FORCE AND NEWTON’S LAWS OF MOTION INTRODUCTION Force is a push or pull which tries to change or successfully changes the state of rest or of uniform motion of a body, i.e., force is the cause of translatory motion. 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.
of nuclear forces. It acts within the nucleus that too upto a very small distance. It does not depends on charge and acts equally between a proton and proton, a neutron and neutron, and proton and neutron, electrons does not experience this force. It acts for very short distance order of 10–15 m. (d) Weak Force : 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 : FN:FEM:FW:FG::1:10–2:10–7:10–38
(a) Gravitational Force : 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 medium. Gravitational force is the weakest force of nature.
On the basis of contact forces are classified into two categories (i)
Contact forces
(ii) Non contact or field forces
(b) Electromagnetic Force :
(a) Contact force :
Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force
Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. e.g. Normal, Tension etc.
(i)
( b ) Field force :
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 experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions and repulsions between atoms/molecules. (c) Nuclear Force :
Force which acts on an object at a distance by the interaction of the object with the field produced by other object is called field force. e.g. Gravitational force, Electro magnetic force etc.
DETAILED ANALYSIS OF CONTACT FORCE (a) Normal force (N) : It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. 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, fission, and fusion, etc. results because of unbalancing
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Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure.
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.
(b) Tension : Tension is the magnitude of pulling force exerted by a string, cable, chain, rope etc. W hen a string is connected to a body and pulled out, the string said to be under tension. It pulls the body with a force T, 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.
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. (c) Force Exerted by spring :
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.
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. 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 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 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
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Note : Spring force is also electromagnetic in nature : (d) Friction force : 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) More about frictional force : (A) About static friction
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. flim= SN. Here s is coefficient of static friction.
(ii) Statics and Kinetic Frictions : •
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 – fs = 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) About kinetic friction : 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 (F2) 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).
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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 = •
fk N
There are two types of kinetic frictions: (i)
Sliding friction : The force of friction when one body slides over the surface of the another body is called sliding friction.
(ii) Rolling friction : 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 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 conservative if the total work done by the 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). 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 the body from A to B. However, when body is released from B, we recover mgh of work. Other examples of conservative forces are spring force, electrostatic force etc. (b) Non-Conservative Force : 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),
flim > fkinetic > frolling.
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
the work done by friction would be greater than FS. Finally, if the book is moved through any closed path, 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 for a conservative force.
greater than k .Theoretical value of can be o to but practical value is 0 < 1.6
SYSTEM Two or more than two objects which interact with each
(a) Conservative Force : 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 in going from A to B. This is illustrated in figure.
other form a system. Classification of forces on the basis of boundary of system : (a) Internal Forces : Forces acting with in a system among its constituents. (b) External Forces : Forces exerted on the constituents of a system by the outside surroundings are called as external forces.
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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. of sphere ‘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’ : (exerted by A)
of the object of system. Contact forces are normal , 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’ 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 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 -
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F = 0 with F =
F
x
=0;
Note : The absolute unit of force remains the same
everywhere, but the gravitational unit of force varies from place to place because it depends on the value of g.
F = 0
F
i
y
=0;
F
z
( b ) Applications of Newton’s 2 nd Law
=0
(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 objects are in equilibrium : 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
(a) Stable equilibrium :
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 several 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 than two unknowns, go back to step 2 and select
O (b) Unstable equilibrium : If on slight displacement from equilibrium position a body moves in the direction 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.
another object and repeat these steps. Eventually at 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’ of mass 10 kg is suspended with string as shown in figure. Find tension in the string. (g = 10 m/s2).
Sol.F.B.D. of block 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).
. (a) Newtons 2 nd law of motion : 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 external force. Any internal force in the system is not to
Sol.F.B.D. of 10 kg block For equilibrium of block along Y axis.
be included in F . In S.I. the absolute unit of force is newton (N) and gravitational unit of force is kilogram weight or kilogram force (kgf.)
F
y
T0
0
T0 = 10 g T0 = 100 N
10g
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N2 = 50 sin 30º + N3 N3 = 100 – 25 = 75 N & N4 = 50 cos 30º + 20 g N4 = 243.30 N
F.B.D. of point ‘A’ y
F
y
T2
0 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 :
1 100 = N. 3 2
F.B.D. of point of ‘B’ y 60º T3
B
T = 10 g = 100 N F.B.D. of pulley :
T4 x
30º T2
F
y
= 0 T4 cos 60º = T2 cos 30º
T4 = 200 N
Since string is massless, so tension in both sides
F
of string is same. So magnitude of force exerted by string on pulley
and
x
= 0 T3 + T2 sin30º = T4 sin 60º
200 T3 = 7.
3
N
Two blocks are kept in contact as shown in figure. Find :-
=
100 2 100 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 :
(ii) Accelerating Objects :
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 all 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. Step 3 : Choose a convenient coordinate system, show
N1 = 10 g = 100 N .......(1) N2 = 100 N .........(2) F.B.D. of 20 kg block
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
x
= max &
F
y
= may.
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.
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Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at
Sol.For calculating the value of F0. F.B.D of whole system 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 in figure. Find the contact force between m1 and m2.
10 g = 100 N
F0 –100 = 10 × 2 F0 = 120 N Sol.Considering both blocks as a system to find the common acceleration. Common acceleration
F a= m1 m2 .......(1)
F
m1
m2
........(1)
(b) According to Newton’s second law, net force on rope. F = ma = 2 × 2 = 4N ............(2) (c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.
a
T–4g=4.2 To find the contact force between ‘A’ and ‘B’ we draw F.B.D. of mass m2. F.B.D. of mass m2
T = 48 N 11. A block of mass 50 kg is kept on another block of 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.
F
x
= max
B A
N = m2 . a N=
m2F m1 m2
F sin ce a m1 m 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 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
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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.
2
=9.8 m/s )
SPRING BALANCE
Sol.Since string is pulled by 100 N force. So tension in the
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. 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 block spring balance hook m
F.B.D. of ‘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 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 calculating the reading, first we draw 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
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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.
0;
or 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 : Along y-axis we have ;
(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 ?
T
spring balance
T
hook
F
y
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
mg . .......(ii) cos sin
T 10kg
T
17. Discuss the direction of friction in the following cases : T
T 10kg
It is clear from above discussion that pull force is smaller than push force.
(b) T
or F =
10kg
(c) Sol. In all the three cases the spring balance reads 10 kg. To understand this let us cut a section inside the spring as shown;
(i)
A man walks 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. (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 is easier than 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
f1 – f 2 = 0
f1 = f2
N1 = N2.
&
N1
f1
N2
f1
f2
f2
Ground N1
N2
(b)
10 10
PAGE # 1010
(ii) When man is gaining the speed : The frictional force on rear leg f1 will be greater than frictional force on front leg f2 (fig. b).
f f acceleration of the man, a = 1 2 . m
Block is to be raised without acceleration, so T = 25 g.
N1 = 25 g + 50 g = In II
(iii) When cycle is gaining speed : In this case torque
75 g = 75 × 9.8 = 735 N nd
case, let the force exerted by the man on the floor
in N2 . Consider the forces inside the dotted box, we
is applied on the rear wheel of the cycle by the
have
chain-gear system. Because of this the slipping
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).
N1
N2
f1
f2 (a)
N1
N2
f1
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 kg respectively.
f2 (b)
(iv) When cycle is slowing 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. In Ist case, let the force exerted by the man on the floor is N1. Consider the forces inside the dotted box, we have N1 = T + 50 g.
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.
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PAGE # 1111
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 speed, 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 a=
5.
(A) 0.49 m/s
(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 = 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 : (A) 2.5 m/s2
(B) 5.0 m/s2
(C) 10 m /s2
(D) none of the above
EXERCISE 6. Normal Force : 1.
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’ 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’ (C) N’> N
(B) N > N’
(D) cannot be determined (A) Same (C) 2 : 1 2.
(B) 1 : 2 (D) 1 : 3
Two forces of 6N and 3N are acting 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.
STATEMENT-1 : Block A is moving on horizontal surface 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
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
3.
