’
INTRODUCTION
of nuclear forces. It acts within the nucleus that too upto a very small distance. It does not depends on
Force is a push or pull which tries to change or successfully changes the state of rest or of uniform
charge and acts equally between a proton and proton,
motion of a body, i.e., force is the cause of translatory motion.
electrons does not experience this force. It acts for very –15 short distance order of 10 –15 m.
a neutron and neutron, and proton and neutron,
It arises due to interaction of the bodies either due to contact (e.g., normal reaction, friction, tension, spring force etc.) or from a distance (e.g., gravitational or electric force). FUNDAMENTAL FORCES
All forces observed in nature such as muscular force, tension, reaction, friction, weight, electric, magnetic, nuclear, etc., can be explained in terms of only following four basic interactions.
The force of interaction which exists between two particles of masses m1 and m2, due to their masses is called gravitational force. The gravitational force acts over long distances and does not need, any intervening
It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range of weak force is very small, in fact much smaller than size of a proton or a neutron. It has been found that for two protons at a distance of 1 fermi : –2 –7 –38 FN:FEM:FW:FG::1:10 –2 :10 –7 :10 –38
On the basis of contact forces are classified into two categories (i)
Contact forces
medium. Gravitational force is the weakest force of nature.
(ii) Non contact or field forces
Force exerted by one particle on the other because of
Forces which are transmitted between bodies by short
the electric charge on the particles is called electromagnetic force. Following are the main
range atomic molecular interactions are called contact
characteristics of electromagnetic force (i)
forces. When two objects come in contact they exert contact forces on each other. e.g. Normal, Tension Tension etc.
These can be attractive or repulsive.
(ii) These are long range forces. (iii) These depend on the nature of medium between the charged particles. (iv) All macroscopic forces (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and
Force which acts on an object at a distance by the interaction of the object with the field produc ed by other object is called field force. e.g. Gravitational force, Electro magnetic force etc. DETAILED ANALYSIS OF CONTACT FORCE
force experienced by a deformed spring are electromagnetic forces. These are manifestations
It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in
of the electromagnetic attractions and repulsions
contact are pressed against each other. It is the electromagnetic force.
between atoms/molecules.
e.g.1 A table is placed on Earth as shown in figure It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, Radioactivity, fission, and fusion, etc. results because of unbalancing
11 1 PAGE # 1
Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure.
Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. When a string is connected to a body and pulled out, the string said to be under tension. It p pulls ulls the body with a force T, whose whose direction is away from the body and along the length of the string. Usually strings are regarded to be massless and unstretchable, known as ideal string.
e.g.2 A boy pushes a block kept on a frictionless surface.
Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force.
A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block.
Note : (i) Tension in a string is an electromagnetic force and it arises only when string is pulled. If a massless string is not pulled, tension in it is zero. (ii) String can not push a body in direct contact.
A spring is made of a coiled metallic wire having a definite length. When it is neither pushed nor pulled then its length is called natural length. length. At natural length the spring does not exert any force on the objects attached to its ends. f the spring is pulled at the ends, its length becomes larger than its natural length, it is known as stretched or extended spring. Extended spring pulls objects attached to its ends. A
Normal force exerted by block on the surface of inclined plane is shown in figure. Here normal force is a component of weight of the body perpendicular to the inclined surface i.e. N = mgsin
B Normal spring
Spring force on A
Spring force on B
A
B Stretched spring Spring force on B
Spring force on A
B
A
Compressed spring
Force acts perpendicular to the surface
1.
Two Two blocks are kept in contact on a smooth surface as shown in figure. Draw normal force exerted by A on B.
If the spring is pushed at the ends, its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the objects attached to its ends.
F = 0 spring in natural length does not exerts any force on its ends
x F
Sol. In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero.
Note : •
Normal is a dependent force it comes c omes in role when one surface presses the other.
F
Fext
F = – kx ;k = spring constant or stiffness constant (unit = N/m) x = extension in spring
x F
F Fext
F = – kx x = compression in spring
22 2 PAGE # 2
Note : Spring force is also electromagnetic in nature :
When a body is moving on a rough surface resistance to the motion occurs because of the interaction between the body and its surroundings. We call such resistance as force of friction. Friction is also considered as component of contact force which acts parallel to the surfaces in contact. (i) Origin of friction : The frictional force arises due to molecular interactions between the surfaces at the points of actual contact. When two bodies are placed one over other, the actual area of contact is much smaller then the total surface areas of bodies. The molecular forces starts operating at the actual points of contact of the surfaces. Molecular bonds are formed at these contact points. When one body is pulled over
(C) Once the motion started, the smaller force is now necessary to continue the motion (F3) and thus frictional force decreases. The force of friction when body is in state of motion over the surface is called kinetic or dynamic friction fk (figure d).
the other, these bonds are broken, and the material get deformed and new bonds are formed. The local deformation sends vibrations into the bodies. These Vibrations ultimately dumps out and energy of vibrations appears as heat. Hence to start or carry on the motion, there is a need of force.
Body 1 Body 2
(iii) (iii) More about about frictio frictional nal force force : (A) (A) About About stati static c frictio friction n
Actual area of contact
1.
The limiting friction depends on the materials of the surfaces in contact and their state of polish.
2.
The magnitude of static friction is independent of the apparent area of contact so long as the normal reaction remains the same.
3.
The limiting friction is directly proportional to the magnitude of the normal reaction between the two surfaces i.e. f lim= SN. Here s is coefficient of static friction.
(ii) (ii) Statics Statics a and nd Kinet Kinetic ic Frictio Frictions ns : •
Experiment : (A) Consider a block placed on a table, and a small force F1 is acted on it. The block does not move. It indicates that the frictional force fs starts acting in opposite direction of applied force and its magnitude is equal of F1(figure b). That is for the equilibrium of the block, we have F1 – f s = 0 or F1 = fs The force of friction when body is in state of rest over
We can write, s =
the surface is called static friction (fs). (B) As the applied force increases the frictional force also increases. When the applied force is increased
(B) (B) About About kinet kinetic ic frict friction ion : 1.
The kinetic friction depends on the materials of the surface in contact.
2.
It is also independent of apparent area of contact as long as the magnitude of normal reaction remains the same.
3.
Kinetic friction is almost independent of the velocity, provided the velocity is not too large not too small.
up to a certain limit (F 2) such that the block is on the verge of motion. The value of frictional force at this
flim N
stage is called limiting friction flim (figure c).
33 3 PAGE # 3
4.
The kinetic friction is directly proportional to the magnitude of the normal reaction between the surfaces. fk = k N. Here k is coefficient of kinetic friction. We can write, k =
•
A force is conservative if the total work done by the
fk N
force on an object in one complete round is zero, i.e. when the object moves around any closed path (returning to its initial position).
There are two types of kinetic frictions: (i) (i)
Slid Slidin ing g fri frict ctio ion n : The force of friction when one body slides over the surface of the another body is called sliding friction.
A force is conservative if there is no change in kinetic energy in one complete round. KE = 0 This definition illuminates an important aspect of a conservative force viz. Work done by a conservative force is recoverable. Thus in figure, we shall have to do mgh amount of work in taking taki ng the body from A to B.
(ii) (ii) Rolli Rolling ng fric frictio tion n : When a wheel rolls without slipping over a horizontal surface, there is no relative motion of the point of contact of the wheel with respect to the plane. Theoretically for a rolling
However, when body is released from B, we recover mgh of work. Other examples of conservative forces are spring force, electrostatic force etc.
wheel the frictional force is zero. This can only possible when bodies in contact are perfectly rigid and contact of wheel with the surface is made only at a point. But in practice no material body is perfectly rigid and therefore bodies get deformed when they pressed each other. The actual area of their contact no longer remains a point, and thus a small amount of friction starts acting between the body and the surface. Here frictional force is called rolling friction. It is clear from above discussion that rolling friction is very much smaller than sliding friction.
A force is non-conservative if the work done by that force on a particle moving between two points depends on the path taken between the points. The force of friction is an example of non-conservative force. Let us illustrate this with an instructive example. Suppose we were to displace a book between two points on a rough horizontal surface (such as a table). If the book is displaced in a straight line between the two points, the work done by friction is simply FS where : F = force of friction ; S = distance between the points. However, if the book is moved along any other path between the two points (such as a semicircular path), the work done by friction would be greater than FS. FS.
flim > fkinetic > frolling.
Finally, if the book is moved through any closed path,
Note : s and k are dimensionless quantities and independent of shape and area of contact . It is a property of two contact surfaces. s will always be greater than k .Theoretical value of can be o to practical value is 0 <
the work done by friction is never zero, it is always negative. Thus the work done by a non-conservative force is not recoverable, as it is i s for a conservative force.
but
1.6
SYSTEM
Two or more than two objects which interact with each other form a system. Classification of forces on the basis of boundary of A force is said to be conservative if the amount of work done in moving an object against that force is independent on the path. One important example of conservative force is the gravitational force. It means that amount of work done in moving a body against gravity from location A to location B is the same whichever path we may follow i n going from A to B. This
system : (a) Inte Intern rnal al Forc Forces es : Forces acting with in a system among its constituents. (b) E xt xt e rn rn a l Fo Fo r ce ce s : Forces exerted on the constituents of a system by the outside surroundings are called as external forces.
is illustrated in figure.
44 4 PAGE # 4
FREE BODY DIAGRAM
B
A free body diagram consists of a diagrammatic representations of single body or a subsystem of bodies isolated from surroundings showing all the forces acting on it.
A
Sol.F.B.D. Sol.F.B.D. of sphere A’ ‘A’ :
Steps for F.B.D. Step 1 : Identify the object or system and isolate it from other objects, clearly specify its boundary. Step 2 : First draw non-contact external force in the diagram, generally it is weight. Step 3 : Draw contact forces which acts at the boundary
F.B.D. of sphere ‘B’ :
of the object of system. Contact forces are normal ,
(exerted by A)
friction, tension and applied force. In F.B.D, internal forces are not drawn only external are drawn. 2.
A block of mass ‘m’ is kept on the ground as shown in figure.
Note : Here NAB and NBA are the action - reaction pair (Newton’s third law).
(i)
4.
Draw F.B.D. of block.
Draw F.B.D. for systems shown in figure below.
(ii) Are forces acting on block forms action- reaction pair. (iii) If answer is no, draw action reaction pair. Sol.(i)
F.B.D. of block
Sol.
(ii) ‘N’ and mg are not action -reaction pair. Since pair act on different bodies, and they are of same nature. (iii) Pair of ‘mg’ mg ’ of block acts on earth in opposite direction.
earth
mg
and pair of ‘N’ acts on surface as shown in figure. N
TRANSLATORY EQUILIBRIUM
When several forces acts on a body simultaneously in such a way that resultant force on the t he body is zero, i.e.,
F = 0 with F = 3.
Two sphere A and B are placed between two vertical walls as shown in figure. Draw the free body diagrams of both the spheres.
Fi the body is said to be in translatory
equilibrium. Here it is worthy to note that : (i)
As if a vector is zero all its components must vanish i.e. in equilibrium as -
55 5 PAGE # 5
F = 0 F = 0 ; F = 0 ; F = 0
F = 0 with F =
i
y
x
Note : The absolute unit of force remains the same everywhere, but the gravitational unit of force varies from place to place p lace because it depe depends nds on the value value of of g. nd
’
z
(i) (i) So in equilibrium forces along x axes must balance each other and the same is true for other directions. If a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is at rest, equilibrium is called static, otherwise dynamic.
When When obje objects cts are in equili equilibri brium um : Steps to solve problem involving objects in equilibrium :
Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the freebody diagram for the object. Label all external forces
Static equilibrium can be divided into following three types :
acting on it. Step 3 : Choose a convenient coordinate system and resolve all forces into rectangular components along x and Y direction.
If on slight displacement from equilibrium position a body has a tendency to regain its original position it is said to be in stable equilibrium. In case of stable equilibrium potential energy is minimum and so center of gravity is lowest.
Step 4 : Apply the equations
F
x
0 and
F
y
0.
Step 5 : Step 4 will give you two equations with se veral unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities. Step 6 : If step 5 produces two equations with more
O
than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at
(b) Unstable equilibrium : If on slight displacement from equilibrium position a body moves in the directio n of displacement, the equilibrium is said to be unstable. In this situation potential energy of body is maximum and so center of gravity is highest.
step 5 you will have enough equations to solve for all unknown quantities.
5.
O
(c) Neutral equilibrium : If on slight displacement from equilibrium position a body has no tendency to come back to its original position or to move in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of body is constant and so center of gravity remains at constant height.
A ‘block’ block’ of mass 10 kg is suspended with string as shown in figure. Find tension in the string. (g = 10 m/s2).
Sol.F Sol.F.B.D. of block bloc k For equilibrium of block along Y axis
F
y
0
T – 10 g = 0 T = 100 N 6.
The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1 , T2, T3 and T4. (g = 10 m/s2).
. nd
The rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction of the applied force.
In relation F = ma the force F stands for the net
Sol.F Sol.F.B.D. of 10 kg block b lock
external force. Any internal force in the system is not to
For equilibrium of block along Y axis.
be included in F .
F
In S.I. the absolute unit of force is newton (N) and gravitational unit of force is kilo gram weight or kilogram force (kgf.)
y
T0
0
T0 = 10 g T0 = 100 N
10g
66 6 PAGE # 6
N2 = 50 sin 30º + N 3 N3 = 100 – 25 = 75 N & N4 = 50 cos 30º + 20 g N4 = 243.30 N
F.B.D. of point A’ ‘A’ y
F
y
0
T2 30º
T2 cos 30º = T0 = 100 N
T1
x
A
8.
200 T2 =
3
F
x
N T0
0
T1 = T2 . sin 30º
200 =
3
Find magnitude of force exerted by string on pulley.
.
Sol B.
F.B.D. of 10 kg block b lock :
1 100 = N. 3 2
F.B.D. of point of ‘B’ y 60º T3
T = 10 g = 100 N F.B.D. of pulley pul ley :
T4 x
B 30º T2
F = 0 T cos 60º = T cos 30º y
and
2
T4 = 200 N
Since string is massless, so tension in both sides
of string is same.
Fx = 0 T3 + T2 sin30º = T4 sin 60º
T3 = 7.
4
So magnitude of force exerted by string on pulley
200 3
=
N
Two Two blocks are kept k ept in contact as shown in figure. Find :-
1002 100 2
= 100 2 N
Note : Since pulley is in equilibrium position, so net
(a) forces exerted by surfaces (floor and wall) on blocks.
forces on it is zero.
(b) contact force between two blocks.
Steps to solve problems involving objects that are in accelerated motion : Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the free - body diagram for that object. Label al l external forces
SolA : F.B.D. of 10 kg block
acting on it. Be sure to include all the forces acting on the chosen body, but be equally careful not to include any force exerted by the body on some other body. Some of the forces may be unknown , label them with algebraic symbols.
N1 = 10 g = 100 N .......(1) N2 = 100 N .........(2) F.B.D. of 20 kg k g block
Step 3 : Choose a convenient coordinate system, show location of coordinate axis explicitly in the free - body diagram, and then determine components of forces with reference to these axis and resolve all forces into x and y components. Step 4 : Apply the equations
F = ma & F = ma . x
x
y
y
Step 5 : Step 4 will give two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities.
77 7 PAGE # 7
Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select
Sol.For Sol.For calculating the value of F0. F.B.D of whole system
another object and repeat these steps. Eventually at F0
step 5 you will have enough equations to solve for all unknown quantities. 2 (a) 2m/s
9.
A force F is applied horizontally on mass m1 as shown
10 g = 100 N
in figure. Find the contact force between m1 and m2. F0 –100 –100 = 10 × 2 F0 = 120 N Sol.Considering Sol. Considering both blocks as a system to find the common acceleration.
(b) According to Newton’s second law, net force on rope. F = ma = 2 × 2 = 4N
Common acceleration
F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.
m1 m2 .......(1)
F
m1
............(2)
(c) For calculating tension at the m iddle point we draw
F a=
........(1)
m2
a
T – 4 g = 4 . 2 To find the contact force between A’ ‘A’ and ‘B’ we draw F.B.D. of mass m2.
T = 48 N
F.B.D. of mass m2
11. A block of mass 50 kg is kept on another block o f mass 1 kg as shown in figure. A horizontal force of 10 N is applied on the 1Kg block. (All surface are smooth). Find : (g = 10 m/s2) (a) Acceleration of blocks A and B.
(b) Force exerted by B on A.
Fx = max
B A
N = m2 . a
m2F N= m1 m2
F sin ce a m m 1 2
Sol.(a)
50 kg 1 kg
F.B.D. of 50 kg
10. A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an external force F0. (a) What is F0 ? (b) What is the net force on rope ?
N2 = 50 g = 500 N along horizontal direction, there is no force aB = 0 (b) F.B.D. of 1 kg k g block : N1 N2
(c) What is the tension at middle point of the rope ? 10 N
(g = 10 m/s2) 1g
along horizontal direction 10 = 1 aA. aA = 10 m/s2 along vertical direction N1 = N2 + 1g = 500 + 10 = 510 N
88 8 PAGE # 8
12. One end of string which passes through pulley and
N = Mg
connected to 10 kg mass at other end is pulled by 100
= 60 × 10
N force. Find out the acceleration of 10 kg mass. (g
N = 600 N.
=9.8 m/s2)
SPRING BALANCE
It does not measure the weight. It measures the force exerted by the object at the hook. Symbolically, it is represented as shown in figure.
Sol.Since Sol.Since string is pulled by 100 N force. So tension in the
A block of mass ‘m’ is suspended at hook. When spring balance is in equilibrium, we draw the F.B.D. of mass m for calculating the reading of balance.
string is 100 N F.B.D. of 10 kg k g block spring balance hook m
F.B.D. of o f ‘m’. 100 – 10 g = 10 a 100 – 10 × 9.8 = 10 a
mg – T = 0 T = mg Magnitude of T gives the reading of spring balance.
a = 0.2 m/s2. WEIGHING MACHINE
A weighing machine does not measure the weight but
14. A block of mass 20 kg is suspended through two light spring balances as shown in figure . Calculate the :
measures the force exerted by object on its upper surface. 13. A man of mass 60 Kg is standing on a weighing machine placed on ground. Calculate the reading of
weighing machine
machine (g = 10 m/s2).
Sol.For Sol.For calculating the reading of weighing machine, we draw F.B.D. of man and machine separately. F.B.D of man
(1) reading of spring balance (1). (2) reading of spring balance (2). Sol.For Sol.For calculating the reading, first fi rst we draw F.B.D.of F.B.D.of 20 kg block.
F.B.D of man taking mass of man as M
F.B.D. of weighing machine
N
N weighing machine
F.B.D. 20 kg
T N = Mg
N1
Mg
Mg 20 g
Here force exerted by object on upper surface is N Reading of weighing machine mg – T = 0 T = 20 g = 200 N
99 9 PAGE # 9
Since both the balances are light so, both the scales will read 200 N.
As the block is in equilibrium along y-axis, so we have
F
y
15. (i)
A 10 kg block is supported by a cord that runs to a spring scale, which is supported by another cord from the ceiling figure (a). What is the reading on the scale ?
(ii) In figure (b) the block is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by cord to a wall. What is the reading of the scale. (iii) In figure (c) the wall has been replaced with a second 10 kg block on the left, and the assembly is stationary. What is the reading on the scale now ?
or
0;
N = mg + F sin
To just move the block along x-axis, we have F cos = N = (mg + F sin )
or
F=
mg .......(i) cos – sin
Pull Pull : Along y-axis we have ;
T
F
y
spring balance
T
hook
0;
N = mg – F sin
10 kg
To just move the block along x-axis, we have
(a)
F cos = N = (mg – F sin )
T
T T 10kg
F=
mg cos sin
. .......(ii)
It is clear from above discussion that pull force is smaller than push force.
(b) T
or
T
17. Discuss the direction of friction in the following cases :
T
T
10kg
10kg
(c) Sol. In all the three cases the spring balance reads 10 kg. To understand this let us cut a section sectio n inside the spring as shown;
(i)
A man walks slowly, slowly, without change in speed.
(ii) A man is going with increasing speed. (iii) When cycle is gaining speed. (iv) When cycle is slowing down . Sol. Sol. (i)
Consider a man walks slowly without acceleration, and both the legs are touching the ground as shown in figure (a). The frictional force on rear leg is in forward direction and on front leg will be on
As each part of the spring is at rest, so F= T. As the block is stationary, so T= 10g = 100N. 16. Pull Pull is easie easierr than than push push
Push : Consider a block of mass m placed on rough horizontal surface. The coefficient of static friction between the block and surface is . Let a push force F is applied at an angle with the horizontal.
backward direction of motion. As a = 0, Fnet = 0
or or
f1 – f 2 = 0
f1 = f2
&
N1 = N2.
N1
f1
N2
f1
f2
f2
Ground N1
N2
(b)
10 10 10 PAGE # 10
(ii) (ii) When When man is gaini gaining ng the the speed speed : The frictional force on rear leg f1 will be greater than frictional force on front leg f2 (fig. b).
T = 25 g.
f f acceleration of the man, a = 1 2 . m
N1 = 25 g + 50 g
=
75 g = 75 × 9.8 = 735 N
Block is to be raised without acceleration, so
In IInd case, let the force exerted by the man on the floor
(iii) (iii) When cycle cycle is gaining gaining speed speed : In this case torque is applied on the rear wheel of the cycle by the chain-gear system. Because of this the slipping
in N2 . Consider the forces inside the dotted box, we have N2 = 50 g – T
tendency of the point of contact of the rear wheel is
and T = 25 g
backward and so friction acts in forward direction.
N2 = 50 g – 25 g
The slipping tendency of point of contact of front
= 25 g = 25 × 9.8 = 245 N.
wheel is forward and so friction acts in backward
As the floor yields to a downward force of 700 N, so the
direction. If f1 and f2 are the frictional forces on rear
man should adopt mode
.
and front wheel, then acceleration of the cycle a =
f1 – f2 , where M is the mass of the cycle together M with rider (fig. a).
19. Figure shows a weighing machine kept in a lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45
N1
N2
f1
f2 (a)
N1
N2
f1
kg respectively respect ively..
f2 (b)
(iv) (iv) When When cycle cycle is s slowing lowing down : When torque is not applied (cycle stops pedaling), the slipping tendency of points of contact of both the wheels are forward, and so friction acts in backward direction (fig. b). If f1 and f2 are the frictional forces on rear and front wheel, then retardation
f1 f2 a= M 18. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in fig.. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding.
Answer the following questions. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion. (take g = 10 m/s2) (i) Find the mass of the object in kg and the normal force acting on the block due to weighing machine? (ii) Find the acceleration of the lift such that weighing machine shows its true weight ?
Sol. (i)
50g 50g
Sol. The FBD for the two cases are shown in figure. b y the man on the floor is In Ist case, let the force exerted by
T + N – Mg = Ma 45 g + 15 g = M(g + a) 450 + 150 = M(10 + 5) M = 40 kg Normal force is the reaction applied by weighing machine i.e. 15 × 10 = 150 N.
N1. Consider the forces inside the dotted box, we have N1 = T + 50 g.
11 11 11 PAGE # 11
4.
A dish of mass 10 g is kept horizontally in air by firing bullets of mass 5 g each at the rate of 100 per second. If the bullets rebound with the same s peed, what is the velocity with which the bullets are fired :
(ii)
T + N – Mg = Ma 45 g + 40 g = 40(g + a) 450 + 400 = 400 +40 a
5.
(B) 0.098 m/s
(C) 1.47 m/s
(D) 1.96 m/s
A block of metal weighing 2 kg is resting on a frictionless plank. If struck by a jet releasing water at a
450 45 a= = m/s2 40 4
rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be :
6. Normal Force :
1.
(A) 0.49 m/s
(A) 2.5 m/s2
(B) 5.0 m/s2
(C) 10 m /s2
(D) none of the above
A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ N ’ then
Two blocks are in contact on a frictionless table. One has mass m and the other 2m.A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be :
(A) N= N’
(B) N > N’
(C) N’> N (D) cannot be determined (A) Same (C) 2 : 1 2.
(B) 1 : 2 (D) 1 : 3
Two Two forces of 6N and 3N are acting a cting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ?:
ASSERTION / REASON
7.
STA STATEME TEMEN NT-1 : Block A is moving on horizontal ho rizontal surface surfac e towards right under action of force. All surface are smooth. At the instant shown the force exerted by block A on block B is equal to net force on block B.
6N 2kg 1kg (A)1N (C) 4N 3.
•
3N
STATEMENT-2 : From Newtons’s third law, the force
(B) 2N (D) 5N
exerted by block A on B is equal in magnitude to force exerted block B on A
There are two forces on the 2.0 kg k g box in the overhead view of figure but only one is shown. The second force is nearly :
(A) statement-1 is true, Statement 2 is true, t rue, statement-2 is correct explanation for statement-1.
y
(B) statement-1 is true, Statement 2 is true, t rue, statement-2 is NOT a correct explanation for statement-1. (C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True
F1 = 20 20 N x
8. 30º a = 12 m/s
A certain force applied to a body A gives it an acceleration –2 of 10 ms –2 . The same force applied to body B gives it –2 an acceleration of 15 ms –2 . If the two bodies are joined
2
(A) –20 –20 jˆ N
(B) – 20 ˆi + 20 jˆ N
(C) –32 –32 ˆi – 12 3 jˆ N
(D) –21 –21 ˆi – 16 jˆ N
together and same force is applied to the combination, the acceleration will be : (IJSO/Stage-I/2011) –2 (A) 6 ms –2
–2 (B) 25 ms –2
–2 (C) 12.5 ms –2
–2 (D) 9 ms –2
12 12 12 PAGE # 12
9. Four blocks are kept in a row on a smooth horizontal table with their centres of mass collinear as shown in the figure. An external force of 60 N is applied from left on the 7 kg block to push all of them along the table. The forces exerted by them are :(IAO/Sr./Stage-I/2008) :(IAO/Sr./Stage-I/2008) 60N
P
Q
R
S
7 kg
5 kg
2 kg
1 kg
(A) 32 N by P on Q (C) 12 N by Q on R
13. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination and . If M2 > M1 then the acceleration of block M2 down the inclined will be :
(B) 28 N by Q on P (D) 4 N by S on R
Tension :
(A)
10. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled hori horizontally zontally with a force F. F. If the rope AB makes an angle with the vertical in equilibrium,then the tension in the string AB is :
M2 (sin ) M1 M2
g
M2 sin M1 sin g M1 M2
(C)
(B)
M1g(sin ) M1 M2
(D) Zero
14. Three masses of 1 kg, 6 kg and 3 kg are connected to each other by threads and and are placed on table as shown in figure. What is the acceleration with which –2 the system is moving ? Take Take g = 10 m s –2 :
(A) F sin (C) F cos
(B) F /sin (D) F / cos
11. In the system shown in the figure, the acceleration of the 1kg mass and the tension in the string connecting between A and B is :
(A) Zero
–2 (B) 1 ms –2
–2 (C) 2 m s –2
–2 (D) 3 m s –2
15. The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant (A)
(C)
g 8g downward, 4 7
(B)
g 6 downward, g 7 7
(D)
g g upward, 4 7
downward force F= 2 mg, where g is acceleration due to gravity. The acceleration of mass in case I is :
g upward, g 2
12. A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a –2 string with an acceleration of 2.2 m s –2 . The tension T1 and T2 will be respectively :(Use g =9.8 m/s2)
(A) Zero (B) More than that in case II (A) 200 N, 80 N (C) 240 N, 96 N
(B) 220 N, 90 N (D) 260 N, 96 N
(C) Less than that in case II (D) Equal to that in case II
13 13 13 PAGE # 13
16. A 50 kg person stands on a 25 kg platform. He pulls massless rope which is attached to the platform via the frictionless, massless pulleys as shown in the figure. The platform moves upwards at a steady velocity if the force with which the person pulls the rope is :
20. Two blocks of mass m each is connected with the string which passes over fixed fi xed pulley, as shown in figure. The force exerted by the string on the pulley P is :
(A) mg
(B) 2 mg
(C)
(D) 4 mg
2 mg
21. One end of a massless rope, which passes over a (A) 500 N (C) 25 N
(B) 250 N (D) 50 N
17. Figure shows four blocks that are being pulled along a smooth horizontal surface. The mssses of the blocks and tension in one cord are given. The pulling force F is :
massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that rope can bear is 360 N, with what minimum safe acceleration (in m/s2) can a monkey of 60 kg move down on the rope :
F 60º
4kg
30N 3kg
(A) 50 N (C) 125 N
P 2kg
1kg
(B) 100 N (D) 200 N C
18. A10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. The magnitude of the least
(A) 16
(B) 6
acceleration the monkey must have if it is to lift the
(C) 4
(D) 8
package off the ground is :
22. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :
(A) 4.9 m/s2 (C) 9.8 m/s2
(B) 5.5 m/s2 (D) none of these
(A)
(B)
(C)
(D) None of these
19. Two blocks, each of mass M, are connected by a massless string, which passes over a smooth
massless pulley pulle y. Forces F act on the blocks as shown. The tension in the string is :
23. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is : (A) 15 kg (A) Mg
(B) 2 Mg
(C) Mg + F
(D) none of these
(B)
15 kg 2
(C) 5 kg (D) Infinitely large (or not possible)
14 14 14 PAGE # 14
24. In the figure, the blocks A, B and C of mass each have
29. Two bodies of masses M1 and M2 are connected to
acceleration a1 , a2 and a3 respectively . F1 and F2 are
each other through a light spring as shown in figure. If
external forces of magnitudes 2 mg and mg
we push mass M1 with force F and cause acceleration
respectively then which of the following relations is
a1 in mass M1 what will be the acceleration in M2 ?
correct :
(A) F/M2
(B) F/(M1 + M2)
(C) a1
(D) (F –M –M1a1) /M /M2
30. A spring balance is attached to 2 kg trolley and is used to pull the trolly along a flat surface as shown in the fig. (A) a1 = a2 = a3
(B) a1 > a2 > a3
The reading on the spring balance remains at 10 kg
(C) a1 = a2 , a2 > a3
(D) a1 > a2 , a2= a3
during the motion. The acceleratio n of the trolly is (Use
25. A weight is supported by two strings 1.3 and 2.0 m
–2 g= 9.8 m –2 ):
long fastened to two points on a horizontal beam 2.0 m apart. The depth of this weight below the beam is : (IAO/Jr./Stage-I/2007) (A) 1.0 m (C) 0.77 m
(B) 1.23 m (D) 0.89 m
26. A fully loaded elevator has a mass of 6000 kg. The tension in the cable as the elevator is accelerated –2 downward with an acceleration of 2ms –2 is (Take g = I0 –2 – 2 ms ) (KVPY/2007) (A) 7·2 × 104 N (C) 6 × 104 N
(B) 4.8 × 104 N (D) 1.2 × 104 N
–2 (A) 4.9 ms –2
–2 (B) 9.8 ms –2
–2 (C) 49 ms –2
–2 (D) 98 ms –2
31. A body of mass 32 kg is suspended by a spring balance from the roof of a vertically operating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance bal ance showed 30 kg & 36 kg respectively. The velocity of the lift is : (A) Decreasing at 20 m & increasing at 50 m
27. A light string goes over a frictionless pulley. At its one end hangs a mass of 2 kg and at the other end hangs a mass of 6 kg. Both the masses are supported by hands to keep them at rest. When the masses are released, they being to move and the string gets taut. –2 (Take g = 10 ms –2 ) The tension in the string during the motion of the masses is : (KVPY/2008) (A) 60 N (B) 30 N (C) 20 N (D) 40 N
(B) Increasing at 20 m & decreasing at 50 m (C) Continuously decreasing at a constant rate throughout the journey (D) Continuously increasing at constant rate throughout the journey Friction Force : 32. A ship of mass 3 × 107 kg initially at rest is pulled by a
Force Exerted by Spring :
force of 5 × 10 4 N through a distance of 3m. Assume that the resistance due to water is negligible, the spee d
28. In the given figure. What is the reading of the spring
of the ship is : (A) 1.5 m/s
(B) 60 m/s
(C) 0.1 m/s
(D) 5 m/s
balance:
33. When a horse pulls a cart, the force needed to move the horse in forward direction is the force exerted by : (A) The cart on the horse (A) 10 N
(B) 20 N
(B) The ground on the horse (C) The ground on the cart
(C) 5 N
(D) Zero
(D) The horse on the ground
15 15 15 PAGE # 15
34. A 2.5 kg block is initially at rest on a horizontal surface.
A 6.0 N horizontal force and a vertical force P are applied to the block as shown in figure. The coefficient of static friction for the block and surface is 0.4. The magnitude of friction force when P = 9N : (g = 10 m/s 2)
40. A bock of mass 5 kg is held against wall by applying a horizontal force of 100N. If the coefficient of friction between the block and the wall is 0.5, the frictional force acting on the block is : (g =9.8 m/s2)
100N
(A) 6.0 N (C) 9.0 N
(B) 6.4 N (D) zero
35. The upper half of an inclined plane with inclination is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is : (A) 2 tan (B) tan (C) 2 sin (D) 2 cos 36. Minimum force required to pull the lower block is (take g = 10 m/s2) :
(A) 1 N (C) 7 N
(B) 5 N (D) 10 N
37. N bullets each of mass m are fired with a velocity v m/ s at the rate of n bullets per sec., upon a wall. If the bullets are completely stopped by the wall, the reaction offered by the wall to the bullets is : (A) N m v / n (B) n m v (C) n N v / m (D) n v m / N 38. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be , then the stopping distance is :
P2 (B) 2 mg
P (A) 2 mg
P2
P (C)
(D)
2 m2g
(A) 20 N (C) 12N
(A)100 N (C) 49 N
(B) 50 N (D) 24.9 N
41. A heavy roller is being pulled along a rough road as shown in the figure. The frictional force at the point of contact is : (IAO/Jr./Stage-I/2007)
F
(A) parallel to F (C) perpendicular to F
(B) opposite to F (D) zero
42. When a motor car of mass 1500 kg is pushed on a road by two persons, it moves with a small uniform velocity. On the other hand if this car is pushed on the same road by three persons, it moves with an acceleration of 0.2 m/s2. Assume that each person is producing the same muscular force. Then, the force of friction between the tyres of the car and the surface of the road is : (IAO/Jr./Stage-I/2009) (A) 300 N (C) 900 N
(B) 600 N (D) 100 N
43. A block of mass M is at rest on a plane surface incl ined at an angle to the horizontal The magnitude of force exerted by the plane on the block is : (KVPY/2009) (A) Mg cos (B) Mg sin (C) Mg tan (D) Mg 44. A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The correct representation of the variation of the frictional forces, ƒ, exerted by the table on the block with time t is given by : (KVPY/2010)
2 m2g
39. What is the maximum value of the force F such that the block shown in the arrangement, does not move : F 60º
5kg
(A)
(B)
(C)
(D)
1 2 3 m = 3kg
(B) 10 N (D) 15 N
16 16 16 PAGE # 16
45. A small child tries to move a large rubb er toy placed on the ground. The toy does not move but gets deformed
Weighing Machine : 49. The ratio of the weight of a man in a stationary lift and
under her pushing force (F) which is obliquely upward as shown . Then
when it is moving downward with with uniform acceleration ‘a’ 3:2. The value of ‘a’ is : (g = acceleration, due to
(KVPY/2011)
gravity) (A) (3/2)g (C) (2/3) g
(B) g (D) g/3
50. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time t1 when elevator is stationary and in time t2 if it is moving
(A) The resultant of the pushing force (F) , weight of
uniformly. Then (A) t1 = t2
the toy, normal force by the ground on the toy and the (B) The normal force by the ground is equal and oppo-
(B) t1 > t2 (C) t1 < t2
site to the weight of the toy.
(D) t1 < t2 or t1 > t2 depending
frictional force is zero.
(C) The pushing force (F) of the child is balanced by
•
ASSERTION / REASON
the equal and opposite frictional force 51. STA STATEMENT TEMENT-1 -1 : A man standing in a lift l ift which is moving
(D) The pushing force (F) of the child is balanced by
upward, will feel his weight to be greater than when
the total internal force in the toy generated due to
the lift was at rest.
deformation
STATEMENT-2 : If the acceleration of the lift is ‘a’ upward
46. On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each eac h other by means of a rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. How far has the girl travelled when she meets the box ? (KVPY/2011)
then the man of mass m shall feel his weight to be equal to normal reaction (N) exerted by the lift given N = m(g+a) (where g is acceleration due to gravity (A) statement-1 is true, Statement 2 is true, statement2 is correct explanation for statement 1. (B) statement-1 is true, Statement 2 is true, statement-
(A) 10 m
2 is NOT a correct explanation for statement-1.