There are two forces on the 2.0 kg 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, statement-2 is correct explanation for statement-1.
y
(B) statement-1 is true, Statement 2 is true, 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 N x
8. 30º a = 12 m/s
A certain force applied to a body A gives it an acceleration of 10 ms–2 . The same force applied to body B gives it an acceleration of 15 ms–2 . If the two bodies are joined
2
(A) –20 ˆj N
(B) – 20 ˆi + 20 ˆj N
(C) –32 ˆi – 12 3 ˆj N
(D) –21 ˆi – 16 ˆj N
together and same force is applied to the combination, the acceleration will be : (A) 6 ms
–2
(C) 12.5 ms–2
(IJSO/Stage-I/2011) (B) 25 ms–2 (D) 9 ms–2
12 12
PAGE # 1212
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) 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 horizontally with a force F. If the rope AB makes an angle with the vertical in equilibrium,then the tension in the string AB is :
M2 (sin ) g M1 M2
(B)
M2 sin M1 sin g M1 M2
(D) Zero
(C)
M1g(sin ) M1 M2
14. Three masses of 1 kg, 6 kg and 3 kg are connected to each other by threads and are placed on table as shown in figure. What is the acceleration with which the system is moving ? 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
(B) 1 ms–2
(C) 2 m s–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
g 8g (A) downward, 4 7 (C)
g 6 downward, g 7 7
g g (B) upward, 4 7 (D)
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 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
PAGE # 1313
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 pulley, as shown in figure. The force exerted by the string on the pulley P is :
(A) mg
(B) 2 mg
(C) 2 mg
(D) 4 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 30N 60º
4kg
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 acceleration the monkey must have if it is to lift the package off the ground is :
(A) 16
(B) 6
(C) 4
(D) 8
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. 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
PAGE # 1414
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–M1a1)/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 acceleration of the trolly is (Use
25. A weight is supported by two strings 1.3 and 2.0 m
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
(B) 1.23 m
(C) 0.77 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 downward with an acceleration of 2ms–2 is (Take g = I0 ms –2 ) (KVPY/2007) (A) 7·2 × 104 N (C) 6 × 104 N
(B) 4.8 × 104 N (D) 1.2 × 104 N
(A) 4.9 ms–2
(B) 9.8 ms–2
(C) 49 ms–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 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. (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 Force Exerted by Spring : 28. In the given figure. What is the reading of the spring balance:
(A) 10 N
(B) 20 N
(C) 5 N
(D) Zero
(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 of 5 × 104 N through a distance of 3m. Assume that the resistance due to water is negligible, the speed of the ship is : (A) 1.5 m/s (B) 60 m/s (C) 0.1 m/s (D) 5 m/s 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 (B) The ground on the horse (C) The ground on the cart (D) The horse on the ground
15 15
PAGE # 1515
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/s2)
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 : (A)
P 2 mg
(B)
P2 (D) 2 m2g
P (C)
P2 2 mg
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
(A) 20 N (C) 12N
5kg
(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 inclined 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)
(A)
(B)
(C)
(D)
1 2 3 m = 3kg
(B) 10 N (D) 15 N
16 16
PAGE # 1616
45. A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed under her pushing force (F) which is obliquely upward as shown . Then
(KVPY/2011)
(A) The resultant of the pushing force (F) , weight of
the toy, normal force by the ground on the toy and the frictional force is zero. (B) The normal force by the ground is equal and opposite to the weight of the toy. (C) The pushing force (F) of the child is balanced by the equal and opposite frictional force (D) The pushing force (F) of the child is balanced by the total internal force in the toy generated due to deformation 46. On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each 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)
Weighing Machine : 49. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ 3:2. The value of ‘a’ is : (g = acceleration, due to gravity) (A) (3/2)g (B) g (C) (2/3) 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 uniformly. Then (A) t1 = t2 (B) t1 > t2 (C) t1 < t2 (D) t1 < t2 or t1 > t2 depending
•
ASSERTION / REASON
51. STATEMENT-1 : A man standing in a lift which is moving upward, will feel his weight to be greater than when the lift was at rest. STATEMENT-2 : If the acceleration of the lift is ‘a’ upward 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) 16 m
(C) statement-1 is true, Statement 2 is false
(D) 4m 47. Which of the following does NOT involve friction ?
(D) statement-1 is False, Statement 2 is True 52. A beaker containing water is placed on the platform of
(IJSO/Stage-I/2011) (A) Writing on a paper using a pencil
a digital weighing machine. It reads 900 g. A wooden
(B) Turning a car to the left on a horizontal road. (C) A car at rest parked on a sloping ground
the beaker (without touching walls of the beaker). Half
(D) Motion of a satellite around the earth.
the reading of weighing machine will be :
block of mass 300 g is now made to float in water in the wooden block is submerged inside water. Now, (IAO/Jr./Stage-I/2009)
48. In the two cases shown below, the coefficient of kinetic 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
PAGE # 1717
54. In which of the following cases the net force is not zero ? (A) A kite skillfully 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 time. The force acting on the particle is zero in the region.
(A) AB (C) CD
(B) BC (D) DE
58. A force of magnitude F1 acts on a particle so as to accelerate if from rest to velocity 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 59. In a imaginary atmosphere, the air exerts a small force F on any particle in the direction of the 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 t1 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. W hich of the following represents the stone’s position ? (A) x = 5t – 3 (B) x = 5t2 + 8t – 3 2 (C) x = –5t + 5t – 3 (D) x = 5t3 + 4t2 – 3 61. Three forces act on a particle that moves with
unchanging velocity v = (3 ˆi – 4 ˆj ) m/s. Two of the forces are F1 = (3 ˆi + 2 ˆj – 4 kˆ ) N and F2 = (–5 ˆi + 8 ˆj + 3 kˆ ) N. The third force is : (A) (–2 ˆi + 10 ˆj – 7 kˆ ) N (B) (2 ˆi – 10 ˆj + kˆ ) N (C) (7 ˆi – 2 kˆ + 10 ˆj ) N (D) none of these
(A) – ˆi m/s
(B) – 1.5 ˆi m/s
(C) 6.5 ˆi m/s
(D) 13 ˆi m/s
57. Figure shows the displacement of a particle going
62. An 80 kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2 . The mass of the parachute is 5.0 kg. The upward force on the open parachute from the air is : (A) 620 N (C) 800 N
(B) 740 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
PAGE # 1818
64. A body of mass 5 kg starts from the origin with an initial velocity u = 30 ˆi + 40 ˆj ms–1 . If a constant force F = –( ˆi + 5 ˆj ) N acts on the body, the time in which the
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 : (B) 20 s
(C) 40 s
(D) 80 s
65. STATEMENT-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 for statement 1. (B) statement-1 is true, Statement 2 is true, statement2 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
3
F(N)
(A) 5 s
3
1
0,0
2
4
6
8
10
x(m)
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 : (A) 3.16 ms–1
(B) 9.3 ms–1
(C) 8 ms
(D) 6.8 ms–1
–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 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
1x 2 2 x1 (B) x x 2 1
1x 2 2 x1 (C) x x 1 2
1x1 2 x 2 (D) x x 1 2
19 19
PAGE # 1919
CARBON INTRODUCTION
VERSATILE NATURE OF CARBON
The compounds like urea, sugars, fats, oils, dyes, proteins, vitamins etc., which are isolated directly or indirectly from living organisms such as animals and plants are called organic compounds.The branch of chemistry which deals with the study of these compounds is called ORGANIC CHEMISTRY.
VITAL FORCE THEORY OR BERZELIUS HYPOTHESIS Organic compounds cannot be synthesized in the laboratory because they require the presence of a mysterious force (called vital force) which exists only in living organisms.
WOHLER’S SYNTHESIS
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.
In 1828, Friedrich Wohler synthesized urea (a well known organic compound) in the laboratory by heating ammonium cyanate.
(iii) Tendency to form multiple bonds : Due to small size of carbon it has a strong tendency to form multiple bonds (double & triple bonds).
(NH 4)2 SO4 + Ammonium sulphate
(iv) Isomerism : It is a phenomenon by the virtue of which two compounds have same molecular formula but different physical and chemical properties.
2 KCNO Potassium cyanate
2NH 4CNO + K2SO 4 Ammonium Potassium sulphate cyanate
CLASSIFICATION OF ORGANIC COMPOUNDS
Note : Urea is the first organic compound synthesized in the laboratory.
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 ORGANIC 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
(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.
Aromatic compounds
Alicyclic compounds
Organic chemistry is treated as a separate branch because of following reasons-
(a) Open Chain Compounds : 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
Toluene
Phenol n-Butane is a straight chain alkane while 2-Methylpropane is branched alkane.
Ethyl benzene
Aniline
Note : Benzene is the parent compound of majority of aromatic organic compounds.
HYDROCARBONS (b) Closed Chain or Cyclic Compounds : Apart from the open chains, the organic compounds can have cyclic or ring structures. 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 of Hydrocarbons : The hydrocarbons can be classified as :
e.g.
or
or
Cyclopropane
Cyclobutane
(i) Saturated hydrocarbons : (A) Alkanes : Alkanes are saturated hydrocarbons containing only carbon - carbon and carbon - hydrogen single covalent bonds. General formula- 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
CnH2n C2H 4 C3H 6
(Ethene) (Propene)
(ii) Aromatic compounds : Organic compounds which
(B) Alkynes : They are also unsaturated hydrocarbons which contain carbon-carbon triple bond. They contain four hydrogen atoms less than the corresponding alkanes.
contain one or more fused or isolated benzene rings are called aromatic compounds.
General formula e.g.
CnH 2n–2 C2H 2 (Ethyne) C3H 4
(Propyne)
21 21
PAGE # 21
Examples :
NOMENCLATURE OF ORGANIC 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) (a) Basic rules of IUPAC nomenclature of organic compounds :
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.
M E P B E Propene
Eth -
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 may be divided into two parts(A) Word root
C
(B) Primary suffix
(A) Word roots for carbon chain lengths : Chain length
Word root
Chain length
Word root
C1 C2 C3 C4 C5
MethEth Prop But Pent-
C6 C7 C8 C9 C10
HexHeptOctNonDec-
(B)
–H
H
H
H
C
C
H H Ethyl
(B) Primary suffix :
(C)
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PAGE # 22
A branched chain hydrocarbon is named using the
e.g.
following general IUPAC rules : Rule1: Longest chain rule : Select the longest possible
2–Methylpentane
4–Methylpentane
(Correct)
continuous chain of carbon atoms. If some multiple
(Wrong)
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 . (ii) Saturation or unsaturation determines the primary suffix (P. suffix). (iii) Alkyl substituents are indicated by prefixes.
CH3
e.g.
Prefix : Methyl
CH3 – CH – CH2 – CH – CH3
Word root : Hept-
CH3
CH2 – CH2 – CH3
e.g. CH3 – CH2 – C – CH3 CH2
e.g. CH3 – CH 2– CH – CH2– CH 3 CH – CH3
e.g.
P. Suffix : -ane
Prefix : Methyl Word root : ButP. Suffix : –ene
Prefixes : Ethyl, Methyl Word root : PentP. Suffix : -ane
4
CH3 |
3
2
(Wrong)
1
1
3-Methylbut-1-yne (Correct)
Prefix : Methyl Word root : pentP. Suffix: - ane
e.g. CH3 – CH2 – CH – CH2 – CH3
2–Methylbut – 3 – ene
(Correct)
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 prefix di, tri etc. , is attached to the name of the substituents. The positions of the substituents are separated by commas. CH3 5 4 3 2 1 CH3 – CH2– C – CH – CH3
e.g.
CH3 CH3
2,3 - Dimethylpentane
2,3,3 - Trimethylpentane
e.g.
2,3,5 -Trimethylhexane
CH3
CH3 |
2
2,2,4 - Trimethylpentane
Rule 4 : Alphabetical arrangement of prefixes: If there are different alkyl substituents present in the compound their names are written in the alphabetical order. However, the 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 such a way that the alkyl group which comes first in alphabetical order gets the lower position.
e.g.
2–Methylbutane (Correct)
3–Methylbutane (Wrong)
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 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. (a) Main Functional Groups : Word root : Hex Primary suffix : - ane Substituent : two methyl & one ethyl groups IUPAC name : 4-Ethyl - 2, 4 - dimethylhexane
(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), Ethanal (CH3CHO) etc.
Some other example :
(i)
Word root : Prop P. Suffix : -ane Substituent : two methyl groups IUPAC name : 2, 2 - Dimethylpropane
(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)
Word root : But P. Suffix : - ene Substituent : two methyl groups IUPAC name : 2, 3 - Dimethylbut - 1 - ene (iii)
Word root : Hex P. Suffix : - yne Substituent : one methyl group IUPAC name : 4 - Methylhex - 2 - yne
(b) Nomenclature of Compounds Containing Functional Group : In case functional group (other 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
Prefix
Carboxylic acid
Carboxy
Suffix
IUPAC Name
- oic acid
Alkanoic acid
(R = CnH2n+1) Ester
Aldehyde
Carbalkoxy
– CHO
Formyl or oxo
R – CHO
Ketone
oxo
Alcohol
– OH
R – OH
Alkenes
C=C
CnH2n
Alkynes
C C
Halides
–X (X = F,Cl,Br,I)
Hydroxy
alkyl (R’) - oate - al - one
Alkyl alkanoate
Alkanal Alkanone
- ol
Alkanol
–
- ene
Alkene
CnH2n–2
–
- yne
Alkyne
R–X
Halo
–
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 functional group does not have the carbon atom, then atoms as parent chain. The selected chain must the carbon atom of the parent chain attached to the include the carbon atoms involved in the functional functional group should get the lowest possible groups like – COOH, – CHO, – CN etc, or those which
number.
carry the functional groups like – OH, – NH2,– Cl, Step 5 :
– NO2 etc.
The name of the compound is written as -
The number of carbon atoms in the parent chain
Prefixes - word root - primary suffix - secondary suffix
decides the word root.
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.
Compound
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 alcohol
Ethyl carbinol
1- Propanol
Structure
H
4
H
5
CH3 – CH2 – CH2 – CH2 – OH
n-Butyl alcohol
6
HCOOH
Formic acid
–
Methanoic acid
Acetic acid
–
Ethanoic acid
CH3COOH
7
n-Propyl carbinol
methyl acetic acid
8
CH3 – CH2 – COOH
Propionic acid
9
CH3 – CH2 – CH2 – COOH
Butyric acid
ethyl acetic acid
10
CH3 – CH2 – CH2 – CH2 – COOH
Valeric acid
n-Propyl acetic acid
H
H – C – C – O – H
Isopropyl alcohol Dimethyl carbinol 2 - Propanol
C H3
1- Butanol
Propanoic acid
Butanoic acid
O
Some more examples :
(iii) CH3 – CH2 – CH2 – NH2 Word root : Prop Primary suffix : - ane Secondary suffix : - amine IUPAC name : Propan - 1 - amine
(i)
Word root : HeptPrimary suffix : – ane Functional group : – OH Secondary suffix : – ol IUPAC Name :
(iv)
2, 5-Dimethylheptan–1– ol
(ii) Word root : Pent Primary suffix : – ene Secondary suffix : – oic acid Position of double : 2nd bond IUPAC name : Pent-2-en-1-oic acid/Pent-2-enoic acid
Pentanoic acid
Word root Primary suffix Substituent IUPAC name
: Prop: - ane : nitro(prefix) : 1 - Nitropropane
Word root Primary suffix Prefix IUPAC name
: But : – ane : – chloro : 2 - Chlorobutane
(v)
26 26
PAGE # 26
ISOMERISM (vi) Word 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. (a) Characteristics of Homologous Series : (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 by a common difference of – CH 2 and differ in molecular mass by 14.
(a) Chain Isomerism : 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. (b) Homologues : 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 of n n=1 n=2 n=3
(iii) C4H8
Molecular formula IUPAC name CH 4 Methane C2H 6 Ethane C3H 8 Propane
(ii) Homologous series of alkenes General formula :CnH2n Value of n n=2 n=3 n=4
Molecular formula C2H 4 C3H 6 C4H 8
IUPAC name Ethene Propene But-1-ene
Common name Ethylene Propylene - Butylene
(iii) Homologous series of alkynes General formula : CnH2n–2 Value of n n=2 n=3 n=4
Molecular formula C2H 2 C3H 4 C4H 6
IUPAC Common name name Ethyne Acetylene Propyne Methyl acetylene But -1-yne Ethyl acetylene
CH3 – CH2 – CH = CH2 , But - 1 - ene
Methylpropene
(b) Position Isomerism : In this type of isomerism, isomers differ in the structure due to difference in the position of the multiple bond or functional group. e.g. (i) C4H8 CH3 – CH2 – CH = CH2 , CH3 – CH = CH – CH3 But -1 - ene But -2 - ene (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.
(c) Functional Group Isomerism : In this type of isomerism, isomers differ in the structure due to the presence of different functional groups. e.g.
(a) Physical Properties : (i) C3H8O CH3 – CH2 – O – CH3
CH3 – CH2 – CH2 – OH
Methoxy ethane
Propan-1-ol
(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.
(ii) C4H6 CH3 – CH2 – C CH But - 1- yne
CH2 = CH – CH = CH2 Buta - 1, 3 - diene [or 1, 3 - Butadiene ]
ALKANES Alkanes are aliphatic hydrocarbons having only C – C single covalent bonds. These are also known as saturated hydrocarbon as they contain single bond
Note : Alkanes are unaffected by most chemical reagents and hence are known as paraffins (parum-little, affinis affinity).
SOME COMMON EXAMPLES OF ALKANES M olcula r Form ula
Structure
CH 4
CH 4
C2H6
CH 3–CH 3
C3H8
CH 3–CH 2 –CH 3
C 4 H 10
Methane
C H 3 – C H 2 – CH 2 – C H 3
CH 3 – CH – CH 3 | CH 3
C 5 H 12
CH
3
– CH
2
– CH
2
– CH
3
CH 3 – CH – C H 2 – C H 3 | CH 3
CH
C 6 H 14
Trivia l Na m e
3
–
CH | 3 C – CH 3 | CH 3
CH 3 – CH 2 – CH 2 – CH 2 – CH 2 – CH 3 C H 3 – CH – CH 2 – CH 2 – CH 3
IUPAC Na m e Methane
Ethane
Ethane
Propane
Propane
n-Butane Isobutane
Butane 2–Methy lpropane
n-P entane
Pentane
Isopentane
2–Methy lbutane
Neopentane
2,2–Dimethylpropane
n-Hexane
Hexane
Isohexane
2–Methy lpentane
–
3–Methy lpentane
–
2,3–Dimethylbutane
Neohexane
2,2–Dimethylbutane
CH3 CH 3 – CH2 – CH – CH 2 – CH 3 CH 3
CH 3 – CH – CH – CH 3 CH 3 CH 3 CH 3 CH 3 – C – CH 2 – CH3 CH 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 acetate 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.