(B) Since there is no friction, the girl will not move
(C) statement-1 is true, Statement 2 is false
(C) 16 m
(D) statement-1 is False, Statement 2 is True
(D) 4m 47. Which of the following does NOT involve friction ? (IJSO/Stage-I/2011)
52. A beaker containing water is placed on the platform of a digital weighing machine. It reads 900 g. A wooden
(A) Writing on a paper using a pencil
block of mass 300 g is now made to float in water in
(B) Turning a car to the left on a horizontal road.
the beaker (without touching walls of the beaker). Half
(C) A car at rest parked on a sloping ground
the wooden block is submerged inside water. Now,
(D) Motion of a satellite around the earth.
the reading of weighing machine will be : (IAO/Jr./Stage-I/2009)
below, the coefficient of kinetic 48. In the two cases shown below, friction between the block and the surface is the same, and both the blocks are moving with the same uniform speed. Then, (IAO/Sr./Stage-I/2008)
(A) 750 g
(B) 900 g
(C) 1050 g
(D) 1200 g
Miscellaneous : F1
(A) F1 = F2 (B) F1 < F2 (C) F1 > F2 (D) F1 = 2F2 if sin = Mg/4F2
F2
53. An object will continue accelerating until : (A) Resultant force on it begins to decreases (B) Its velocity changes direction (C) The resultant force on it is zero (D) The resultant force is at right angles to its direction of motion
17 17 17 PAGE # 17
54. In which of the following cases the net force is not zero ? (A) A kite skillful ly held stationary in the sky (B) A ball freely falling from a height (C) An aeroplane rising upward at an angle of 45° with the horizontal with a constant speed (D) A cork floating on the surface of water.
55.
Figure shows the displacement of a particle going
along the X-axis as a function of ti me. The force acting
58. A force of magnitude F1 acts on a particle so as to accelerate if from rest to velocity v. v. The force F1 is then replaced by another force of magnitude F2 which decelerates it to rest. (A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these
on the particle is zero in the region.
(A) AB
(B) BC
(C) CD
(D) DE
59. In a imaginary atmosphere, the air exerts a small force F on any particle in the direction of th e particle’s motion. A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then (A) t1 < t2 (B) t1 > t2 (C) t1 = t2 (D) The relation between t 1 and t2 depends on the mass of the particle
56. A 2 kg toy car can move along x axis. Graph shows force Fx, acting on the car which begins to rest at time t = 0. The velocity of the car at t = 10 s is :
60. A single force F of constant magnitude begins to act on a stone that is moving along x axis. The stone continues to move along that axis. Which of the following represents the stone’s position ? (A) x = 5t – 3 (B) x = 5t2 + 8t – 3 (C) x = –5t –5t2 + 5t – 3 (D) x = 5t3 + 4t2 – 3 61. Three forces act on a particle that moves with unchanging velocity v = (3 ˆi – 4 jˆ ) m/s. Two of the forces are F1 = (3 ˆi + 2 jˆ – 4 kˆ ) N and F2 = ( –5 –5 ˆi + 8 jˆ
+ 3 kˆ ) N. The third force is :
(A) ( –2 –2 ˆi + 10 jˆ – 7 kˆ ) N (B) (2 ˆi – 10 jˆ + kˆ ) N (C) (7 ˆi – 2 kˆ + 10 jˆ ) N (D) none of these
62. An 80 kg person is parachuting and experiencing a (A) – ˆi m/s
(B) – 1.5 ˆi m/s
(C) 6.5 ˆi m/s
(D) 13 ˆi m/s
downward acceleration of 2.5 m/s2 . The mass of the parachute is 5.0 kg. The upward force on the open
57.
Figure shows the displacement of a particle going
parachute from the air is : (A) 620 N
(B) 740 N
(C) 800 N
(D) 920 N
along the x-axis as a function of time : 63. A block of mass m is pulled on the smooth horizontal surface with the help of two ropes, each of mass m, connected to the opposite faces of the block. The forces on the ropes are F and 2F. The pulling force on the block is : (A) The force acting on the particle is zero in the region AB (B) The force acting on the particle is zero in the region BC (C) The force acting on the particle is zero in the region CD (D) The force is zero no where
(A) F (C) F/3
(B) 2F (D) 3F/2
18 18 18 PAGE # 18
64. A body of mass 5 kg starts from the origin with an initial –1 velocity u = 30 ˆi + 40 jˆ ms –1 . If a constant force –( ˆi + 5 jˆ ) N acts on the body, the time in which the F = –(
68. A body of 0.5 kg moves along the positive x - axis under
the influence of a varying force F (in Newtons) as shown
below :
(KVPY/2011)
y-component of the velocity becomes zero is : (A) 5 s
(B) 20 s
(C) 40 s
(D) 80 s 3
65. STA STATEMENT TEMENT-1 -1 :According to the newton’s third law of motion, the magnitude of the action and reaction force is an action reaction pair is same only in an inertial frame of reference. STATEMENT-2 : Newton ’ s laws of motion are applicable in every inertial reference frame. (A) statement-1 is true, Statement 2 is true, statement2 is correct explanation explanation for statement 1. (B) statement-1 is true, Statement 2 is true, statement2 is NOT a correct explanation for statement-1.
) N ( F
3
1
0,0
2
4
6
8
10
x(m)
(C) statement-1 is true, Statement 2 is false (D) statement-1 is False, Statement 2 is True –1 If the speed of the object at x = 4m is 3.16 ms –1 then its
66. A body of mass 10 g moves with constant speed 2 m/ s along a regular hexagon. The magnitude of change in momentum when the body crosses a corner is : (IAO/Sr./Stage-I/2007) (A) 0.04 kg-m/s
(B) zero
(C) 0.02 kg-m / s
(D) 0.4 kg-m/s
67. An object with uniform density is attached to a spring that is known to stretch linearly with applied force as shown below
speed at x = 8 m is : –1 (A) 3.16 ms –1
–1 (B) 9.3 ms –1
–1 (C) 8 ms –1
–1 (D) 6.8 ms –1
69. A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bullet is 49 gm, what is the weight of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8 m/s2 . (KVPY/2010) (A) 50 kg (B) 75 kg (C) 100 kg (D) 125 kg
When the spring object system is immersed in a liquid of density 1 as shown in the figure, the spring stretches by an amount x1 ( > 1). When the experiment experiment is repeated in a liquid of density 2 < 1 . the spring is stretched by an amount x2. Neglecting any buoyant force on the spring, the density of the object is: (KVPY/2011)
1x1 2 x 2 (A) x x 1 2
1x2 2 x1 (B) x x 2 1
1x2 2 x1 (C) x x 1 2
1x1 2x 2 (D) x x 1 2 19 19 19 PAGE # 19
INTRODUCTION
VERSATILE VERSAT ILE NATURE OF CARBON
About 3 million organic compounds are known today. The main reasons for this huge number of organic compounds are (i) Catenation : The property of self linking of carbon atoms through covalent bonds to form long straight or branched chains and rings of different sizes is called catenation.Carbon shows maximum catenation in the periodic table due to its small size, electronic configuration and unique strength of carboncarbon bonds. (ii) Electronegativity and strength of bonds : The electronegativity of carbon (2.5) is close to a number of other elements like H (2.1) , N(3.0) , P (2.1), Cl (3.0) and O (3.5). So carbon forms strong covalent bonds with these elements. (iii) Tendency to form multiple bonds : Due to small size of carbon it has a strong tendency to form multiple bonds (double & triple bonds). (iv) Isomerism : It is a phenomenon by the virtue of which two compounds have same molecular formula but different physical and chemical properties.
CLASSIFICATION OF ORGANIC COMPOUNDS
Note : Urea is the first organic compound synthesized in the laboratory l aboratory..
The organic compounds are very large in number on account of the self -linking property of carbon called catenation. These compounds have been further classified as open chain and cyclic compounds. Organic compounds
MODERN DEFINITION OF ORG ORGANIC ANIC CHEMISTRY Organic compounds may be defined as hydrocarbons and their derivatives and the branch of chemistry which deals with the study of hydrocarbons and their derivatives is called ORGANIC CHEMISTRY.
Open chain compounds
Closed chain compounds
Aromatic compounds
Alicyclic compounds
Organic chemistry is treated as a separate branch because of following reasons(i) Organic compounds are large in number. (ii) Organic compounds generally contain covalent bond. (iii) Organic compounds are soluble in non polar solvents. (iv) Organic compounds have low melting and boiling points.
These compounds contain an open chain of carbon atoms which may be either straight chain or branched chain in nature. Apart from that, they may also be saturated or unsaturated based upon the nature of bonding in the carbon atoms. For example.
(v) Organic compounds show isomerism . (vi) Organic compounds exhibit homology.
,
,
20 20
PAGE # 20
,
e.g. Benzene
Tolue oluene ne
Phenol
n-Butane is a straight chain alkane while
Ethyl hyl benzene
Aniline
Note : Benzene is the parent compound of majority of aromatic organic compounds.
2-Methylpropane is branched alkane. HYDROCARBONS
Apart from the open chains, the organic compounds can have cyclic or ring structures. s tructures. A minimum of three atoms are needed to form a ring. These compounds have been further classified into following types. (i ) Alicyclic compounds : Those carbocyclic compounds which resemble to aliphatic compounds in their properties are called alicyclic compounds .
The organic compounds containing only carbon and hydrogen are called hydrocarbons. These are the simplest organic compounds and are regarded as parent organic compounds. All other compounds are considered to be derived from them by the replacement of one or more hydrogen atoms by other atoms or groups of atoms. The major source of hydrocarbons is petroleum. Types Types of Hydrocarbons : The hydrocarbons can be classified as :
e.g.
or
or
Cyclopropane
Cycl Cyclob obut utan ane e
(i) Saturated hydrocarbons : (A) Alkanes : Alkanes are saturated hydrocarbons containing only carbon - carbon and carbon - hydrogen single covalent bonds. General formulaformula- CnH2n + 2(n is the number of carbon atoms) e.g.
or
Cyclopentane
CH4 ( Methane) C2H6 (Ethane)
(ii) Unsaturated hydrocarbons : (A) Alkenes : These are unsaturated hydrocarbons which contain carbon - carbon double bond. They contain two hydrogen less than the corresponding alkanes. General formula e.g.
or
Cyclohexane
(ii) Aromatic compounds : Organic compounds which contain one or more fused or isolated benzene rings are called aromatic compounds.
CnH2n C2 H 4 C3 H 6
(Ethene) (Propene)
(B) Alkynes : They are also unsaturated hydrocarbons hydrocarbons which contain carbon-carbon triple bond. They contain four hydrogen atoms less than the corresponding alkanes. General formula e.g.
CnH 2n C2 H 2
–
C3 H 4
2
(Ethyne) (Propyne)
21 21
PAGE # 21
Examples : NOMENCLATURE NOMENCLATU RE OF ORGAN ORGANIC IC COMPOUNDS
a
Nomenclature means the assignment of names to organic compounds . There are two main systems of nomenclature of organic compounds (1) Trivial system (2) IUPAC system (International Union of Pure and Applied Chemistry)
M E P B E Propene
Eth -
Note : The name of the compound, in general , is written in
For naming simple aliphatic compounds, the normal saturated hydrocarbons have been considered as the parent compounds and the other compounds as their derivatives obtained by the replacement of one or more hydrogen atoms with various functional groups.
the following sequence(Position of substituents )-(prefixes ) (word root)-(p suffix). (iii) Names of branched chain hydrocarbon : The carbon atoms in branched chain hydrocarbons are
(i) Each systematic name has two or three of the following parts-
present as side chain . These side chain carbon atoms constitute the alkyl group or alkyl radicals. An alkyl group
(A) Word root : The basic unit of a series is word root which indicate linear or continuous number of carbon atoms.
is obtained from an alkane by removal of a hydrogen. General formula of alkyl group = CnH2n+1
(B) Primary suffix : Primary suffixes are added to the word root to show saturation or unsaturation in a carbon chain. (C) Secondary suffix : Suffixes added after the primary suffix to indicate the presence of a particular functional group in the carbon chain are known as secondary suffixes.
An alkyl group is represented by R. e.g. H
(A)
H
–
H
H Methyl
(ii) Names of straight chain hydrocarbons : The name of straight chain hydrocarbon hydrocarbon may be divided into two parts(A) W ord root
C
(B) Primary suffix
(A) Word roots for carbon chain lengths : Chain length
Word root root
Chain length
Word root
C1 C2 C3 C4 C5
Meth-Meth Eth Et hProp Pr op Butt Bu Pent-
C6 C7 C8 C9 C10
HexHexHept He pt-OctOc tNonNo nDec ec--
(B)
–
H
H
H
H
C
C
H H Ethyl
(B) Primary suffix :
(C)
22 22
PAGE # 22
A branched chain hydrocarbon hydrocarbon is named using the
e.g.
following general ge neral IUPAC rules : Rule1: Longest chain rule : Select the longest possible
2 –Methylpentane
continuous chain of carbon atoms. If some multiple
4 –Methylpentane
(Correct)
(W rong)
bond is present , the chain selected must contain the e.g.
multiple bond. (i) The number of carbon atoms in the selected chain
3 –Methylbut –1 – ene
determines the word root .
(Correct)
(ii) Saturation or unsaturation determines the primary suffix (P. suffix). (iii) Alkyl substituents are indicated by prefixes.
CH3
e.g.
e.g.
Prefix : Methyl
CH3 – CH – CH2 – CH – CH3 CH2 – CH2 – CH3
CH3
CH3 – CH2 – C – CH3 CH2
e.g. CH3 – CH – 2 CH – CH2 – CH 3 CH – CH3
e.g.
Word root : HeptP. Suffix : -ane
Prefix : Methyl Word root : ButP. Suffix : –ene
Prefixes : Ethyl, Methyl Word root : PentP. Suffix : -ane
4
CH 3 |
3
2
(W rong)
1
3-Methylbut-1-yne (Correct)
Prefix : Methyl Word root : pentP. Suffix: - ane
e.g. CH3 – CH2 – CH – CH2 – CH3
2 –Methylbut – 3 – ene
1
Rule 2 : Lowest number rule: The chain selected is numbered in terms of arabic numerals and the position of the alkyl groups are indicated by the number
3
4
2-Methylbut-3-yne (Wrong)
Rule 3 : Use of prefixes di, tri etc. : If the compound contains more than one similar alkyl groups,their positions are indicated separately and an appropriate numerical numerical prefix prefix di, tri etc. , is attached to the name of the substituents. The positions of the substituents are separat separated ed by by commas commas.. CH3 5 4 3 2 1 CH3 – CH2– C – CH – CH3
e.g.
CH3 CH3
2,3 - Dimeth Dimethyl ylpen pentan tane e
2,3,3 2,3,3 - Trimeth Trimethylp ylpent entane ane
e.g.
2,3,5 -Trimethylhexane
CH3
CH 3 |
2
2,2,4 - Trimethylpentane
Rule 4 : Alphabetical Alphabetical arrangement of prefixes: If there are different alkyl substituents present in the compound their names are written in the alphabetical order. However, However, the the numerical numerical prefixes such as di , tri etc. , are not considered for the alphabetical order.
of the carbon atom to which alkyl group is attached . (i) The numbering is done in such a way that the
e.g.
substituent carbon atom has the lowest possible number. 3-Ethyl - 2,3-dimethylpentane (ii) If some multiple bond is present in the chain, the carbon atoms involved in the multiple bond should get lowest possible numbers.
Rule 5 : Naming of different alkyl substituents at the equivalent positions : Numbering of the chain is done in s uch a way that the alkyl group which comes first in alphabetical order gets the lower position.
e.g.
2 –Methylbutane (Corre ct )
3 –Methylbutane (W rong)
e.g. 3-Ethyl-4-methylhexane
23 23
PAGE # 23
Rule - 6 : Lowest sum rule According to this rule numbering of chain is done in such a way that the sum of positions of different substituents gets lower value. e.g.
FUNCTIONAL GROUP
An atom or group of atoms in an organic compound or molecule that is responsible for the compound’s characteristic reactions and determines its properties is known as functional group. An organic compound generally consists of two parts (i) Hydrocarbon radical (ii) Functional group
(i)
e.g. Hydrocarbon Hydrocarbon radical •
Functional group
Functional group is the most reactive part of the
molecule. •
Functional group mainly determines the chemical
properties of an organic compound. •
Hydrocarbon radical mainly determines the physical
properties of the organic compound.
W ord root : He Hex Primary suffix : - ane Substituent Substituent : two two methy methyll & one ethyl ethyl grou groups ps IUPAC IUP AC name : 4-Ethyl - 2, 4 - dimethylhexane
(i)
W ord root : Prop P. Suffix : -ane Substit Substituen uentt : two two met methy hyll groups groups IUPAC IUP AC name name : 2, 2 - Dimethylpropane
(i) Hydroxyl group ( – OH) : All organic compounds containing - OH group are known as alcohols . e.g. Methanol (CH3OH) , Ethanol (CH3 – CH2 – OH) etc . (ii) Aldehyde group ( –CHO) : All organic compounds containing –CHO group are known as aldehydes. e.g. Methanal (HCHO), (HCHO), Ethanal (CH (CH3CHO) etc. (iii) Ketone group ( –CO –) : All organic compounds containing –CO – group are known as ketones. e.g. Propanone (CH 3 COCH 3 ), Butanone (CH3COCH2CH3) etc. (iv) Carboxyl group ( – COOH) : All organic compounds containing carboxyl group are called carboxylic acids. e.g. CH3COOH (Ethanoic acid) CH3CH2COOH(Propanoic acid) (v) Halogen group (X = F, Cl, Br, I) : All organic compounds containing –X (F, Cl, Br or I) group are known as halides. e.g. Chloromethane (CH3Cl), Bromomethane (CH3Br) etc .
(ii)
W ord root : But P. Suffix : - ene Substit Substituen uentt : two two met methy hyll groups groups IUPAC IUP AC name name : 2, 3 - Dimethylbut - 1 - ene (iii)
W ord root : Hex P. Suffix : - yne Substit Substituen uentt : one one methy methyll grou group p IUPAC IUP AC name name : 4 - Methylhex - 2 - yne
In case functional group (other (othe r than C = C and C C) is present, it is indicated by adding secondary suffix after the primary suffix. The terminal ‘e’ of the primary suffix is removed if it is followed by a suffix beginning with ‘ a ’ , ‘ e ’ , ‘ i ’ , ‘ o ’ , ‘ u ’ . Some groups like –F, – Cl, – Br and – are considered as substituents and are indicated by the prefixes.
O Some groups like – CHO, – C – , – COOH, and
–
OH
are considered as functional groups and are indicated by suffixes.
24 24
PAGE # 24
Class
Functional Group
General Formula
Carboxylic acid
Prefix Carboxy
Suffix - oic acid
IUPAC Name Alkan kanoic acid
(R = CnH2n+1) Ester
Aldehyde
Carbalkoxy
–
CHO
R – CHO
Ketone
oxo
Alcohol
–
OH
R – OH
Alkenes
C=C
CnH2n
Alkynes
C C
Halides
Formyl or oxo
–
CnH2n
X
(X = F,Cl,Br,I)
2
–
R – X
Hydroxy –
–
Halo
alkyl (R’) - oate
- al - one
Alkyl alkanoate
Alkanal Alkanone
- ol
Alkanol
- ene
Alkene
- yne
Alkyne
–
Haloalkane
Steps of naming of an organic compound
Step 4 :
containing functional group :
The carbon atoms of the parent chain are numbered in such a way so that the carbon atom of the functional
Step 1:
group gets the lowest possible number . In case the
Select the longest continuous chain of the carbon atoms as parent chain. The selected chain must include the carbon atoms involved in the functional groups like – COOH, – CHO, – CN etc, or those which carry the functional groups like – OH, –
–
NH2, – Cl,
NO2 etc.
The number of carbon atoms in the parent chain decides the word root.
functional group does not have the carbon at om, then the carbon atom of the parent chain attached to the functional group should get the lowest possible number. Step 5 : The name of the compound is written as Prefixes - word root - primary suffix - secondary suffix Note :
Step 2 : The presence of carbon - carbon multiple bond decides the primary suffix.
The number of carbon atoms in the parent chain decides the word root.
Step 3 : The secondary suffix is decided by the functional group.
25 25
PAGE # 25
S.No.
C ompound Co
Common name Derived name
IUPAC Name
1
CH3 – OH
Methyl alcohol or Wood spirit
Carbinol
Methanol
2
CH3 – CH2 – OH
Ethyl alcohol
Methyl carbinol
Ethanol
3
CH3 – CH2 – CH2 – OH
n-Propyl al alcohol
Ethyl ca carbinol
1- Pr Propanol
Structure
H
4
Isopropyl alcohol Dimethyl carbinol
H
2 - Propanol
–
C H
5
CH3 – CH2 – CH2 – CH2 – OH
n-Buty n-Butyll alcoh alcohol ol
6
HCOOH
Form ic acid
–
Methanoic acid
Acetic acid
–
Ethanoic acid
CH3COOH
7
n-Prop n-Propyl yl carbino carbinoll
methyl acetic acid
8
CH3 – CH2 – COOH
Propionic acid
9
CH3 – CH2 – CH2 – COOH
Butyric acid
ethyl acetic acid
Valeric acid
n-Propyl acetic acid
H –
C
–
O – H
C H3
1- Butan Butanol ol
Propanoic acid
Butanoic acid
O
10
CH3 – CH2 – CH2 – CH2 – COOH
(iii) CH3 – CH2 – CH2 – NH2 W ord root : Prop Primary suffix : - ane Secondary suffix : - amine IUPAC name : Propan - 1 - amine
(i)
W ord root : HeptPrimary suffix : – ane Functional group : – OH Secondary suffix : – ol IUPAC IUP AC Name :
(iv)
2, 5-Dimethylheptan –1 – ol
(ii) W ord root : Pe P en t Primary suffix : – ene Secondary suffix : – oic acid Position of double : 2nd bond IUPAC name : Pent-2-en-1-oic Pent-2-en-1-oic acid/Pent-2-enoic acid/Pent-2-enoic acid
Pentanoic acid
W ord root Primary suffix Su b s t i t u e n t IUPAC name
: Prop: - ane : nitro(prefix) : 1 - Nitropropane
W ord root Primary suffix Prefix IUPAC name
: B ut : – ane : – chloro : 2 - Chlorobutane
(v)
26 26
PAGE # 26
ISOMERISM
(vi) W ord root Primary suffix Secondary suffix Prefix IUPAC name
: But : – ane : – one : Methyl : 3 - Methylbutan - 2- one
Such compounds which have same molecular formula but different physical and chemical properties are known as isomers and the phenomenon is known as isomerism.
HOMOLOGOUS SERIES
Homologous series may be defined as a series of similarly constituted compounds in which the members possess similar chemical characteristics and the two consecutive members differ in their molecular formula by – CH2.
(i) All the members of a series can be represented by the same general formula. e.g. General formula for alkane series is CnH2n+2 . (ii) Any two consecutive members differ in their formula formu la by a common difference of – CH 2 and differ in molecular mass by 14.
The isomerism in which the isomers differ from each other due to the presence of different carbon chain skeletons is known as chain isomerism. e.g. (i) C4H10 ,
(iii) Different members in a series have a common functional group. e.g. All the members of alcohol family have –OH group . (iv) The members in any particular family have almost identical chemical properties. Their physical properties such as melting point, boiling point, density etc, show a regular gradation with the increase in the molecular mass.
2 - Methylpropane (Isobutane) (ii) C5H12
2 - Methylbutane (Isopentane)
(v) The members of a particular series can be prepared almost by the identical methods.
The different members of a homologous series are known as homologues. 2, 2 -Dimethylpropane (neo - pentane)
e.g. (i) Homologous series of alkanes General formula : CnH2n+2 Value alue of n n=1 n=2 n=3
(iii) C4H8
Molec Mo lecula ularr form formula ula IUP IUPAC nam name e CH4 Methane C2 H 6 Ethane C3 H 8 Propane
(ii) Homologous series of alkenes General formula :CnH2n Value alue of n n=2 n=3 n=4
Molecu Mole cula larr form formul ula a C2 H 4 C3 H 6 C4 H 8
IUP IUPAC name name Ethene Propene But-1-ene
Comm Common on name name Ethylene Propylene - Butylene
(iii) Homologous series of alkynes General formula : CnH2n
CH3 – CH2 – CH = CH2 , But - 1 - ene
Methylpropene
In this type of isomerism, isomers i somers differ in the structure due to difference in the position of the multiple bond or functional group. e.g. (i) C4H8 CH3 – CH2 – CH = CH 2 , CH3 – CH = CH – CH3 But -1 - ene B u t - 2 - e ne
2
–
Value alue of n n=2 n=3 n=4
Molecu Mole cula larr form formul ula a C2 H 2 C3 H 4 C4 H 6
IUP IUPAC Comm Common on name name Ethyne Acetylene Propyne Methyl acetylene But -1-yne Ethyl acetylene
(ii) C3H8O CH3 – CH2 – CH2 – OH , Propan-1-ol
CH3 – CH – CH3 OH Propan-2-ol
27 27
PAGE # 27
only. These compounds are open chain compounds which are also addressed as Acyclic compounds. Alkanes have the general formula CnH2n+2 .The carbon atoms in alkanes are in a state of sp3 hybridization, i.e. the carbon atoms have a tetrahedral geometry.
In this type of isomerism, isomers differ in the structure due to the presence of different functional groups. e.g. (i) C3H8O CH3 – CH2 – O – CH3
(i) Alkanes of no. of carbon atoms C1 to C4 are gases. Carbon atoms C5 to C17 are liquids and C18 & onwards are solids. (ii) Alkanes are colourless and odourless. (iii) They are non-polar in nature, hence they dissolve only in non-polar solvents like benzene, carbon tetrachloride etc. (iv) Boiling point of alkanes increases as their molecular weight increases.
CH3 – CH2 – CH2 – OH
Methoxy ethane
Propan-1-ol
(ii) C4H6 CH3 – CH2 – C CH But - 1- yne
CH2 = CH – CH = CH 2 Buta - 1, 3 - diene [or 1, 3 - Butadiene ]
Note :
ALKANES
Alkanes are unaffected by most chemical reagents
Alkanes are aliphatic hydrocarbons having only C – C single covalent bonds. These are also known as saturated hydrocarbon as they contain single bond
M o l c u l a r F o rm u l a
S tr u c tu r e
CH 4
CH 4
C2 H6
CH 3 –CH 3
C3 H8
CH 3 –CH 2 –CH 3
C4 H10
CH 3
CH
–
CH 3
C6 H14
–
–
CH
–
2
CH
CH
3
–
C H – CH
CH
2
–
CH
3
–
CH
–
CH
3
–
–
3
CH
CH
–
CH 3
–
CH CH
C H – CH
CH CH
E t ha ne P ro p a n e
n-Butane Isobutane
Butane 2 –Methylpropane
3
n-Pentane
Pentane
Isopentane
2 –Methylbutane
Neopentane
2,2 –Dimethylpropane
2–
CH
2 –
–
CH
3
2
CH
n-Hexane
Hexane
Isohexane
2 –Methylpentane
–
3 –Methylpentane
–
2,3 –Dimethylbutane
Neohexane
2,2 –Dimethylbutane
3
2
–
CH
3
3
–
CH
3
3
3
C – CH CH
M et h an e
3
2
CH
–
2
2
CH
CH
–
CH | 3 C – CH 3 | CH 3
–
–
2
CH – CH 2 | CH 3
CH
I UP A C Na m e
3
3
3
CH 3
–
CH 3
CH
CH
P ro p a n e
CH 2
–
CH
–
3
E t ha n e
CH 2
–
3
CH
T ri v i a l Na m e M e t ha ne
CH 3 – CH | CH
C5 H12
and hence are known as paraffins (parum-little, affinis affinity).
2
–
CH
3
3
28 28
PAGE # 28
Note : The C – C bond distance in alkanes is 1.54 Å and the bond energy is of the order of 80 Kcal per mole.
(vi) Laboratory Method : Methane is prepared in the laboratory by heating a mixture of dry sodium acetat e and soda lime in a hard glass tube as shown in figure. It is a decarboxylation reaction.
METHANE
It is a product of decomposition of organic matter in absence of oxygen. It is found in coal mines (hence the name damp fire), marshy places (hence the name marsh gas) and the places where petroleum is found. Note :
Hard glass tube Delivery tube
Sodium acetate and soda lime
Gas jar Cork Bubbles of methane gas
Burner
Methane is a major constituent of natural gas. Gas
Methane is a colourless gas with practically no smell and is almost insoluble in water. It melts at – 183º C and boils at – 162ºC. Methane has tetrahedral geometry in which H – atoms are situated at four corners of the regular tetrahedron. Bond angle is 109º28’. It has sp 3 hybridisation.
Trough Beehive shelf Water
Iron stand
Preparation of methane gas
Methane, so formed is collected by downward displacment of water. This gas contains some hydrogen, ethylene etc. as impurities which can be removed by passing the impure gas through alkaline potassium permanganate solution. (i) Combustion : (A) Methane burns with explosive violence in air forming carbon dioxide and water. CH4 + 2O2 CO2 + 2H2O + Heat (B) In the presence of insufficient supply of oxygen. 2CH4 + 3O2 2CO + 4H2O + Heat
Tetrahedral
(ii) Halogenation : (A) In direct sunlight (i) Direct synthesis :
C
+
Ni 500ºC
2H2 Hydrogen
Carbon
CH4 + 2Cl2
CH4
CO2 + 4H2
Ni powder 300ºC
Methane
CH4 + H2O
Ni powder 300ºC 12H2O 12H Water
+
2H
Methane
3CH4 Methane
+
4Al(OH) 4Al (OH)3 Aluminium hydroxide
Zn –Cu Couple CH4 + H2O Methane
H Hydrogen iodide
(v) Reduction of methanol or formaldehyde or formic acid with H CH3OH
+
2H
Red P
HCHO Methanal
C H4
+
+
H2 O
Methane
Methanol +
4H
Red P
Cl2 CHCl 3 Methyl chloride
Cl2 CHCl 2 2 Methylene dichloride
Cl2 CHCl3 Chloroform
CCl4 Carbon tetrachloride
Fluorine forms similar substitution products in the presence of nitrogen which is used as a diluent because of high reactivity of fluorine. Bromine vapours react very sluggishly while iodine vapours do not react at all.
CH4
CH4 + 2H2O
(iv) Reduction of methyl iodide : CH3 – Methyl iodide
Cl2 CH4 Methane
(iii) Nitration :
(iii) Hydrolysis of aluminium carbide : Al4C3 + Aluminium carbide
C + 4HCl
(B) In diffused light
(ii) Sabatier and Senderens reductive method : Methane can be prepared by passing carbon monoxide or carbon dioxide and hydrogen over finely powdered nickel catalyst at 300ºC. CO + 3H2
h
CH 4 Methane
+
2
+
H 2O
+
HO –NO 2 Nitric ac acid
400 ºC 10 atm.
CH3–NO2
+
H2O
Nitromethane
(iv) Catalytic Air oxidation : This is a method for commercial production of methanol. When a mixture of methane and oxygen in a ratio of 9: 1 by volume is passed through a heated heate d copper tube at 20 0 ºC and at a pressure of 100 atmospheres, methanol is formed. CH4 + 1/2 1/ 2 O2 CH 3OH Methane Methanol (i) Alkanes are used directly as fuels . (ii) Certain alkanes, such as methane, are used as a source of hydrogen. (iii) The carbon obtained in decomposition of alkanes is in very finely divided state and is known as carbon black. This is used in making printer’s ink, paints, boot polish and blackening of tyres. (iv) Alkanes are used as starting materials for a number of other organic compounds e.g. methanol, methyl chloride, methylene dichloride etc.
29 29
PAGE # 29
ALKENES
Alkenes are the simplest unsaturated aliphatic hydrocarbons with one carbon - carbon double bond. Alkenes have general formula CnH2n. The carbon atoms connected by the double bond are in a state of sp2 hybridisation and this part of molecule is planar. A double bond is composed of sigma () and a pi ( ) bond. Alkenes are also called olefines (oil forming) becuase they form oily products with halogens. R – CH = CH CH2 + Br 2
R – CH – CH2
(i) Ethylene is mainly used in the manufacture of ethanol, ethylene oxide and higher 1-alkenes. Ethylene is used for ripening of fruits. It is also used for preparation of mustard gas. [Cl – CH2 – CH2 – S – CH2 – CH2 – Cl] (ii) Polythene from ethylene, teflon from tetra fluoroethylene and polystyrene from styrene are used as plastic materials. Acrilon or orlon obtained from vinyl cyanide is used for making synthetic fibres. ETHENE
Br Br (Oily liquid)
Ethene occurs in natural gas, coal gas and wood gas. It is also formed during the cracking of high boiling petroleum fractions.
(i) Alkenes of C2 to C 4 are gases. Alkenes of carbon atoms C5 to C 14 are liquids and C14 and onwards are solids. (ii) Ethene is colourless gas with faint sweet smell. All other alkenes are colourless and odourless. (iii) Alkenes are insoluble in polar solvents like water, but fairly soluble in non-polar solvents like benzene, carbon tetrachloride etc.
Ethene is a colourless gas (B.P. = –105ºC). It is very sparingly soluble in water but dissolves in acetone, alcohol etc. It burns with smoky flame. Ethene has trigonal planar geometry. Bond angle is 120º. It has sp2 hybridisation.
(iv) Boiling point of alkenes increases with increase in molecular mass. Bond length of C = C is 1.34 Å . The energy of the double bond is 142 Kcal mol 1, which is less than twice the energy of a single bond i.e. 80 Kcal mole-1. –
This indicates that a pi ( ) bond is weaker than a sigma ( ) bond. (c) Preparation of Ethene :
(i) By dehydration of alcohol (Lab. method) : Molecular formula
Structure
Trivial Name
IUPAC Name
C2H4
H2C = CH2
Ethylene
Ethene
C3H6
CH3 – CH = CH2
Propylene
Propene
C4C8
CH3 – CH2 – CH = CH2
-Butylene
1-Butene
-Butylene (cis)
2-Butene (cis)
-Butylene (trans)
2-Butene (trans)
CH3 – C – H CH3 – C – H CH3 – C – H H – C – CH3 CH3 – C = CH2
Isobutylene
2-Methyl propene
CH3 – CH2 – OH
Conc.. H2SO4 Conc
CH2 = CH2 + H2O
165 – 170ºC
Ethene
Ethanol (ii) By cracking of kerosene : Cracking
(CH H2)4– CH3 CH3 – (C
CH2 = CH CH2 CH3 – CH2 – CH2 – CH3 + CH Butane
n-Hexane
Ethene
(iii) From alkyl halides (Dehydrohalogenation) : KOH CH2 – CH2 + KOH (Alcoholic) H X
CH2 = CH2 + KX KX + H2O (Here X = Halogen) Ethene
Ethyl halide
CH3 C5H10
CH2 = CH CH (C (CH2)2 CH3 CH3CH = CH CHCH CH2CH3
CH3 – CH – CH = CH2
–
1-Pentene
–
2-Pentene (cis and trans)
–
3-Methyl-1-butene
–
2-Methyl-2-butene
(i) Addition of halogens : CH2 = CH2 Ethene
+
Cl2 Chlorine
CCl4
Cl Cl 1,2-Dichloroethane (Ethylene dichloride)
CH3 CH3 – C = CH – CH3 CH3 CH2 = C – CH2 – CH3 CH3
–
2-Methyl-1-butene
CH2 = CH2 Ethene
CH2– CH2
CCl4 Br2 Bromine (red-brown colour) +
CH2 – CH2 Br Br 1,2-Dibromoethane (colourless)
30 30
PAGE # 30
Note : Addition of bromine on alkenes in presence of CCl4 is the test for unsaturation. (ii) Addition of halogen acids (Hydrohalogenation) : CH2 = CH CH2 + HCl HCl
CH2 – CH2
Ethene
Cl H Chloroethane
ALKYNES
Alkynes are unsaturated aliphatic hydrocarbons having a carbon-carbon triple bond. Alkynes have general formula CnH2n 2. Thus, they have two hydrogen atoms less than an alkene and four hydrogen atoms less than an alkane with same number of carbon atoms. A triple bond is composed of one sigma () and two pi () bonds. The carbon atoms connected by a triple bond are in state of sp hybridisation. –
(iii) Hydrogenation :
CH2 = CH2 + H2 Ethene
Ni or Pt High T& P
CH3 – CH3
(i) Alkynes of carbon atoms C2 to C4 are gases. Alkynes of carbon atoms C5 to C12 are liquids.Alkynes of C13 & onwards are solids.