Trough Beehive shelf Water
Iron stand
Gas
(a) Properties : 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 sp3 hybridisation.
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.
(b) Structure :
(d) General Reactions :
H
(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
C H
H
H
Tetrahedral
(ii) Halogenation : (A) In direct sunlight
(c) Preparation of Methane : (i) Direct synthesis :
C
+
Ni 500ºC
2H2 Hydrogen
Carbon
CH4 + 2Cl2
CH4
CO2 + 4H2
Ni powder 300ºC
CH4 + H2O
Ni powder 300ºC 12H2O Water
+
2H
+
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
HCHO Methanal
CH4
+
+
H2O
2+
H 2O
Methane
Methanol +
4H
Red P
CH3Cl Methyl chloride
Cl2
Cl2 CH2Cl2 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.
Methane
3CH4 Methane
Red P
Cl2 CH4 Methane
CH4
CH4 + 2H2O
(iv) Reduction of methyl iodide : CH3 – Methyl iodide
C + 4HCl
(iii) Nitration :
(iii) Hydrolysis of aluminium carbide : Al4C3 + Aluminium carbide
h
(B) In diffused light
Methane
(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
Preparation of methane gas
CH4 Methane
+
+
HO–NO2 Nitric acid
400ºC 10 atm.
CH3–NO2
+
H 2O
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 copper tube at 200ºC and at a pressure of 100 atmospheres, methanol is formed. CH4 + 1/2 O2 CH 3OH Methane Methanol (e) Uses : (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
(b) Uses :
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 = CH2 + Br2
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.
(a) Properties : (i) Alkenes of C2 to C4 are gases. Alkenes of carbon atoms C5 to C14 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.
(a) Properties : 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. (b) Structure :
(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 : (b) Some common examples of alkenes Molecular formula
Structure
Trivial Name
IUPAC Name
CH3 – CH2 – OH
Ethene
Ethanol
C2H4
H2C = CH2
Ethylene
C3H6
CH3 – CH = CH2
Propylene
Propene
-Butylene
1-Butene
-Butylene (cis)
2-Butene (cis)
-Butylene (trans)
2-Butene (trans)
C4C8
CH3 – CH2 – CH = CH2 CH3 – C – H CH3 – C – H CH3 – C – H H – C – CH3 CH3 – C = CH2
(i) By dehydration of alcohol (Lab. method) :
Isobutylene
2-Methyl propene
Conc. H2SO4
CH2 = CH2 + H2O
165 – 170ºC
Ethene
(ii) By cracking of kerosene : Cracking
CH3 – (CH2)4– CH3
CH3 – CH2 – CH2 – CH3 + CH2 = CH2 Butane
n-Hexane
Ethene
(iii) From alkyl halides (Dehydrohalogenation) : CH2 – CH2 + KOH (Alcoholic) H X
CH2 = CH2 + KX + H2O (Here X = Halogen) Ethene
Ethyl halide
CH3 C5H10
CH2 = CH (CH2)2CH3
–
1-Pentene
CH3CH = CHCH2CH3
–
2-Pentene (cis and trans)
CH3 – CH – CH = CH2
–
3-Methyl-1-butene
–
2-Methyl-2-butene
–
2-Methyl-1-butene
(d) General Reactions : (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
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
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.
is the test for unsaturation. (ii) Addition of halogen acids (Hydrohalogenation) : CH2 = CH2 + HCl Ethene
CH2 – CH2 Cl H Chloroethane
(iii) Hydrogenation : (a) Properties : Ni or Pt
CH2 = CH2 + H2 CH3 – CH3 High T& P Ethene Ethane
(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.
(iv) Combustion : C2H4 + 3O2 Ethene
2CO2
+
(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 :
nCH2 = CH2
High T & High P
– (CH2– CH2 –)n Polyethene
Ethene
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.
SOME COMMON EXAMPLES OF ALKYNES : M ole cula r form ula
Structure
C2H2
H –C
C3H4
CH 3 – C
C4H6
C5H8
De rive d Na m e
IUPAC na m e
Acetylene
Ethyne
M ethyl acetylene
Propyne
Ethyl acety lene
1–Butyne
Dimethyl acety lene
2– Buty ne
n-Propyl ac etylene
1–Pentyne
Ethyl m ethyl acetylene
2-Pentyne
Isopropyl acety lene
3-M ethy l- 1-buty ne
C–H C –H
CH3 –CH 2 – C CH
CH 3 – CH 2 – CH 2 – C CH
C H 3 – CH – C C H C H3
ETHYNE
(a) Structure :
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
180º H
C
C
H
carbon-carbon hydrogen bond angle is 180º. Linear
31 31
PAGE # 31
(b) Properties : It is a colourless gas which is slightly soluble in water. Pure ethyne has ethereal odour. Acetylene burns with luminous flame like aromatic compounds. This is a highly exothermic reaction.
TEST FOR ALKANES, ALKENES AND ALKYNES (a) Alkanes : (i) Bromine water test: It does not decolourise the bromine water.
Note : The temperature of oxyacetylene flame is about 3000ºC and is used for welding and cutting steel.
(ii) Baeyer’s test: It does not, react with Baeyer’s reagent (alkaline solution of KMnO4).
(c) Preparation :
(b) Alkenes:
(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.
(i) Bromine water test: It decolourises the orange colour of Bromine water.
1200ºC
2C + H2
(iii) Silver nitrate Test: No reaction
H
C2H2
H
Ca(OH)2 + C2H2
H
CCl
4 + Br2 Bromine water (red-brown colour)
H Ethene
(ii) From calcium carbide (Lab. Method) :
CaC2 + 2H2O Calcium carbide
H C=C
H C–C
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
H
H2O + [O]
C=C H
H
H–C–C–H
H
OH OH Ethylene glycol
Ethene
(iii) Silver nitrate Test: No reaction (c) Alkynes :
(d) Chemical Properties :
(i) Bromine water test : It decolourises the Br2 water.
(i) Addition of halogens :
H – C C – H + Br2 Ethyne
Cl HC CH + HCl
H2C = CH
Ethyne
Cl Chloroethene (Vinyl chloride)
HC CH Ethyne
HC CH
H2C = CH2 Ethene
H2/Ni
1,2-Dibromoethene
H
H
Br – C – C – Br Br Br 1,1,2,2-Tetrabromoethane (Colourless)
Cl 1,1-Dichloroethane (Gem dihalide)
(iii) Silver nitrate Test : It gives white precipitate H – C C – H + 2 [Ag (NH3)2 ]NO3 Ethyne Tollen's reagent
Ag – C C – Ag + 2NH4NO3 + 2NH3 White ppt.
CH3 – CH3 Ethane
EXERCISE H2
H2C = CH2 Ethene
Pd/BaSO4
Ethyne
1.
4CO2 + 2H2O + Heat
(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
(v) Polymerisation : H 3HC CH
Fe
The IUPAC name of the compound having the formula (CH3)3 CCH = CH2 is -
(iv) Combustion : 2C2H2 + 5O2 Ethyne
Ethyne
Br2
Br
H3C – CH
(iii) Hydrogenation : H2 Ni
H C=C
(ii) Baeyer’s test : It also decolourises the purple colour of alkaline KMnO4 .
(ii) Addition of Halogen acid : HCl
Bromine water
H Br
H
H
H
H H Benzene
or (C6H6)
2.
Which of the following is not an example of aromatic compound ? (A) Benzene (B) Naphthalene (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) glucose (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) Wohler (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 name for CH3 – C – H is (A) Acetal (C) Ethanal
(B) Methanal (D) Acetaldehyde
12. The 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(CH3) (OH) CH2 – CH(CH3) CH3 is (A) 2,4-Dimethylpentan -2-ol (B) 2,4- Dimethylpentan-4-ol (C) 2,2-Dimethylbutane (D) Isopentanol 18. The IUPAC name of (CH3)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) have similar chemical properties . (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 (C)CH3–CH2–CH2–CH2–CH3 &
32. Ethyne on passing through a red hot iron tube gives (A) mesitylene (C) butenyne
33. The general molecular formula of alkynes is (A) CnH2n (C) CnH2n–2
(D) All of the above 25. Which of the following is a functional isomer of CH3 – COOH ?