Ethane
(iv) Combustion : C2H4 + 3O 3O2 Ethene
2C O2
(ii) Alkynes are colourless and odourless, but ethyne has characteristic odour.
2H2 O + Heat
+
(iii) Boiling point and solubilities in water are relatively higher than those of alkanes and alkenes.
(v) Addition of oxygen :
(iv) Alkynes are weakly polar in nature. (v) Alkynes are lighter than water and soluble in nonpolar solvents. (vi) Boiling point of alkynes increases with the increase in molecular mass.
(vi) Polymerisation :
nCH nC H2 = CH CH2
High T & High P
–
Note : The bond energy of a triple bond is 190.5 Kcal per mole, which is less than thrice the energy of a single () bond.
(CH2– CH2 –)n Polyethene
Ethene
M o l e c u l a r fo r m u l a
S tr u ctu r e
C2 H2
H
C3 H4
CH 3 – C
C4 H6
CH 3 –CH 2 – C
C5 H8
–
C
C – H C – H
CH
CH 3 – CH 2 – CH 2 – C
CH3
–
CH
–
C
CH
CH
D e r i v e d Na m e
I UP AC n a m e
A c et y l e ne
E t hy ne
M et hy l ac et y lene
P ro p y n e
Ethyl acety lene lene
1 –Butyne
Dimethyl acetylene
2 – Butyne
n-Propyl acetylene
1 – Pentyne
Ethyl methyl acetylene
2-Pentyne
Isopropyl Isopropyl acety lene
3-Methyl- 1-butyne
C H3
ETHYNE
It is also known as acetylene. Acetylene is the first member of alkyne series and has a linear geometry. It has sp hybridisation.The carbon-carbon triple bond distance and carbon-hydrogen bond distance have been found to be 1.20 Å and 1.06 Å respectively. The carbon-carbon hydrogen bond angle is 180º. Linear
31 31
PAGE # 31
It is a colourless gas which is slightly solub le in water. water. Pure ethyne has ethereal odour. od our. Acetylene burns with luminous flame like aromatic compounds. This is a highly exothermic reaction.
TEST FOR ALKANES, ALKENES AND ALKYNES
(i) Bromine water test: It does not decolourise the bromine water.
Note : The temperature of oxyacetylene flame is about
(ii) Baeyer’s test: It does not, react with Baeyer’s reagent (alkaline solution of KMnO4).
3000ºC and is used for welding and cutting steel.
(iii) Silver nitrate Test: No reaction
(i) From carbon and hydrogen (Direct synthesis ) : When an electric arc is struck between carbon (graphite) rods in an atmosphere of hydrogen, acetylene is formed.
1200ºC
2C + H2
(i) Bromine water test: It decolourises the orange colour of Bromine water. H
C2H2
H
Ca( a(OH OH))2 + C2H2
H
CCl4
+ Br2 Bromine water (red-brown (red-brow n colour)
H Ethene
(ii) From calcium carbide (Lab. Method) :
CaC2 + 2H2O Calcium carbide
H C=C
C – C
H
H Br
Br H 1,2-Dibromoethane (Colourless)
(ii) Baeyer’s test: It decolourises the purple colour of Baeyer’s reagent.
Calcium Ethyne hydroxide
H
(iii) Dehydrohalogenation of dihaloalkanes :
H C=C
H
H
H2O + [O [O]]
H
H – C – C – H
H
OH OH Ethylene glycol
Ethene
(iii) Silver nitrate Test: No reaction
(i) Bromine water test : It decolourises the Br2 water.
(i) Addition of halogens :
H
H – C C – H + Br 2 Ethyne
Cl H2 C = CH
Ethyne
HCl
Cl Chloroethene (Vinyl chloride)
H2 Ni
Ethyne
HC CH
H2C = CH CH2 Ethene
H2
H2/Ni
Ethyne
(iii) Silver nitrate Test : It gives white precipitate H – C C – H + 2 [Ag (NH3 )2 ]NO3 Ethyne Tollen's reagent
CH3 – CH3 Ethane
1.
4CO2 + 2H2O + Heat
Ethyne
The IUPAC IUPAC name of the compound having the t he formula
(A) 3,3,3-Trimethyl -1-propane (B) 1,1,1-Trimethyl-1-butene (C) 3,3-Dimethyl-1-butene (D) 1,1 –Dimethyl- 3 -butene
H
C – Ag + 2NH 2NH4NO3 + 2NH 2NH3
EXERCISE
(v) Polymerisation :
Fe
Ag – C
White ppt.
(CH3)3 CCH = CH 2 is -
(iv) Combustion : 2C2H2 + 5O2 Ethyne
3HC CH
Br Br 1,1,2,2-Tetrabromoethane (Colourless)
1,2-Dibromoethene
Cl 1,1-Dichloroethane (Gem dihalide)
H2C = CH2 Ethene
Pd/BaSO4
Br
H
Br – C – C – Br
H3 C – CH
(iii) Hydrogenation : HC CH
Br
Bromine water
H
Br2
(ii) Baeyer’s test : It also decolourises the purple colour of alkaline KMnO4 .
(ii) Addition of Halogen acid : HC CH + HCl
H C=C
H
H
H
H H Benzene
or (C6H6)
2.
Which of the following is not an example of aromatic compound ? (A) Benzene
(B) Naphthal ene
(C) Cyclobutane
(D) Phenol
32 32
PAGE # 32
3.
4.
5.
The IUPAC name of the following compound is CH2 = CH – CH (CH3)2 (A) 1,1-Dimethyl -2-propene (B) 3-Methyl-1-butene (C) 2-Vinylpropane (D) 1-Isopropylethylene Which of the following is an alkyne ? (A) C4H8 (B) C5H8 (C) C7H19 (D) None of these The IUPAC name of the following compound is -
C2H5 CH3 – CH – C = CH2
(A) 3-Ethyl-2-methylbut-3-ene (B) 2-Ethyl-3- methylbut -1-ene (C) 2-Methyl-3-ethylbut-3-ene (D) 3-Methyl-2-ethylbut-1-ene 6.
The first organic compound synthesized in the laboratory was (A) urea (B) gl ucose (C) alcohol (D) None of these
7.
Propane is an (A) acyclic compound (B) open chain compound (C) alipthatic compound (D) All of these
8.
The scientist who gave vital force theory was (A) Berzelius (B) Avogadro (C) W ohler (D) Lavoisier Which one of the following is not an organic compound ? (A) Hexane (B) Urea (C) Ammonia (D)Ethyl alcohol
10. Vast number of carbon compounds is due to the fact that carbon has (A) variable valency (B) property of catenation (C) great chemical affinity (D) None of these
O ||
11. The IUPAC IUPAC name for fo r CH3 – C – H is (A) Acetal (C) Ethanal
(B) Methanal (D) Acetaldehyde
12. The IUPAC IUPAC name of compound H CH3 – C – CH2 – CH3 is COOH
(A) Butan -3- oic acid (B) Butan -2- oic acid (C) 3-Methylbutanoic acid (D) 2-Methylbutanoic acid
14. How many chain isomers (non- cyclic aliphatic) could be obtained from the alkane C6H14 ? (A) 6 (B) 5 (C) 4 (D) 3 15. The isomerism exhibited by n-propyl alcohol and isopropyl alcohol is (A) chain isomerism (B) position isomerism (C) functional isomerism (D) None of these 16. Any two consecutive members in a homologous series differ in molecular mass by(A) 8 (B) 14 (C) 24 (D) 12
CH3
9.
13. The functional group, present in CH3COOC2H5 is (A) ketonic (B) aldehydic (C) ester (D) carboxylic
17. The IUPAC name ofCH3 – C(CH 3) (OH) CH2 – CH(CH 3) CH 3 is (A) 2,4-Dimethylpentan -2-ol (B) 2,4- Dimethylpentan-4-ol (C) 2,2-Dimethylbutane (D) Isopentanol 18. The IUPAC name of (CH 3)2 CHCH2 CH2 Br is(A) 1-Bromopentane (B) 2-Methyl-4-bromopentane (C) 1-Bromo -3- methylbutane (D) 2-Methyl-3-bromopentane 19. Which of the following does not belong to homologous series of alkanes ? (A) C2H6 (B) C3H4 (C) C4H10 (D) C5H12 20. Isomers have (A) same molecular formula & same structure. (B) different molecular formula & different structure. (C) same molecular formula & different structure. (D) different molecular formula & same structure. 21. Which of the following properties is not true regarding organic compounds ? (A) They are generally covalent compounds. (B) They have high melting and boiling points. (C) They are generally insoluble in water. (D) They generally show isomerism. 22. The nature of linkage in organic compounds is generally (A) ionic bond (B) covalent bond (C) co-ordinate covalent (D) metallic bond 23. Which of the following statements is incorrect ? The members of the homologous series of alkanes (A) are all straight chain compounds. (B) have the general formula CnH2n+2. (C) (C) hav have e simi similar lar chemic chemical al prop proper ertie ties s . (D) show a regular gradation of physical properties.
33 33
PAGE # 33
24. Which of the following pairs is an example of chain isomer ? (A) CH3 – CH2 – OH & CH3OCH3 (B) CH3 – CH2 – CHO & CH3 – CO-CH3
(A) mesitylene
(B) benzene
(C) butenyne
(D) None
33. The general molecular formula of alkynes is -
(C)CH3 –CH2 –CH2 –CH2 –CH3 & (D) All of the above
(A) CnH2n
(B) CnH2n
(C) CnH2n
(D) CnH2n+2
4
–
2
–
25. Which of the following is a functional isomer of CH3 – COOH ? (A) CH3 – CH2 – OH
32. Ethyne on passing through a red hot iron tube gives -
34. Which of the following is not a pair of homologues ? (A) Ethylacetylene and Dimethylacetylene (B) Methylacetylene and Ethylacetylene (C) 1-Butyne and 2-Pentyne
(B)
(D) 1-Pentyne and 3-Hexyne 35. Ethyne is obtained by dehydrobromination of -
(C)
(D) All of these
26. Which of the following forms a homologous series ? (A) Ethane, Ethylene, Acetylene (B) Ethane, Propane, Butanol (C) Methanal, Ethanol, Propanoic acid (D) Butane, 2-Methylbutane, 2,3-Dimethylbutane 27. Homologous have the same (A) empirical formulae (B) molecular formulae (C) chemical properties (D) physical properties
(A) CHBr = CHBr
(B) CH 3CHBr 2
(C) CH3 –CH2Br
(D) None of these
36. The temperature of oxy-acetylene flame is (A) 200ºC
(B) 1600ºC
(C) 2000ºC
(D) > 2500º
37. Ethyne is isoelectronic with-
28. Write down the general formula of homologous series whose third homologue is C4H6 ? (A) CnH2n 2 (B) Cn H2n + 2 (C) Cn + 1 H2n 2 (D) Cn H2n –
–
29. The general formula of saturated hydrocarbons is (A) CnH2n (B) CnH2n+2 (C) CnH2n 2 (D) CnH2n+1 –
30. The third member of methyl ketone homologous series is (A) Acetone (B) 2 –Butanone (C) 2 –Pentanone (D) 3 –Pentanone. 31. The values of bond energies of single, double and triple bonds are in the order (A) C – C > C = C > C C (B) C = C > C – C > C C (C) C C > C = C > C – C (D) C = C > C C > C – C
(A) chlorine
(B) oxygen
(C) nitrogen gas
(D) CO2
38. C C bond length is (A) 1.54 Å
(B) 1.20 Å
(C) 1.34 Å
(D) 1.39 Å
39. Which of the following gives silver nitrate test ? (A) Methane
(B) Ethene
(C) Ethyne
(D) All
40. Which of the following does not decolourise bromine water ? (A) Alkanes on only
(B) Alkenes only
(C) Alkynes only
(D) (B) and (C) both
41. Unsaturated fatty acids contain [IJSO-State-I/2011] (A) atleast one double bond (B) two double bonds (C) more than two double bonds (D) no double bond
34 34
PAGE # 34
Pre-requisite : Before going through this chapter, cha pter, you you should be thorough with the basic concepts of the same chapter explained in IX NCERT. NCERT.
Logarithm to any base a (where a > 0 and a (i)
LOGARITHM
If a is a positive real number, other than 1 and x is a rational number such that ax = N, then x is the logarithm of N to the base a. If ax = N then loga N = x. [ Remember N will be +ve] ‘
Fundamental Laws of Logarithm :
’
1 ).
loga a = 1
(ii) loga 0 = not defined
[As an = 0 is not possible, where n is any number] (iii) loga ( –ve no.) = not defined.
[As in loga N, N will always be (+ ve)] (iv) loga (mn) = loga m + logan
There are two systems of logarithm which are generally used. (i) Common logarithm : In this system base is always taken as 10. (ii) Natural logarithm : In this system the base of the logarithm is taken as e . Where e is an irrational number lying between 2 and 3. (The approximate value of e upto two decimal places is e = 2.73) ‘
’
‘
[Where m and n are +ve numbers] m (v) loga = logam – logan n (vi) loga(m)n = n logam
’
(vii) logam
Some Useful Results : (i) If a > 1 then (a) loga x < 0 [for all x satisfying satisfying 0 < x < 1] (b) loga x = 0 for x = 1 (c) loga x > 0 for for x > 1
(d) x > y function.
log m b log a b
(viii) logam . logma = 1 (ix) If a is a positive real number and n is a positive ‘
’
‘
’
rational number, then
loga x > loga y i.e. logax is an increasing
aloga n n (x) If a is a positive real number and n is a positive ‘
Graph of y = loga x, a > 1
’
’
rational number, then
y
logaq np p log n a
q
y = log logax, a > 1 0
x'
‘
(1,0)
(xi) ploga q qloga p
x
(xii) logax = logay x = y y'
(ii) If 0 < a < 1, then (a) loga x < 0 for all x > 1 (b) loga x = 0 for x = 1 (c) logax > 0 for all all x satisfy satisfying ing 0 < x < 1
(d) x > y function.
logax < loga y i.e. loga x is a decreasing
Graph of y = loga x, 0 < a < 1. y
Sol.
(1,0) 0
x
log3a = 4 a = 34
a = 81. Ex.2 Find the value of log
9 27 3 – log log 8 32 4
Sol. Given :
log
y = logax, 0 < a < 1.
x'
Ex.1 If log3a = 4, find value of a.
3 9 27 3 9 27 – log log log log 4 8 32 4 8 32
9 32 3 log 8 27 4 = log1 = 0.
[ loga1 = 0]
y' PAGE # 3535
Ex.3 If 2log4x = 1 + log4(x – 1), find the value of x. Sol. Given
FACTORS FA CTORS AND A ND MUL M ULTIPLES TIPLES
2log4x = 1 + log4(x – 1) Factors : a is a factor of b if there exists a relation ‘
log4x2 – log4(x – 1) = 1 x2 log4 =1 x – 1
’
‘
’
such that a × n = b, where n is any natural number. ‘
’
x2 41 = x – 1
1 is a factor of all numbers as 1 × b = b.
x2 = 4x – 4
x2 – 4x + 4 = 0
(x – 2)2 = 0
x = 2.
Factor of a number cannot be greater than the number (in fact the largest factor will be the number itself). Thus factors of any number will lie between 1 and the number itself (both inclusive) and they are limited.
Ex.4 Evaluate : 3 2
–
Sol. Given 3 2
–
log 3 5
log3 5
.
= 3 2. 3
= 9. 3
–
log 3 5
‘
log3 5 –
1
9 . 5
’
‘
’
of the type b × n = a . Thus the multiples of 6 are 6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.
1
–
=9×5 =
Multiples : a is a multiple of b if there exists a relation
[ am + n = am.an]
The smallest multiple will be the number itself and the number of multiples would be infinite.
NOTE :
To understand what multiples are, let’s just take an Ex.5 If A = log 27625 + 7
log11 13
log 7 and B = log9125 + 13 11 ,
then find the relation between A and B. Sol. A = log27625 + 7
log11 13
4
= log 33 5 + 7
log11 13
example of multiples of 3. The multiples are 3, 6, 9, 12, .... so on. We find that every successive multiples appears as the third number after the previous. So if one wishes to find the number of multiples of 6 less than 255, we could arrive at the number through
or, A =
4 log 13 log35 + 7 11 3
....(i)
log 7 and,B = log9125 + 13 11
or, B = log 32 5 3 + 7
A –
3 4 log35 = B – log35 2 3 4
3
log35 <
3 2
If one wishes to find the multiples of 36, find
log11 13
3 log 13 or, B = log35 + 7 11 2 By (i) and (ii) we have,
log35
255 = 42 (and the remainder 3). The remainder is of 6 no consequence to us. So in all there are 42 multiples. 255 =7 36
(and the remainder is 3). Hence, there are 7 multiples of 36. ...(ii)
Ex.7 How many numbers from 200 to 600 are divisible by
4, 5, 6 ? Sol. Every such number must be divisible by L.C.M. of (4, 5, 6) = 60. 600 60
–
200 60 = 10 – 3 = 7.
Such numbers are 240, 300, 360, 420, 480, 540 and 600.
A < B. Ex.6 Find the value of log25125 – log84 Sol. Given, log25125 – log84
= log 5 2 5 3 – log 23 2 2
Clearly, there are 7 such numbers. Factorisation : It is the process of splitting any number
into a form where it is expressed only in terms of the most basic prime factors. For example, 36 = 22 × 3 2. It is expressed in the
=
3 2 log55 – log2 2 2 3
factorised form in terms of its basic prime factors. Number of factors : For any composite number C,
3 2 = – 2 3 5 = . 6
[ loga a 1 ]
which can be expressed as C = ap × bq × cr ×....., where a, b, c ..... are all prime factors and p, q, r are positive integers, the number of factors is equal to (p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 3 2. So the factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.
PAGE # 3636
Ex.8 If N = 123 × 34 ×52, find the total number of even factors of N. Sol. The factorised form of N is (22 × 31)3 × 34 × 52 26 × 37 × 52. Hence, the total number of factors of N is (6 + 1) (7 + 1) (2 + 1) = 7 × 8 × 3 = 168. Some of these are odd multiples and some are even. The odd multiples are formed only with the combination of 3s and 5s. So, the total number of odd factors is (7 + 1) (2 + 1) = 24. Therefore, the number of even factors is 168 – 24 = 144.
(i) No. of closed lockers = No. of non-perfect square numbers from 1 to 1000 = 1000 – 31 = 969. (ii) Upto 500 students they can go to two or more than two lockers, while the rest 500 can go to only one locker. (iii) The 31 perfect squares ( the last being 312 = 961) will be open while the lockers from 971 to 1000 is yet to be accessed last time so they all are open. The total being = 31 + 30 = 61 (iv) The no. of students that have gone to locker no. 840 is same as the no. of factors of 840. 840 = 23 × 3 × 5 × 7. So, the no. of factors = (3 + 1) (1 + 1) (1 + 1) (1 + 1) = 32. HCF AND LCM
Ex.9 A number N when factorised can be written N = a4 × b3 × c7. Find the number of perfect squares which are factors of N (The three prime numbers a, b, c > 2). Sol. In order that the perfect square divides N, the powers of ‘a’ can be 0, 2 or 4, i.e. 3. Powers of ‘b’ can be 0, 2, i.e. 2. Power of ‘c’ can be 0, 2, 4 or 6, i.e. 4. Hence, a combination of these powers given 3 × 2 × 4 i.e. 24 numbers. So, there are 24 perfect squares that divides N. Ex.10 Directions : (i to iv) Answer the questions based on the given information. There are one thousand lockers and one thousand students in a school. The principal asks the first student to go to each locker and open it. Then he asks the second student go to every second locker and close it. The third student goes to every third locker, and if it is closed, he opens it, and it is open, he closes it. The fourth student does it to every fourth locker and so on. The process is completed with all the thousand students. (i) How many lockers are closed at the end of the process ? (ii) How many students can go to only one locker ? (iii) How many lockers are open after 970 students have done their job ? (iv) How many student go to locker no. 840 ? Sol. Sol. (i to iv) : Whether the locker is open or not depends on the number of times it is accessed. If it is accessed odd number of times, then it is open while if it is accessed even number of times then it is closed. How many times a locker will be accessed depends on the locker no. If it contains odd number of factors, then it will be open and if it contains even number of factors. Then it will be closed. We know that a perfect square contains odd number of factors while a non-perfect square contains even number of factors. Thus the lockers with perfect square number will be open and the number of these perfect squares from 1 to 1000 determines the no. of open lockers.
LCM (least Common Multiple) : The LCM of given numbers, as the name suggests is the smallest positive number which is a multiple of eac h of the given numbers HCF (Highest Common factor) : The HCF of given numbers, as the name suggests is the largest factor of the given set of numbers.
Consider the numbers 12, 20 and 30. The factors and the multiples are Given numbers
Factors
1, 2, 3, 4, 6, 12 1, 2, 4, 5, 10, 20 1, 2, 3, 5, 6, 10, 15, 30
12 20 30
Multiples
12, 24, 36, 48, 60, 72, 84, 96, 108, 120.... 20, 40, 60, 80, 100, 120..... 30, 60, 90, 120....
The common factors are 1 and 2 and the common multiples are 60, 120... Thus the highest common factor is 2 and the least common multiple meaning of HCF it is the largest number that divides all the given numbers. Also since a number divides its multiple, the meaning of LCM is that it is the smallest number which can be divided by the given numbers.
HCF will be lesser than or equal to the least of the numbers and LCM will be greater than or equal to the greatest of the numbers.
Ex.11 Find a number greater than 3 which when divided by 4, 5, and 6 always leaves the same remainder 3. Sol. The smallest number which, when divided by 4, 5 and 6, leaves leaves the remainder 3 in each case is LCM (4, 5 and 6) + 3 = 60 + 3 = 63. Ex.12 In a school 437 boys and 342 girls have been divided into classes, so that each class has the same number of students and no class has boys and girls mixed. What is the least number of classes needed? Sol. We should have the maximum number of students in a class. So we have to find HCF (437, 342) = 19. HCF is also the factor of difference of the number.
437 342 + 19 19 = 23 + 18 = 41 classes.
Number of classes = `
PAGE # 3737
For any two numbers x and y, x × y = HCF (x, y) × LCM (x, ( x, y) .
DIVISIBLITY
Division Algorithm : General representation of result
HCF and LCM of fractions :
is, is ,
LCM of numerators LCM of fractions = HCF of deno min ators
Dividend Re mainder Quotient Divisor Divisor
HCF of numerators HCF of fractions = LCM of deno min ators Make sure the fractions are in the most reducible form. Ex.13 Find the H.C.F. H.C.F. and L.C.M. of
H.C.F. of given gi ven fractions = Sol. H.C.F.
Dividend = (Divisor × Quotient ) + Remainder Ex.18 On dividing 15968 by a certain number, the quotient
8 16 2 10 , , and . 9 27 81 3
H.C.F. of ( 2, 8,16,10) 2 = , L.C.M. of (3, 9, 81, 27) 81
is 89 and the remainder is 37. Find the divisor. Dividend Re mainder 15968 37 Quotient 89 = 179.
Sol. Divisor =
L.C.M. of ( 2, 8,16,10) 80 L.C.M. of given fractions = = . H.C.F.of (3, 9, 81, 27) 3
(i) (xn – an) is divisible by (x – a) for all the values of n. (ii) (xn – an) is divisible by (x + a) and (x – a) for all the
numbe r which when divided by 6, 7, 8, 9 Ex.14 Find the least number and 10 leaves remainder 1. Sol. As the remainder is same Required number = LCM of divisors + Remainder = LCM (6, 7, 8, 9, 10) +1
= 2520 + 1 = 2521. Ex.15 Six bells start tolling together and they toll at intervals
of 2, 4, 6, 8, 10, 12 sec. respectively, find (i) after how much time will all six of them toll together ? (ii) how many times will they toll together in 30 min ? Sol. The time after which all six bells will toll together must be multiple of 2, 4, 6, 8, 10, 12. Therefore, required time = LCM of time intervals. = LCM (2, 4, 6, 8, 10, 12) = 120 sec. Therefore after 120 s all six bells will toll together. After each 120 s, i.e. 2 min, all bell are tolling together. 30 1 Therefore in 30 min they will toll together 2 = 16 times 1 is added as all the bells are tolling together at the start also, i.e. 0th second. Ex.16 LCM of two distinct natural numbers is 211. What is their HCF ? Sol. 211 is a prime number. So there is only one pair of distinct numbers possible whose LCM is 211,
i.e. 1 and 211. HCF of 1 and 211 is 1. Ex.17 An orchard has 48 apple trees, 60 mango trees and 96 banana trees. These have to be arranged in rows
such that each row has the same number of trees and all are of the same type. Find the minimum number of such rows that can be formed. Sol. Total number of trees are 204 and each of the trees are exactly divisible by 12. HCF of (48, 60, 96).
204 = 17 such rows are possible. 12
NOTE :
even values of n. (iii) (xn + an) is divisible by (x + a) for all the odd values of n. Test of Divisibility : No .
D i vi si b l i ty T e st
2
U n i t d i gi gi t s h o u l d b e 0 or or e ve ve n
3
The sum of digits digits of no. should be divisible divisible by 3
4
The no formed formed by last last 2 digits digits of given given no. should be divisible divisible by 4.
5
U n itit d ig ig it it s h ou ou ld ld b e 0 o r 5. 5.
6
N o s h o u ld ld b e d iv ivi s i b le le b y 2 & 3 b ot ot h
8
The number formed formed by last 3 digits digits of given given no. should be divisible divisible by 8.
9
Sum of digits digits of given given no. should be divisible divisible by 9
11
Th e d ififfe re re nc nc e b et etw ee ee n s um u m s o f th e d ig ig itits a t e ve ve n & a t o dd dd p la la ce ce s should b e zero or multiple of 11.
25 Last 2 digits digits of the number number should be 00, 25, 50 or 75.
Rule for 7 : Double the last digit of given number and
subtract from remaining number the result should be zero or divisible by 7. Ex.19 Check whether 413 is divisible by 7 or not. Sol. Last digit = 3, remaining number = 41, 41 – (3 x 2) = 35
(divisible by 7). i.e. 413 is divisible by 7. This rule can also be used for number having more than 3 digits. Ex.20 Check whether 6545 is divisible by 7 or not. Sol. Last digit = 5, remaining number 654, 654 – (5 x 2)
= 644; 64
–
(4 x 2) = 56 divisible by 7. i.e. 6545 is
divisible by 7. Rule for 13 : Four times the last digit and add to
remaining number the result should be divisible by 13. Ex.21 Check whether 234 is divisible by 13 or not .
divi sible Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible by 13.
PAGE # 3838
Rule for 17 : Five times the last digit of the number and subtract from previous number the result obtained should be either 0 or divisible by 17.
REMAINDERS
The method of finding the remainder without actually performing the process of division is termed as
Ex.22 Check whether 357 is divisible by 17 or not. Sol. 357, (7 x 5) – 35 = 0, i.e. 357 is divisible by 17. Rule for 19 : Double the last digit of given number and add to remaining number The result obtained should
remainder theorem. Remainder
should always be positive. For example if
we divide –22 by 7, generally we get an and d
be divisible by 19.
–
–
3 as quotient
1 as remainder. But this is wrong because
remainder is never be negative hence the quotient
Ex.23 Check whether 589 is divisible by 19 or not.
should be –4 and remainder is +6. We can also get
Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number is divisible by 19.
remainder 6 by adding –1 to divisor 7 ( 7 –1 = 6).
Ex.24 Find the smallest number of six s ix digits which is exactly
Two numbers, x and y, are such that tha t when divided by Ex.28 Two
divisible by 111. 111. Sol. Smallest number of 6 digits is 100000. On dividing 100000 by 111, we get 100 as remainder. b e added = (111 (111 – 100) = 11. Number to be Hence, required number = 100011.
6, they leave remainders 4 and 5 respectively. Find the remainder when (x2 + y2) is divided by 6. Sol. Suppose x = 6k1 + 4 and y = 6k2 + 5
x2 + y2 = (6k1 + 4)2 + (6k2 + 5)2 = 36k12 + 48k1 + 16 + 36k22 + 60k2 + 25 = 36k12 + 48k1 + 36k22 + 60k2 + 41
Ex.25 Find the largest four digit number which when reduced by 54, is perfectly divisible by all even natural numbers less than 20. Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18.
Their LCM = 2 × LCM of first 9 natural numbers = 2 × 2520 = 5040. This happens to be the largest four-digit number divisible by all even natural numbers less than 20. 54 was subtracted from our required number to get this number. Hence, (required number – 54) = 5040 Required number = 5094.
Obviously when this is divided by 6, the remainder will be 5. Ex.29 A number when divided by 259 leaves a remainder
139. What will be the remainder when the same number is divided by 37 ? num ber be P. Sol. Let the number So, P – 139 is divisible by 259. Let Q be the quotient then,
Ex.26 Ajay multiplied 484 by a certain number to get the result 3823a. Find the value of ‘a’. Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.
4 is a factor of 484 and 11 is also a factor of 484. Hence, 3823a is divisible by both 4 and 11. To be divisible by 4, the last two digits have to be divisible by 4. ‘
P = 259Q + 139
P 259Q 139 = 37 37
259 is divisible by 37,
P 139 =Q 259
When 139 divided by 37, leaves a remainder of 28. Ex.30 A number being successively divided by 3, 5 and 8
leaves remainders 1, 4 and 7 respectively. Find the
a’ can take two values 2 and 6.
respective remainders if the order of divisors be
38232 is not divisible by 11, but 38236 is divisible by 11. Hence, 6 is the correct choice.
reversed. Sol.
3 x 5 y 1 8 z 4 1 7
Ex.27 Which digits should come in place of and $ if the number 62684$ is divisible by both 8 and 5 ?
divisibl e by 5, so 0 or 5 must Sol. Since the given number is divisible come in place of $. But, a number ending with 5 in never divisible by 8. So, 0 will replace $. Now, the number formed by the last three digits is 40, which becomes divisible by 8, if is replaced by 4 or 8. Hence, digits in place of and $ are (4 or 8 or 0) and 0 respectively.
z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79 ; x = (3y + 1) = (3 × 79 + 1) = 238. Now, 8 238 5 29 6 3 5 4 1 2
Respective remainders are 6, 4, 2.
PAGE # 3939
Ex.31 A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. Then find out the number. Sol. 4 x
5 y 2 6 z 3 1 4
z = (6 × 1 + 4) = 10 y = (5 × z + 3) = (5 × 10 + 3) = 53 x = (4 × y + 2) = (4 × 53 + 2) = 214 Hence, the required number is 214. Ex.32 In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors of 585) and got the remainders 4, 8 and 12. If he had divided number by 585, then find out the remainder. Sol. 5 x 9 y 4 13 z 8 1 12
Now, 1169 when divided by 585 gives remainder = 584.
NOTE :
Ex.36 What is the remainder when 1415 16
15 Sol. 14
16
is divided by 5 ?
= (15 –1)odd = 15n + ( –1)odd, i.e. a (multiple of 5)
–
1. Thus when divided by 5 the remainder will be ( –1),
i.e. 4. Ex.37 What is the remainder when 357 + 27 is divided by 28? Sol. 357 = (33)19 357 + 27 = (27)19 + 27 = (28 – 1)19 + 27 = 28M + ( –1)19 + 27 [Expand by binomial theorem] = 28M – 1 + 27 = 28M + 26 When 28M + 26 divided by 28, the remainder is 26. Hence, the required remainder is 26. Ex.38 What is the remainder when 82361 + 83 361 + 84361 + 85361 + 86361 is divided by 7? Sol. 82 361 + 83 361 + 84 361 + 85 361 + 86 361 = [(84 – 2) 361 + (84 – 1)361 + 84361 + (84 + 1)361 + (84 + 2)361] Since, 84 is a multiple of 7, then the remainder will be when, ( – 2)361 + ( –1)361 + 1361 + 2361 is divided by 7 is ( – 2)361 + ( –1)361 + 1 361 + 2 361 = 0. So the remainder is zero. CYCLICITY
(i) Binomial Expansion : (a + b)n = an + (a – b)n = an –
n 1!
n
an 1b + –
an 1b + –
n(n 1) 2!
n(n 1)
an 2b2 + .... + bn, or –
an 2b2 – ......+ ( – 1)nbn. –
1! 2! Hence, first term is purely of a i.e an and last digit is purely of b, i.e. bn.
(ii) Total Total number of terms in the expansion of (a + b)n is (n + 1). Ex.33 What is the remainder when 738 is divided by 48. Sol.
2 19
19
19
49 7 48 1 7 = = = so by using 48 48 48 48 binomial expansion, we can say that 18 terms are completely divisible by 48 but the last term which is 38
119 48
We are having 10 digits in our number systems and some of them shows special characteristics like they, repeat their unit digit after a cycle, for example 1 repea t its unit digit after every consecutive power. So, its cyclicity is 1 on the other hand digit 2 repeat its unit digit after every four power, hence the cyclicity of 2 is four. The cyclicity of digits are as follows : Digit
Cyclicity
0, 1, 5 and 6
1
4 and 9
2
2, 3, 7 and 8
4
So, if we want to find the last digit of 245, divide 45 by 4. The remainder is 1 so the last digit of 245 would be same as the last digit of 21 which is 2.
is not divisible. So, 119 = 1 is the remainder..
Ex.34 What is the remainder if 725 is divided by 4? an d Sol. 725 can be written (8 –1)25. There are 26 terms in all and first 25 terms are divisible by 8, hence also by 4. The last term is ( –1) 25. Hence, (8 –1) 25 can be written 8X – 1 or 4Y –1 ( where Y = 2X). So, 4Y – 1 divided by 4 leaves the remainder 3. Ex.35 What is the remainder if 345 is divided by 8 ? Sol. 345 can be written as 922 × 3. 9 can be written as (8 + 1). Hence, any power of 9 can be written as 8N + 1. In other words, any power of 9 is 1 more than a multiple of 8. Hence, (8N + 1) × 3 leaves remainder 3 when divided by 8.
(i) When there is any digit of cyclicity 4 in unit s place. Since, when there is 2 in unit’s place then in 2 1 unit digit is 2, in 22 unit digit is 4, in 23 unit digit is 8, in 24 unit digit is 6, after that the unit’s digit repeats. e.g. unit digit (12)12 is equal to the unit digit of, 24 i.e.6. ’
Ex.39 In (32)33 unit digit is equal to the unit digit of 21 i.e. 2. Ex.40 In (23)15 unit digit is equal to the unit digit of 33 i.e. 7. Ex.41 In (57)9 unit digit is equal to the unit digit of 71 i.e. 7.
digi t of 82 i.e. 4. Ex.42 In (678)22 unit digit is equal to the unit digit
PAGE # 4040
(ii) When there is any digit of cyclicity 2 in unit s place. Since, when there is 4 in unit’s place then in 4 1 unit digit is 4, in 42 unit digit is 6 and so on. ’
HIGHEST POWER DIVIDING A FACTORIAL
Factorial n : Product of n consecutive natural numbers
is known as factorial n it is denoted by n! . ‘
Ex.43 In (34)33 unit digit is 4.
Ex.45 In (49)18 unit digit is 1. (iii) When there is any digit of cyclicity 1 in unit s place. ’
Since, when there is 5 in unit’s place then in 5 1 unit digit is 5, in 52 unit digit is 5 and so on.
(ii) 1359
57 gives the remainder 4 1. So, the last digit of 357 is same as the last digit of 31, i.e. 3.
(ii) The number of digits in the base will not make a difference to the last digit. It is last digit of the base which decides the last digit of the number itself. 59 For 1359, we find which gives a remainder 3. So 4 the last digit of 1359 is same as the last digit digi t of 33, i.e. 7. Ex.49 Find unit’s digit in y = 7 17 + 734 Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6 Ex.50 What will be the last digit of (73 )75 76
= (73)x where x = 75
6476
y y y y! is 2 3 ......., where [ ] represents just x x x
6476
Therefore, the last digit of (73)75
= (75)even power
will be 31 = 3.