(B) benzene (D) None
(B) CnH2n–4 (D) CnH2n+2
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
(A) CH3 – CH2 – OH
(B)
(C)
(D) All of these
(D) 1-Pentyne and 3-Hexyne 35. Ethyne is obtained by dehydrobromination of -
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 (C) CH3–CH2Br
(B) CH3CHBr2 (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 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
NUMBER
Pre-requisite : Before going through this chapter, you should be thorough with the basic concepts of the same chapter explained in IX NCERT.
SYSTEM Fundamental Laws of Logarithm : Logarithm to any base a (where a > 0 and a 1 ). (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] Systems of Logarithm :
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 [Where m and n are +ve numbers]
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)
m (v) loga = logam – logan n
(vi) loga(m)n = n logam
(vii) logam
Some Useful Results :
log m b log a b
(i) If a > 1 then (a) loga x < 0 [for all x satisfying 0 < x < 1] (b) loga x = 0 for x = 1 (c) loga x > 0 for x > 1 (d) x > y loga x > loga y i.e. logax is an increasing function.
(viii)logam . logma = 1
Graph of y = loga x, a > 1
rational number, then
(ix) If ‘a’ is a positive real number and ‘n’ is a positive rational number, then
a loga n n (x) If ‘a’ is a positive real number and ‘n’ is a positive
y
logaq np p log n a
q
y = logax, a > 1 x'
0
(1,0)
x
(xi) ploga q qloga p (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 x satisfying 0 < x < 1 (d) x > y logax < loga y i.e. loga x is a decreasing function.
Graph of y = loga x, 0 < a < 1. y
a = 34
a = 81. Ex.2 Find the value of log
(1,0) 0
x
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. Sol. log3a = 4
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 AND MULTIPLES
2log4x = 1 + log4(x – 1) Factors : ‘a’ is a factor of ‘b’ if there exists a relation such that a × n = b, where ‘n’ is any natural number.
log4x2 – log4(x – 1) = 1 log4
x2 =1 x –1
x2 = 4x – 4 2
(x – 2) = 0
x2 x –1
41 =
x2 – 4x + 4 = 0
x = 2.
1 is a factor of all numbers as 1 × b = b.
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 – log3 5 . Sol. Given 3 2 – log3 5 = 3 2.3 – log3 5 = 9. 3
Multiples : ‘a’ is a multiple of ‘b’ if there exists a relation
[ am + n = am.an]
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.
log3 5 –1
= 9 × 5–1
The smallest multiple will be the number itself and the number of multiples would be infinite.
9 . 5
NOTE : To understand what multiples are, let’s just take an
=
Ex.5 If A = log27625 + 7
log11 13
and B = log9125 + 13
then find the relation between A and B. Sol. A = log27625 + 7
log11 13
4 log 13 = log 3 3 5 + 7 11
log11 7
,
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
4 log 13 or, A = log35 + 7 11 3
and,B = log9125 + 13
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, 3 4 log35 = B – log35 2 3
4 3 log35 < log35 3 2
A < B.
Ex.6 Find the value of log25125 – log84 Sol. Given, log25125 – log84 = log 5 2 5 3 – log 2 3 2 2
=
255 = 42 (and the remainder 3). The remainder is of 6 no consequence to us. So in all there are 42 multiples.
log11 7
or, B = log 3 2 5 3 + 7
A–
....(i)
3 2 log55 – log2 2 2 3
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. 600 200 60 – 60 = 10 – 3 = 7.
Such numbers are 240, 300, 360, 420, 480, 540 and 600. 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 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 × 32. 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. (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 each 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 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, y).
DIVISIBLITY Division Algorithm : General representation of result
HCF and LCM of fractions :
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
Dividend = (Divisor × Quotient ) + Remainder
Make sure the fractions are in the most reducible form. Ex.18 On dividing 15968 by a certain number, the quotient 10 8 16 2 Ex.13 Find the H.C.F. and L.C.M. of , , and . 9 27 81 3
H.C.F. of (2, 8,16,10 ) 2 Sol. H.C.F. of given fractions = = , 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 Ex.14 Find the least number which when divided by 6, 7, 8, 9 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 : (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 even values of n. (iii) (xn + an) is divisible by (x + a) for all the odd values of n. Test of Divisibility :
No.
Divisiblity Te st
2
U nit digit s hould be 0 or even
3
The s um of digits of no. s hould be divis ible by 3
4
The no form ed by las t 2 digits of given no. s hould be divis ible by 4.
5
U nit digit s hould be 0 or 5.
6
N o s hould be divis ible by 2 & 3 both
8
The num ber form ed by las t 3 digits of given no. s hould be divis ible by 8.
9
Sum of digits of given no. s hould be divis ible by 9
11
The difference betw een s um s of the digits at even & at odd places s hould be zero or m ultiple of 11.
25 Las t 2 digits of the num ber s hould 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 . 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 be divisible by 19.
remainder theorem. Remainder should always be positive. For example if we divide –22 by 7, generally we get –3 as quotient and –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. Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number is divisible by 19. Ex.24 Find the smallest number of six digits which is exactly divisible by 111. Sol. Smallest number of 6 digits is 100000. On dividing 100000 by 111, we get 100 as remainder. Number to be added = (111 – 100) = 11. Hence, required number = 100011. 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. 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. ‘a’ can take two values 2 and 6. 38232 is not divisible by 11, but 38236 is divisible by 11. Hence, 6 is the correct choice. Ex.27 Which digits should come in place of and $ if the number 62684$ is divisible by both 8 and 5 ? Sol. Since the given number is divisible by 5, so 0 or 5 must 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.
should be –4 and remainder is +6. We can also get remainder 6 by adding –1 to divisor 7 ( 7–1 = 6). Ex.28 Two numbers, x and y, are such that when divided by 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 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 ? Sol. Let the number be P. So, P – 139 is divisible by 259. Let Q be the quotient then,
P = 259Q + 139
P 259 Q 139 = 37 37
P 139 =Q 259
259 is divisible by 37, 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 respective remainders if the order of divisors be reversed. Sol.
3 x 5 y 1 8 z 4 1 7 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. To find the remainder of big number
16
Ex.36 What is the remainder when 1415
is divided by 5 ?
16
15 Sol. 14 = (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 + 83361 + 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 + 1361 + 2361 = 0. So the remainder is zero.
NOTE :
CYCLICITY (i) Binomial Expansion : (a + b)n = an +
n 1!
n
(a – b)n = an –
an–1b +
n(n 1) 2!
an – 2b2 + .... + bn, or
n(n 1)
an–1b + 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 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. 19
19
19
49 48 1 7 38 72 Sol. = = = 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
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 repeat 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? 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.
To Fi n d t h e U n i t Di g i t i n Ex p o n e n t i a l Ex p r e s s i o n s : (i) When there is any digit of cyclicity 4 in unit’s place. Since, when there is 2 in unit’s place then in 21 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. Ex.42 In (678)22 unit digit is equal to the unit digit of 82 i.e. 4.
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 41 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!’.
33
Ex.43 In (34) unit digit is 4.
So, n! = n(n – 1)(n – 2)...321. e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.
Ex.44 In (29)15 unit digit is 9. 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 51 unit digit is 5, in 52 unit digit is 5 and so on.
Ex.52 What is the highest power of 2 that divides 20! (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.
Sol. (i) The cyclicity of 3 is 4. Hence,
(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 of 33, i.e. 7.
Ex.49 Find unit’s digit in y = 717 + 734 Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6 Ex.50 What will be the last digit of (73 )75 6476
= (73)x where x = 75 64
6476
Therefore, the last digit of (73 )75
6476
76
= (75)even power
will be 31 = 3. 55
Ex.51 What will be the unit digit of (87 ) 6355
75 63
= (87)x where x = 75 63
55
. = (75)odd
Cyclicity of 7 is 4. To find the last digit we have to find the remainder 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
completely? Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20 = 1 × (21) × 3 × (22) × 5 × (21 × 31) × 7 × (23) × ..... so on. In order 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 20
product, i.e
= 5. Hence, 5 numbers contribute an 22 extra 1 to exponents. Similarly, there are 2 numbers that are divisible by 23 and 1 number that is divisible by 24. Hence, the total 1s contributed to the exponent
Cyclicity of 3 is 4 To find the last 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.
75 Sol. Let (87 )
y y y y! is 2 3 ......., where [ ] represents just x x x
part.
Ex.47 In (46)13 unit digit is 6.
Sol. Let (73 )75
factorial one. Hence 0! = 1 = 1! The approach to finding the highest power of x dividing
the integral part of the answer and ignoring the fractional
Ex.46 In (25)15 unit digit is 5.
Ex.48 Find the last digit of (i) 357
The value of factorial zero is equal to the value of
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. 5 52 Hence, the total contributions to the power of 5 is 24, or the number 100! is divisible by 524. 1
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 and also 4 × 9 = 36, can further be divided by 6. Thus, when the divisor is a composite number find the
8 122 8 15 2 8 1 7 0 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.
Ex.56 What is the largest power of 12 that would divide 49! ?
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.
Ex.59 Convert (169)10 in base 7.
Highest power of 3 in 49! = 22 Highest power of 2 in 49! = 46 46 = 23 2 Highest power of 12 will be 22. (Since the common
Highest power of 4 in 49! =
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.
31 3 0
0 7 3
Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B, (1987)10 = (3703)8
C, D, E, F.
After 9, we use the letters to indicate digits. For 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 + 81 × 3 + 82 × 2)10 (231)8 = (1 + 24 + 128)10 (231)8 = (153)10 Ex.63 Convert (0.03125)10 to base 16. Sol. 16 0.03125 = 0.5 0 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 product
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
of two consecutive natural numbers. Therefore, R.H.S.
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
numbers have been summed up and their sum is equal
ALPHA NUMERICS NUMBERS
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 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-digit 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 if 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 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.
How many numbers between 200 and 600 are divisible by 14?
of the multiplicand, we can see that 4 x D gives the unit digit of 2 and 4 x 2 gives the first digit of D. Yes, you got
(A) 28
(B) 29
(C) 27
(D) None of these
it right D = 8. Writing the multiplication again with the value of D we get 2BC8 x 4 = 8CB2.
7.
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
The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25
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 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 digit of 1.
(C) 120 sec, 15 times
(D) None of these
Hence, C is 2 or 7. You can easily 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.
Convert 0.225 in to form p/q. 3 (A) 10 (C)
2.
3.
4.
5.
9 50
9 (B) 40
(D)
9 400
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 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 : (B) 2100 (A) 21000 2000 (C) 2 (D) 210000 Find the value of ‘x’ if 2logx 7 + log7x 7 + 3log49x 7 = 0 (A) x =
4 3
(C) x = 7–4/3
(A) P(P +1)
(B) (P + 2)P
(C) (P + 1)(P – 1)
(D) None of these
12. Find out (A + B + C + D) such that AB x CB = DDD, where AB and CB are two-digit numbers and DDD is a threedigit number. (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?
(B) x = 7–1/2 (D) Either (B) or (C)
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) (B) 216 + 1 (C) 296 + 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 (C) 5, 4
(B) 5, 6 (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 (C) 1 or 5
(B) 1 (D) None of these
26. What is the remainder when 3040 is divided by 17? (A) 1 (B) 16 (C) 13
(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. W hen 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. W hen 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 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
(B) 11
(D) Smallest 3-digit prime number 37. How many numbers between 400 and 600 begin with or end with a digit of 5 ? (A) 40
(B) 100
(C) 110
(D) 120
38. If (12 + 22 + 32 + .....+ 102) = 385, then the value of (22 + 42 + 62 +...... + 202). (A) 770
(B) 1155
(C) 1540
(D) (385 × 385)
39. Find the total number of prime factors in the expression (4)11 × (7)5 × (11)2. (A) 37
(B) 33
(C) 26
(D) 29
40. The largest number which exactly divides the product 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
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
(D) 15
43. Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (A) 1050 (C) 1440
(B) 540 (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 (C) 41
(D) 48
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 (B) 1086 (C) 213 (D) 691 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 7 . What was the number erased? 17 (A) 7 (B) 8 (C) 9 (D) None of these
is 35
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 following numbers necessarily produces an integer, when multiplied by D? (A) 18 (B) 108 (C) 198 (D) 288 48. W hat 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 49. Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12? (A) 0 (B) 9 (C) 3 (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) 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8
53
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 (B) 78 and 60 (C) 63 and 81 (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
58. The number of positive n in the range 12 n 40 such that the product (n –1) (n – 2).... 3.2.1 is not divisible by n is : (A) 5 (B) 7 (C) 13 (D) 14 59. A rich merchant had collected many gold coins. He did not want any body to know about him. One 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
6
is :
(B) 6 (D) 10
(A) 6 17
(B)
(C)
(D)
1 1 1 62. log 10 1 + log 10 1 + log 10 1 + ... + 2 3 4 1 . When simplified has the value equal log10 1 1999
(B) 3 (D) 100
63. Arrange the following rational number in ascending order
3 4 7 1 , , , . 7 5 9 2
(A)
4 7 3 1 , , , 5 5 9 2
(B)
3 1 7 4 , , , 7 2 9 5
(C)
4 7 1 3 , , , 5 9 2 7
(D)
1 3 7 4 , , , 2 7 9 5
3
4
(A) 3 (C) 5 (D) cannot be determine
12
25 2
(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
72. If the decimal number 2111 is written in the octal system, then what is its unit place digit ? (A) 0 (C) 2 73. If
to : (A) 1 (C) 10
17 , 2,12 25 , 3 4 .
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
(B) 1 (D) 3
1 log 3 M 3 log 3 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 such 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
32 is :
[NSTSE-2009] (A) 5 (C) 7
(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 c2 + 6a = – 14, then the value of (a2 + b2 + c2) is : [IJSO-2009] (A) 14 (B) 25 (C) 36 (D) 47 81. Let N = 28, the sum of All distinct factors of N is : [IJSO-2009] (A) 27 (B) 28 (C) 55 (D) 56 82. The units digit of (1 + 9 + 92 + 93 + --------- + 92009) is : [IJSO-2009] (A) 0 (B) 1 (C) 9 (D) 3 83. The biggest among the following is : (A) 21/2 (B) 31/3 (C) 61/6 (D) 81/8
[IJSO-2009]
84. If a, b 1, ab > 0, a b and logba = logab, then ab = ? [IAO- 2009] (A) 1/2 (B) 1 (C) 2 (D) 10
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 (C) 51 (D) 2009 86. If HCF (p, q) = 12 and p × q = 1800 n then LCM (p, q) is : [NSTSE 2010] (A) 3600 (B) 900 (C) 150 (D) 90
87. If
x y +
be : (A) 16 (C) 2
y 10 = and x + y = 10, then 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. [IAO 2008] (A) 0 (B) 2 (C) 2.5 (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 (B) 7 (C) 0 (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 ANGLE An angle is the amount of rotation of a revolving line with respect to a fixed line. If the rotation is in anticlock-wise sense, then the angle measured is positive and if the rotation is in clock-wise sense, then the angle measured is negative.
(ii) Centesimal system : In this system a right angle is divided into 100 equal parts, called grades. Each grade is sub divided into 100 minutes, and each minute into 100 seconds. Thus, 1 right angle = 100 grades (100g) 1 grade = 100 minutes (100’) 1 minute = 100 seconds (100”) (iii) Circular system : In this system the unit of measurement is radian. One radian, written as 1c, is 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
The number of radians is an angle subtended by an
plane. These lines divide the plane into four equal parts are known as quadrants. The lines X’OX and
arc of a circle at the centre is equal to
YOY’ are known as X-axis and Y-axis respectively. These two lines taken together are known as the co-
=
length of arc . radius
s r
ordinate axes. The regions XOY, YOX’, X’OY’ and Y’OX are known as first, second, third and fourth quadrants
Where, = angle in radian, s = arc length and r = radius.
respectively.
(b) Relation Between Three Measurement of Angles :
System
of
D G 2R 90 100
Where, D = number of degrees, G = number of grades, and R = number of radians. NOTE : (i) The angle between two consecutive digits in a clock is 30º = (/6 radians). (ii) The hour hand rotates through an angle of 30º in one hour, i.e. (1/2)º in one minute. (a) Systems of measurement of angles : (i) Sexagesimal system
(ii) Centesimal system
(iii) Circular system (i) Sexagesimal system : In this system a right angle is divided into 90 equal parts called degrees. Each degree is divided into 60 equal parts called minutes and each minute is divided into 60 equal parts called seconds. Thus, 1 right angle = 90 degrees ( 90º) 1º = 60 minutes (60’) 1’ = 60 seconds (60”)
(iii) The minute hand rotates through an angle of 6º in one minute. Ex.1 Express in radians 47º 25’ 36”. '
36 3 Sol. 47º 25’ = 47º 25 60 5 '
'
º
128 128 1 32 = 47º = 47 = 47º 5 5 60 75
º
º
3557 3557 c 3557 c = = = . 13500 75 180 75
PAGE # 4949
Ex.2 Express in degrees : 2 (a) 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 that in the second ; how many sides do the polygons
(b) (– 2)c .
c
have ? Sol. Suppose the second regular polygon has number of
º
2 2 180 = 24º Sol. (a) 15 15 180 (b) (– 2)c = 2
side = x. The first regular polygon will have number of side = 2x.
º
º
6 180 7 ( 2) = 114 = 11 22 '
8 6 = – 114º 60 = – 114º 32 11 11
8 = – 114º 32’ 60 11
( 4 x 4) right angle 2x And each angle of the second polygon
Each angle of the first polygon =
º
=
'
''
4 x 4 2x 4 : =3:2 2x x
4 x 4 6 x 12 x x
4x – 4 = 6x – 12
= – 114º 32’ 44”. Ex.3 Express in radians 345g 25’ 36”. Sol. 345g 25’ 36” = 345.2536g =
3452536 c = 1.726268 c 2000000
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.