Ex.51 What will be the unit digit of (87)75
6355
= (87)x where x = 75 63
55
. = (75)odd
Cyclicity of 7 is 4. l ast digit we have to find the remainder To find the last when x is divided by 4. x = (75)odd power = (76 – 1)odd power where x is divided by 4 so remainder will be –1 or 3, but remainder should be always positive.
Therefore, the last digit of (87)75 Hence, the last digit is of (87)75
6355
6355
× 5 × (2 1 × 31) × 7 × (2 3) × ..... so on. In o rder to find the
highest power of 2 that divides the above product, we need to find the sum of the powers of all 2 in this expansion. All numbers that are divisible by 21 will contribute 1 to the exponent of 2 in the product 20
= 10. Hence, 10 numbers contribute 21 to the 21 product. Similarly, all numbers that are divisible by 22 will contribute an extra 1 to the exponent of 2 in the product, i.e
20
= 5. Hence, 5 numbers contribute an 22 extra 1 to exponents. Similarly, there are 2 numbers
by 24. Hence, the total 1s contributed to the exponent
Cyclicity of 3 is 4 To find the last l ast digit we have to find the remainder when x is divided by 4. x = (75)even power = (76 – 1)even power , where n is divided by 4 so remainder will be 1. 6476
completely? Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 2 0 = 1 × (2 1) × 3 × (2 2)
that are divisible by 23 and 1 number that is divisible
6355
The approach to finding the highest power of x dividing
Ex.52 What is the highest power of 2 that divides 20!
Sol. (i) The cyclicity of 3 is 4. Hence,
75 Sol. Let (87)
The value of factorial zero is equal to the value of factorial one. Hence 0! = 1 = 1!
part.
Ex.47 In (46)13 unit digit is 6.
Sol. Let (73 )
’
the integral part of the answer and ignoring the fractional
Ex.46 In (25)15 unit digit is 5.
75 64
‘
So, n! = n (n – 1) (n – 2)...3 2 1.e.g. 1.e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.
Ex.44 In (29)15 unit digit is 9.
Ex.48 Find the last digit of (i) 357
’
will be 73 = 343. is 3.
of 2 in 20! is the sum of ( 10 + 5 +2 +1) = 18. Hence, group of all 2s in 20! gives 218 x (N), where N is not divisible by 2. If 20! is divided by 2x then maximum value of x is 18. Ex.53 What is the highest power of 5 that divides of
x = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1. Sol. Calculating contributions of the different powers of 5,
we have
100
= 20,
100
= 4. 51 52 Hence, the total contributions to the power of 5 is 24, or the number 100! is divisible by 524.
Ex.54 How many zeros at the end of first 100 multiples
of 10. Sol. First 100 multiple of 10 are = 10 × 20 × 30 × ......× 1000 = 10100 (1 × 2 × 3 × .......× 100) = 10100 × 1024 × N = 10124 × N Where N is not divisible by 10 So, there are 124 zero at the end of first 100 multiple of 10. PAGE # 4141
Ex.55 What is the highest power of 6 that divides 9!
9 9 = 1 and 2 = 0. Thus 6 6 answers we get is 1 which is wrong. True there is just
Sol. By the normal method.
(i) Conversion from base 10 to any other base : Ex.58 Convert (122)10 to base 8 system. Sol.
one multiple of 6 from 1 to 9 but the product 2 × 3 = 6
8 122 8 8
and also 4 × 9 = 36, can further be divided by 6. Thus, when the divisor is a composite number find the
15 1 0
2 7 1
highest power of its prime factors and then proceed. In The number in decimal is consecutively divided by the
this case, 9! can be divided by 27 and 34 and thus by 64 (In this case we need not have checked power of 2 as
number of the base to which we are converting the
it would definitely be greater than that of 3).
decimal number. Then list down all the remainders in the reverse sequence to get the number in that base.
power of 12 that would divide 49! ? Ex.56 What is the largest power
So, here (122)10 = (172)8.
Sol. To check the highest power of 12 in 49!, we need to
check the highest powers of 4 and 3 in it. Highest power of 3 in 49! = 22 Highest power of 2 in 49! = 46 46 Highest power of 4 in 49! = = 23 2 Highest power of 12 will be 22. (Since the common
Ex.59 Convert (169)10 in base 7.
7 169 7 24 7 3 Sol. 0
Sol. Highest power of 5 in 36! is 8.
So, there will be 8 zeros at the end of 36!. So, at the end of 36!36! , there will be 8 × 36! zeros. BASE SYSTEM
Remainder
(169)10 =(331)7
power between 3 and 4 is 22). Ex.57 How many zeros will be there at the end of 36!36! ?
1 3 3
Ex.60 Convert (0.3125)10 to binary equivalent. Sol.
Integer
2 0.3125 = 0.625
0
2 0.625 = 1.25
1
2 0.25 = 0.50
0
2 0.50 = 1.00
1
Thus The number system that we work in is called the
(0.3125)10 = (0.1010)2
decimal system . This is because there are 10 digits
‘
’
in the system 0-9. There can be alternative system that
Ex.61 Convert (1987.725)10 (........)8
can be used for arithmetic operations. Some of the
Sol. First convert non-decimal part into base 8.
most commonly used systems are : binary, octal and
8 1987 8 248 3
hexadecimal.
8 8
These systems find applications in computing. Binary system has 2 digits : 0, 1. Octal system has 8 digits : 0, 1,..., 7. Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,
C, D, E, F.
31 3 0
0 7 3
(1987)10 = (3703)8
After 9, we use the letters to indicate digits. For instance, instance ,
Now we have to convert (0.725)10 (........)8
A has a value 10, B has a value 11, C has a value 12,...
Multiply
0.725 × 8 = [5.8] 0.8 × 8 = [6.4] 0.4 × 8 = [3.2] 0.2 × 8 = [1.6] 0.6 × 8 = [4.8]
so on in all base systems. The counting sequences in each of the systems would be different though they follow the same principle. Conversion : Conversion of numbers from (i) decimal
system to other base system. (ii) other base system to decimal system.
...5 ...6 ...3 ...1 ...4
Keep on accomplishing integral parts after multiplication with decimal part till decimal part is zero.
(0.725)10 = (0.56314...)8
(1987.725)10 = (3703.56314...)8
PAGE # 4242
(ii) Conversion from any other base to decimal system : Ex.62 Convert (231)8 into decimal system. Sol. (231)8 , the value of the position of each of the numbers ( as in decimal system) is : 1 = 80 × 1 3 = 81 × 3 2 = 82 × 2 Hence, (231)8 = (80 × 1 + 8 1 × 3 + 82 × 2)10 (231)8 = (1 + 24 + 128)10 (231)8 = (153)10 Ex.63 Convert (0.03125)10 to base 16. 0 Sol. 16 0.03125 = 0.5 16 0.5 = 8.0 8 So (0.03125)10 = (0.08)16
Ex.68 The sum of first n natural numbers is a three-digit
number, all of whose digits are the same. What is the value of n? Sol. In 5 seconds, you can solve the equation
n(n 1) = aaa (111, 222, etc) . How do you proceed 2 next ? If you think it's hit-and-trial from this point, you are wrong. Here goes the simple logic. It might strike you instantly if you have been working with numbers:
n(n 1) = aaa = a × 111 = a × 3 × 37 2 n(n + 1) = 6a x 37
Look at the L.H.S. of the equation, n(n + 1) is a produc t
Ex.64 Convert (761.56)8 (......)16 Sol. In such conversion which are standard form conversions, it is easier to (761.56)8 (.....)2 (.....)16 Converting every digit in base 8 to base 2, (111110001.101110)2 (1F1.B8)16 Ex.65 Convert (3C8.08)16 to decimal Sol. (3C8.08)16 = 3 162 + C 161 + 8 16 + 0 16 1 + 8 16 2 = 768 + 192 + 8 + 0 + 0.03125 = (968.03125)10 So, (3C8.08)16 = (968.03125)10 –
of two consecutive natural numbers. Therefore, R.H.S. should also be a product of two consecutive natural numbers. One of the numbers is 37. Therefore, what could the other number 6a, consecutive to 37 be? It can only be 36, giving a = 6 and n = 36. Therefore, 36 numbers have been summed up and their sum is equal
–
ALPHA NUMERICS NUMBERS
to 666. Ex.69 If ABC x CBA = 65125, where A, B and C are single
digits, then A + B + C = ? Sol. As the unit digit of the product is 5, therefore, the unit
digit of one of the numbers is 5 and the unit digit of the other number is odd. Therefore, AB5 x 5BA = 65125,
aa Ex.66 If a – b = 2, and
b b
where A = 1, 3, 5, 7 or 9. then find the value of a, b and c.
cc 0 Sol. These problems involve basic number (i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers . Hence, their sum cannot exceed 198. So, c must be 1. (iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6 and b = 4. Such problems are part of a category of problems called alpha numerics. a 3b
a c Ex.67 If _____ a a 9
As the product of two three-digit numbers is a five-digi t number, and not a six-digit number, A can only be equal to 1. IB5 x 5B1 = 65125. The digit sums of both numbers, 1B5 and 5B1 will be same. Therefore, the product would give digit sum of a perfect square. The digit sum on the R.H.S. is 1. Therefore, the digit sum of each number can be 1or 8. Correspondingly B will be 4 or 2 (as digit sum cannot be equal to 1).
then find a, b and c if each of them is
distinctly different digit. Sol. (i) since the first digit of (a 3 b) is written as it is after subtracting ac carry over from a to 3. (ii) there must be a carry over from 3 to b, because i f no carry over is there, it means 3 – a = a. 3 2a = 3 a = 2 which is not possible because a is a digit. For a carry over 1, 2 – a = a a=1 (iii) it means b and c are consecutive digit (2, 3), (3, 4),.... (8, 9)
Keeping B = 2, we can see that 125 x 521 = 65125. Ex.70 Find the four-digit number ABCD such that
ABCD x 4 = DCBA. Sol. Any number multiplied by 4 will give us an even number.
Hence, the digit D when multiplied by 4 will give us an even number. Since A is the unit digit of the product produc t it is even. Hence, A = 2, 4, 6 or 8 (It cannot be 0). A is also the first digit of the multiplicand and if A = 4, 6 or 8 the product ABCD x 4 will become a 5 digit number. Hence A = 2. Writing the value of A we get 2BCD x 4 = DCB2. PAGE # 4343
Now for the value of D looking at the first and last digits
6.
by 14?
of the multiplicand, we can see that 4 x D gives the unit digit of 2 and 4 x 2 gives the fi rst digit of D. Yes, you got it right D = 8. Writing the multiplication again with the 7.
value of D we get 2BC8 x 4 = 8CB2.
(A) 28
(B) 29
(C) 27
(D) None of these
The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25 is :
Now for the value of B. A number is divisible by 4 if the number formed by the last two digits is divisible by 4. Since the number 8CB2 is a multiple of 4, the number
How many numbers between 200 and 600 are divisible
8.
(A) 47
(B) 60
(C) 72
(D) 94
How many three-digit numbers would you find, which
B2 should be divisible by 4. Or, the number B2 = 12,
when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2,
32, 52, 72 or 92. Hence the original number ABCD is
3, 4, and 5 respectively ?
21C8, 23C8, 25C8, 27C8 or 29C8. But the last 4
(A) 4
(B) 3
numbers when multiplied by 4 will not n ot give you the first
(C) 2
(D) 1
digit of 8 in the product. Therefore B = 1 and the original number is 21C8. We write the multiplication again
9.
Six strings of violin start vibrating simultaneously and they vibrate at 3, 4, 5, 6,10 and 12 times in a minute, find :
21C8 x 4 = 8C12. Now for the value of C notice that when you multiply 8,
i. After how much time will all six of them vibrate together ?
the unit digit of 21C8, by 4 you write 2 in the unit digit of
ii. How many times will they vibrate together in 30 min ?
the product and carry 3. The tenth digit of the product is
(A) 60 min, 31 times
(B) 60 sec, 31 times
1. Therefore, 4 x C + 3 (carry over) gives a unit uni t digit of 1.
(C) 120 sec, 15 times
(D) None of these
Hence, C is 2 or 7. You You can easily easil y check by the hundreds
10. The HCF of 2 numbers is 11 and their LCM is 693.
digit in the product (which is C again) that C = 7.
If their sum is 176, find the numbers.
Therefore, our answer is 2178 x 4 = 8712.
(A) 99,77
(B) 110, 66
(C) 88,77
(D) 121, 44
11. If P is a prime number, then the LCM of P and (P + 1) is 1.
2.
3.
4.
5.
Convert 0.225 in to form p/q. 3 (A) 10
9 (B) 40
9 (C) 50
9 (D) 400
(C) (P + 1)(P – 1)
(D) None of these
AB and CB are two-digit numbers and DDD is a threedigit number.
There are four prime numbers written in ascending order. The product of the first three is 385 and that of the last three is 1001. The last number is : (A) 11 (B) 13 (C) 17 (D) 19 If logxy = 100 and log2x = 10, then the value of y is : (A) 21000 (B) 2100 (C) 22000 (D) 210000 Find the value of ‘x’ if 2logx 7 + log7x 7 + 3log49x 7 = 0
(A) 21
(B) 19
(C) 17
(D) 18
13. Three pieces of cakes of weights 4
1 3 Ibs, 6 Ibs and 2 4
1 Ibs respectively are to be divided into parts of equal 5 weights. Further, each must be as heavy as possible. 7
If one such part is served to each guest, then what is the maximum number of guests that could be entertained ? (A) 54
(B) 72
(C) 20
(D) 41
14. How many natural numbers between 200 and 400 are
there which are divisible by i. Both 4 and 5?
4 (A) x = 3
(B) x = 7
(C) x = 7
(D) Either (B) or (C)
4/3
(B) (P + 2)P
12. Find out (A + B + C + D) suc h that AB x CB = DDD, where
When (55)10 is represented in base 25 then the expression is : (A) (25)25 (B) (35)25 (C) (55)25 (D) none of these
–
(A) P(P +1)
–
1/2
ii. 4 or 5 or 8 or 10 ? (A) 9, 79
(B) 10, 80
(C) 10, 81
(D) None of these
PAGE # 4444
15. 461 + 462 + 463 + 464 is divisible by :
(A) 3
(B) 10
(C) 11
(D) 13
16. If x is a whole number, then x2 (x2 – 1) is always divisible
by : (A) 12
(B) 24
(C) 12 – x
(D) Multiple of 12
17. If 653 xy is exactly divisible by 80, then the find the value
of (x + y). (A) 2
(B) 3
(C) 4
(D) 6
18. Find the unit digit of (795 – 358).
(A) 6 (C) 3
(B) 4 (D) None of these
19. When a number P is divided by 4 it leaves remainder
3. If the twice of the number P is divided by the same divisor 4 than what will be the remainder ? (A) 0 (B) 1 (C) 2
(D) 6
20. If (232 +1) is divisible by a certain number then which of
the following is also divisible by that number. (A) (216 – 1) (C) 296 + 1
(B) 216 + 1 (D) None of these
21. If the number 357y25x is divisible by both 3 and 5, then
find the missing digit in the unit’s place and the thousand place respectively are : (A) 0, 6 (B) 5, 6 (C) 5, 4
(D) None of these
22. A number when divided by 342 gives a remainder 47.
When the same number is divided by 19, what would be the remainder ? (A) 3
(B) 5
(C) 9
(D) None of these
23. What is the remainder when 9875347 × 7435789 × 5789743 is divided by 4 ?
(A) 1
(B) 2
(C) 3
(D) None of these
24. What is remainder when 784 is divided by 2402?
(A) 1 (C) 2401
(B) 6 (D) None of these
25. P is a prime number greater than 5. What is the
remainder when P is divided by 6? (A) 5 (B) 1 (C) 1 or 5
(D) None of these
26. What is the remainder when 3040 is divided by 17?
(A) 1 (C) 13
(B) 16 (D) 4
27. What is the remainder when 650 is divided by 215? (A) 1 (B) 36 (C) 5 (D) 214 28. What is the remainder when 7413 – 4113 + 7513 – 4213 is divided by 66? (A) 2 (B) 64 (C) 1 (D) 0 29. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be : (A) 1, 2 (B) 2, 3 (C) 3, 2 (D) 4, 1 30. When Sholey screened on the TV there was a commercial break of 5 min after every 15 min of the movie. If from the start of the movie to the end of the movie there was in all 60 min of commercials that was screened what is the duration the movie ? (A) 180 min (B) 195 min (C) 169 min (D) 165 min Directions : (31 to 35) Read the following information carefully and answer the questions given below. In a big hostel, there are 1,000 rooms. In that hostel only even numbers are used for room numbers, i.e. the room numbers are 2, 4, 6, ...., 1998, 2000. All All the rooms have one resident each. One fine morning, the warden calls all the residents and tells them to go back to their rooms as well as multiples of their room numbers. When a guy visits a room and finds the door open, he closes it, and if the door is closed, he opens it, All 1,000 guys do this operation. All the doors were open initially. 31. The last room that is closed is room number ? (A) 1936 (B) 2000 (C) 1922 (D) None of these 32. The 38th room that is open is room number : (A) 80 (B) 88 (C) 76 (D) None of these 33. If only 500 guys, i.e. residents of room number 2 to 1000 do the task, then the last room that is closed is room number (A) 2000 (B) 1936 (C) 1849 (D) None of these 34. In the case of the previous question, how many rooms will be closed in all ? (A) 513 (B) 31 (C) 13 (D) 315 35. If you are a lazy person, you would like to stay in a room whose number is : (A) more than 500 (B) more than 1000 (C) 500 (D) 2000
PAGE # 4545
36. A 4-digit number is formed by repeating a 2-digit
number such as 2525, 3232 etc. Any number of this from is exactly divisible by : (A) 7 (C) 13
45. In a number system, the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes :
(A) 406 (C) 213
(B) 11
(D) Smallest 3-digit prime number 37. How many numbers between 400 and 600 begin with
46. A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came and erased
one number. The average of the remaining numbers
or end with a digit of 5 ? (A) 40
(B) 100
(C) 110
(D) 120
7 . What was the number erased? 17 (A) 7 (B) 8 (C) 9 (D) None of these
is 35
38. If (12 + 22 + 32 + .....+ 102) = 385, then the value of
(22 + 42 + 62 +...... + 202). (A) 770
(B) 1155
(C) 1540
(D) (385 × 385)
47. Let D be a recurring decimal of the form D = 0. a1 a2 a1 a2 a1 a2 ....., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the
39. Find the total number of prime factors in the expression 11
5
following numbers necessarily produces an integer, when multiplied by D? (A) 18 (B) 108 (C) 198 (D) 288
2
(4) × (7) × (11) . (A) 37
(B) 33
(C) 26
(D) 29
40. The largest number which exactly divides the product
48. What is the value of the following expression
1 1 1 1 (2 2 1) ( 4 2 1) (6 2 1) ..... (20 2 1) ? 9 10 (A) (B) 19 19 10 11 (C) (D) 21 21
of any four consecutive natural numbers is : (A) 6 (B) 12 (C) 24
(D) 120
41. The largest natural number by which the product of
three consecutive even natural numbers is always divisible, is : (A) 6 (C) 48
(B) 24 (D) 96
49. Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
(A) 0 (C) 3
42. A 3-digit number 4a3 is added to another 3-digit
number 984 to give the four-digit number 13b7, which is divisible by 11. Then ,(a + b) is : (A) 10 (B) 11 (C) 12
one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (A) 1050 (B) 540 (C) 1440 (D) 1590 44. Three friends, returning from a movie, stopped to eat
at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a monetary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl? (A) 38 (B) 31 (D) 48
(B) 9 (D) 6
50. Let N = 553 + 173 – 723, then N is divisible by : (A) both 7 and 13 (B) both 3 and 13
(C) both 17 and 7
(D) 15
43. Anita had to do a multiplication. Instead of taking 35 as
(C) 41
(B) 1086 (D) 691
(D) both 3 and 17
51. Convert the number 1982 from base 10 to base 12. The results is : (A) 1182 (B) 1912
(C) 1192
(D) 1292
52. If n2 = 12345678987654321, find the value of n ? (A) 12344321 (B) 1235789 (C) 11111111 (D) 111111111
53
1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8
is equal to : (A) 9 (C) 4.5
(B) 8 (D) None of these
54. The LCM of two numbers is 567 and their HCF is 9. If the difference between the two numbers is 18, find the
two numbers : (A) 36 and 18 (C) 63 and 81
(B) 78 and 60 (D) 52 and 34 PAGE # 4646
55. If a, a + 2, and a + 4 are prime numbers, then the number of possible solution for a is : (A) three (B) two (C) one (D) more than three 56. Find the square root of 7 – 4 3 .
(A) 2 – 3
(B) 5 – 3
(C) 2 – 5
(D) None of these
1 2 n 40 such 58. The number of positive n in the range 12 that the product (n –1) (n – 2).... 3.2.1 is not divisible by n is : (A) 5 (B) 7 (C) 13 (D) 14 gol d coins. He did 59. A rich merchant had collected many gold not want any body to know about him. One One day, day, his wife asked, “How many gold coins do we have? ” After pausing a moment he replied, “Well ! if divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the square of the two numbers. “ The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins the merchant has ? (A) 96 (B) 53 (C) 43 (D) 48 60. 76n – 66n, where n is an integer > 0, is divisible by : (A) 13 (B) 127 (C) 559 (D) All of these
10 25 108 154 225
(A) 4 (C) 8
is :
(B) 6 (D) 10
1 1 1 62. lo g 10 1 + log 10 1 + log10 1 + ... + 2 3 4 1 . When simplified has the value equal log 10 1 1999
to : (A) 1 (C) 10
(B) 3 (D) 100
63. Arrange the following rational number in ascending
order
3 4 7 1 , , , . 7 5 9 2
4 7 3 1 (A) , , , 5 5 9 2
(C)
4 7 1 3 , , , 5 9 2 7
3 1 7 4 (B) , , , 7 2 9 5
(D)
1 3 7 4 , , , 2 7 9 5
6
17 , 2,12 25 , 3 4 .
(A) 6 17
(B)
(C)
(D)
3
4
12
25 2
65. If log 10N 2.5 then, find out total number of digits in N.
57. How many even integers n, where 100 n 200, are divisible neither by seven nor by nine ? (A) 40 (B) 37 (C) 39 (D) 38
61. The value of
64. Which of the following surds is greatest in magnitude
(A) 3 (C) 5 (D) cannot be determine
(B) 4
66. If log x = n then 2n is equal to : (A) log (x2) (B) (logx)2 (C) log (x+2) (D) log 2x 67. Given log2 = 0.3010, then log 16 is : (A) 2.4080 (B) 1.2040 (C) 0.2408 (D) 1.9030 68. The value of [log10 (5 log10 100)]2 is : (A) 0 (B) 1 (C) 2 (D) 10 69. If log10 [log10(log10x)] = 0. (A) x = 103 (C) x = 155
(B) x = 1010 (D) None
70. If n = 67 then find the unit digit of [3n + 2n ]. (A) 1 (B) 10 (C) 5 (D) None 71. What is the decimal equivalent of the 25 digits of
hexadecimal number (100.....001)16 ? (A) 223 + 1 (C) 292 + 1
(B) 224 + 1 (D) 296 + 1
111 is written in the octal system, 72. If the decimal number 2111 then what is its unit place digit ?
(A) 0 (C) 2 73. If
(B) 1 (D) 3
1 log3 M 3 log3 N = 1+ log 5, then : 0.008 3
9 N 3 3 (C) M N
9 (A) M
9 M 3 9 (D) N M
9 (B) N
74. The value of x, when log3(log2 x) + 2 log9(log7 8) = 2, is : (A) 243 (B) 27
(C) 343
(D) 64
75. Find x if log10 1250 + log1080 = x. (A) 5 (B) 4 (C) 8 (D) 7 76. P, Q and R are three natural numbers suc h that P and Q are primes and Q divides PR. Then out of the following the correct statement is : [IJSO-2008] (A) Q divides R (B) P divides R (C) P divides QR (D) P divides PQ
PAGE # 4747
77. It is required to decide if 1107 is a prime number or not. The number of trials of division necessary is : [IJSO-2008] (A) 10 (B) 11 (C) 12 (D) 235 78. The number of integers between
–
8 and
(B) 6 (D) 8
79. When expanded, the number of zeros in 100010 is : [NSTSE-2009] (A) 13 (B) 30 (C) 4 (D) 10 80. If a2 + 2b = 7, b2 + 4c = – 7 and c 2 + 6a = – 14, then the value of (a2 + b2 + c2) is : [IJSO-2009] (A) 14 (B) 25 (C) 36 (D) 47
distinct factors factors of N is : 81. Let N = 28, the sum of All distinct [IJSO-2009] (A) 27 (B) 28 (C) 55 (D) 56 82. The units digit of (1 + 9 + 9 + 9 + --------- + 9 ) is : [IJSO-2009] 2
(A) 0 (C) 9
3
(C) 61/6
2009
(B) 1 (D) 3
83. The biggest among the following is : (A) 21/2 (B) 31/3
[IJSO-2009]
(D) 81/8
84. If a, b 1, ab > 0, a b and logba = logab, then ab = ? [IAO- 2009]
(A) 1/2 (C) 2
(B) 1 (D) 10
(C) 51
(D) 2009
86. If HCF (p, q) = 12 and p × q = 1800 n then LCM (p, q) is :
32 is :
[NSTSE-2009]
(A) 5 (C) 7
85. If 2009 = pa.qb, where "p" and "q" are prime numbers, then find the value of p + q. [NSTSE 2009] (A) 3 (B) 48
[NSTSE 2010]
(A) 3600 (C) 150 87. If
x + y
be : (A) 16 (C) 2
(B) 900 (D) 90 10 y = and x + y = 10, then t hen the value of xy will 3 x [NSTSE 2010]
(B) 9 (D) 10
88. The value of log10(3/2) + log10 (4/3) + ......... up to 99
terms. (A) 0 (C) 2.5
[IAO 2008]
(B) 2 (D) None of the above
89. In the familiar decimal number system the base is 10. In another number system using base 4, the counting
proceeds as 1, 2, 3, 10, 11, 12, 13, 20, 21 .... The twentieth number in this system will be : [IJSO-2010] (A) 40 (B) 320 (C) 210 (D) 110 90. If the eight digit number 2575d568 is divisible by 54 and 87, the value of the digit ‘d’ is : [IJSO-2011]
(A) 4 (C) 0
(B) 7 (D) 8
91. If x < 0 and log7 (x2 – 5x – 65) = 0, then x is : [IJSO-2011] (A) –13 (B) –11 (C) – 6 (D) – 5
PAGE # 4848
TRIGONOMETRY
(ii) Centesimal system : In this system a right angle is
ANGLE
divided into 100 equal parts, called grades. grades . Each grade An angle is the amount of rotation of a revolving line
is sub divided into 100 minutes, minutes, and each minute into
with respect to a fixed line. If the rotation is in
100 seconds. seconds. Thus, 1 right angle = 100 grades (100 g) 1 grade = 100 minutes (100 )
anticlockanticlock-wise sense, sense, then the angle measured is positive and if the rotation is in clock-wise sense, sense,
’
then the angle measured is negative. negative.
1 minute = 100 seconds (100 ) ”
(iii) Circular system : In this system the unit of radian, written as 1c, is measurement is radian. One radian, the measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.
QUADRANTS Let X OX and YOY be two lines at right angles in a ’
’
plane. These lines divide the plane into four equal parts are known as quadrants. quadrants. The lines X OX and ’
YOY are known as X-axis and Y-axis respectively. ’
These two lines taken together are known as the coordinate axes. axes. The regions XOY, YOX , X OY and Y OX ’
’
’
’
The number of radians is an angle subtended by an arc of a circle at the centre is equal to
.
s r
=
Where, = angle in radian, s = arc length and r = radius.
are known as first, first, second, second, third and fourth quadrants (b )
respectively.
Where, D = number of degrees, G = number of grades, R = number of radians.
and NOTE :
(i) The angle between two consecutive digits in a clo ck is 30º = ( /6 radians). radians). (ii) The hour hand rotates through an angle of 30º in one hour, hour, i.e. (1/2)º in one minute. minute. (i) Sexa exagesi gesima mall sy syste stem
(ii) Cente ntesima imal system
(iii) Circular system (i) Sexagesimal system : In this system a right angle is divided into 90 equal parts called degrees. degrees. Each degree is divided into 60 equal parts called minutes
(iii) The minute hand rotates through an angle of 6º in one minute. minute. Ex.1 Express in radians 47º 25 ’ 36”.
36 3 = 47º 25 60 5
Sol. 47º 25 ’
and each minute is divided into 60 equal parts called º
seconds. seconds. Thus, 1 right angle = 90 degrees ( 90º) 1º = 60 minutes (60 ) ’
”
1 = 60 seconds (60 ) ’
128 128 1 32 = 47 = 47º = 47º 5 5 60 75
º
º
3557 c 3557 3557 c . = = = 13500 75 180 75
PAGE # 4949
Ex.2 Express in degrees :
2 15
Ex.6 The angles in one regular polygon is to that in another
c
as 3 : 2, also the number of sides in the first is twice
(a)
(b) ( – 2)c .
c
have ?
º
2 2 180 = 24º Sol. (a) 15 15 180 (b) ( – 2)c = 2
Sol. Suppose the second regular polygon has number of side = x.
º
º
6 180 7 ( 2) = 114 = 11 22
The first regular polygon will have number of side = 2x. ( 4 x 4) Each angle of the first polygon = right angle
º
2x And each angle of the second polygon
6 8 = – 114º 60 = – 114º 32 11 11
8 = – 114º 32’ 60 11
that in the second ; how many sides do the polygons
=
4 x 4 2x 4 : =3:2 2x x
4 x 4 6 x 12 x x
''
= – 114º 32 44 . ’
”
Ex.3 Express in radians 345g 25’ 36 . Sol. 345g 25’ 36 = 345.2536g ”
”
=
3452536 2000000
c = 1.726268 c
Ex.4 One angle of a triangle is
2x grades another is 3
x 3x degrees, whilst the third is radians ; express 75 2 all angles in degrees. Sol.
x 12 x º And 180º = 75 75 5 3 12 3 But xº + xº + xº = 180º 5 5 2
6xº + 15xº + 24xº = 1800 45xº = 1800 x = 40º Hence, three angles of the triangle are 24º, 60º and 96º.
Ex.5 The angles of a triangle are in A.P. A.P. and the number of degrees in the least is to the number of radians in the greatest is 60 to c. Find the angles in degrees. Sol. The three angles in A.P. A.P. ; if y is common difference, let these angles be (x – y)º, xº and (x + y)º. x + y + x + x – y= 180º x = 60º According to the question. ( x y) 60 (x y) or
180
(x – y) = (x + y)
3 (x – y) = x + y 4y = 2x y = x/2
Ex.7 The radius of a certain circle circle is 30 cm, find the approximately length of an arc of this circle ; if the length of the chord of the arc be 30 cm. Sol. Let ABC be the circle whose centre is O and AC is chord.
AOC = 60º =
3
Hence, arc AC = radius ×
3
180 º
= 30 ×
= 10 = 31.4159 cm.
3
Trigonometry means, the science which deals with the measurement of triangles.
A right angled triangle is shown in Figure. Figure.
C as
B is of
B is called hypotenuse. hypotenuse. There
are two other angles i.e. × 60º
TRIGONOMETRY
90º. Side opposite to
60º = 30º. 2 Hence three angles are 30º, 60º and 90º.
y=
respectively are 8 and 4.
x c
4x – 4 = 6x – 12 2x = 8 x = 4. The number of sides in the first and second polygons
In AOC, AO = OC = AC = 30 cm.
2 g 2 9 º 3 x = x xº 3 3 10 5
c
( 2x 4 ) right angle x
A and
C. If we consider
, then opposite side to this angle is called
perpendicular and side adjacent to is called base. base. (i) Six Trigonometric Ratios are : sin =
Perpendicular = Hypotenuse
P H
=
AB AC
PAGE # 5050
Hypotenuse cosec = = Perpendicular Base cos = = Hypotenuse
sec =
tan =
B H
Hypotenuse = Base
P
13 , then prove that : 5 tan2 A – sin2 A = sin4A sec2 A.
Ex.9 If cosec A =
AC = AB
BC = AC
H B
Perpendicu lar = Base
H
=
P B
Sol. We have cosec A =
So, we draw a right triangle ABC, right angled at C such that Hypotenuse AB = 13 units and perpendicular BC = 5 units
AC BC
=
By pythagoras theorem, (13)2 = (5)2 + AC2 AB2 = BC BC2 + AC2
AB BC
AC2 = 169 – 25 = 144 AC =
cot =
BC Base B = = AB Perpendicular P
1 cot
cos =
1 sec
1 sin = cos ec
We also observe that tan =
sin cos and cot = cos sin
m Ex.8 If tan = , then find sin n Sol. Let P = m and B = n
tan =
P B
=
m n
cot =
144 = 12 units
BC 5 5 BC = and sin A = = AC 12 13 AB AB 13 B and sec A = = AC 12
tan A =
(ii) Interrelationship in Basic Trigonometric Ratios : tan =
13 Hypotenuse = . 5 Perpendicular
1 tan
sec =
cosec
1 cos 1 = sin
L.H.S tan2 A – sin2 A
5 = 12 25 = 144
2 –
–
5 13
2
13
5
25 169
C
=
25(169 144 ) 144 169
=
25 25 144 169
A
12
R.H.S. sin4A × sec2 A 4
5 13 = × 13 12 4 2 5 13 = 13 4 12 2 =
54 2
2
=
13 12 So, L.H.S = R.H.S
2
25 25 144 169
Hence Proved.
Ex.10 In ABC, right angled at B, AC + AB = 9 cm and BC = 3 cm. Determine the value of cot C, cosec C, sec C. Sol. In ABC, we have (AC)2 = (AB)2 + BC BC2
(9 – AB)2 = AB2 + (3)2 [ AC + AB = 9cm AC = 9 – AB] 81 + AB2 – 18AB = AB 2 + 9 72 – 18 AB = 0
H2 = P2 + B2 H2 = m22 + n22 H = m 2 n2
sin = sin =
m P = H m2 n 2 m m2 n2
72 AB = = 4 cm. 18 Now, Now, AC + AB = 9 cm
AC = 9 – 4 = 5 cm
A
C
5cm
3cm
4cm
B
BC 3 AC 5 , cosec C = , AB 4 AB 4 AC 5 . sec C = BC 3
So, cot C = .
PAGE # 5151
TRIGONOMETRIC
TRIGONOMETRIC TAB TABLE LE
RATIOS
OF
COMPLEMENTARY ANGLES
sin (90 – ) = cos
cos (90 – ) = sin
tan (90 – ) = cot
cot (90 – ) = tan
sec (90 – ) = cosec cosec (90 – ) = sec cot 54 º tan 20 º tan 36º cot 70 º
Ex.14
Sol.
Ex.11 Given that cos (A – B) = cos A cos B + sin A sinB, find
cot 54 º tan 20 º tan 36º cot 70 º
cos 15º = cos 15º =
2
1 1 3 + 2 2 2
3 1 2 2
cot( 90º 36 º ) tan( 90º 70 º ) + tan 36 º cot 70º
=
tan 36 º cot 70 º + tan 36 º cot 70 º
.
each. Find the length of diagonals and also find its area. Sol. Let ABCD be a rhombus of side 10 cm and
BAD = BCD = 60 º. Diagonals of parallelogram bisect each other.
BD = 10 cm cos 30º =
2
2
2 sin 68º 2 cot 15º cos 22º 5 tan 75 º
Ex.15.
Ex.12 A Rhombus of side of 10 cm has two angles of 60 º
OB sin 30º = AB 1 OB = 10 2 OB = 5 cm A BD = 2(OB) BD = 2 ( 5 )
–
–
= 1 + 1 – 2 = 0.
–
So, AO = OC and BO = OD In right triangle AOB
2
[cot (90 – ) = tan and tan (90 – ) = cot ]
We get cos (45º – 30º) = cos 45º cos 30º + sin 45º sin 30º
–
2.
=
the value of cos15º. Sol. Putting A = 45º and B = 30º
1
–
D
Sol.
2 sin 68 º 2 cot 15º cos 22º 5 tan 75 º –
3 tan 45 º tan 20º tan 40º tan 50 º tan 70º 5
=
2 sin( 90 º 22º ) 2 cot( 90 º 75 º ) cos 22º 5 tan 75 º
–
3(1)(tan 20 º tan 70 º )(tan 40º tan 50º ) 5
=
2 cos 22º 2 tan 75º cos 22º 5 tan 75º
–
3[tan( 90 º 70º ) tan 70 º ][tan( 90 º 50 º ) tan 50º ] 5
C
O 30º B
OA AB
3 OA = 10 2
OA = 5 3 AC = 2(OA) AC = 2 ( 5 3 ) = 10 3 cm So, the length of diagonals AC = 10 3 cm & BD = 10 cm. 1 Area Area of Rhombu Rhombus s= × AC × BD 2 1 = × 10 3 × 10 = 50 3 cm2. 2
3 tan 45 º tan 20º tan 40º tan 50º tan 70º . 5
2 3 (cot 70º tan 70º) (cot 50º tan 5 0º) 5 5 t a n ( 9 0 º – ) = c o t , c o t ( 9 0 º – ) = t a n
= 2 – [
&
sin (90º – ) = cos] = 2 –
2 3 5 5
= 2 – 1 = 1.
Ex.16 If sin 3A = cos (A – 26º) where 3A is an acute angle, find the value of A. Sol. sin 3A = cos (A – 26º)
cos (90º – 3A) = cos (A – 26º) [ sin = cos(90º – ]
90º – 3A = A – 26º 4A = 116º A = 29º
PAGE # 5252
AREA OF TRIANG TRIANGLE LE In a
ABC,, a & b are the length of 2 sides of triangle ABC
and
is the included angle between them.
Then, Area of triangle =
1 ab sin 2
Proof : Ex.17 A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60º and the angle of depression of the base of the hill as 30º. Calculate the distance of the hill from the ship and the height of the hill. Sol. Let x be distance of hill from man and h + 8 be height of hill which is required. Const. : Draw a line from B perpendicular to AC, i.e.BD i.e.BD AC
BD BD = a sin a 1 Area of triangle = base height 2 1 b BD = 2 1 1 b a sin = = ab sin 2 2 sin =
ANGLE OF ELEVA ELEVATION TION
In order to see an object which is at a higher level
compared to the ground level we are to look up. The line joining the object and the eye of the observer is known as the line of sight and the angle which this line
h x In right triangle BCD, =
tan 30º =
of sight makes with the horizontal drawn through the eye of the observer is known as the angle of elevation. elevation. Therefore, the angle of elevation of an object (Figure). helps in finding out its height (Figure).
In right triangle ACB, AC h tan 60º = BC x
1 3
CD 8 BC x
8 x=8 x
3
Height of hill = h + 8 =
3 .x + 8 =
3 8 3 + 8
= 32 m.
Distance of ship from hill = x = 8 3 m. Ex.18 A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angle of elevation of the bottom and the top of the flag staff are respectively 30º and 60º. Find the height of tower. Sol. Let AB be the tower of height h metre and BC be the
ANGLE OF DEPRE DEPRESSION SSION When the object is at a lower level than the observer ’s eyes, he has to look downwards to have a view of the object. In that case, the angle which the line of sight makes with the horizontal through the observer’s eye is known as the angle of depression (Figure).
height of flag staff surmounted on the tower. Let the point on the plane be D at a distance x meter from the foot of the tower In ABD tan 30º =
x= 3h
AB AD
1 3
=
h x
.....(i)
PAGE # 5353
1
3000 = xy 3
( AB = CD)
x + y = 3000 3
.......(ii)
From equation (i) and (ii)
x + 3000 = 3000 3 x = 3000 3
–
3000
x = 3000 ( 3
–
1)
x = 3000 × (1.732 – 1) x = 3000 × 0.732 x = 2196 m. Dis tan ce covered Time taken 2196 = m/sec 15 2196 18 = Km/hr 15 5 = 527.04 Km/hr Hence, the speed of aeroplane is 527.04 Km/hr.
Speed of Aeroplane = In ADC AC AD 5h 3 = x 5h x= 3 From (i) and (ii) tan 60º =
3 h=
......(ii)
Ex.20 If the angle of elevation of a cloud from a point h metres above a lake is and the angle of depression
5h 3
3h = 5 + h
2h = 5
of its reflection in the lake is , prove that the distance of the cloud from the point of observation is 2h sec . tan tan
5 = 2.5 m 2 So, the height of tower = 2.5 m
h=
Ex.19 The angle of elevation of an aeroplane from a point on the ground is 45º. After a flight of 15 sec, the elevatio n changes to 30º. If the aeroplane is flying at a height of
Sol. Let AB be the surface of the lake and let C be a point of observation such that AC = h metres. Let D be the position of the cloud and D’ be its reflection in the lake. Then BD = BD ’. In DCE
3000 metres, find the speed of the aeroplane. Sol. Let the point on the ground is E which is y metres from point B and let after 15 sec flight it covers x metres distance. In AEB
tan 45º =
AB EB tan =
3000 1= y
y= 3000 m In CED tan 30º =
CD ED
.......(i)
CE =
DE CE
H tan
......(i)
In CED’ tan =
ED' EC
PAGE # 5454
CE =
hHh tan
4.
hand when the time is 7 : 20 am is :
2h H CE = tan
.......(ii)
From (i) & (ii)
5.
H 2h H = tan tan H tan = 2h tan + H tan
H tan – H tan = 2h tan H (tan – tan ) = 2h tan 2h tan H = tan tan ........(iii) In DCE
6.
DE sin
7.
8.
2h tan tan tan sin sin 2h cos CD = tan tan sin CD =
CD =
9.
Hence Proved.
703 (B) 720
705 720
(D)
710 720
c
3.
Degree measure of
7 is : 6
(A) 210°
(B) 240°
(C) 270°
(D) None
If tan =
a a sin b cos then value value of is = b a sin – b cos
a2 b2
(B)
a 2 – b 2 a
a2 b2 1
(D)
a 2 b2
If tan 15º = 2 –
a 2 – b2
a 2 b2
3 , then the value of cot2 75º is :
(A) 7 + 3
(B) 7 – 2 3
(C) 7 – 4 3
(D) 7 + 4 3
4 3 cot2 30º + 3 sin2 60º – 3 cosec2 60º – tan2 30º 3 4 2 and b = 3 tan 45º + cos 0º – cot 90º then logb(a) is :
If a =
1 2
(B) –1 (D)
1 2
The angles of the triangles ABC and DEF are given as follows : A = 90 0, B = 300, D = 900 and E = 300. If the side BC is twice the side EF, EF, which of the following statement statem ent is true? (A) Sin B = 2 Sin E
(B) Sin E = 2 Sin B
(C) Sin B = Sin E
(D) Sin A = Sin B
10. The value of the expression
Radian measure of 175º 45 ’ is : 700 (A) 720
2.
(D) 140º
(C)
2h sec tan tan
2h sec observation is . tan tan
(C)
(C) 120º
(A) 2
Hence, the distance of the cloud from the point of
1.
(B) 100º
(C)
H CD = sin Substituting the value of H from (iii)
(A) 80º
(A)
DE sin = CD
CD =
Angle between the minute hand of a clock and hour
1 Degree measure of is : 4
4 3 cot2 30º + 3sin2 60º – 2cosec 2 60º – tan2 30º is : 3 4
(A) 1 (C)
10 3
(B)
–
20 3
(D) 5
11. The value of the expression
(A) 15º 19’5”
(B) 14º 19’ 5”
(C) 15º 18º 6”
(D) 14º 18º 6”
A horse is tied to a post by a rope. If the horse moves along a circular path always keep the rope tight and describes 88 metres when it has traced out 72º at the centre, then the length of rope is :
5 sin2 30º cos 2 45 º 4 tan 2 60º is : 2 sin 30º cos 60º tan 45º (A) 4 (C)
53 12
(B) 9 (D)
55 6
12. The value of tan 5º tan 10º tan 15º tan 20º...... tan 85º is :
(A) 60 m
(B) 65 m
(A) 1
(B) 2
(C) 70 m
(D) 72 m
(C) 3
(D) None
PAGE # 5555
13. If +
1 = and sin = , then sin is : 3 2
2 3
(A)
21. If each of , and is a positive acute angle such that
(B)
2 (C) 3
tan (
2 2 3
3 (D) 4
14. If 7 sin = 24 cos ; 0 < < 14 tan – 75 cos – 7 sec (A) 1 (C) 3
, then value of 2 is equal to : (B) 2 (D) 4
tan is equal to : 15. If tan = 4, then sin3 sin cos cos (A) 0
2
x sin
6
cos 2
4
6
sec
cos ec 2
4
3
tan
cos ec
6
(C) – 2
(D) 0
17. The area of a triangle is 12 sq. cm. Two Two sides are 6 cm and 12 cm. The included angle is :
1 (A) cos 1 3
1 (B) cos 1 6
1 (C) sin 1 6
1 (D) sin 1 3
–
–
–
18. If + = 90º and = 2, then cos + sin equals to : 1 2
(B) 0
(C) 1
(D) 2
(A)
1 1 (C) 37 , 45° & 52 2 2
(D) none
22. If tan (A
1
and tan (A + B) = 3 , 3 0º < A + B 90º, A > B. Then the value of A and B is : (A) 45º, 30º (B) 45º, 15º (C) 60º, 30º (D) none –
B) =
C 2 C 2
(B) sec
C 2
(D) sin
C 2
24. An aeroplane when flying at a height 2500 m from the ground, passes vertically above another aeroplane. At an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 45º and 30º respectively, respectively, then the vertical distance between the two aeroplanes at that instant is : (A) 1158 m (B) 1058 m (C) 1008 m (D) none 25. The shadow of a tower is 30 metres when the sun ’s altitude is 30º. When the sun ’s altitude is 60º, then the length of shadow will be : (A) 60 m (B) 15 m (C) 10 m (D) 5m
–
2
(B) 60º, 45º & 75º
(C) cosec
4
(B) 6
(A) 45º, 45º & 90º
(A) cos
(A) 4
=
A B equals to : 2
16. Find the value of x from the equation :
)
cos
(D) 1
cot 2
–
23. If A, B, C are the interior angl es of a triangle triangl e ABC, then
(B) 2 2
(C)
1 1 , cos( + – ) = and 2 2 + – ) = 1, then the values of , and is :
sin ( +
2
26. The angles of elevation of the top of a vertical tower from two points 30 metres apart, and on the same straight line passing through the base of tower, are 300 and 600 respectively. The height of the tower is : (A) 10 m (B) 15 m (C) 15 3 m
(D) 30 m
19. The difference between two angles is 19º and their 890 sum is grades. Find the greater angle. 9 (A) 63º (B) 35º
(C) 27º
(D) 54º
20. If and are angles in the first quadrant, tan = sin =
1
1 , 7
, then using the formula fo rmula sin (A + B) = sin A 10 cos B + cos A sin B, one can find the value of ( + 2) to
be : (A) 0º (C) 60º
(B) 45º (D) 90º
27. If the angle of elevation of a cloud from a point 200 metres above a lake is 30º and the angle of depression of its reflection in the lake is 60º, then the height of the cloud (in metres) above the lake is : (A) 200 (B) 300 (C) 500 (D) none 28. The angle of elevation of the top of tower from the top and bottom of a building h metre high are and , then the height of tower is : (A) h sin cos / / sin ( + ) (B) h cos cos / / sin( –) (C) h cos sin / / sin( –) (D) None of these
PAGE # 5656
29. When a eucalyptus tree is broken by strong wind, its top strikes the ground at an angle of 30º to the ground and at a distance of 15 m from the foot. What is the height of the tree? (A) 15 3 m
(B) 10 3 m
(C) 20 m
(D) 10 m
30. A man at the top of a vertical lighthouse, observes a boat coming directly towards it.If it takes 20 minutes for the angle of depression to change from 30º to 60º,
35. The trigonometric expression
sec – 1 sin – 1 cot2 + sec2 has the value 1 sin 1 sec [IJSO-2009] (A) –1
(B) 0
(C) 1
(D) 2
36. (1 + tan2 ) / (1 + cot2 ) = (A) tan2 (C) sec
[IJSO-2009] (B) cot 2
(D) cosec2
2
the time taken by the boat to reach the lighthouse from the point when the angle of depression was was 30º, is : (A) 30 minutes (B) 20 minutes (C) 10 minutes
37. If cos + sin =
[NSTSE 2009]
(D) 5 minutes
2 tan
(A) 31. In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and A = 60º, then the length of AD is : (A) 2 3
(B)
15 3 (C) 8
12 3 7
(C)
2 cos sin
(B)
2 sin
(D) none of these.
38. The tops of two poles of heights 20m and 14m are connected by a wire. If the wire makes an angle of 30 º
(D) None of these
32. The expression (1 – tan A + sec A) (1 – cot A + cosec A) has value : (A) – 1 (C) + 1
2 cos, then cos – sin = ?
[IJSO-2008] (B) 0 (D) + 2
with the horizontal, then the length of the wire is : [NSTSE 2009] (A) 40 m (B) 12 m (C) 28 m
(D) 68 m
x 1 2 3 4 = 1, 0º < x < 100º, then the 39. If sin 2 . . ...... x 2 1 2 3
33. A person on the top of a tower observes a scooter moving with uniform velocity towards the base of the tower. He finds that the angle of depression changes from 30 º to 60º in 18 minutes. The scooter will reach the base of the tower in next : (A) 9 minutes (B) 18 / ( 3 (C) 6
–
value of x is equal to : (A) 91º
(B) 80º
(C) 49º
(D) 46º
[IJSO-2008] 40. If p =
1) minutes
[NSTSE 2010]
1 – sin x 1 – sin x cos x ,q= ,r= , then 1 sin x cos x 1 sin x
Which one of the following statement is correct ?
3 minutes
(D) the time depends upon the height of the tower 34. In the diagram, PTR and QRS are straight lines. Given 4 and "T" is the midpoint of PR, calcula te 3 the length of PQ, in cm. [NSTSE 2009]
that, tan xº =
(A) p = q r
[NSTSE-2010] (B) q = r p
(C) r = p q
(D) p = q = r
41. If sin + cosec = 2, then [sin8 + cosec8 ] will have the value : [IJSO-2010] (A) 2 (C) 26
(B) 24 (D) 28
S
42. An aeroplane is flying horizontally at a height of
xº
m c 3
R
3150 m above a horizontal plane ground. At a particular instant it passes another aeroplane verti-
T
P
cally below it. At this instant, the angles of elevation of the planes from a point on the ground are 30º and 60º. Hence, the distance between the two planes at that
m c 6
instant is : [IJSO-2011] Q
(A)
8
(B) 9
(C)
59
(D) 10
(A) 1050 m.
(B) 2100 m.
(C) 4200 m.
(D) 5250 m.
PAGE # 5757
PROTOPLASM •
INTRODUCTION
All the living organisms are essentially formed of numerous coordinated compartments called as cells. Every cell basically formed of two functional regions as plasma membrane and protoplasm. The ground substance of protoplasm, after removing nucleus, all the cell organelles and cell inclusions, is called hyaloplasm/ cytoplasm. It consists of high water contents containing various compounds of biological importance, some of which are soluble in water water e.g. glucose, amino acids, minerals etc. while some of these are insoluble in water e.g. lipids.
•
•
•
•
A compound which releases H+ ion when dissolved in water is called as an acid, e.g., HCl, H2SO4 etc. and base releases OH- ion, e.g., NaOH, KOH etc. Salt is a compound formed, when an acid and a base react with each other. A cell has many salts of Na+, K+, Ca++ and Cl-, HCO3-, PO4-3 etc. A large amount of minerals also occur as hard deposits as crystals within the cell. The salt concentration in cells and in body fluids is of great importance for normal cell functioning. (B) Gase Gases: s: Oxygen, carbon dioxide , nitrogen and other gases are also present in protoplasm.
Physical Properties :
(ii) Its specific gravity is slightly above that of water.
(C) Water : Water is not an organic molecule because it does not contain carbon. The bonding properties of water account for some of its characteristics, which are very important to living organisms.
(iii) Its viscosity has been found to be like that of glycerin.
(i) Water is the main component of cell contents and body fluid.
(iv) It has power of responding to external stimuli, like heat, electric shocks, application of chemicals etc.This property of protoplasm is called irritability. irritability.
(ii) It is neutral with pH 7. It ionises to H+ and OH- ion. Phospholipids, nucleic acids and proteins by accepting or donating H+ ions from water attain specific ionic state.
•
(i) Protoplasm is a polyphasic colloidal system.
(v) I t exhibits streaming movement e.g. rotatory movements in the leaves of aquatic plan ts like Hydrilla and Vallisneria.
(iii) It forms an average 55 to 60% of living material. (iv) Water dissolves more substances in it than any other liquid due to it's highest known dielectric constant (the measure of capacity to neutralize the attraction between electric charges).Only polar molecules dissolve in water.
(vi) Amoeboid movement of the protoplasm can also be noticed in myxomycetes and Amoeba. (vii) In general, the pH of cytoplasm is slightly acidic i.e. 6.8, however pH of the nucleoplasm is 7.6 to 7.8. •
(v) It is generally non toxic to the cell. Colloids like starch, glycogen and protein remain dispersed in water in cell cytoplasm.
Chemical Properties :
The collection of various types of biomolecules of a cell collectively form cellular pool. pool. Elements do not occur in free form but combines to form organic molecules and inorganic molecules. Organic and inorganic compounds occur in a ratio of 9 : 1. Cellular pool is mainly constituted by : •
•
(vi) It is a medium of heat exchange and transfer. (vii) It participates in chemical reactions both as a reactant and a product. It forms an ideal medium for chemical reactions, because dissolved molecules can make intimate contact.
Inorganic materials include salts, minerals and water. These materials generally occur in aqueous phase which contains molecules and ions d issolved in water.
(viii) It acts as a lubricating and protective fluid.
Organic compounds as carbohydrates, lipids, amino acids, proteins, nucleic acids and vitamins. These molecules usually occur in aqueous and non aqueous phase.
A large series of covalent compounds are formed with the help of carbon, hydrogen and some other elements. These are called as organic compounds. Their special properties distinguish them from inorganic compounds found in living bodies.
–
•
The inorganic substances include salts, minerals and water. (A) Minerals: These occur in ionic state and form only 1-3 % of cellular pool. Cellular functions fail to occur in the absence of proper ionic balance in the cell cytoplasm and extracellular fluid.
(i) Carbohydrates : •
Carbohydrates can be chiefly composed of carbon, hydrogen and oxygen. In this hydrogen and oxygen atoms are present in a ratio of 2 : 1. As in carbohydrates hydrogen and oxygen are present therefore these are also termed as hydrates of carbon. PAGE # 59
•
•
•
•
Carbohydrates have general formula as Cn H2n On.
•
Carbohydrates are widely distributed in plant tissues and in animal tissues. In animals they are in the form of glucose and glycogen. In plants they are in the form of cellulose and starch. One gram of carbohydrate yields about 4 kilocalories of energy ene rgy..
(B) Disaccharides : The disaccharides are sugars composed of two molecules of the same or different monosaccharides, united by a glycosidic glycosidic linkage. They have a general formula Cn (H 2O) n 1 . These include maltose, lactose , sucrose.
Carbohydrates can be defined chemically as aldehyde or ketone derivatives or the poly hydric (more th an one –
–
OH group) alcohol and their derivatives.
Carbohydrates can be further divided as : (A) Monosaccharides (B) Disaccharides (C) Polysaccharides (A) Monosaccharides : These sugars cannot be hydrolyzed into simpler forms. They have the general formula C n H 2n O n . The simplest types of monosaccharides ar e glyceraldehyde and dihydroxyacetone. Depending upon the number of carbon atoms present, these can be further subdivided into trioses (e.g. (e.g. Glyceraldehyde), Glyceraldehyde ), tetrose (e.g. Erythrose), Erythrose), pentoses (e.g. (e.g. Ribose, Deoxyribose ) etc. Generally if free H is present at carbon 1 the sugar is an aldose but if a CH2OH group is substituted, the sugar is a ketose. They have reducing property due to the presence of aldehyde or ketone group present in them. Some examples are as follows :
•
•
•
–
•
Glucose : It is a hexose. It 's formula is C6 H12 O6 . It is normally found in fruit juice and formed in the body by the hydrolysis of starch, cane sugar, maltose and lactose. Glucose is said to be the sugar of the body. It is a principal sugar in blood, serving the tissue as a major metabolic fuel. Normal level of blood glucose is 80 120 mg / 100 ml of blood. When the blood sugar level exceeds the threshold value i.e. 180 mg / 100 ml, glucose begins to appear in the urine. This condition is called as glycosuria.
•
•
Glucose simple chain •
•
•
•
•
C H 2O H O H OH
H OH
H H
OH
H OH Ring structure of glucose
•
Note : The simple ring structure of glucose is given by Haworth. Haworth. Fructose : Fructose or fruit sugar is also known as levulose. Similar to glucose it is a ketohexose and less readily absorbed by tissue cells. It is obtained by the hydrolysis of cane sugar.
Maltose: They consist of two glucose residues. It's occurrence have been reported in germinating cereals and malt. It is the major product of enzymatic hydrolysis of starch. Lactose: It is found in milk to the extent of about 5%. Upon hydrolysis it yields a mixture of galactose and glucose. Sucrose : A single molecule of sucrose consists of one glucose and one fructose molecule. It is the common sugar of commerce and the kitchen. It is derived commercially from either cane sugar or occurs in varying amount in a variety of fruits, seeds, leaves , flowers, roots and in maple sugar. On hydrolysis it yields an equimolar mixture of glucose and fructose. (C) Polysaccharides : Polysaccharides (Glycans (Glycans)) are those which yield more than six molecules of monosaccharides on hydrolysis. It's general formulae is (C6 H10 O5)n. It's examples are as follows :
–
Structure of glucose O II C – H I H – C – OH I HO – C – H I H – C – OH I H – C – OH I C H 2O H
Galactose : It is found in milk sugar or lactose along with glucose. It is synthesized in the mammary glands and combines with glucose to make the lactose of milk. In the liver it can be changed to glucose and thus used in the body. It is a part of glycolipids and glycoproteins.
Cellulose : It is the chief constituent of the frame work of plants, constituting 50% or more of all carbon in vegetation. It is a linear and unbranched homopolysaccharide of about 6000 to 10,000 - D Glucose molecules. Mammals do not have cellulase enzyme and therefore cannot digest wood & vegetable fibers. Purest form of cellulose is found in cotton which is about 90%. Starch : It is the most important food source of carbohydrates and is found in cereals, potatoes, legumes and other vegetables. Chemicall y, the starch is formed of two glucose polymers : -Amylose (an unbranched but spiral chain of about 200-2000 Glucose molecules) and Amylopectin (a branched chain of about 2000-20,000 -Glucose molecules.). Natural starch is insoluble in water and gives a blue colour with iodine solution. Glycogen : The counterpart of starch in the animal body is glycogen thats why it is also called as animal starch, which occurs in significant amount in liver and muscles. Glycogen is non reducing sugar which gives red colour with iodine. It is a branched homopolysaccharide formed of about 30,000 D-Glucose molecules. –
•
Note : Glycosidic linkage: The linkage between the hydroxyl groups of two monosaccharide molecules with the release of one molecule of water. PAGE # 60
•
•
•
•
•
Biological significance of carbohydrates : Carbohydrates serve as an important structural material in some animals and in all plants, where they constitute the cellulose framework. Carbohydrates are essential for life. Almost all animals use them as respiratory fuel. In animal cells, carbohydrates are in the form of glucose and glycogen, which serve as an important source of energy for the vital activities. Carbohydrates play a key role in the metabolism of amino acids and fatty acids.
•
Is o l e u c i n e
Al a n i n e
Ar g i n i n e
L e u c in e
As p a r g i n e
H i s ti d i n e
Me th i o n i n e
As pa p a r ti c a c i d
–
P h e n yl a l a n i n e
C ys tte e in e
–
Th r e o n i n e
G l u ta m ic ic ac a ci d
–
Tr yp to p h a n
G l u ta m in in e
–
Va l i n e
G l yc i n e
–
L ys i n e
P r o li n e
–
–
Serine
–
–
Tyrosine
–
Essential amino acids : They are 8 in number. They are not synthesized in a human body and are obtained from food etc. are called as essential amino acids. Non essential amino acids : They are 10 in number. They are synthesized in a human body and are termed as non essential amino acids. –
–
Semi essential amino acids : They are two in number and needed by growing children and lactating and pregnant women. –
H I
N
H O I II —C—C— I R1 H I
H I
N
—
C
— I R1
OH
O II
C
—
H I
H—
N
H I —C— I R2
H O H I I II N—C—C— I R2
O II C —
OH
OH
Peptide bond formation
•
The R group is variable in different amino acids. Amino acids can react with acid and base both, this is due to the presence of carboxyl and amino groups in them. There are about 20 amino acids that take part in the formation of proteins. The 20 amino acids are further divided into three groups :
Note : Peptide bond : Polypeptide and simple protein consist entirely of long chain of amino acids linked together by peptide bonds formed between the carboxyl group of one amino acid and the amin o group of other amino acid. A molecule of water is released out during bond formation.
H—
•
•
Semi-Essential Amino acids
H—
•
•
Non-Essential Amino acids
Some carbohydrates have highly specific functions e.g. ribose in the nucleoprotein of the cells, galactose in certain lipids and the lactose of milk. (ii) Protein : The name protein is derived from the Greek word proteios, which means"Of means"Of the first rank". This was coined by Berzelius in 1838. Proteins are the complex nitrogenous substances found in the cells of animals and plants. Chemically proteins are polymers of molecular units called as amino acids. These polymers contain carbon, oxygen, nitrogen and hydrogen atoms. Usually sulphur atoms are also present .Certain proteins contain phosphorus or some trace metal elements, such as copper, iron etc. in addition to the other elements. The proteins have high molecular weight. One gram of protein yields 4 kilocalories of energy. The amino acids found in a molecule of protein are linked together by peptide bonds. The general structure of a amino acid is represented by the following formula :
•
Essential Amino acids
•
•
Biological significance of proteins : They act as a structural components of cell. They are essential for growth and repair of the body. All the enzymes are made up of protein s. They help to catalyze various reactions occurring in our body. body. They play important roles as hormones, antibodies, etc. Haemoglobin , the respiratory pigment of animals is a conjugated protein composed of colourless basic protein the globin and haem. (iii) Lipid : Term lipid was coined by Bloor. Bloor. Fats and their derivatives are collectively known as lipids (In greek Lipas = fat ). The principal component associated with most lipids are the fatty acids. acids. The lipids are a heterogenous groups of substanc es which have the common property of being relatively relatively insoluble in water and soluble in non polar solvents such as ether, ether , chloroform and benzene. They consist of comparatively less oxygen. One gram of fat yields 9 kilocalories of energy. Similar or different fatty acids participate in the composition of a fat molecule. The lipids include fats, oils, ghee, waxes and related compounds. –
PAGE # 61
•
•
•
•
•
•
•
Note : Lipids generally consist of a single molecule of glycerol and three molecules of fatty acids joined acids joined together by ester bonds. Therefore these are also termed as triglycerides. Three molecules of water are released during the formation of triglycerides.
stranded i.e. it has two polynucleotide chains. •
1.
Biological significance of lipids : They takes part in the synthesis of steroids, hormones, vitamin D, bile salts etc. They act as a solvent s olvent for fat soluble vitamins i.e. vitamin A , D, E and K. They act as storage compounds in animals, in the fruits and seeds of plants and in other organism. They act as structural cellular components particularly in cell membranes. They are found in the form of phospholipids, glycolipids and sterols. sterols. They act as insulators. They provide electrical and thermal insulation. They are deposited beneath the skin and other internal organs to reduce the heat loss. They also work as shock absorbers and other mechanical impacts.
2.
Transf ansfer er RNA (t RNA) :
•
It
is 10
•
It
is synthesized in the nucleus by DNA.
•
It
is also known as soluble RNA.
•
It
is also known as adapter RNA.
•
RNA.. At the time of protein synthesis It is the smallest RNA
•
•
3.
•
•
•
place of nitrogen base, thymine present in DNA there is a nitrogen base uracil in RNA. RNA is made up of only one polynucleotide chain i.e. RNA is single stranded.
•
In
RNA, polynucleotide chain runs in 3' 5' direction.
•
Exception : RNA found in Reo virus is double –
has the most complex structure.
Messen Mess enge gerr RN RNA (m RNA) : The m RNA is 1 5 % of the cells total RNA. The m RNA is produced by genetic DNA in the nucleus. This process is called as transcription, m RNA is also called as template RNA. It acts as the template for protein synthesis. –
–
–
•
General properties of enzymes :
•
They remain unaltered at the end.
•
They are required in small quantities.
•
They accelerate the rate of reaction.
•
They are proteinaceous in nature.
•
Enzymes are highly specific towards substrate.
•
Certain enzymes exhibit the property of reversibility.
In
In
It
Enzymes are protein catalysts for biochemical reactions in the living cells. Th e substance which increases reaction rate is called as catalyst and the phenomenon is called as catalysis. The term enzyme is derived from Greek word which means 'in yeast' because the yeast cells were the first to reveal enzyme activity in living organisms. Enzyme was first introduced by W. Kuhne in 1878. Berzelius was the first to define and recognize the nature of catalyst. In 1926 J.B Sumner isolated the enzyme urease as a crystalline protein for the first time. Enzymes could be intracellular and extracellular enzymes. enzymes. When the enzymes remain and function inside the cells, they are called as endoenzymes or intracellular enzymes. The enzymes which leave the cell and function outside the cell are called extra cellular enzymes.
Nitrogen bases : There are two types of purines which include adenine (A), and guanine (G) and pyrimidines which include thymine (T), uracil (U) and cytosine (C). In DNA adenine, thymine, guanine and cytosine present while in RNA uracil is present in place of thymine.
place of deoxyribose sugar of DNA, there is a presence of ribose sugar in RNA.
15% of total RNA
–
Pentose sugar : It is a 5 - carbon containing sugar which is ribose is RNA and deoxyribose in DNA.
(B) RNA : (Ribonucleic acid) Structure of RNA is fundamentally the same as DNA but there are some differences. The differences are as follows.
–
–
Nucleotides : A single nucleotide consist of following parts :
Phosphate group : PO4-3 group in the form of H3PO4
–
it acts as a carrier of amino acids.
Å. One turn of spiral has a distance of 34 Å and distance between two adjacent nucleotides is 3.4 Å.
•
–
–
(A) DNA (Deoxyribose (Deoxyribose nucleic aci ds) : DNA is coiled macromolecule made of two antiparallel chains held together by hydrogen bonds. DNA has diameter of 20
•
Ribos ibosom omal al RNA (r RNA) : This RNA is 80% of the cell's total RNA. It is the most stable form of RNA. It is found in ribosomes and it is produced in nucleolus. They are present as 80 S type of ribosome in S type of ribosome in eukaryotic cells and 70 prokaryotic cells. It is the site of protein synthesis. –
(iv) Nucleic acids : These are the hereditary materials of living organisms. There are two types of nucleic acids :
•
Types Types of RNA- A cell contains three types of o f RNA :
(vi) Pigments :
The coloured substance found in the living being is called as pigment. The beauty of nature is due to animals, birds and flowers having different pigments. The living beings de pend on sun for fo r energy. The green pigment in nature is called as chlorophyll, chlorophyll, can only PAGE # 62
store light energy obtained from the sun, in the form of chemical energy. Thus, chlorophyll is the nutritional basis of life on earth. The colour of our skin is due to the pigment melanin. melanin. Haemoglobin & haemocyanin pigments play an important role in transportation of oxygen in the body of living beings. Pigments belong to the group carotenoid are found in both plants and animals. BIOCHEMICAL REACTIONS
The reactions undergoing inside a living cell to sustain life are called as biochemical reactions. The biological system can't use heat liberated in biological reactions directly as they are isothermic so the biological systems use chemical energy (ATP) to perform various living processes. Biochemical reactions are catabolic ( breakdown/exergonic reactions) reactions ) and anabolic (synthetic reactions), reactions), collectively they are called as metabolic reactions. reactions.
EXERCISE 1.
Which of the following is a disaccharide ? (A) Galactose (B) Fructose (C) Maltose (D) Dextrin
2.
Nucleic acids are made up of (A) amino acids (B) pentose sugars ( C) n u c l e o s i d e s (D) nucleotid es
3.
Which of the following is not a carbohydrate ? (A) Starch (B) Glycogen (C) W ax (D) Glucose
4.
5.
To get quick energy one should use (A) carbohydrates (B) fats (C) vitamins (D) proteins Circular and double stranded DNA occurs in (A) golgi body (B) mi tochondria (C) nucleus (D) cytoplasm
6.
The most abundant protein in human body is (A) collagen (B) myosin (C) actin (D) albumin
7.
Which is not a polysaccharide ? (A) Sucrose (B) Starch (C) Glycogen (D) Cellulose
8.
9.
The decreasing order of the amount of organic compounds, present in an animal body is (A) carbohydrates, proteins, fats, and nucleic acid (B) proteins, fats, nucleic acid and carbohydrates (C) proteins, fats, carbohydrates and nucleic acid (D) carbohydrates, fats, proteins and nucleic acid Term protoplasm was introduced by (A) Purkinje (B) Schultze (C) Sutton and Boveri (D) Van Mohl
10. Which of the following is a monosaccharide ? (A) pentose sugar (B) hexose sugar (C) glucose (D) all of the above 11. The process of m-RNA synthesis on a DNA template is known as (A) translation (B) transcription (C) transduction (D) transformation 12. Which amino acid is non essential for a human body ? (A) Glycine (B) Phenylalanine (C) Arginine (D) Methionine 13. Double helix model of DNA was proposed by (A) Watson and Crick (B) Schleiden and Schwann (C) Singer and Nicholson (D) Kornberg and Khurana 14. Which of the following nitrogen base is not found in DNA ? (A) Thymine (B) Cytosine (C) Guanine (D) Uracil 15. Glycogen is a / an (A) polymer of amino acids (B) polymer of fatty acids (C) unsaturated fat (D) polymer of glucose units 16. Carbohydrate is a (A) polymer of fatty acids (B) polymer of amino acids (C) polyhydric aldehyde or ketone (D) none of the above 17. In which form, food stored in animal body ? (A) Glucose (B) Glycogen (C) Cel lulose (D) ATP 18. Chemically enzymes are (A) fats (C) hydrocarbons
(B) carbohydrates (D) proteins
19. Long chain molecules of fatty acids are obtained by (A) polymerisation of two carbon compounds (B) decomposition of fats (C) polymerisation of glycogen (D) conversion of glycogen 20. The amino acids which are not synthesized in the body are called as (A) non essential (B) essential (C) deaminated (D) all of them are correct –
21. Fats in the body are formed when (A) glycogen is formed from glucose (B) sugar level becomes stable in blood (C) extra glycogen storage in liver and muscles is stopped (D) all of the above
PAGE # 63
22. Which element is not found in nitrogen base ? (A) Nitrogen (B) Hydrogen (C) Carbon (D) Phosph orous 23. Proteins are the polymers of (A) amino acids (C) enzymes
(B) natural protein (D) nucleic acids
29. Similarity in DNA and RNA is (A) both are polymers of nucleotides (B) both have similar pyrimidine (C) both have similar sugar (D) None of the above 30. In which stage of cell cycle, DNA replication occurs (A) G1 phase (B) S Phase (C) G2 phase (D) M phase c31. Bacteria cannot survive in a highly salted pickle because [IJSO-Stage-I/2011] (A) they become plasmolysed and consequently die. (B) they do anaerobic respiration. (C) water is not available to them. (D) of all the reasons mentioned above. –
24. DNA polymerase is needed for (A) replication of DNA (B) synthesis of DNA (C) elongation of DNA (D) all of the above 25. Duplication of DNA is called as (A) replication (B) transduction (C) transcription (D) translation 26. Ligase enzyme is used for (A) denaturation of DNA (B) splitting of DNA into small fragments (C) joining fragments of DNA (D) digestion of lipids
(D) glycogen & fructose
–
–
32. Maximum vitamin A content is likely to be found in the extract of [IJSO-Stage-I/2011] (A) sprout of pulse (B) cod liver (C) white muscles (D) rose petals
27. Orange juice contains plenty of (A) vitamin C (B) vitamin A (C) vitamin D (D) vitamin E 28. Sucrose is composed of (A) glucose & fructose (B) glucose & glycogen (C) two molecules of glucose
–
33. The ointment prescribed for burns usually contains, among other ingredients, [IJSO-Stage-I/2011] (A) vitamin A (B) vitamin B (C) vitamin D (D) vitamin E 34. Unsaturated fatty acids contain [IJSO-Stage-I/2012] (A) atleast one double bond (B) two double bonds (C) more than two double bonds (D) no double bond
PAGE # 64
SERIES COMPLETION Series completion problems deals with numbers, Ex 5. alphabets and both together. While attempting to solve the question, you have to check the pattern of the series. Series moves with certain mathematical Sol. operations. You have to check the pattern. Type of questions asked in the examination : (i) Find the missing term(s). (ii) Find the wrong term(s).