2 g 2 9º 3 x = x xº 3 3 10 5
And
x c x 12 x º 180º = 75 75 5
But
3 12 3 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. 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. ; 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 c (x y) 180 or (x – y) = (x + y) × 60º 180 º 3 (x – y) = x + y 4y = 2x y = x/2 60 º y= = 30º. 2 Hence three angles are 30º, 60º and 90º.
( 2x 4) right angle x
2x = 8 x = 4. The number of sides in the first and second polygons respectively are 8 and 4. Ex.7 The radius of a certain 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. In AOC, AO = OC = AC = 30 cm.
AOC = 60º =
3
Hence, arc AC = radius ×
= 30 × = 10 = 31.4159 cm. 3 3
TRIGONOMETRY Trigonometry means, the science which deals with the measurement of triangles. Trigonometric Ratios :
A right angled triangle is shown in Figure. B is of 90º. Side opposite to B is called hypotenuse. There are two other angles i.e. A and C. If we consider C as , then opposite side to this angle is called perpendicular and side adjacent to is called base. (i) Six Trigonometric Ratios are : Perpendicular P AB sin = Hypotenuse = = H AC PAGE # 5050
cosec =
13 , then prove that : 5 2 2 tan A – sin A = sin4A sec2 A.
Hypotenuse H AC = = Perpendicular P AB
Ex.9 If cosec A =
BC Base B = = AC Hypotenuse H
cos =
sec =
Hypotenuse H AC = = Base B BC
tan =
Perpendicu lar AB P = = Base BC 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 By pythagoras theorem, AB2 = BC2 + AC2 (13)2 = (5)2 + AC2 2 AC = 169 – 25 = 144 AC =
cot =
BC Base B = = AB Perpendicular P
1 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 =
1 tan
cot =
1 sec = cos
L.H.S tan2 A – sin2 A 2
1 cos = sec
1 sin = cos ec
cosec =
1 sin
5 5 = – 12 13
sin cos and cot = cos sin
Ex.8 If tan =
m , then find sin n
Sol. Let P = m and B = n
tan =
P m = B n
H2 = P2 + B2 H2 = m22 + n22 H = m 2 n2
sin = sin =
m P = H m2 n 2 m 2
m n2
.
2
25 25 = – 169 144
=
25(169 144 ) 144 169
=
25 25 144 169
We also observe that tan =
13 Hypotenuse = . 5 Perpendicular
13
5
C
12
A
R.H.S. sin4A × sec2 A 4
2
5 13 = × 13 12 5 4 13 2 = 13 4 12 2 54 25 25 = = 2 144 169 13 12 2 So, L.H.S = R.H.S
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 + BC2 (9 – AB)2 = AB2 + (3)2 C [ AC + AB = 9cm AC = 9 – AB] 81 + AB2 – 18AB = AB2 + 9 72 – 18 AB = 0 5cm 3cm 72 AB = = 4 cm. 18 Now, AC + AB = 9 cm A B 4cm AC = 9 – 4 = 5 cm BC 3 AC 5 , cosec C = , So, cot C = AB 4 AB 4 AC 5 . sec C = BC 3
PAGE # 5151
TRIGONOMETRIC TABLE
TR IG O N O M E T R I C RA TIO S O F COMPLEMENTARY ANGLES sin (90 – ) = cos
cos (90 – ) = sin
tan (90 – ) = cot
cot (90 – ) = tan
sec (90 – ) = cosec cosec (90 – ) = sec Ex.14
Sol.
Ex.11 Given that cos (A – B) = cos A cos B + sin A sinB, find the value of cos15º. Sol. Putting A = 45º and B = 30º We get cos (45º – 30º) = cos 45º cos 30º + sin 45º sin 30º
cos 15º =
cos 15º =
1 2
1 1 3 + 2 2 2
3 1 2 2
cot 54 º tan 20 º –2 tan 36 º cot 70 º
=
cot( 90 º 36 º ) tan( 90 º 70 º ) + –2 tan 36 º cot 70 º
=
tan 36 º cot 70 º + –2 tan 36 º cot 70 º
[cot (90 – ) = tan and tan (90 – ) = cot ] = 1 + 1 – 2 = 0. Ex.15. –
.
Ex.12 A Rhombus of side of 10 cm has two angles of 60º each. Find the length of diagonals and also find its area. Sol. Let ABCD be a rhombus of side 10 cm and BAD = BCD = 60º. Diagonals of parallelogram bisect each other. So, AO = OC and BO = OD In right triangle AOB D C OB sin 30º = AB 1 OB O = 2 10 OB = 5 cm 30º A BD = 2(OB) B BD = 2 ( 5 ) BD = 10 cm OA cos 30º = 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 × AC × BD Area of Rhombus = 2 1 = × 10 3 × 10 = 50 3 cm2. 2
cot 54 º tan 20 º – 2. tan 36 º cot 70 º
Sol.
2 sin 68 º 2 cot 15 º cos 22º 5 tan 75 º
3 tan 45 º tan 20º tan 40º tan 50 º tan 70º . 5
2 sin 68 º 2 cot 15 º cos 22º 5 tan 75 º
3 tan 45 º tan 20º tan 40º tan 50 º tan 70º 5 2 sin( 90 º 22 º ) 2 cot( 90 º 75 º ) = cos 22º 5 tan 75 º –
–
3(1)(tan 20 º tan 70 º )(tan 40 º tan 50 º ) 5
=
2 cos 22 º 2 tan 75 º cos 22º 5 tan 75 º
–
3[tan( 90 º 70 º ) tan 70 º ][tan( 90 º 50 º ) tan 50 º ] 5
2 3 (cot 70º tan 70º) (cot 50º tan 50º) 5 5 [ t an ( 90 º – ) = c ot , c ot ( 90 º – ) = ta 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 TRIANGLE In a ABC, a & b are the length of 2 sides of triangle 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 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 ELEVATION 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 of sight makes with the horizontal drawn through the eye of the observer is known as the angle of elevation. Therefore, the angle of elevation of an object helps in finding out its height (Figure).
In right triangle ACB, AC h tan 60º = BC x h x In right triangle BCD,
3 =
tan 30º = 1
3
CD 8 BC x
8 x x=8
3
Height of hill = h + 8 =
3 .x + 8 =
3 8 3 + 8
= 32 m.
Distance of ship from hill = x = 8 3 m.
ANGLE OF DEPRESSION 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).
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 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 1 AB h tan 30º = = 3 AD x
x=
3h
.....(i)
PAGE # 5353
1
3
3000 = xy
x + y = 3000
( AB = CD) 3
.......(ii)
From equation (i) and (ii)
x + 3000 = 3000 3
x = 3000 3 – 3000
x = 3000 × (1.732 – 1) x = 3000 × 0.732 x = 2196 m.
x = 3000 ( 3 – 1)
......(ii)
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.
2h = 5
Ex.20 If the angle of elevation of a cloud from a point h metres above a lake is and the angle of depression of its reflection in the lake is , prove that the distance of the cloud from the point of observation is
Speed of Aeroplane = In ADC AC AD 5h 3 = x 5h x= 3 From (i) and (ii)
tan 60º =
5h
3 h=
3h = 5 + h
3
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 elevation changes to 30º. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane. Sol. Let the point on the ground is E which is y metres from
2h sec . tan tan 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
point B and let after 15 sec flight it covers x metres distance. In AEB
tan 45º =
AB EB
3000 1= y y= 3000 m In CED
tan 30º =
CD ED
DE CE H CE = tan In CED’
tan =
.......(i)
tan =
......(i)
ED' EC
PAGE # 5454
hHh tan
CE =
2h H CE = tan
4.
hand when the time is 7 : 20 am is : .......(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
H=
6.
2h tan ........(iii) tan tan In DCE sin = CD =
DE sin
7.
CD =
8.
2h tan CD = tan tan sin sin 2h cos CD = tan tan sin
2h sec observation is . tan tan
700 720
Hence Proved.
(B)
703 720
710 (D) 720
(D) None
If tan =
a a sin b cos then value of is = b a sin – b cos
a2 b2
(B)
a2 – b2 a
9.
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 If a = cot2 30º + 3 sin2 60º – 3 cosec2 60º – tan2 30º 3 4 and b = 3 tan2 45º + cos 0º – cot 90º then logb(a) is :
1 2
(B) –1 (D)
1 2
The angles of the triangles ABC and DEF are given as follows : A = 900, B = 300, D = 900 and E = 300. If the side is true? (A) Sin B = 2 Sin E
(B) Sin E = 2 Sin B
(C) Sin B = Sin E
(D) Sin A = Sin B
4 3 cot2 30º + 3sin2 60º – 2cosec2 60º – tan2 30º is : 3 4
(A) 1 (C)
10 3
(B) –
20 3
(D) 5
11. The value of the expression
(B) 14º 19’ 5” (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 : (A) 60 m (C) 70 m
(C) 270°
10. The value of the expression
c
3.