Ex 6.
Sol. NUMBER SERIES
(i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression) 2
(ii) a, ak, ak , ak , ................(Geometric Progression) Ex 7.
Sol. (iv) Series of prime numbers – i.e. 2, 3, 5, 7, 11, ...... (v) Series of composite numbers – i.e. 4, 6, 8, 9, 10, 12, ................. Directions : (1 to 10) Find the missing numbers : Ex 1.
Sol.
16, 19, 22, 25, ? (A) 27 (B) 28 (C) 29 (D) 25 (B) As per series a, a + d, a + 2d,......... a = 16 d=3 a + 4d = 16 + 4 × 3
Ex 8.
Sol.
Ex 9. Ex 2.
Sol.
Ex 3.
Sol. Ex 4.
Sol.
9, 18, 36, ?, 144 (A) 70
(B) 56
(C) 54 (D) 72 2 (D) As As per series, a, ak, ak , ak3, ........ a = 9, k = 2 ak3 = 9 × 23 = 72 2, 6, 14, 26, ? (A) 92 (B) 54 (C) 44 (D) 42 (D) The pattern is +4, +8, +12, +16, ....... 240, ? , 120, 40, 10, 2 (A) 120 (C) 40
(B) 240 (D) 10
1 1 1 1 (B) The pattern is ×1, × , × , × , × 3 5 2 4 missing term = 240 × 1 = 240
3, 9, 36, 180, ? ( A ) 1 0 80 (B) 900 (C) 720 (D) None of these (A) Each term is multiplied by 3, 4, 5 and so on respectively. Therefore, the next term would be 180 × 6 = 1080.
A multiple series is a mixture of more than one series :
3
a a a (iii) a, , 2 , 3 , .............(Geometric .............(Geometric Progression) k k k
8, 12, 21, 46, 95, ? ( A ) 1 88 (B) 214 (C) 148 (D) 216 2 2 (D) The pattern is + 2 , + 3 , + 52, + 72, ....... missing number = 95 + 112 = 216
Sol.
Ex 10.
Sol.
4, 7, 3, 6, 2, 5, ? (A) 0 (B) 1 (C) 2 (D) 3 (B) The sequence is a combination of two series I 4, 3, 2, ? II 7, 6, 5 The pattern followed in I is – 1, – 1, – 1 missing number = 2 – 1 = 1 14, 15, 12, 16, 9, 18, 4, 21, ? (A) 2 (B) 3 (C) – 3 (D) – 5 (C) The sequence is a combination of two series. 14, 12, 9, 4, (....) and 15. 16, 18, 21 The pattern followed in is – 2, – 3, – 5, ....... missing number = 4 – 7 = – 3 1, 1, 4, 8, 9, ? ,16, 64 (A) 21 (B) 27 (C) 25 (D) 28 (B) (B) (i) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............] (ii) (ii) 1, 8, __, __, 64 mixe mixed d combin combinat atio ion n 3, 6, 24, 30, 63, 72, ?, 132 (A) 58 (B) 42 (C) 90 (D) 120 (D) The difference between the terms is given below as :
3
6
24 30
Difference 3
18
6
Difference
3
15
63
33 3
72 9 15
?
48
132 ?
?
Therefore, alternate difference between the difference is 3 and 15 respectively. Hence, the next term would be 72 + 48 = 120. PAGE # 65
Directions : (11 to 12) Find the wrong term(s) — Ex 11.
Sol.
Ex 13.
1 2 28 64 14 0 3 7 (P) (Q) ( R) (S) (T) Which number will come in place of (T) ? ( A ) 1 4 12 (B) 164 (C) 696 (D) 78
Sol.
(A)
5, 8, 10, 12, 15, 18, 20, 23 (A) 8 (B) 12 (C) 15 (D) 18 (B)
Therefore, number 12 is wrong and should be Similarly replaced by 13. Ex 12.
Sol.
37
1, 3, 8, 19, 42, 88, 184 (A) 3 (B) 8 (C) 19 (D) 88 (D) 3
1 2
19
8 11
5 3
6
12
(R)
( S)
(T)
78
164
340
696
1412
×2+8
×2 +12
×2 +16
×2+20
Therefore, the number 1412 will come in place of (T). Ex 14. 184
89 47
24
(Q)
×2+4
42 23
(P)
95 48
Hence, number 88 is wrong and should be replaced by 89. or 1 × 2 + 1, 3 × 2 + 2, 8 × 2 +3, 19 × 2 + 4, 42 × 2 + 5, Sol. 89 × 2 + 6 Directions : (13 to 14) In each of the following questions, a number series is given. After the series, below it in the next line, a number is given followed by (P), (Q), (R), (S) and (T). You have to complete the series starting with the number given following the sequence of the given series. Then answer the question given below it.
2 9 57 33 7 3 (P) (Q) (R) (S ) Which number will come in place of (Q) ? (A) 113 (B) 17 (C) 3912 (D) 8065
(T)
(A) Similarly, 3 ×8 –7
(P)
(Q)
17
113 ×7 –6
(R)
(S)
673
3361 ×5 –4
×6 –5
Therefore, the number numbe r 113 will will come in place pla ce of (Q).
ALPHABET SERIES
(i) Position of the letters
(ii) Difference between the alphabets
(i) Position of alphabets : Alphabets in order :
Alphabets in reverse order :
Directions : (15 to 24) Find the missing term(s) : Ex 15.
Sol.
Ex 16.
Sol.
Ex 17.
B, E, H, ? (A) K (B) L Sol. (C) J (D) M (A) In the given series, every letter is moved three steps forward to obtain the corresponding letters of the next term. So, the missing term is K. Q, N, K, ?, E (A) H (B) I (C) J (D) G (A) In the given series, every letter is moved three Ex 18. steps backward to obtain the corresponding letters of the next term. So, the missing term is H.
A, Y, Y, D, W, G, U, J, ? (A) R (B) T (C) S (D) P (C) The given sequence consists of two series : . A, D, G, J in which each letter is moved three steps forward to obtain the next term . Y, Y, W, U, ? in which each letter l etter is moved mov ed two steps backward to obtain the next term. So, the missing term would be S. AG, AG, LR, WC, W C, HN, ? (A) SX (C) SY
(B) RY (D) TX PAGE # 66
Sol.
(C) The first letter of each group and the second Ex 24. letter of each group differs by 11 letters between them. L A W H Sol. 12
1
11
23
11
11 R 18
G 7
Similarly,
Alphabetical positions
8
11
Difference in Alphabetical positions C N 3 14 11
11
D E F 15 , G H I 24 A B C 6 , 4 5 6 1 2 3 7 8 9
N
S
So, the missing term would be J K L 33 11 12 10
Alphabetical positions Difference in Alphabetical positions
Therefore, the next group of letter would be SY. SY. H
Directions : (25 to 27) Find the wrong term (s) : Ex 25.
Y
And 11
Ex 19.
Sol.
Ex 20.
Sol.
Ex 21.
Sol.
Ex 22.
Sol.
Ex 23.
Sol.
(ABC) – 6, (DEF) – 15, (GHI) – 24, ? (A) (IJK) – 33 (B) (JKM) – 33 (C) (IJK) – 32 (D) (JKL) – 33 (D) In a given series Let A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, and so on
11
Sol.
AD, EI, JO, PV, ? (A) VD (B) W C (C) W D (D) VE (C) The first letter of subsequent groups have a difference of 4, 5 and 6 places respectively, respect ively, whereas the second letter of the subsequent groups has a difference of 5, 6, and 7 places respectively, Therefore, on following the same pattern, we get Ex 26. ‘WD ’ as the next term which would replace the question mark. Sol. AB, BA, ABD, DBA, PQRS, ? (A) SRQP (B) SRPQ (C) SQRP (D) RSQP (A) The first term is reversed to get second term. The third term is reversed to get the fourth term. Similarly, to get the sixth term, we reverse the fifth Ex 27. term. So, the missing term would be SRQP. HEJ, JGL, LIN, NKP, ? (A) MOR (B) PNS Sol. (C) PMR (D) NPT (C) First letter of each group differs by 2 letters. Second letter of each group differs by 2 letters. Third letter of each group differs di ffers by 2 letters. All the letters differ in the forward direction. Hence, the next choice would be PMR. XYQ, ZAR, BCS, DET, DET, ? (A) GFU (B) FUG (C) FZU (D) FGU (D) Here, first two terms of every group of letters are in continuation, like XY, ZA, BC, DE, and the third letter of each group is again in forward continuation, i.e. Q, R, S, T. Hence, the term replacing the question mark would be FGU.
ABD, DGK, HMS, NTB, SBL, ZKW (A) NTB (B) DGK (C) SBL (D) ZKW (A) First letter of first, second, third,.........terms is moved three, four, five, ........steps forward respectively. Similarly Similarly,, second letter is moved five, six, seven,......steps forward respectively and third letter is moved seven, eight, nine,........steps forward respectively. Hence, NTB is the wrong term and should be replaced by MTB. EPV, FQW, GRX, HTY, HTY, ITZ (A) FQW (B) GRX (C) HTY (D) ITZ (C) In every term, first second and third letter is in alphabetical order to its next term respectively. Fourth term is not following the same rule. Hence, HTY is the wrong term and should be replaced by HSY. D4V, G10T, G10T, J20R, J2 0R, M43P M43 P, P90N P90 N (A) P90N
(B) G10T
(C) J20R
(D) D4V
(B) First letter of every term is moved three steps forward in each next term. Second number of every term of the pattern × 2 + 1, × 2 + 2, × 2 + 3,............and third letter of every term is moved two steps backward. Hence, G10T is the wrong term and should be replaced by G9T.
LETTER REPEATING SERIES
Pattern of such questions is that some letters in sequence are missing. (i) The letters may be in cyclic order (clockwise or anti-clockwise). (ii) To solve a problem, we have to select sel ect one of the alternative from the given alternatives. The alternative which gives a sequence form of letters is the choice.
17Z5, 15X4, 13V3, ?, 9R1 (A) 11S2 (B) 11T2 (C) 11U2 (D) 11T3 Directions : (28 to 32) Find the missing term(s) : (B) The first number & second letter of every term is moved two steps backward & the third number Ex 28. a a _ b a a _ b b b _ a of every term is moved one step backward. ba ckward. So, the ( A ) b aa (B) abb missing term would be 11T2. (C) bab (D) aab
PAGE # 67
Sol.
(A) we proceed step by step to solve the above Ex 35. series: Steps :
1.
The first blank space should be filled in by 'b 'b' so that we have two a's a's followed by two b's. b's.
2.
Second blank space should be filled in by 'a 'a' so that we have three a's followed by three b' s.
Sol.
3.
a _ h _ _ c _ n e _ h _ e a c _ _ _ _ _ 2 1 _ 4 3 _ 5 _ _ 2 5 4 _ _ _ _ _ _ _ _ The last five terms in the series are ( A) 3 2 524 (B) 43215 (C) 25314 (D) 32541 (B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, the numbers series runs as 21543 15432 54321 43215. If first digit of a group of five digits is placed as the last digit, we obtain the second group of five digits and so on.
The last blank space must be filled in by 'a 'a' to keep the series in sequence. Ex 29. _ bca _ ca _ c _ b _ Ex 36. _ m y e _ _ y l x _ y l m _ _ l _ _ _ _ (A) aabbc (B) abbbc 4 6 _ 5 8 6 _ _ _ 5 7 _ 6 5 8 _ _ _ _ _ (C) aabcc (D) abbac The last five terms of the number series are ( A) 4 6 758 (B) 74658 (C) 76485 (D) 46785 Sol. (D) Sol. (D) By taking e = 5, l = 4, m = 6, y = 7 and x = 8 the number series runs as 46758 67485 74658 46785. By taking the digits in the groups of five, we find Series is abc/ abc/ abc/ abc. So, pattern abc is that first digit of the first group (i.e. 4) is the third repeated. digit of the second group and the last two digits Ex. 30 a _ abb _ aa _ ba _ a _ b have interchanged their positions. The same rule (A) ababa (B) aabba applies in others groups also. (C) aabab (D) aaabb Sol(C) Series is aaabb/ aaabb/ aaabb. So, pattern Direction : (37) In the following question, three sequences aaabb is repeated. of letter/numbers are given which correspond to each other in some s ome way. way. In the given question, quest ion, you Ex 31. a _ c _ abb _ ca _ a have to find out the letter/numerals that come in (A) baca (B) bbca the vacant places marked by (?). These are given (C) bacc (D) bacb as one of the four alternatives under the question. Sol(A) Series is abc/ aabbcc/ aaa Mark your answer as instructed. Ex 32. bc _ b _ c _ b _ ccb Ex 37. C B _ _ D _ B A B C C B (A) cbcb (B) bbcb (C) cbbc (D) bcbc _ _ 2 3 5 4 _ _ ? ? ? ? Sol(A) Series is bccb / bccb / bccb. So, pattern bccb is p _ p q _ r _ q _ _ _ _ repeated ( A) 4 5 5 4 (B) 4 3 3 4 (C) 4 2 2 4 (D) 2 5 5 2 Directions : (33 to 34) The question given below is based (C) Comparing the positions of the capital letters, on the letter series, In series, some letters are Sol. numbers and small letters, we find p corresponds missing. Select the correct alternative. If more than to C and 2 corresponds to p. So, p and 2 correspond five letters are missing, select the last five letters of the series. to C. q corresponds to A and 3 corresponds to q. So, q and 3 corresponds to A. Also, 5 corresponds Ex 33. xyzu _ yz _ v _ _ uv _ _ _ _ _ _ _ to D. So, the remaining number i.e., 4 corresponds (A) uvxyz (B) vuzyx to B. So, BCCB corresponds to 4, 2, 2, 4. (C) uvzyx (D) vuxyz Sol. (A) The series is x y z u v / y z u v x/ z u v x y/u v x y z MISSING TERMS IN FIGURES Thus the letters are written in a cyclic order. EX 34.
Sol.
abcd _ bc _ e _ _ de _ _ _ _ _ _ _ (A) deabc (B) edcba (C) decba (D) edabc (A) The series is a b c d e / b c d e a / c d e a b / b e a b c Thus the letters are written in a cyclic order.
Directions : (38 to 47) Find the missing number(s) :
Ex 38.
Direction : (35 to 36) There is a letter series in the first row and a number series in the second row. Each number in the number series stands for a letter in the letter series. Since in each of that series some term are missing you have to find out as to what Sol. those terms are, and answer the questions based on these as given below in the series.
6
9
15
8
12
20
4
6
?
(A) 5 (B) 10 (C) 15 (D) 21 (B) In the first row, row, 6 + 9 = 15 In the second row, 8 + 12 = 20 In the third row, missing number = 4 + 6 = 10. PAGE # 68
3 Ex 39.
10 2
6
Ex 43.
30 3
9
4 (A) 11 (C) 3
Sol.
(B) 6 (D) 2 Sol. 6 4 8 (C) Clearly, in the column, 3 18 3 27 We take x in place of ? In the column, 2 Similarly in the column,
15 x 5
9 ,x
95 15
Ex 40.
27D
9E
7I
21K
3M
4D
?
7J
Sol.
5
2
5
]
]
]
5
26
4
6
3
?
8
4
6
3C
?
6
(A) 15 (B) 20 (C) 25 (D) 40 (B) Clearly In first figure 6 × 3 – 4 × 2 = 18 – 8 = 10 In second figure 9 × 5 – 5 × 3 = 45 – 15 = 30 In third figure 6 × 5 – 2 × 5 = 30 – 10 = 20
3 Ex 44.
5
5
29 3 5
(A) 32 (B) 22 (C) 18 (D) 27 (B) In first figure 5 × 4 + 6 = 26 In second figure 8 × 3 + 5 = 29 missing number in third figure 6 ]
]
Sol.
(A) 11E (B) 28G (C) 35I (D) 48F (B) The letters in the first row form a series C, D, E (a series consecutive letters). The letters in the second row form a series I, K, M (a series of Ex 45. alternate letters). Similarly, the letters in the third row will form the series D, G, J (a series in which each letter is three steps ahead of the previous one). So, the missing letter is G. Also, the number in the second column is equal to the product of the numbers in the first and third Sol. columns. So, missing number is (4 × 7) i.e. 28. Thus, the answer is 28G.
]
5
8
3
7
6
174
5
× 3 + 4 = 22
3
2
5
?
336
3 2 9 2 7 9 6 4 5 ( A ) 1 40 (B) 150 (C) 200 (D) 180 (B) In first figure 8 × 5 × 3 + 3 × 2 × 9 = 120 + 54 = 174 In second figure 6 × 7 × 5 + 2 × 7 × 9= 210 + 126 = 336 missing number in third figure 3 × 2 × 5 + 6 × 4 × 5 = 30 + 120 = 150 ]
]
]
4 Ex 41.
7
9
11
41 ? 5
6
Sol.
5 2
15
25 21
7
Ex 46.
40
1
(A) 16 (B) 9 (C) 85 (D) 112 (C) Hint ; 42 + 52 = 16 + 25 = 41 12 + 22 = 1 + 4 = 5 62 + 72 = 36 + 49 = 85
Sol.
176
?
( A ) 1 84 (B) 210 (C) 241 (D) 425 (A) The number at the bottom is the difference of squares of two numbers given at top In first figure 112 – 92 = 121 – 81 = 40 In second figure 152 – 72 = 225 – 49 = 176 In third figure 252 – 212 = 625 – 441 = 184 ]
]
Ex 42.
84 14 12
81 18
9
(A) 16 (C) 61 Sol.
88 ? 11 (B) 21 (D) 81
(A) In first figure, 12 In second figure, 9
3
14
2 18 2
Ex 47.
= 84. = 81.
Let the missing number In third figure be x.
x Then, 11 2
]
88 2 = 88 or x =
11
= 16.
Sol.
5
33 6 3
4
7 48
5
5
3 ?
4
5
(A) 47 (B) 45 (C) 37 (D) 35 (D) In first figure, 6 × 3 + 3 × 5 = 33 In second figure, 5 × 4 + 4 × 7 = 48 In third figure, 5 × 4 + 3 × 5 = 35 PAGE # 69
4
17.
Directions : (1 to 25) Find the missing numbers : 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
2, 8, 18, 32, ? (A) 62 (C) 50 16, 54, 195, ? (A) 780 (C) 816 14, 316, 536, 764, ? (A) 981 (C) 8110
(B) 60 (D) 46
3, 15, 35, ?, 99, 143 (A) 63 (C) 69
(B) 77 (D) 81
9, 16, 30, 58, ? ( A ) 1 04 (C) 116
(B) 114 (D) 118
3, 12, 27, 48, 75, 108, ? ( A ) 1 92 (C) 162
(B) 183 (D) 147
1, 4, 12, 30, ? (A) 60 (C) 64
(B) 62 (D) 68
94, 166, 258, ?, 4912 ( A ) 3 6 10 (C) 1026
(B) 1644 (D) 516
(B) 180 (D) 452
22.
7,19, 55, 163, ? (A) 387 (C) 527
(B) 329 (D) 487
1, 2, 9, 4, 25, 6, ? (A) 51 (C) 50
(B) 49 (D) 47
24.
16, 33, 67, 135, ? (A) 371 (C) 271
(B) 175 (D) 287
25.
101, 100, ?, 87, 71, 46 (A) 92 (C) 89
(B) 88 (D) 96
13.
6, 24, 60, 120, 210, 336, ?, 720 (A) 496 (B) 502 (C) 504 (D) 498
16.
(B) 581 (D) 481
2, 3, 6, 18, ?, 1944 (A) 154 (C) 108
100, 50, 52, 26, 28, ? 16, 8 (A) 30 (B) 36 (C) 14 (D) 32
15.
137, 248, 359, 470, ? ( A ) 5 82 (C) 571
21.
(B) 210 (D) 258
3, 1, 4, 5, 9, 14, 23, ? (A) 32 (C) 41
(B) 37 (D) 28
3, 6, 18, 72, 360, ? (A) 720 (C) 1600
(B) 1080 (D) 2160
78, 79, 81, ?, 92, 103, 119 (A) 88 (B) 85 (C) 84 (D) 83
(B) 195 (D) 103
20.
8, 11, 15, 22, 33, 51, ?, 127, 203 (A) 80 (B) 53 (C) 58 (D) 69
2, 12, 36, 80, 150, ? (A) 194 (C) 252
2, 9, 28, 65, ? ( A ) 1 21 (C) 126
1, 11, ?, 11, 11, 11, 11, 16, 11 11 (A) 1 (B) 11 (C) 6 (D) 192
(B) 1048 (D) 9100
8, 24, 16, ?, 7, 14, 6, 18, 12, 5, 5, 10 (A) 14 (B) 10 (C) 7 (D) 5
(B) 112 (D) 108
19. (B) 802 (D) 824
12.
14.
18.
0, 6, 20, 42, 72, ? ( A ) 1 06 (C) 110
23.
Directions : (26 to 28) In each of the following questions, a number series is given. After the series, below it in the next line, a number is given followed by (P), (Q), (R), (S) and (T). You have to complete the series starting with the number given following the sequence of the given series. Then answer the question given below it. 26.
2 3 8 27 5 (P) (Q) (R) (S ) (T) Which of the following numbers will come in place of (T) ? (A) 184 (B) 6 (C) 925 (D) 45
27.
5 18 48 11 2 7 (P) (Q) (R) (S ) Which number will come in place of (S) ? ( A ) 1 72 (B) 276 (C) 270 (D) 376
28.
(T)
15 15 9 25 9 32 3 7 (P) (Q) (R) (S ) (T) Which of the following numbers will come in place of (R) ? ( A ) 2 51 (B) 315 (C) 176 (D) 151 PAGE # 70
Directions : (29 to 35) Find the wrong term(s) — 29.
30.
31.
32.
33.
34.
35.
9, 11, 15, 23, 39, 70, 135 (A) 23 (B) 39 (C) 70 (D) 135 3, 9, 36, 72, 216, 864, 1728, 3468 (A) 3468 (B) 1728 (C) 864 (D) 216 2, 5, 11, 20, 30, 47, 65 (A) 5 (C) 30 121, 143, 165, 186, 209 (A) 143 (C) 186
(B) 20 (D) 47
(B) 165 (D) 209
8.
9.
10.
11.
(B) 24
(C*) 34 (A) 15 (C) 34
(D) 51 (B) 24 (D) 51
(B) 37
(C) 69
(D) 132
105, 85, 60, 30, 0, – 45, – 90 (A) 85 (B) – 45 (C) 105 (D) 0
3.
4.
5.
6.
7.
CYD, FTH, IOL, LJP, LJP, ? (A) PET (C) OEY
(B) OET (D) PEV
ZGL, XHN, VIQ, TJU, ? (A) RKX (C) RLZ
(B) RKY (D) RKZ
ZSD, YTC, XUB, WVA, ? (A) VZZ (C) VWZ
(B) ZVX (D) VZX
RML, VIJ, ZFH, Z FH, DDF, ? (A) HDC (C) HCD
(B) CHI (D) DIC
LRX, DJP, VBH, NTZ, ? (A) ELS (C) GKS
(B) FMR (D) FLR
MAD, OBE, SCH, YDM, ? (A) HET (C) GET
(B) HES (D) UAE
2B, 4C, 8E, 14H, ? ( A ) 2 2L (C) 22K
(B) 24L (D) 2M
14.
15.
16.
Directions : (1 to 24) Find the missing term(s) :
2.
(B) ELS (D) DLS
13.
17.
1.
JXG, HTJ, FPN, ?, BHY (A) EKS (C) DLR
MTH, QRK, UPN, YNQ, ? (A) CKT (B) ELT (C) CLT (D) EKT
9, 13, 21, 37, 69, 132, 261 (A) 21
(B) KBO (D) LBN
12.
9, 15, 24, 34, 51, 69, 90 (A) 15
DFK, FEL, HDM, JCN, ? (A) KBN (C) LBO
X, U, S, P, N, K, I, ? (A) J (C) M
(B) K (D) F
18.
1 BR, 2 EO, 6 HL, 15 K I, ? (A) 22 NF (B) 31 NF (C) 31 NE (D) 28 NF
Z, X, U, Q, L, ? (A) F (C) G
(B) K (D) E
19.
P3C, R5F, T8I, V12L, ? (A) Y17O (C) X17O
A, H, N, S, W, ? (A) A (C) B
(B) Y (D) Z
20.
Z 15 A, W 13 C, ?, Q 9 G, N 7 I ( A) T 1 2 E (B) R 11F (C) T 11E (D) R 13 D
Q, T, V, Y, Y, A, ? (A) B (C) D
(B) C (D) F
21.
B3M, E7J, H15G, K31D, ? (A) N65A (B) O63A (C) N63A (D) N63Z
X, A, D, G, J, ? (A) N (C) M
(B) O (D) P
22.
5X9, 8U12, 11R15, 14O18, ? ( A ) 1 7 L21 (B) 17K21 (C) 17M21 (D) 17L23
Z, L, X, J, V, H, T, F, F, ?, ? (A) R, D (C) S, E
(B) R, E (D) Q, D
23.
6C7, 8F10, 11J14, 15O19, ? (A) 19U24 (B) 20U25 (C) 19U25 (D) 20U24
AZ, YB, CX, WD, ? (A) VE (C) EU
(B) UE (D) EV
24.
B2E, D5H, F12K, H27N, ? (A) J58Q (B) J56Q (C) J57Q (D) J56P
(B) X17M (D) X16O
PAGE # 71
Directions : (25 to 30) Find the wrong term(s) : 25.
26.
27.
28.
29.
30.
ECA, JHF JH F, OMK, TQP, YWU (A) ECA (B) JHF (C) TQP (D) YW U
9.
10.
a _ cab _ a _ c _ b c (A) bbac (C) abba
(B) abab (D) bcba
ba _ cb _ b _ bab _ (A) acbb (C) cabb
(B) bcaa (D) bacc
DKY, DKY, FJW, FJW , HIT, JHS, LGQ (A) FJW (B) LGQ (C) JHJ (D) HIT
11.
DVG, DVG, FSI, HPK, JNM, LJO L JO (A) DVG (B) JNM (C) HPK (D) LJO
a _ bc _ a _ bcda _ ccd _ bcd _ ( A ) a b d dbd (B) acbdbb (C) adbbad (D) bbbddd
12.
CDF, CDF, DEG, EFH, FHI (A) CDF (C) FHI
cc _ ccdd _ d _ cc _ ccdd _ dd (A) dcdcc (B) dcddc (C) dccdd (D) None of these
(B) DEG (D) EFH
13.
ZLA, BMY, CNW, FOU, HPS (A) ZLA (B) BMY (C) FOU (D) CNW
a _ b a a _ b a a _ b a (A ) a a b (C) b b a
(B) b a b (D) b b b
14.
babbb_b_b_bb (A ) b b a (C) a b a
(B) b a a (D) a a a
m _ l _ ml _ m _ llm ( A) l m m m (C) lmm l
(B) lmlm (D) mllm
G4T, J10R, M20P, P43N, S90L (A) G4T
(B) J10R
(C) M20P
(D) P43N
15.
Directions : (16 to 19) The questions given below are based on the letter series, In each of these series, some Directions : (1 to 15) Which sequence of letters when placed letters are missing. Select the correct alte rnative. If at the blanks one after the other will complete the more than five letters are missing, select the last given letter series ? five letters of the series. 1.
2.
3.
4.
a_baa_ aa__ab (A) a a a a (C) b b a a
(B) b a a a (D) a b b a
_ a a b b _ a _ a b _ b (A) b b a a (C) b a a b
(B) b a b a (D) a b a b
a a b _ a a a _ b b a _ (A) b a a (C) b a b
(B) a b b (D) a a b
a__b_a _ab_aa (A) a b a a b (C) b b a b b
(B) b b a b a (D) b a a b a
5.
abc _ d _ bc _ d _ b _ cda (A) bacdc (B) cdabc (C) dacab (D) dccbd
6.
a _ bbc _ aab _ cca _ bbcc (A) bacb (B) acba (C) abba (D) caba
7.
_ b c _ _ b b _ a a b c (A) acac (B) babc (C) abab (D) aacc _ b c c _ ac _ a a b b _ a b _ c c (A) aabca (B) abaca (C) bacab (D) bcaca
8.
16.
_ _ r _ ttp _ _ s _ tp _ _ _ s _ _ _ (A) rstqp (B) tsrqp (C) rstpq (D) None
17.
_ _ x _ zbxazyxabyz zbxazyxabyz _ _ _ _ _ (A) abxzy (B) abzxy (C) abxyz (D) bxayz
18.
x _ xxy _ x _ xy _ yxx _ _ yy _ y (A) xyyyy (B) xxyyx (C) yxxyx (D) xyxyx
19.
_ _ r _ tqrptsrpqst _ _ _ _ _ (A) pqrts (B) pqtrs (C) pqrst (D) qrpst
Directions : (20 to 23) There is a letter series in the first row and a number series in the second row. Each number in the number series stands for a letter in the letter series. Since in each of that series some term are missing you have to find out as to what those terms are, and answer the questions based on these as given below in the series. 20.
a b _ c d _ a _ a b d _ d b a _ 1 _ 3 _ 3 2 _ 1 _ _ _ 4 _ _ _ _ The last four terms in the series are ( A ) 1 2 34 (B) 3112 (C) 3211 (D) 4312 PAGE # 72
21.
22.
23.
_ b n t _ _ n a m _ n a b _ _ a _ _ _ _ 1 3 _ 2 5 3 _ _ 5 2 4 _ 3 2 5 _ _ _ _ _ The last five terms in the series are (A) 13425 (B) 41325 (C) 34125 (D) 13452 n _ g f _ t _ f h t n _ _ t _ b _ f 1 3 _ 2 4 5 0 _ 4 _ _ 3 _ _ _ _ _ _ The last five terms of the number series are (A) 50123 (B) 40321 (C) 40231 (D) 51302
3.
_ m i a x _ i r x a _ _ m a _ _ _ _ _ _ 4 _ 5 _ 7 3 _ _ _ 6 _ _ _ _ _ _ _ _ _ _ The last five term of the letter series are (A) r m x i a (B) x m r a i (C) x r m a i (D) r m i x a
25.
_ A C _ B D _ C D C D 2 _ 4 1 _ 1 4 _ _ _ _ r s _ q r _ p ? ? ? ? (A) p q p q (B) p r p r (C) r q r q (D) r s r s
6.
( A ) 2 35 (C) 144
(B) 141 (D) 188
12
18
30
16
6
(A) 18 (C) 9
5
12
_ A D A C B _ _ B D C C 2 4 _ _ 2 3 5 3 _ _ _ _ p _ _ q _ _ r s ? ? ? ? (A) p r s s (B) p s r r (C) r p s s (D) s r p p
36
18 27
?
(B) 12 (D) 6
6
6
21
7
4
5
(A) 14 (C) 32
A _ BA C _ D _ B C D C _ 4 _ 3 _ 2 _ 5 ? ? ? ? d c _ _ b a c b _ _ _ _ (A) 2 4 5 4 (B) 2 5 4 5 (C) 3 4 5 4 (D) 4 5 2 5
32 40
8
4
7. 26.
(B) 92 (D) 102
4.
Directions : (24 to 26) In each of the following questions, three sequences of letter/numbers are given which 5. correspond to each other in some way. In each question, you have to find out the letter/numerals that come in the vacant places marked marke d by (?). These are given as one of the four alternatives under the question. Mark your answer as instructed. 24.
(A) 112 (C) 82
?
8
10 (B) 22 (D) 320
5
9
8
5
15
?
3
5
6
(A) 12 (C) 16
(B) 11 (D) 26
(A) 72 (C) 9
(B) 18 (D) 19
(A) 1 (C) 90
(B) 18 (D) 225
(A) 20 (C) 24
(B) 22 (D) 12
8. Directions : (1 to 39) Find the missing term in the given figures
1. 9. (A) 36 (C) 25
(B) 9 (D) 64
2.
10. (A) 14 (C) 11
(B) 18 (D) 13
PAGE # 73
11.
7
11
49 49
12
8
54
15
4
?
(A) 36 (C) 25
12.
18 24 32 12 14 16 3 ? 4 72 112 128
(A) 28 (C) 81
(B) 3 (D) 5
2 4
10
3
26
5
70
90
27 80
39 45
29
33
42
20.
43
43
(A) 69 (C) 50
30 70
59
31 44
43
1 3 5 35 (A) 1 (C) 3
(B) 142 (D) 198
7 3 4 74
3
6 4
1
140 4 10 7
3 (B) 14 (D) 22
4
3
8
9
4
20
9
24
11
?
13
(A) 117 (C) 32
(B) 36 (D) 26
(A) 26 (C) 27
(B) 25 (D) 30
20
34 184
12 18 30
16 32 40
36 18 34
30
44
?
22.
6 ? 8 104 (B) 2 (D) 4
5
(A) 48 (C) 44
23.
(B) 9 (D) 64
5
21
16 109 2
2 2 53 19
6
15
(A) 25 (C) 7
17.
(A) 33 (C) 135
8
?
2
21.
48 56
(A) 127 (C) 158
12
6
10
?
15
35
80
? 39
38
15.
40
(B) 49 (D) 60
101
8
6 (B) 3 (D) 5
29
72
5
2
(A) 16 (C) 20
?
J
4
(A) 1 (C) 4
16.
6
6 4
8
H
5
14.
19.
4
C
(B) 36 (D) 49
6
(A) 2 (C) 4
13.
18.
(B) 7 (D) 0
(B) 145 (D) 18
24.
17
? 13
(B) 129 (D) 49
2
3
3 33 2
4 54 2
4
5
(A) 78 (C) 94
51
6 3
?
5
4 (B) 82 (D) 86 PAGE # 74
48
2 25.
4
5
28 5
1
7 38 4
3
2
3
(A) 14 (C) 11
?
34.
3
3
8
10
2
?
1
6
56
90
2
20
0
7
(A) 0
(B) 3
(B) 18 (D) 26
(C) 5
(D) 7
2
15 35. 26.
(A) 9 (C) 10 27.
HCA - 138 FED - 456 E?H - 87?
BIG - 792
64
9 21
81
25
2
64
25 ?
144
6
5
? 6
4
8
11
(A) 48 (C) 35
(B) 72 (D) 120
(A) 38 (C) 4
(B) 64 (D) 16
3 8
4
6
101
36
6
12 ?
8
? 15
56
( A ) 1 27 (C) 158
184
(B) 142 (D) 198
8
4
(A) 3 (C) 5
34
38 35
6
1
48
43
37.
(B) 23 (D) 31
2 3
25
16
(A) 19 (C) 25
29.
65
36.
(B) I, 9 (D) I, 5
36 26
49
80
16
13
(B) 11 (D) 12
(A) G, 6 (C) G, 5
28.
7
9
(B) 4 (D) 6
12 18 30
16 32 40
6
8
36 18 27
38.
29 30.
39
27 80
45 (A) 69 (C) 50
33 43
29 42 43
30 70
31 44
59
40 80
? 39
(B) 49 (D) 60
(A) 18 (C) 9
10 20 39.
4
9
9 6
31. (A) 0 (C) 3
(B) 2 (D) 1
(B) 12 (D) 6
16
16 12
(A) 60 (C) 21 40.
?
20 (B) 50 (D) 25
Find the value of X in the following figure :
15
32.
4 33
(A) 12 (C) 14 33
?
27
(B) 9 (D) 10
(B) JNS (D) KRS
2 36
Find the missing letters from left to right.
(A) JSN (C) JRS
2
(A) 3 (C) 8
8 32
X
18
9
22
11
12
3
(B) 4 (D) 12 PAGE # 75
PUZZLE TEST Directions : (1 to 5) Read the following information carefully and answer the questions given below it. (i). Five professors (Dr. Joshi, Dr. Davar, Dr. Natrajan, Dr. Choudhary and Dr. Zia) teach five different subjects subjec ts (zoology, physics, botany botan y, geology and history) in four universities ( Delhi, Gujarat, Mumbai, and Osmania). Do not assume any specific order. (ii). Dr. Choudhary teaches zoology in Mumbai University . (iii). Dr. Natrajan is neither in Osmania University nor in Delhi University and he teaches neither geology nor history. (iv). Dr. Zia teaches physics but neither in Mumbai University nor in Osmania University. (v). Dr. Joshi teaches history in Delhi University. (vi). Two professors are from Gujarat University. (vii). One professor teaches only one subject and in one University only. Ex 1.
Ex 2.
Ex 3.
Ex 4.
Ex 5.
Sol. :
Who teaches geology ? (A) Dr Natrajan (C) Dr. Davar
(B) Dr. Zia (D) Dr. Joshi
Which university is Dr. Zia from ? (A) Gujarat (B) Mumbai (C) Delhi (D) Osmania Who teaches botany ? (A) Dr. Zia (C) Dr. Joshi
(B) Dr. Davar (D) Dr. Natrajan
Names Dr. Joshi Dr. Davar Dr. Natrajan Dr. Choudhary Dr. Zia
University Delhi Osmania Gujarat Mumbai
Subject History Geology Botany Zoology
Gujarat
Physics
On the basis of the above table, rest of the questions can be solved very easily. 1.
(C) Dr. Davar teaches geology.
2.
(A) Dr. Zia is from Gujarat university.
3.
(D) Dr. Natrajan teaches botany. botany.
4.
(B) Dr. Davar is from Osmania University.
5.
(D) Dr. Natranjan - Gujarat University is the correc t combination.
Ex 6.
Ramesh is taller than Vinay who is not as tall as Karan. Sanjay is taller than Anupam but shorter than Vinay. Who among them is the tallest ? (A) Ra Ramesh (B) Karan (C) Vinay (D) Cannot be determined (D) In this question ranking of Karan is not de fined. Consequently, either Ram or Karan occupies the top position wi th th re re ga ga rd rd to to he he ig ig ht ht . He nc nc e, e, option (d) is the correct choice.
Sol.
Directions : (7 to 11) Read the following information carefully and answer the questions given below it : There are five men A, B, C, D and E a nd six women P, Q, R, S, T and U. A, B and R are advocates; C, D, Which of the following combinations is correct ? P, Q and S are doctors and the rest are teachers. (A) Delhi University - Dr. Zia Some teams are to be selected from amongst (B) Dr. Choudhary - geology these eleven persons subject to the following (C) Dr. Davar - Mumbai University conditions : (D) Dr. Natranjan - Gujarat University A, P and U have to be together. (1 to 5) B cannot go with D or R. From the given information in the question : E and Q have to be together. From II, we get Dr. Choudhary teaches zoology in C and T have to be b e together. Mumbai University. D and P cannot go together. From III, We get Dr. Natrajan is neither in Osmania C cannot go with Q. nor in Delhi University. Therefore, he will be either If the team is to consist consis t of two male advocates, two at Mumbai or Gujarat University. Similarly, as he Ex 7. lady doctors and one teacher, the members of the teaches neither geology nor history, therefore, he team are must be teaching physics or botany. ..........(1) (A ) A B P Q U (B) A B P U S From IV, Dr. Zia Physics but as he is not teaching (C) A P R S U (D) B E Q R S in either Mumbai or Osmania University, he must Sol. (B) The male advocates are A and B, lady doctors be teaching either in Delhi or Gujarat University...(2) University...(2) are P, P, Q and S ; teachers are a re E, T and U. Form V, we get Dr Joshi teaches history in Delhi Now, Now, A and B will be select ed. University Form (1) and (2), we conclude that Dr Natarajan teaches botany. And from (1), (2) and VI, A, P and U have to be together. Now, we have to we get both Natarajan and Zia teach in Gujarat select one lady doctor more. It can be Q or S. But Q University. Finally, On summarisation we can and E have to be together. Since E is not selected, prepare the following table. so S will be selected. Thus, the team is A B P U S. Who is from Osmania University ? (A) Dr. Natrajan (B) Dr. Davar (C) Dr. Joshi (D) Dr. Zia
PAGE # 76
Ex 8.
Sol.
Ex 9.
If the team is to consist of one advocate, two Directions : (12 to 15) Read the following paragraph carefully : doctors, three teachers and C may not go with T, Four women A, B, C and D and three men E, F and the members of the team are : G play bridge, a game for four players. (A) A E P Q S U (B) A E P Q T U (i) The group consists of three married couples and a widow. (C) B E Q S T U (D) E Q R S T U (ii) Spouses are never partners in a game. (B) The advocates are A, B and R ; doctors are (iii) No more than one married couple ever plays i n C, D, P, Q, S ; teachers are E, T and U. The team the same game. (iv) (iv) One day they played played four four games as follows. follows. consists of 3 teachers i.e. E, T, U. Now, Now, A, P and U A and E versus B and F. have to be together. E and Q have to be together. A and G versus D and F. Thus, the team is A E P Q T U. B and C versus F and G. C and E versus D and G. If the team is to consist of one male advocate, one male doctor, one lady doctor and two teach ers, the Ex 12. members of the team are :
Sol.
(A) A C P T U
(B) A D E P T
(C) A D E P U
(D) B C E Q U
Ex 13.
(A) The male advocates are A and B ; male doctors are C and D ; l ady doctors are P, Q and S ; teachers are E, T and U. If A is selected, P and U will be Ex 14. selected. D and P cannot go together. So, a male doctor C will be selected. C and T have to be together. Thus, the team is A C P T U. If B is Ex 15. selected, D will not be selected. So, male doctor C will be chosen. C and T have to be together. Now, the second teacher to be selected selecte d is E or U. But, U cannot go without A. So, E will be selected. E and Q have to be together. Thus, the team can also be B C E Q T.
Ex 10.
If the team is to consist of one advocate, three doctors and one male teacher, the members of the team are:
Sol.
(A) A D P S U
(B) C D R S T
(C) D E Q R S
(D) D E Q R T
(B) B (D) D
Whom is G married to ? (A) A (C) C
(B) B (D) D
Which of the following is a widow ? (A) A (B) B (C) C (D) D
(D) E is married to D.
13.
(A) F is married to A.
So, two other doctors D and S will be selected. P is not selected, so A will not be selected. D is selected, so B cannot be selected. Thus, the team is D E Q R S.
Sol.
Whom is F married to ? (A) A (C) C
12.
(C) The advocates are A, B and R ; the doctors are 14. C, D, P, P, Q and S ; male teacher is E. Clearly, E will 15. be selected. E and Q have to be together. togethe r. C and Q cannot be together. So, C will not be selected. P Ex 16.
If the team is to consist of two advocates, two Sol. doctors, two teachers and not more than three ladies, the members of the team are :
(B) B (D) D
Sol. : (12 to 15) From (iv), is married either to A or to C. If F is married to A, then G is married to B or to C. If G is married to B, then E is married to D ; if G is married to C, then E is married to B or to D. If F is married to C, then G is married to B ; then E is married to D. Hence, the married couples are : FA, GB, ED or FA, GC, EB or FA, GC, ED or FC, GB, ED. Of these, only FA, FA, GB, ED does not contradict any of the statements.
also cannot be selected because U is not selected.
Ex 11.
Whom is E married to ? (A) A (C) C
(B) G is married to B. (C) C is a widow. A vagabond runs out of cigarettes. He searches for the stubs, having learnt that 7 stubs can make a new cigarette, good enough to be smoked, he gathers 49 stubs, If he smokes 1 cigarette every three - quarters of an hour, how long will his supp ly last ? (A) 5.25 hr (B) 6 hr (C) 4.5 hr (D) 3 hr (B) He has got =
49 7
7 cigarettes.
The duration of time he will take to smoke these
3 hr = 5.25 hr (i.e. 5 hr and 15 4
(A) A B C P T U
(B) A C P R T U
7 cigarettes = 7
(C) A E P Q R T
(D) B C E Q R T
min). Now note that after he has smoked these 7 cigarettes, he will collect 7 more stubs (one form each), form which he will be able to make another
(A) A C P R T U and A E P Q R T are wrong wron g because bec ause each of these combinations consists of four ladies. B C E Q R T is incorrect because B and R cannot go together.
cigarette. This will take him another
3 hr (45 min) 4
to smoke. Therefore, total time taken = 6hr.
PAGE # 77
Directions : (17 to 18) Read the following information and 1. answer the questions that follow. There are 70 clerks working with M/s. Jha Lal Khanna & Co. chartered accountants, of which 30 are female. 2. (i) 30 clerks are married. (ii) 24 clerks are above 25 years of age (iii) 19 Married clerks are above 25 years of age; among them 7 are males. 3. (iv) 12 males are above 25 years of age (v) 15 males are married. Ex 17.
Ex 18.
How many unmarried girls are there ? (A) 12 (B) 15 (C) 18 (D) 10
4.
Who stays in locality Q ? (A) A (C) C
(B) B (D) E
What is E s occupation ? (A) Busi ness (B) Engineer (C) Lawyer (D) Doctor ’
Agewise who among the following lies between A and C ? (A) Lawyer (B) Doctor (C) Cloth merchant (D) En Engineer What is B s occupation ? (A) Busi ness (B) Engineer (C) Lawyer (D) Doctor ’
How many of these unmarried girls are above 25 ? (A) 12 (B) 15 (C) 4 (D) 0 5. What is C s occupation ? (17 to 18) : From the given data, we can make the (A) Doctor (B) Lawyer following table with the help of which rest of the (C) Engineer (D) Busin ess questions can be solved very easily. Directions : (6 to 10) Study the given information carefully Ma le le ((4 40) Fe m ma a le le ((3 30) and answer the questions that follow. Above 25 There are four people sitting in a row : one each from India, Japan, USA and Germany, but not in Married 7 12 that order, Unmarried 5 0 wearing caps of different colours colou rs - green, . They are wearing Below 25 yellow, red and white, not necessarily in that order. married 8 3 II. One is wearing a kurta and one a T-shirt. III. The Indian is wearing a green cap and a jacket. unmarri ed 20 15 IV. The American is not seated at either end. 40 30 Total V. The persons with kurta and T-shirt are sitting next to each other. There are 15 unmarried girls. VI. The persons with kurta wears a red cap and sits next to the Japanese. In these 15 unmarried girls no one is above 25. VII. The Japanese wears a shirt and is not seated at either end. VIII. The man with white cap wears T-shirt and is seated at one end. ’
Sol.
17. 18.
Directions : (1 to 5) Study the following information carefully and answer the questions given below it : There are five friends A, B, C, D and E. Two of them are businessmen while the other three belong to different occupations viz. medical, engineer and legal. One businessman and the lawyer stay in the same locality S, while the other three stay in three different localities P, Q and R. Two of these five persons are Hindus while the remaining three come from three different communities viz. Muslim, Christian and Shikh. The lawyer is the oldest in age while one of the businessmen who runs a factory is the youngest. The other businessman is a cloth merchant and agewise lies between the doctor and the lawyer. D is a cloth merchant and stays in locality S while E is a Muslim and stays in locality R. The doctor is a Christian and stays in locality P, P, B is a Shikh while A is a Hindu and runs a factory factor y.
6.
Who wears the T-shirt ? (A) Indian (C) American
(B) Japanese (D) German
7.
Who is wearing a kurta ? (A) Indian (B) Japanese (C) American (D) German
8.
What is the colour colo ur of the cap worn by the Japanese? (A) Red (B) Green (C) Yellow (D) W hite
9.
Who precedes the man wearing T-shirt ? (A) Indian (B) Japanese (C) American (D) German
10.
Who precedes the man wearing jacket ? (A) Indian (B) German (C) Japanese (D) Cannot say
PAGE # 78
Directions : (11 to 15) Read the following information Directions : (19 to 23) Read the information given below carefully and answer the questions that follow. and answer the questions. I. There are six students ( A, B, C, D, E and F) in a The age and height of six children in a class are as group. Each student can opt for only three choices follows : out of the six which are music, reading, painting, (i) A is taller and older than B but shorter and badminton, cricket and tennis. younger than C. II. A, C and and F lik like e rrea eadi ding ng.. (ii) D is taller than E who is not as tall as B. III. III. D does does not like like badmin badminton, ton, but likes likes music. music. (iii) The oldest is the shortest. IV. Both Both B and and E like like paintin painting g and music music.. (iv) The youngest would be fourth if the children V. A and D do not not like paintin painting, g, bu butt they they like like cricket. cricket. stood in a line according to their height and one VI. All stude student nt except except one one like badmin badminton ton.. started counting from the tallest. VII. II. Two students students like tennis. tennis. (v) D is younger than F but older than E who is VIII. III. F does not like cricket, music and and tennis. older than C. 11.
12.
Which pair of students has the same combination 19. of choices ? (A) A and C (B) C and D (C) B and E (D) D and F 20. Who among the following students likes both tennis and cricket ? (A) A and B (B) C 21. (C) B and D (D) D
13.
How many students like painting and badminton ? (A) 1 (B) 2 (C) 3 (D) 4
14.
Who among the following do not like music ? (A) A , C and D (B) A, B and C (C) A, C and F (D) B, D and F
15.
Which of the following is the most popular choice? (A) Tennis (B) Badminton 23. (C) Reading (D) Painting
22.
Who among them is the tallest ? (A) B (B) E (C) C (D) Data inadequate Who is older than B but younger than C ? (A) F (B) D (C) A (D) Data inadequate Which of the following statements is definitely true? (A) D is the most old person (B) B has the max. height (C) A is older than D (D) F is the shortest Which of the following is the correct order of height in descending order? (A) A, C, D, B, E, F (B) F, D, E, C, A, B (C) D, C, A, B, E, F (D) C, D, A, B, E, F Whose Rank in height cannot be positioned definitely ? (A) B (B) D (C) C (D) E
R earns more than H but not as much as T, T, M earns more than R. Who earns least among them? Directions : (24 to 28) Study the information given below (A) R (B) T and answer the questions that follow. (C) H (D) M (i) Six Plays P, P, Q, R, S, T and U are to be organised from Monday to Saturday i.e. 10 to 15 one play each 17. Harish is taller than Manish but shorter than day. Suresh. Manish is shorter than Anil but taller than (ii) There are two plays between R and S and one Raghu. Who among them is the shortest having play between P and R. regard to height ? (iii) There is one play between U and T and T is to (A) Anil (B) Manish be organised before U. (C) Raghu (D) Cannot be determined (iv) Q is to be organised before P, not necessarily Direction : (18) Examine the following statements : immediately. I. Either A and B are of the same age or A is older (v) The organisation does not start with Q. than B. The organisation would start from which play ? II. Either C and D are of the same age or D is older 24. (A) P (B) S than C. (C) T (D) None III. B is older than C. 16.
18.
Which one of the following conclusions can be 25. drawn from the above statements ? (A) A is older than B (B) B and D are of the same age 26. (C) D is older than C (D) A is older than C
On which date is play T to be organised ? (A) 10th (B) 11th (C) 12th (D) None The organisation would end with which play ? (A) P (B) Q (C) S (D) None
PAGE # 79
27.
28.
Which day is play Q organised ? (A) Tuesday (B) W ednesday (C) Thursday (D) None
31.
Which of the following is the correct sequence of organising plays ? 32. (A) PTRUQS (B) QSTURP (C) SUTRQP (D) None
If it is sure that Henna will go to the fair, then who among the following will definitely go ? (A) Rama
(B) Shamma
(C) Reena
(D) Rama and Reena
If Tina does not go to the fair, which of the foll owing statements must be true ? (i) (i)
Henna enna cann cannot ot go
(ii) (ii) Sham Shamma ma can canno nott go Directions : (29 to 30) Read the following information (iii) (iii) Reen Reena a cann cannot ot go go carefully and answer the questions given below it. (iv) (iv) Rama Rama cann cannot ot go go I. Seven books are placed one above the other in a (A) (i) and (ii) (B) (iii) and (iv) particular way . (C) (i), (iii) and (iv) (D) (i) and (iv) II. The history book is placed directly above the civics book. Directions : (33 to 37) Read the following paragraph III. The geography book is fourth from the bottom carefully and choose the correct alternative. and the English book is fifth from the top. The office staff of XYZ corporation presently IV. There are two books in between the civics and consists of three females A, B, C and five males D, economics books. E, F, G and H. The management is planning to 29.
To find the number of books between the civics and the science books, which other extra piece of information is required, from the following ? (A) There are two books between the geography and the science books. (B) There are two books between the mathematics and the geography books . (C) There is one book between the English and the science books. (D) The civics book is placed before two two books 33. above the economics book.
open a new office in another city usin g three males and two females of t he present staff. To To do so they plan to separate certain individuals who do not function well together. The following guidelines were established I.
Fema Females les A and and C are are not not to to be be toge togethe therr
II.
C and E shou should ld be separ separat ated ed
III III.
D and and G should should be be separ separat ated ed
IV.
D and and F should should not not be part part of of a team. team.
If A is chosen to be moved, which of the following cannot be a team ?
To know which three books are kept above the English book, which of the following additional pieces of information, if any, is required? (A) The economics book is between the English 34. and the science books.
(A) ABDEH
(B) ABDGH
(C) ABEFH
(D) ABEGH
(B) There are two books between the English and the history books. (C) The geography book is above the English book. 35. (D) No other information is required.
(A) 1
(B) 2
(C) 3
(D) 4
Directions : (31 to 32) A five-member team that includes Rama, Shamma, Henna, Reena, and Tina, is planning to go to a science fair but each of them 36. put up certain conditions for going .They are as follows.
(A) B
(B) D
(C) F
(D) G
30.
I. If Rama goes, then at least one amongst Shamma and Henna must go. II. If Shamma goes, then Reena will not go. 37. III. If Henna will go, then Tina must go. IV. If Reena goes, then - Henna must go. V. If Tina goes, then Rama must go but Shamma cannot go. VI. If Reena plans not to go the fair, then Rama will also not go.
If C and F are to be moved to the new office, how many combinations are possible ?
If C is chosen to the new office, which number of the staff cannot be chosen to go with C ?
Under the guidelines, which of the following must be chosen to go to the new office ? (A) B
(B) D
(C) E
(D) G
If D goes to the new office, which of the following is/are true ? I.
C cannot be be chosen
II.
A cann cannot ot be chos chosen en
III.
H mus mustt be be c cho hose sen. n.
(A) I only
(B) II only
(C) I and II only
(D) I and III only
PAGE # 80
Directions : (38 to 42) Study the following information Directions : (45 to 49) Read the following information carefully and answer the questions given below. carefully and answer the questions that follow : (i) There is a family of six persons- L, M, N, O, P A team of five is to be selected from amongst five and Q. They are professor, businessman, boys A, B, C, D and E and four girls P, P, Q, R and S. chartered account, bank manager, engineer and Some criteria for selection are : medical representative, not necessarily in that A and S have to be together order. (ii) There are two married couples in the family. P cannot be put with R. (iii) O, the bank manager is married to the lady D and Q cannot go together. professor. C and E have to be together. (iv) Q, the medical representative, is the son of M R cannot be put with B. and brother of P. Unless otherwise stated, these criteria are (v) N, the chartered accountant, is the t he daughter - in applicable to all the questions below : law of L. (vi) The businessman is married to the chartered If two of the members have to be boys, the team 38. acconuntant. will consist of : (vii) P is an unmarried engineer. (viii) L is the grandmother of Q (A) A B S P Q (B) A D S Q R (C) B D S R Q 39.
(D) C E S P Q
45.
How is P related to Q. (A) Brother (B) Sister (C) Cou Cousi sin n (D) Either brother or sist ister
46.
Which of the following is the profession of M ? (A) Professor (B) Chartered accountant (C) Businessman (D) Medical representative
47.
Which of the following is the profession of L ? (A) Professor (B) Charted accountant ( C) B u s i n e s s m a n (D) Engineer
If R be one of the members, the other members of the team are :
40.
(A) P S A D
(B) Q S A D
(C) Q S C E
(D) S A C E
If two of the members are girls and D is one of the members, the members of the team other than D are :
41.
(A) P Q B C
(B) P Q C E
(C) P S A B
(D) P S C E
If A and C are members, the other members of the 48. team cannot be :
42.
(A) B E S
(B) D E S
(C) E S P
(D) P Q E
49.
If including P at least three members are girls, the members of the team other than P are : (A) Q S A B
(B) Q S B D
(C) Q S C E
(D) R S A D
Directions : (43 to 44) Read the given information carefully and answer the questions that follow : Ratan, Anil, Pinku and Gaurav are brothers of Rakhi, Sangeeta, Pooja and Saroj, not necessarily in that order. Each boy has one sister and the names of bothers and sisters do not begin with the same ’
Sangeeta s brothers. Saroj is not Ratan s sister. 43.
44.
’
Pooja s brother is ’
(A) Ratan
(B) Anil
(C) Pinku
(D) Gaurav
Which of the following are brother and sister ? (A) Ratan and Pooja oja
(B) Anil and Saroj
(C) Pinku Pinku and and Sangeet Sangeeta a
(D) Gaurav Gaurav and and Rakh Rakhii
How is O related to Q? (A) Father (C) Uncle
(B) Grandfather (D) Brother
Directions : (50 to 54) I. There is a group of six persons P,Q, P,Q, R, S, T and U from a family. They are Psychologist, Manager, Lawyer, Jeweller, Doctor and Engineer. II. The Doctor is grandfather of U, who is a Psychologist. III. The Manager S i s married to P. IV. IV. R, the Jeweller i s married to the Lawyer. V. Q is the mother of U and T. VI. There are two married couples in the family.
letter. Pinku and Gaurav are not Saroj s or 50. ’
Which of the following is one of the couples ? (A) QO (B) OM (C) PL (D) None of these
51.
52.
What is the profession of T ? (A) Doctor (B) Jeweller (C) Manager (D) None of these How is P related to T ? (A) Brother (C) Father
(B) Uncle (D) Grandfather
How many male members are their in the family ? (A) One (B) Three (C) Four (D) Data inadequate
PAGE # 81
53.
54.
What is the profession of P ? (A) Doctor (B) Lawyer (C) Jeweller (D) Manager
III. Two persons from the same side of the roads
Which of the following is one of the pairs of couples in the family ? (A) PQ (B) PR (C) PS (D) Cannot be determined
movement, he or she cannot immediately move
cannot move in consecutive movements . “
IV. If one person crosses the road in a particular
Direction : (55) The ages of Mandar, Shivku, Pawan and 56. Chandra are 32, 21, 35 and 29 years, not in order, Whenever asked they lie of their own age but tell the truth abut others. (i) Pawan says, My age is 32 and Mandar s age is not 35 (ii) Shivku says, My age is not 2 9 and Pawan s 57. age in not 21 (iii) Mandar says, My age is 32. “
”
’
back to the other side. V. Jai and Reema did not take part in first 3 movements. What is the minimum possible number of movements that took place in the entire game ? (A) 3
(B) 4
(C) 5
(D) 6
”
“
’
”
“
55.
”
What is Chandra s age ? (A) 32 years (B) 35 years (C) 29 years (D) 21 years
If number of movements are minimised in the game, then which of the following combination of friends can never be together on one particular side of the road during the course of the game ?
’
Directions : (56 to 57) Answer the questions on the basis of the information given below. 5 friends Nitin, Reema, Jai, Deepti and Ashutosh are playing a 58. game of crossing the roads. In the beginning, Nitin, Reema and Ashutosh are on the one side of the road and Deepti and Jai are on the other side. At the end of the game, it was found that Reema and Deepti are on the one side and Nitin, Jai and Ashutosh are on the other side of the road. Rules of the game are as follows : I. One movement means only one person crosses the road from any side to the other side. II. No two persons can cross the road simultaneously from any side to the other side. “
”
(A) Nitin, Reema amd Deepti (B) Nitin, Jai and Deepti (C) Deepti, Jai and Ashutosh (D) Ashutosh, Nitin and Deepti You have 12 similar l ooking coins. 11 of them weigh the same. One of them has a different weight, but you don t know whether it is heavier or lighter. You You ’
also have a scale. You can put coins on both sides of the scale and it ll tell you which side is heavier or ’
will stay in the middle if both sides weigh the same. What is the minimum number of weighing required to find out the odd coin. (A) 3
(B) 4
(C) 5
(D) 6
PAGE # 82
CALENDAR AND CLOCK TEST Similarly, 200 years = 10 odd days = 03 odd days
We are to find the day of the week on a mentioned date. Certain concepts are defined as under.
An ordinary year has 365 days.
In an ordinary year, first and last day of the year are same.
A leap year has 366 days. Every year which is divisible by 4 is called a leap year. year. For example 1200, 1600, 1992, 2004, etc. are all leap years.
For a leap year, if first day is Monday than last day will be Tuesday for the same year.
In a leap year, February is of 29 days but in an ordinary year, it has only 28 days.
Year ending in 00's but not divisiable by 400 is not considered a leap year. e.g., 900, 1000, 1100, 1100, 1300, 1400, 1500, 1700, 1800, 1900, 2100 are not leap years.
After counting the odd days, we find the day according to the number of odd days.
Sunday for 0 odd day, Monday for 1 odd day and so on as shown in the following table. Table : 1 (Odd days for week days) Days
Sunday Monday Tuesd ay W ednesday Thurs Thursday day Friday Friday Saturday
Odd Days
The day on which calendar started (or the very first day ) i.e., 1 Jan, 0001 was Monday. Monday. Calendar year is from 1 Jan to 31 Dec. Financial year is from 1 April to 31 March.
ODD DAYS
The no. of days exceeding the complete no. of weeks in a duration is the no. of odd days during that duration. COUNTING OF ODD DAYS
Every ordinary year has 365 days = 52 weeks +1 day. Ordinary year has 1 odd day.
Every leap year 366 days = 52 weeks + 2 days. Leap year has 2 odd days.
Odd days of 100 years = 5, Odd days of 200 years = 3, Odd days of 300 years = 1, Odd days of 400 years = 0.
124 7
= 5 odd days.
0
1
2
3
4
5
6
Table : 2 (Odd days for months in a year) Ordinary Year
Days
Odd Days
Leap year
Days
Odd Days
January
31
3
January
31
3
F ebruary
28
0
February
29
1
March
31
3
March
31
3
April
30
2
April
30
2
Ma y
31
3
May
31
3
June
30
2
J u ne
30
2
Total
181 days
6
Total
182 d ays
0
July
31
3
July
31
3
August
31
3
August
31
3
September
30
2
Sept ember
30
2
Oct ober
31
3
October
31
3
Novem ber
30
2
November
30
2
Decem ber
31
3
December
31
3
Total
184 days
1
Total
184 d ays
2
Table : 3 (Odd days for every quarter) Months of years
I st three months 1 Jan to 31 March
Total days
90 / 91 Ord. / Leap
Odd days
6 /0 0 Ord. / Leap Odd day
Explanation : 100 years = 76 ordinary years + 24 leap years Ex 1. ( The year 100 is not a leap year) = 76 odd days + 2 × 24 odd days = 124 odd days. Sol. Odd days =
15
= 1 odd day.. 7 20 1 400 years = = 0 odd day (1 is added as 400 7 is a leap year) Similarly, 800, 1200, 1600, 2000, 2400 years contain 0 odd days. 300 years =
II nd three III rd three months months 1 Apr to 1 July to 30 June 30 Sep. 91
92 1 Odd day
Iv th three Total year months 1 Jan to 1 Oct. to 31 Dec. 31 Dec. 92
365 / 366 Ord. / Leap
1 1 /2 Odd day Ord. / Leap
If it was Saturday on 17th December 1982 what will be the day on 22nd December 1984 ? Total number of odd days between between 17 Dec.1982 to 17 Dec.1984 the number of odd days = 1+2 = 3. From 17 to 22 Dec. number of odd days = 5 od d days = 1 odd od d day. 3 + 5 = 8 odd Sunday. Saturday + 1 odd day = Sunday. PAGE # 83
Ex 2. Sol.
Find the day of the week on 16 January, 1969. Ex 6. 1600 years have 0 odd day. .....................(A) 300 years have 1 odd day. day. ......................(B) ......................(B) Sol. 68 years have 17 leap years and 51 ordinary ordin ary years. Thus = (17 × 2 + 51 × 1 ) = 85 odd days ' 01' odd day ...(C) ‘
‘
Ans. Ex 3. Sol.
’
’
16 J Jan anu uary h ha as = ' 02' odd days..(D) Adding (A) + (B) +(C) +(D), We get, get, 0 + 01 +01 +01 +02 +02 = 04 04 odd odd days days Thursday
‘
‘
Ans.
Ex 7.
1996 1997 1998 1999 2000 2001 2002 2003 2
1
1
1
2
Total = 2 + 1 + 1 + 1 + 2 = 7= 0 odd days Hence, year 2001 will have the same calendar as
Prove that last day of a century cannot be Tuesday, Thursday or Saturday. Saturday.
’
’
’
‘
will be -
year 1996.
Find the day of the week on 18 July, 1776 (leap Sol. year). Here 1600 years have 0 odd day.................... day.....................(A) .(A) 100 years have 5 odd days..............................(B) 75 years = (18 leap years + 57 ordinary years) = (18 × 2 + 57 × 1) = 93 odd days = (7 × 13 + 2) = 2 odd days.............................( days.............................(C) C) st Now, the no. of days from 1 January to 18 July, 1776 = 182 + 18 = 200 days = (28 × 7 + 4) days = 4 odd days.....................(D) Adding (A) + (B) +(C) +(D), We get get,, 0 + 5 + 2 + 4 = 04 odd days days Thursday ‘
The year next to 1996 having the same Calendar
’
100 years have = 5 odd days
Last day of st century is Friday
200 years have = 10 odd days
Last day of IInd century is Wednesday
= 3 odd days 300 years have = 15 odd days
Last day of rd century is Monday
= 01 odd day 400 years have = (5 × 4 + 1) Last day of 4th century is Sunday = 21 odd days = 0 odd days Since the order keeps on cycling, we see that the last day of the century centu ry cannot be Tuesday, Thursday or Saturday.
Ex 4. Sol.
On what dates of October, 1975 did Tuesday fall ? For determining the dates, we find the day on 1st Oct, 1975. 1600 years have 0 odd days.....................( days.....................(A). A). 300 years have 01 odd days.....................(B) days.....................(B).. 74 years have (18 leap years + 56 ordinary years) 2 × 18 + 1 × 56 = 92 odd days = 01 odd days.............(C) days.............(C) st st Days from 1 January to 1 Oct., 1975 1st Jan – 30 June + 1 st July to 1st Oct. 181 + 31 + 31 + 30 + 1 = 274 days = 01 odd days......(D) (274/7= 01 days) Adding (A) + (B) +(C) +(D) = 0 + 01 +01 +01 = '03' odd days Wednesday( 1st Oct), hence 7,14,21,28 Oct. will Tuesday fall. ‘
‘
’
‘
‘
Ans.
Ex 5. Sol.
:
’
Minute hand and hour hand coincides once in every hour. They coincide 11 times in 12 hours and 22
’
times in 24 hours.
They coincide only one time between 11 to 1 O’ clock. at 12 O’ clock.
’
Calendar for 1995 will serve for 2006, prove ? The Calendar for 1995 and 2006 will be the same ,if day on 1st January of both the years is the same. This is possible only if the total odd days between 31st Dec. 1994 and 31st Dec.2005 is 0. [one day before both the years as we want to know the day on 1st January of both the years i.e. same] During this period, we have 3 leap years (1996, 2000, 2004) an d 08 ordinary years (1995,1997,1998,1999, 2001, 2002, 2003,2005) Tota otal odd days ays = (2 (2 × 3 + 1 × 8) = 14 = 0 odd days (Thus (Thus Proved) Proved)
Minute hand and hour hand are opposite once in every hour. They do it 11 times in 12 hours and 22 times in 24 hours.
They opposite only one time between 5 to 7 O’ clock. at 6 O’ clock.
Both hands (minute and hour) are perpendicular twice in every hour. 22 times in 12 hours and 44 times in 24 hours.
In one minute, hour hand moves 1/2º and minute hand moves 6º. In one hour, hour hand moves 30º and minute hand moves 360º.
In an hour, minute hand moves 55 minutes ahead of hour hand.
PAGE # 84
Case-II When the time taken (20 + 15) = 35 min.
HANDS COINCIDE
Ex.8 Sol.
At what time between 3 O’Clock and 4 O’Clock will the two hands coincide ? At 3 O’clock the distance between the two hands is 15 minutes when they coincide with each other the distance between the two hands will be 0 min. So, the time taken (15 + 0 ) = 15 minutes. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in 60 min. 55
MIRROR MIRR OR IMAGE OF CLOC CLOCK K
60 40 480 7 = = 43 min. 55 11 11
Hence, the right time is 43
to find the mirror image, time is subtracted from 11 : 60.
23 : 60. Ex.11
The time in the clock is i s 4 : 46, what is the mirror image ?
Sol.
(11 : 60) – (4 : 46) = 7 : 14.
7 min. past 2. 11
At what time between 4 O’clock and 5 O ’clock will the hands are perpendicular ? At 4 O’clock the distance between the two hands is 20 min. When they are at 15 minutes distance, they are perpendicular to each other. When the time taken (20 – 15) = 5 min. Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 5 min. ahead of hour hand in
Both types of angles are 360º in total. If one angle is known, other can be obtained by subtracting from 360º.
Ex.13
At 4 : 30, what is the angle formed between hour hand and minute hand ?
Sol.
Case-I
If the time is between 11 11 O’clock to 1 O ’clock, then to find the mirror image, time is subtracted from
HANDS ARE PERPENDICULAR
Sol.
If the time is between 1 O’clock to 11 O’clock, then
At what time between 2 O’clock and 3 O ’clock will Ex.12 The time in the clock is 12 : 35, then find its mirror the two hands be opposite ? image. At 2 O’clock the distance between the two hands is Sol. (23 : 60) – (12 : 35) = 11 : 25. 10 minutes. When they are at 30 minutes distance, they are opposite to each other. The time taken TO FIND THE ANGLE BETWEEN TWO HANDS (30 + 10 ) = 40 min. Minute hand is 55 min. ahead of hour hand in Angle are of two types : 60 min. Positive angle : It is obtained by moving from hour Minute hand is 1 min. ahead of hour hand in hand to minute hand moving in clockwise direction . 60 min. Negative angle : It is obtained by moving from 55 minute hand to hour hand. Minute hand is 40 minutes ahead of hour hand in
Ex.10
hand is 35 min. ahead of hour hand in
60 35 420 2 = = 38 min. 55 11 11 2 Hence, the right time is 38 min. past 4. 11
4 minute past 3. 11
HANDS ARE OPPOSITE
Sol.
Minute
Minute hand is 15 min. ahead of hour hand in
Hence the right time is 16
Ex.9
60 min.
60 15 180 4 = = 16 min. 55 11 11
Minute hand is 55 min. ahead of hour hand in
At 4 O’ clock angle between hour and min. hand is of 120º. In 30 min. minute hand make an angle of 180º.
So, the resultant angle is 180º – 120º = 60º. But in 30 min. hour hand hand will also cover an an angle of 15º. Hence, the final angle between both hands is 60º – 15º = 45º. Short trick
60 5 60 5 = = 5 min. 55 11 11 5 Hence, the right time is 5 min. past 4. 11
PAGE # 85
9.
If the day before yesterday was Saturday, what day will fall on the day after tomorrow ?
Ex.14
Sol.
A bus for Delhi leaves every thirty minutes from a bus stand. An enquiry clerk told a passenger that the bus had already left ten minutes ago and the next bus will leave at 9.35 A.M. At what what time did the enquiry clerk 10. give this information to the passenger ? Bus leaves after every 30 minutes. The next bus will leave at 9 : 35 A.M. The last bus left at 9 : 35 – 0 : 30 = 9 : 05 A.M. but clerk said that bus had left 10 minutes earlier. 11. 9 : 05 + 0 : 10 = 9 : 15 A.M.
(A) Friday
(B) Thursday
(C) W ednesday
(D) Tuesday
If February 1, 1996 is Wednesday, what day is March 10, 1996 ? (A) Monday
(B) Sunday
(C) Saturday
(D) Friday
If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month ?
1.
2.
3.
4.
Find the day of the week on 26 January, 1950. (A) Tuesday (B) Friday (C) W ednesday (D) Thursday
If it was Saturday on 17th November, 1962 what will be the day on 22nd November, 1964 ? 14.
7.
(A) Thursday
(B) Saturday
(C) Sunday
(D) Tuesday
At what time are the hands of a clock together between 5 and 6 ? (A) 33
3 min. past 5 11
(B) 28
3 min. past 5 11
(C) 27
3 min. past 5 11
(D) 26
3 min. past 5 11
At what time between 9 and 10 will the hands of a
4 minutes past 9 11 6 (C) 16 minutes past 9 11 9 (D) 16 minutes past 9 11
(B) 16
At what time between 5 & 5 : 30 3 0 will the hands of a clock be at right angle ?
Find the day of the week on 15 August, 1947.
10 minutes past 5 11 5 (B) 11 minutes past 5 11 10 (C) 9 minutes past 5 11 9 (D) 10 minutes past 5 11 (A) 10
(B) Friday (D) Thursday
Karan was born on Saturday 22nd March 1982. On what day of the week was he 14 years 7 months and 8 days of age ? (A) Sunday (B) Tuesday (C) W ednesday (D) Monday 16.
8.
Mohini went to the movies nine days ago. She goes
(A) 16 minutes past 9
Sangeeta remembers that her father's birthday was certainly after eighth but before thirteenth of
(A) Tuesday (C) W ednesday
(D) Friday
clock be in the straight line, but not together ?
December. Her sister Natasha remembers that their father's birthday was definitely after ninth but before fourteenth of December. On which date of December was their father's birthday ? (A) 10th (B) 11th (C) 12th (D) Data inadequate 15. 6.
(C) W ednesday
week is today ?
Are the years 900 and 1000 leap years ? (A) Yes (B) No (C) Can't say (D) None of these
(B) Tuesday (D) Sunday
(B) Monday
to the movies only on Thursday. What day of the
Which two months in a year have the same calendar ? (A) June, October (B) April, November 13. (C) April, July (D) October, December
(A) Monday (C) W ednesday 5.
12.
(A) Sunday
Ajay left home for the bus stop 15 minutes earlier
If on 14th day after 5th March be Wednesday, what day of the week will fall on 10th Dec. of the same year ? (A) Friday (B) W ednesday
(A) 8.30 a.m.
(B) 8.45 a.m.
(C) Thursday
(C) 8.55 a.m.
(D) Data inadequate
(D) Tuesday
than usual. It takes 10 minutes to reach the stop. He reached the stop at 8.40 a.m. What time does he usually leave home for the bus stop ?
PAGE # 86
17.
The priest told the devotee, "The temple bell is 26. rung at regular intervals of 45 minutes. The last bell was rung five minutes ago. The next bell is due to be rung at 7.45 a.m." At what time did the priest give this information to the devotee ? (A) 7.40 a.m. (B) 7.05 a.m. (C) 6.55 a.m. (D) None of these
18.
There are twenty people working in an office. The first group of five works between 8.00 A.M. A.M. and 2.00 P.M. The second group of ten works between 10.00 27. A.M. and 4.00 P.M. P.M. And the third group of five works between 12 noon and 6.00 P.M. There are three computers in the office which all the employees frequently use. During which of the following hours the computers are likely to be used most ? (A) 10.00 A.M. –– 12 noon (B) 12 noon –– 2.00 P.M. (C) 1.00 P.M. –– 3.00 P.M. (D) 2.00 P.M. –– 4.00 P.M. 28.
19.
A tired worker slept at 7.45 p. p.m.. If he rose at 12 noon, for how many hours did he sleep ? (A) 5 hours 15 min. (B) 1 16 6 ho hours 1 15 5 mi min. (C) 12 hours (D) 6 hours 45 min.
20.
How many times are the hands of a clocks perpendicular in a day ? (A) 42 (B) 48 (C) 44 (D) 46
21.
If a clock shows 04: 28 then its mirror image will be ? (A) 07: 42 (B) 07: 32 (C) 08: 32 (D) 08: 42
22.
A watch, which gains uniformly, is 3 minutes slow at noon on Monday and is 3 minutes 48 seconds fast at 2 p.m. on the th e following Monday. Monday. What time it was correct ? (A) 2 p.m. On Tuesday 30. (B) 2 p.m. On Wednesday (C) 3 p.m. On Thursday (D) 1 p.m. On Friday.
23.
How many times are the hands of a clocks coincide in a day ? (A) 10 (B) 11 (C) 12 (D) 22
29.
31.
24.
At what time between 2 and 3 O’ clock the hands of a clock will make an angle of 160º ? (A) 20 minu minute tes s p pas astt 2 (B) (B) 30 30 m min inut utes es past past 2 (C) (C) 40 40 min minut utes es past past 2 (D) (D) 50 50 min minut utes es past past 2
25.
Ashish leaves his house at 20 minutes to seven in the morning, reaches Kunal’s house in 25 minutes, they finish their breakfast in another 15 minutes and leave for their office which takes another 35 32. minutes. At what time do they leave Kunal’s house to reach their office ? (A) 7.40 am (B) 7.20 am (C) 7.45 am (D) 8.15 am
The train for Lucknow leaves every two and a half hours from New Delhi Railway Station. An announcement was made at the station that the train for Lucknow had left 40 minutes ago and the next train will leave at 18. 00 hrs. At what time was the announcement made ? (A) 15.30 hrs (B) 17.10 hrs (C) 16.00 hrs (D) None of these A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 a.m., at what what time will he first touch a flag at 120 feet from the ground ? ( A) 4 p . m . (B) 5 p.m. (C) 6 p.m. (D) None of these If the two incorrect watches are set at 12 1 2 : 00 noon at correct time, when will both the watches show the correct time for the first time given that the first watch gains 1 min in 1 hour and second watch loses 4 min in 2 hours : (A) 6 pm, 25 days later (B) 12 : 00 noon, 30 days later (C) 12 noon, 15 days later (D) 6 am 45 days later Rajeev and Sanjeev are too close friends Rajeev's watch gains 1 minute in an hour and Sanjeev's watch loses 2 minutes in an hour. Once they set both the watches at 12 : 00 noon, with my correct watch. When will the two incorrect watches of Rajeev and Sanjeev show the same time together? (A) (A) 8 day days s lat later er (B) (B) 10 10 day days s later later (C) 6 days days later later (D) can't can't be dete determin rmined ed At a railway station a 24 hour watch loses 3 minutes in 4 hours. If it is set correctly on Sunday noon when will the watch show the correct time ? (A) 6 pm after 40 days (B) 12 noon after 75 days (C) 12 pm after 100 days (D) 12 noon after 80 days A swiss watch is being shown in a museum which has a very peculiar property. It gains as much in the day as it loses during night between 8 pm to 8 am. In a week how many times will the clock show the correct time ? (A) 6 times (B) 14 times (C) 7 times (D) 8 times A wrist watch which is running 12 minutes late on a Sunday noon is 16 minutes ahead of the correct time at 12 noon on the next Sunday. When is the clock 8 minutes ahead of time ? (A) Thursday 10 10 am (B) Friday noon (C) Friday 8 pm (D) Tuesday noon PAGE # 87
33.
34.
35.
36
37.
A clock loses 2 minutes in a hour and another clock 38. gains 2 minutes in every 2 hours. Both these cloc ks are set correctly at a certain time on Sunday and both the clocks stop simultaneously on the next day with the time shown being 9 am and 10 : 06 AM. What is the correct time at which they stopped? (A) 9 : 54 am (B) 9 : 44 pm (C) 9 : 46 am (D) 9 : 44 am David sets his watch at 6 : 10 am on Sunday, Sunday, which gains 12 minutes in a day. On Wednesday if this watch is showing 2 : 50 pm. What is the correct time ? 39. (A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm
Kumbhakarna starts sleeping between 1 pm and 2 pm and he wakes up when his watch shows such a time that the two hands (i.e., hour-hand and minute-hand) interchange the respective places. He wakes up between 2 pm and 3 PM on the same night. How long does he sleep ? (A) 55
5 min 13
(B) 110
10 min 13
(C) 54
6 min 13
(D) None of these
A clock loses 3% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what
Ramu purchased a second hand Swiss watch which is very costly. In this watch the minute-hand
will be the time that the clock will show exactly 14
3 and hour hand coincide after every 65 minutes. 11
(A) 1 : 36 : 48
Out of the following four choices which does not 41. show the coinciding of the hour hand and minutehand : (A) 3 : 16 : 2 (B) 6 : 32 : 43 (C) 9 : 59 : 05 (D) 5 : 27 : 16
If the actual time is 10 : 10, what is the position of
days from the time it was set right ? (B) 1 : 40 : 48
(C) 1 : 41 : 24 (D) 10 : 19 : 12 How much time does the watch lose or gain per day ? Direction : (40 to 41) A 12 dial clock has its minute hand (A) 4 min (B) 5 min defective. Whenever it touches dial 12, it (C) 4 min, 20 sec (D) none of these immediately falls down to 6 instead of running My watch was 8 minutes behind at 8 pm on Sunday smoothly (the hour hand remains unaffected during but within a week at 8 pm on Wednesday it was 7 that fall). It was set right at 12 ‘O’ clock in the noon. minutes ahead of time. During this period at which time this watch has shown the correct time : 40. What was the actual time when the minute hand of (A) Tuesday 10 : 24 am the clock touched dial 9 for the 5th time? (B) Wednesday 9 : 16 pm ( A ) 2 : 15 (B) 3 : 00 (C) It cannot show the correct time during this period (C) 5 : 15 (D) 6 : 45 (D) None of the above
the hour hand in that defective clock ? (A) Between 2 and 3
(B) Be Between 4 and 5
(C) (C) Bet Betwe ween en 10 and and 11 11
(D) (D) Bet Betwe ween en 3 and and 4
PAGE # 88
CUBE AND DICE-TEST CUBES
A cube is three dimensional figure, having 8 corners, 6 surfaces and 12 edges. If a cube is painted on all of its surfaces with any colour and further divided into various smaller cubes, we get following results. Smaller cubes with three surfaces painted will be present on the corners of the big cube. He r e 2
3
2 2 3
3 2 2 3
1 1
2
2 1
1 2
3
2 2
3
2 2 3 1 1 2 1 1 2 2 2 3
3
2 2 3
2 1 1 2
side of big cube side of small cube
4 1
4
3
2 1 1
2
Ex 1.
How many smaller cubes have three surfaces
2
painted ?
3
(A) 4
(B) 8
(C) 16
(D) 24
2
Sol. Smaller cubes with two surface painted will be present on the edges of the big cube. Smaller cubes with one surface painted will be present on Ex 2. the surfaces of the big cube. Smaller cubes with no surface painted will be present inside the big cube. If a cube is painted on all of its surfaces with a Sol. colour and then divided into smaller cubes of equal size then after separation, number of small er cubes so obtained will be calculated as under : Ex 3. Number of smaller cubes with three surfaces painted = 8 Number of smaller cubes with two surfaces painted = (n – 2) × 12 Number of smaller cubes with one surfaces Sol. painted = (n – 2)2 × 6 Number of smaller cubes with no surfaces painted = (n – 2)3 Ex 4. Where n = No of divisions on the surfaces of the bigger cube =
n=
length of edge of big cube length of edge of one smaller cube
TYPE I
Sol.
(B) Number of smaller cubes with three surfaces pain painte ted d =8 How many smaller cubes have two surfaces painted ? (A) 4
(B) 8
(C) 16
(D) 24
(D) Number of smaller cubes with two surfaces pain painte ted d = (n – 2) × 12 = (4 – 2) × 12 = 24 How many smaller cubes have only one surface painted ? (A) 8
(B) 16
(C) 24
(D) 32
(C) Number of smaller cubes with one surface pain painte ted d = (n – 2)2 × 6 = (4 – 2)2 × 6 = 4 × 6 = 24 How many smaller cubes will have no side painted ? (A) 18
(B) 16
(C) 22
(D) 8
(D) Number of smaller cubes with no surface painted = (n – 2)3 = (4 – 2)3 = (2)3 = 8
TYPE II
If a cube is painted on all of its surfac es with single colour and then divided into various smaller cubes of equal size.
If a cube is painted on all of its surfaces with different colours and then divided into various
smaller cubes of equal size. Directions : ( 1 to 4) A cube of side 4 cm is painted black on all of its surfaces and then divided into various smaller cubes of side 1 cm each. The smaller Directions : ( 5 to 7 ) A cube of side 4 cm is painted black on the pair of one opposite surfaces, blue on the pair cubes so obtained are separated.
444 Total cubes of obtained = 1 1 1
of another opposite surfaces and red on remainin g
64
pair of opposite surfaces. The cube is now divided into smaller cubes of equal side of 1 cm each.
PAGE # 89
Ex 9.
Sol.
Ex 5.
Sol.
Ex 6.
Sol.
How many smaller cubes have two surfaces painted ? (A) 4 (B) 8 (C) 16 (D) 24 (C) Number of smaller cubes with two surfaces painted = Number of cubes present at the corners + Numbers of cubes present at 4 edges = 8 + (n – 2) × 4 = 8 + 8 = 16
How many smaller cubes have three surfaces painted ? (A) 4 (B) 8 (C) 16 (D) 24 (B) Number of smaller cubes with three surfaces painted = 8 (These smaller cubes will have all three surfaces painted with different colour blue, black and red.) How many smaller cubes have two surfaces painted ? Ex 10. (A) 4 (B) 8 (C) 16 (D) 24 (D) Number of smaller cubes with two surfaces painted = 24. And out of this (a) Number Number of cubes cubes with with two surfaces surfaces painted painted Sol. with black and blue colour = 8. (b) Number Number of cubes cubes with with two surfaces surfaces painted painted with blue and red colour = 8. (c) Number Number of cubes cubes with with two two surfaces surfaces painte painted d with black and red color = 8. Ex 11.
How many smaller cubes have only one surface painted ? (A) 8 (B) 16 (C) 24 (D) 32 (D) Number of smaller cubes with one surface painted = Number Number of of cubes present present at the 8 edges + number of cubes present at the four surfaces= (n – 2) × 8 + (n – 2)2 × 4 = 2 × 8 + 4 × 4 = 16 + 16 = 32 How many smaller cubes will have no side painted ? (A) 18 (B) 16 (C) 22 (D) 8 (B) Number of smaller cubes with no side painted = Number of cubes on the two unpainted surfaces + number of cubes present inside the cube. = (n – 2)2 × 2 + (n – 2)3 = 4 × 2 + (2) 3 = 8 + 8 = 16.
How many smaller cubes have only one surface painted ? (A) 8 (B) 16 Sol. (C) 24 (D) 32 Sol. (C) Number of smaller cubes with one surface painted = 24. And out of this (a) Number of cubes cubes with with one one surface surface painted with black colour =8. TYPE IV (b) Number of cubes cubes with with one one surface surface painted with blue colour = 8. If a cube is painted on its surfaces in such a way (c) (c) Number Number of cubes cubes with with one one surface surface painte painted d that one pair of adjacent surfaces is left unpainted. with red colour = 8. Directions : (12 to 15 )A ) A cube of side 4 cm is painted red on TYPE III the pair of one adjacent surfaces, green on the pair of other adjacent surfaces and two adjacent If a cube is painted on its surfaces in such a way surfaces are left unpainted. Now the cube is divided that one pair of opposite surfaces is left unpainted. into 64 smaller cubes of side 1 cm each. Directions : ( 8 to 11 ) A cube of side 4 cm is painted red on the pair of one opposite surfaces, green on the pair of another opposite surfaces and one pair of opposite surfaces is left unpainted. Now the cube is divided into 64 smaller cubes of side 1 cm each.
Ex 7.
Ex 8.
Sol.
How many smaller cubes have three surfaces painted ? (A) 0 (B) 8 (C) 16 (D) 20 (A) Number of smaller cubes with three surfaces Ex 12. painted = 0 (Because each smaller cube at the corner is attached to a surface which is unpainted.)
How many smaller cubes have three surfaces painted ? (A) 2 (B) 4 (C) 8 (D) 6 PAGE # 90
Sol.
Ex 13.
Sol.
Ex 14.
Sol.
Ex 15.
(A) Number of smaller cubes with three surfaces Type-II painted = Number of smaller cubes at two corners =2 Ex 17. The figures given below show the two different positions of a dice. Which number will appear How many smaller cubes have two surfaces opposite to number 2 ?. painted ? (A) 4 (B) 8 (C) 16 (D) 14 (D) Number of smaller cubes with two surfaces painted = Number of smaller cubes at four corners (A) 3 (B) 4 + Number of smaller cubes at 5 edges. (C) 5 (D) 6 = 4 + (n – 2) × 5 = 4 + 2 × 5 Sol. (C) The above question, = 4 + 10 = 14 where only two positions of a dice are given, can easily How many smaller cubes have only one surface be solved with the painted ? following method. (A) 8 (B) 16 Step I. The dice, when unfolded, will appear as shown in (C) 24 (D) 30 the figure given on the right side. (D) Number of smaller cubes with one surface painted = Number of smaller cubes at four Step II. Write the common number to both the dice in the middle block. Since common number is 4, hence surfaces + Number of smaller cubes at 6 edges + number 4 will appear in the central block. Number of smaller cubes at two corners. = (n – 2)2 × 4 + (n – 2) × 6 + 2 Step III. Consider the figure (i) and write the first nu mber in = 4 × 4 + 2 × 6 + 2 = 16 + 12 = 28 + 2 = 30 the anti-clockwise direction of number 4, (common number) in block I and second number How many smaller cubes will have no side painted in block II. Therefore, numbers 3 and 2 being the ? first and second number to 4 in anticlockwise directions respectively, will appear in block I & II (A) 18 (B) 16 respectively. (C) 22 (D) 8
(A) Number of smaller cubes with no surfaces Step IV I V. Consider figure (ii) and wire first and second number in the anticlock-wise direction to number painted = Number of smaller cubes from inside 4, (common number) in block (III) & (IV). Hence the big cube + Number of cubes at two surfaces + numbers 6 and 5 will appear in the blocks III and IV Number of cubes at one edge. respectively. = (n – 2)3 + (n – 2)2 × 2 + (n – 2) Step V. Write remaining number in the remaining block. = (2)3 + (2)2 × + 2 Therefore, number 1 will come in the remaining = 8 + 8 + 2 = 18 block. Now, from the unfolded figures we find that DICES number opposite to 6 is 3, number opposite to 2 is 5 and number opposite to 4 is 1. Therefore, option (C) is our answer. Type-I ( Short Trick : From the given dice, we will take the General Dice : In a general dice the sum of numbers common number as the base and then in its on the any two adjacent faces is ‘7’. respect move clockwise direction and write as follows : 4 – 2 – 3 Standard Dice : In a standard dice the sum of 4 – 5 – 6. numbers on the opposite faces is '7'. Here,we find that number opposite to 6 is 3, number Ex 16. Which number is opposite 4 in a standard dice opposite to 2 is 5 and number opposite to 4 is given below ? remaining number 1. Therefore, option (C) is our answer. ) Sol.
1 5 4 (A) 1 (C) 5 Sol.
Ex 18. (B) 3 (D) Can ’t be determined
Clearly , from the standard dice the sum of numbers on the opposite faces is '7', so number opposite to 4 is 3.
On the basis of two figures of dice, you have to tell what number will be on the opposite face of number 5?
(A) 1 (C) 4
(B) 2 (D) 6
PAGE # 91
Sol.
(D) The above question where only two positions Type-V of a dice are given, can easily be solved with the following method : If in the given dice, there are two numbers common, Ex 21. Which of the following dices is identical to the unfolded figure as shown here ? then uncommon numbers will always be opposite of each other. Therefore, option (D) is our answer.
Type-III Ex 19.
From the following figures of dice, find which number will come in place of ‘?’ (X) (X)
Sol.
(A) 4 (B) 5 (C) 2 (D) 3 (D) If the above dice is unfolded, it will look like as the figure (i) given below.
Sol.
Figure (i)
Now the number in place of ‘?’ can be obtained by making a slight change in the figure as given here. Now comparing figure (ii) with third dice as above, we get that number in place of ? is 3.
Figure (ii)
Sol.
(B)
(C)
(D)
(A) From the unfolded figure of dice, we find that number opposite to 2 is 4, for 5 it is 3 and for 1 it is 6. From this result we can definitely say that figure (B), (C) and (D) can not be the answer figure as numbers lying on the opposite pair of surfaces are present on the adjacent surfaces.
Directions : (1 to 5) A cube is coloured orange on one face, pink on the opposite face, brown on one face and silver on a face adjacent to the brown face. The other two faces are left uncoloured. It is then cut into 125 smaller cubes of equal size. Now answer the following questions based on the above statements. 1.
Type-IV Ex 20.
(A)
A dice has been thrown four times and produces following results. 2.
How many cubes have at least one face coloured pink ? (A) 1 (B) 9 (C) 16 (D) 25 How many cubes have all the faces uncoloured ? (A) 24 (B) 36 (C) 48 (D) 64
3.
How many cubes have at least two faces coloured ? (A) 19 (B) 20 (C) 21 (D) 23
4.
How many cubes are coloured orange on one face and have the remaining faces uncoloured ? (A) 8 (B) 12 (C) 14 (D) 16
Which number will appear opposite to the number 3? 5. (A) 4 (B) 5 (C) 6 (D) 1 (A) From the figures (i), (ii) and (iv) we find that numbers 6, 1, 5 and 2 appear on the adjacent surfaces to the number 3. Therefore, number 4 will be opposite to number 3.
How many cubes one coloured silver on one face, orange or pink on another face and have four uncoloured faces ? (A) 8 (B) 10 (C) 12 (D) 16
PAGE # 92
Directions : (6 to 11) A cube is painted red on two adjacent Directions : (17 to 21) The outer border of width 1 cm of a surfaces and black on the surfaces opposite to cube with side 5 cm is painted yellow on each side red surfaces and green on the remaining faces. and the remaining space enclosed by this 1 cm Now the cube is cut into sixty four smaller cubes of path is painted pink. This cube is now cut into 125 equal size. smaller cubes of each side 1 cm. The smaller 6. How many smaller cubes have only one surface cubes so obtained are now seperated. painted ? (A) 8 (B) 16 17. How many smaller cubes have all the surfaces (C) 24 (D) 32 uncoloured ? 7. How many smaller cubes will have no surface (A) 0 (B) 9 painted ? (C) 18 (D) 27 (A) 0 (B) 4 (C) 8 (D) 16 18. How many smaller cubes have three surfaces 8. How many smaller cubes have less than three coloured ? surfaces painted ? (A) 2 (B) 4 (A) 8 (B) 24 (C) 28 9.
10.
11.
(D) 48
How many smaller cubes have three surfaces 19. painted ? (A) 4 (B) 8 (C) 16 (D) 24
(C) 8
(D) 10
How many cubes have at least two surfaces coloured yellow ? (A) 24
(B) 44
How many smaller cubes with two surfaces painted have one face green and one of the 20. adjacent faces black or red ? (A) 8 (B) 16 (C) 24 (D) 28
(C) 48
(D) 96
(A) 0
(B) 1
How many smaller cubes have at least one surface painted with green colour ? 21. (A) 8 (B) 24 (C) 32 (D) 56
(C) 2
(D) 4
How many cubes have one face coloured pink and an adjacent face yellow ?
How many cubes have at least one face coloured ? (A) 27
(B) 98
(C) 48 (D) 121 Directions : (12 to 16) A cube of 4 cm has been painted on its surfaces in such a way that two opposite Directions : (22 to 31) A solid cube has been painted yellow, surfaces have been painted blue and two adjacent blue and black on pairs of opposite faces. The surfaces have been painted red. Two remaining cube is then cut into 36 smaller cubes such that surfaces have been left unpainted. Now the cube 32 cubes are of the same size while 4 others are is cut into smaller cubes of side 1 cm each. of bigger sizes. Also no faces of any of the bigger cubes is painted blue. 12. How many cubes will have no side painted ? (A) 18 (B) 16 22. How many cubes have at least one face painted (C) 22 (D) 8 blue ? 13. How many cubes will have at least red colour on (A) 0 (B) 8 its surfaces ? (C) 16 (D) 32 (A) 20 (B) 22 (C) 28 (D) 32 23. How many cubes have only one faces painted ? 14.
15.
16.
How many cubes will have at least blue colour on its surfaces ? (A) 20 (B) 8 24. (C) 24 (D) 32 How many cubes will have only two surfaces painted with red and blue colour respectively ? 25. (A) 8 (B) 12 (C) 24 (D) 30 How many cubes will have three surfaces coloured ? (A) 3 (B) 4 26. (C) 2 (D) 16
(A) 24 (C) 8
(B) 20 (D) 12
How many cubes have only two faces painted ? (A) 24 (B) 20 (C) 16 (D) 8 How many cubes have atleast two faces painted ? (A) 36 (B) 34 (C) 28 (D) 24 How many cubes have only three faces painted ? (A) 8 (B) 4 (C) 2 (D) 0 PAGE # 93
27.
28.
29.
How many cubes do not have any of their faces 37. painted yellow ? (A) 0 (B) 4 (C) 8 (D) 16 38. How many cubes have at least one of their faces painted black ? (A) 0 (B) 8 (C) 16 (D) 20 39. How many cubes have at least one of their faces painted yellow or blue ? (A) 36 (B) 32 (C) 16 (D) 0 40.
30.
How many cubes have no face painted ? (A) 8 (B) 4 (C) 1 (D) 0
31.
How many cubes have two faces painted yellow and black respectively ? (A) 0 (B) 8 41. (C) 12 (D) 16
Directions : (32 to 35) Some equal cubes are arranged in the form of a solid block as shown in the adjacent figure. All the visible sufaces of the block (except the bottom) are then painted. 32.
How many cubes have one face painted ? (A) 9 (B) 24 (C) 22 (D) 20
34.
How many cubes have only two faces painted ? (A) 0 (B) 16 (C) 20 (D) 24
How many cubes have only three faces painted ? (A) 4 (B) 12 (C) 6 (D) 20 Directions : (36 to 40) A cuboid of dimensions (6 cm 4 cm 1 cm) is painted black on both the surfaces of dimensions (4 cm 1 cm), green on the surfaces of dimensions (6 cm 4 cm). and red on the surfaces of dimensions (6 cm 1 cm). Now the block is divided into various smaller cubes of side 1 cm. each. The smaller cubes so obtained are separated.
35.
36.
How many cubes will have all three colours black, green and red each at least on one side? (A) 16 (B) 12 (C) 10 (D) 8
If cubes having only black as well as green colour are removed then how many cubes will be left? (A) 4 (B) 8 (C) 16 (D) 30 How many cubes will have 4 coloured sides and 2 sides without colour? (A) 8 (B) 4 (C) 16 (D) 10 How many cubes will have two sides with green colour and remaining sides without any colour? (A) 12 (B) 10 (C) 8 (D) 4
Which alphabet is opposite D ?
(A) E (C) F 42.
How many cubes do not have any of the faces painted ? (A) 27 (B) 8 (C) 10 (D) 12
33.
How many cubes will be formed? (A) 6 (B) 12 (C) 16 (D) 24
(B) C (D) A
What should be the number opposite 4 ?
(i)
(ii)
(iii)
(A) 5 (C) 3
(B) 1 (D) 2
(i)
(ii)
43.
(iii) (iv) Which letter will be opposite to letter D ? (A) A (B) B (C) E (D) F Directions : (44 to 45) The figure (X) given below is the unfolded position of a cubical dice. In each of the following questions this unfolded figure is followed by four different figures of dice. You have to select the figure which is identical to the figure (X). PAGE # 94
50.
44.
Which symbol will appear on the opposite surface to the symbol x?
(X)
(A)
(A)
(B)
(B) =
(C) 51. (B)
(D)
(D) O
Three positions of the same dice are given below. Observe the figures carefully and tell which number will come in place of ‘?’
1 6
3
3
5
(i) 45.
52.
(A)
(iii) (B) 6 (D) 5
6
4
1
2
(i) (C)
?
6
1
(ii)
(A) 2 (C) 6
(D)
?
On the basis of the following figures you have to tell which number will come in place of ‘?’
3
(B)
2
(ii)
(A) 1 (C) 3
(X)
4
4
5
(iii) (B) 3 (D) 4
Directions : (53 to 55) Choose from the alternatives, the boxes that will be formed when figure (X) is folded: Directions : (46 to 48) In each of the following questions, select the correct option for the question asked.
53.
(i) 46.
47.
48.
(X)
(ii)
Which number will come opposite to number 2? (A) 5 (B) 1 (C) 6 (D) 3 Which number will come opposite to number 6? (A) 1 (B) 5 (C) 4 (D) 3
(A)
(B)
(C)
(D)
Which number will come opposite to number 4? (A) 3 (B) 5 (C) 1 (D) 2 +
49.
On the basis of two figures of dice, you have to tell what 54. number will be on the opposite face of number 5?
(i)
(ii)
(A) 1 (C) 4
(B) 2 (D) 6
(X)
(A)
(C)
(B)
+
+
(D)
PAGE # 95
55.
59.
(X)
(i)
(A)
(C)
(ii)
(B) (iii) (iv) Which number is opposite to number 5? (A) 6 (B) 5 (C) 1 (D) 3
(D)
Directions : (60 to 64) Choose the cube from the options that will unfold to give the figure on the left Direction : (56) The six faces of a cube have been marked with numbers 1, 2, 3, 4, 5 and 6 respectively. This cube is rolled down three times. The three 60. positions are given. Choose the figure that will be formed when the cube is unfolded.
X
M
56. M
X
M (A)
(A)
M
X
(B)
(C)
(D)
(E)
(B) 1
4
8
3
61.
7 9
(C)
(D) 9
7 8
1
57.
Which number is opposite 3 in a standard dice given below ?
(A) 1 (C) 5 58.
(B) 4 (D) Can ’t be determined
(A)
4
8 1
7
(B)
(C)
7 (D)
8
7 4 (E)
62. 8 D
Which number is opposite 4 ?
(A) 5 (C) 2
(B) 3 (D) 1
Directions : (59) In the following question four positions of the same dice have been shown. You have to see these figures and select the number opposite to the number as asked in each question.
8
8
(A)
(B)
D (C)
(D)
(E)
(D)
(E)
63. B B
(A)
(B)
(C)
PAGE # 96
66.
Which number/letter is opposite 2 ?
J
3
64.
J (A)
(B)
(D)
Which letter is opposite Q ?
Q O P L N M (A) L (C) N
(B) C (D) 3
(E)
Directions : (65 to 68) In each of the following questions, a diagram has been given which can be folded into a cube. The entries given in the squares indicate the entries on the face of the cube. In each question a number or a letter has been given . Of the four alternatives given below it, you have to find the one that would appear on the face opposite to it in the cube. 65.
(A) A (C) 1
J (C)
I C A B 2
67.
Which number/letter is opposite O? L N M 2 I O (A) L (C) N
68.
(B) M (D) 2
Which letter is opposite R? Q R S P U T
(B) M (D) P
(A) P (C) T
(B) S (D) U
PAGE # 97
FORCE
AND
NEWTON S
LAW
OF
Q ue .
1
2
3
4
5
6
7
8
Ans.
B
C
C
B
A
B
D
A
Q ue .
16
17
18
19
20
21
22
23
24
Ans.
B
D
A
A
A
C
C
D
Q ue .
31
32
33
34
35
36
37
Ans.
B
C
B
A
A
C
Q ue .
46
47
48
49
50
Ans.
D
D
CD CD
D
Q ue .
61
62
63
Ans.
B
A
C
Ques. 1 C Ans. Ques. 1 6 B Ans. Ques. 3 1 C Ans.
11
12
13
14
15
C
C
C
C
C
25
26
27
28
29
30
B
B
B
B
A
D
C
38
39
40
41
42
43
44
45
B
D
A
C
A
B
D
A
A
51
52
53
54
55
56
57
58
59
60
A
D
D
C
B
A, A, C
C
A , B, C
B
B
B
64
65
66
67
68
69
C
D
C
B
B
C
2 C 17 A 32 B
3 B 18 C 33 C
4 B 19 B 34 A
5 B 20 C 35 B
6 A 21 B 36 D
7 D 22 B 37 C
8 A 23 A 38 B
9
10
MOTION(PHYSICS)
A CD CD B
9 10 10 C B 24 25 C C 39 40 C A
11 C 26 D 41 A
12 D 27 C
13 C 28 A
14 B 29 B
15 B 30 C
Q.
1
2
3
4
5
6
7
8
9
10
Ans.
B
A
B
A
D
A
D
C
B
A
Q.
11
12
13
14
15
16
17
18
19
20
Ans.
A
A
D
A
B
A
D
B
C
C
Q.
21
22
23
24
25
26
27
28
29
30
Ans.
B
C
A
C
C
A
B
D
B
B
Q.
31
32
33
34
35
36
37
38
39
40
Ans.
C
B
A
A
B
D
C
C
D
C
Q.
41
42
43
44
45
46
47
48
49
50
Ans.
C
A
D
D
A
A
C
C
C
D
Q.
51
52
53
54
55
56
57
58
59
60
Ans.
C
D
C
C
C
A
C
B
D
D
Q.
61
62
63
64
65
66
67
68
69
70
Ans.
A
B
B
A
A
A
B
B
B
C
Q.
71
72
73
74
75
76
77
78
79
80
Ans.
D
A
B
C
A
A,D
B
D
B
A
Q.
81
82
83
84
85
86
87
88
89
90
Ans.
D
A
B
B
B
B
B
D
D
B
Q.
91
Ans.
C
PAGE # 9898
Q.
1
2
3
4
5
6
7
8
9
10
Ans.
B
B
C
B
A
A
C
C
C
C
Q.
11
12
13
14
15
16
17
18
19
20
Ans.
D
A
B
B
D
B
D
A
D
B
Q.
21
22
23
24
25
26
27
28
29
30
Ans.
C
B
D
B
C
C
D
C
A
A
Q.
31
32
33
34
35
36
37
38
39
40
Ans.
B
D
A
D
B
A
B
B
D
D
Q.
41
42
Ans.
A
B
Q.
EXERCISE-1 (Number Series) Que .
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
C
D
D
A
C
D
B
C
C
C
D
C
C
B
D
Que .
16 16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
B
C
C
C
B
A
B
D
D
A
C
B
B
C
A
Que .
31 31
32
33
34
35
Ans.
C
C
C
D
D
EXERCISE- 2 (Alphabet Series) Qu e.
1
2
3
4
5
6
7
8
9
10
11 11
12 12
13 13
14 14
15 15
Ans.
D
A
D
C
C
A
D
C
D
B
D
C
C
C
D
Qu e.
16 16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
C
A
B
C
C
C
A
B
A
C
D
B
C
D
B
EXERCISE- 3 (Letter Repeating Series) Q u e.
1
2
3
4
5
6
7
8
9
10
11 11
12 12
13 13
14 14
15 15
Ans.
D
D
A
A
C
B
A
C
D
D
C
B
D
C
B
Que.
16 16
17
18
19
20
21
22
23
24
25
26
Ans.
C
A
A
A
C
D
D
D
A
B
D
PAGE # 9999
EXERCISE- 4 (Missing Term In Figure) Q ue .
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
B
D
B
D
C
C
C
D
A
D
D
B
C
A
B
Q ue .
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans .
B
C
A
B
B
A
C
A
D
D
A
C
D
B
A
Q ue .
31
32
33
34
35
36
37
38
39
40
Ans .
B
B
A
C
A
C
B
C
D
B
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
A
B
D
C
A
D
C
C
C
C
C
D
C
C
B
Que.
16 16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans. Que. Ans. Que. Ans.
C
C
D
D
C
D
D
B
B
C
A
A
D
C
D
31 31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
D
C
B
A
B
A
D
A
D
C
D
A
C
B
D
46 46
47
48
49
50
51
52
53
54
55
56
C
A
D
B
D
D
D
A
C
A
A
57 D
58 B
Que.
1
2
3
4
5
6
7
8
9
10
11 11
12 12
13 13
14 14
15 15
Ans.
D
C
B
D
D
B
C
B
C
C
A
B
C
B
A
Que. Ans.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
B
B
B
B
C
B
C
D
C
B
D
C
B
B
D
Que. Ans.
31
32
33
34
35
36
37
38
39
40
41
D
B
D
B
A
A
C
A
D
A
C
Que.
1
2
3
4
5
6
7
8
9
10
11 11
12 12
13 13
14 14
15 15
Ans.
D
C
C
D
A
C
C
D
B
B
C
A
C
D
B
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
C
D
C
B
A
B
D
D
A
D
C
A
D
C
B
Que.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans. Que.
C
D
C
D
C
A
D
C
B
C
B
B
A
D
B
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans. Que.
D 61
A 62
B 63
C 64
D 65
A 66
B 67
D 68
B
D
C
B
A
C
C
Ans.
A
D
E
D
C
A
B
B
PAGE # 100 100