(B) 240°
BC is twice the side EF, which of the following statement
1 Degree measure of is : 4
(A) 15º 19’5” (C) 15º 18º 6”
(D) 140º
7 Degree measure of is : 6 (A) 210°
(C)
Radian measure of 175º 45’ is :
705 (C) 720
2.
(C) 120º
(A) 2
2h sec tan tan Hence, the distance of the cloud from the point of
(A)
(B) 100º
(C)
DE CD
CD =
1.
(A) 80º
(A)
H sin Substituting the value of H from (iii)
Angle between the minute hand of a clock and hour
(B) 65 m (D) 72 m
5 sin2 30 º cos 2 45 º 4 tan 2 60º is : 2 sin 30 º cos 60 º tan 45 º (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) 1
(B) 2
(C) 3
(D) None
PAGE # 5555
13. If + =
21. If each of , and is a positive acute angle such that
1 and sin = , then sin is : 3 2
(A)
2 3
(B)
2 2 3
(C)
2 3
(D)
3 4
1 1 , cos( + – ) = and 2 2 tan ( + – ) = 1, then the values of , and is :
sin ( + – ) =
, then value of 2 14 tan – 75 cos – 7 sec is equal to : (A) 1 (B) 2 (C) 3 (D) 4
14. If 7 sin = 24 cos ; 0 < <
tan is equal to : 15. If tan = 4, then sin 3 sin cos cos (A) 0 (B) 2 2
2
(C)
(B) 6 (D) 0
17. The area of a triangle is 12 sq. cm. Two sides are 6 cm and 12 cm. The included angle is :
1 (C) sin–1 6
1 (D) sin–1 3
1 2
(B) 0
(C) 1
(D) 2
1 22. If tan (A – B) =
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
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, 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
18. If + = 90º and = 2, then cos2 + sin2 equals to : (A)
(D) none
(C) cosec
cot 2 sec tan 2 6 3 4 x sin cos 6 4 cos ec 2 cos ec 4 6
1 (B) cos–1 6
1 1 (C) 37 , 45° & 52 2 2
A B equals to : cos 2 C (A) cos 2
16. Find the value of x from the equation :
1 (A) cos–1 3
(B) 60º, 45º & 75º
23. If A, B, C are the interior angles of a triangle ABC, then
(D) 1
(A) 4 (C) – 2
(A) 45º, 45º & 90º
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 grades. Find the greater angle. 9 (A) 63º (B) 35º (C) 27º (D) 54º
sum is
20. If and are angles in the first quadrant, tan = 1
1 , 7
, then using the formula sin (A + B) = sin A 10 cos B + cos A sin B, one can find the value of ( + 2) to be : (A) 0º (B) 45º (C) 60º (D) 90º sin =
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, the time taken by the boat to reach the lighthouse from the point when the angle of depression was 30º, is : (A) 30 minutes (B) 20 minutes (C) 10 minutes (D) 5 minutes
35. The trigonometric expression sec – 1 sin – 1 cot2 + sec2 has the value 1 sin 1 sec
[IJSO-2009] (A) –1 (C) 1 36. (1 + tan2 ) / (1 + cot2 ) = (A) tan2 (C) sec2 37. If cos + sin =
(A) 2 3 (C)
(B)
15 3 8
12 3 7
(D) None of these
32. The expression (1 – tan A + sec A) (1 – cot A + cosec A) has value : [IJSO-2008] (A) – 1 (B) 0 (C) + 1 (D) + 2 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 30º to 60º in 18 minutes. The scooter will reach the base of the tower in next : [IJSO-2008] (A) 9 minutes (B) 18 / ( 3 – 1) minutes (C) 6
3 minutes
(D) the time depends upon the height of the tower 34. In the diagram, PTR and QRS are straight lines. Given 4 that, tan xº = and "T" is the midpoint of PR, calculate 3 the length of PQ, in cm. [NSTSE 2009]
[IJSO-2009] (B) cot2 (D) cosec2
2 cos, then cos – sin = ?
[NSTSE 2009] (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 :
(B) 0 (D) 2
(C)
2 tan
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 30º 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 39. If sin 2 1 . 2 . 3 ...... x 2 = 1, 0º < x < 100º, then the
value of x is equal to : (A) 91º (C) 49º 40. If p =
[NSTSE 2010] (B) 80º (D) 46º
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 ? [NSTSE-2010] (A) p = q r (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 3150 m above a horizontal plane ground. At a
3 cm
xº
R
T
P
particular instant it passes another aeroplane vertically below it. At this instant, the angles of elevation of
6 cm
the planes from a point on the ground are 30º and 60º. Hence, the distance between the two planes at that instant is : [IJSO-2011] Q
(A) (C)
8
(B) 9
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 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) Gases: 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.
(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) It exhibits streaming movement e.g. rotatory movements in the leaves of aquatic plants 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. 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 dissolved in water.
(viii) It acts as a lubricating and protective fluid. (b) Organic Compounds :
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.
(a) Inorganic Compounds : •
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.
•
Carbohydrates can be defined chemically as aldehyde or ketone derivatives or the poly hydric (more than 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 are glyceraldehyde and dihydroxyacetone. Depending upon the number of carbon atoms present, these can be further subdivided into trioses (e.g. Glyceraldehyde), tetrose (e.g. Erythrose), pentoses (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. Structure of glucose O II C–H I H – C – OH I HO – C – H I H – C – OH I H – C – OH I CH2OH Glucose simple chain
•
(B) Disaccharides : The disaccharides are sugars composed of two molecules of the same or different monosaccharides, united by a glycosidic linkage. They have a general formula Cn (H2O)n–1 . These include maltose, lactose , sucrose. •
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) 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 :
•
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. Chemically, 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.
CH2OH 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.
•
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.
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.
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.
•
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 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 :
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 : •
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.
•
Essential Am ino acids
Non-Essential Am ino acids
Sem i-Essential Am ino acids
Is oleucine
Alanine
Arginine
Leucine
As pargine
His tidine
Methionine
As partic acid
–
Phenylalanine
Cys teine
–
Threonine
Glutam ic acid
–
Tryptophan
Glutam ine
–
Valine
Glycine
–
Lys ine
Proline
–
–
Serine
–
–
Tyros ine
–
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 amino group of other amino acid. A molecule of water is released out during bond formation. H H O H H O I I II I I II H — N — C — C — OH H — N — C — C — OH I I R1 R2 O H H I II I H—N—C—C— I R1
H O H I I II N — C — C — OH I R2
Peptide bond formation •
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 proteins. They help to catalyze various reactions occurring in our 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. Fats and their derivatives are collectively known as lipids (In greek Lipas = fat). The principal component associated with most lipids are the fatty acids. The lipids are a heterogenous groups of substances which have the common property of being relatively insoluble in water and soluble in non – polar solvents such as 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 together by ester bonds. Therefore these are also termed as triglycerides. Three molecules of water are released during the formation of triglycerides.
•
Biological significance of lipids :
•
They takes part in the synthesis of steroids, hormones, vitamin D, bile salts etc.
•
They act as a solvent 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.
•
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.
stranded i.e. it has two polynucleotide chains. •
Types of RNA- A cell contains three types of RNA :
1.
Ribosomal 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 eukaryotic cells and 70 – S type of ribosome in prokaryotic cells. It is the site of protein synthesis.
2.
Transfer RNA (t – RNA) :
•
It is 10 – 15% of total RNA
•
It is synthesized in the nucleus by DNA.
•
It is also known as soluble RNA.
•
It is also known as adapter RNA.
•
It is the smallest RNA. At the time of protein synthesis it acts as a carrier of amino acids.
•
It has the most complex structure.
3.
Messenger 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.
(iv) Nucleic acids : These are the hereditary materials of living organisms. There are two types of nucleic acids :
(v) Enzymes : Enzymes are protein catalysts for biochemical reactions in the living cells. The 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. 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.
(A) DNA (Deoxyribose nucleic acids) : DNA is coiled macromolecule made of two antiparallel chains held together by hydrogen bonds. DNA has diameter of 20 Å. One turn of spiral has a distance of 34 Å and distance between two adjacent nucleotides is 3.4 Å. •
Nucleotides : A single nucleotide consist of following parts :
•
Pentose sugar : It is a 5 - carbon containing sugar which is ribose is RNA and deoxyribose in DNA.
•
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.
•
Phosphate group : PO4-3 group in the form of H3PO4
•
General properties of enzymes :
(B) RNA : (Ribonucleic acid) Structure of RNA is fundamentally the same as DNA but there are some differences. The differences are as follows.
•
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 place of deoxyribose sugar of DNA, there is a presence of ribose sugar in RNA. In 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
(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 depend on sun for energy. The green pigment in nature is called as 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. 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) and anabolic (synthetic reactions), collectively they are called as metabolic 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) nucleosides (D) nucleotides
3.
Which of the following is not a carbohydrate ? (A) Starch (B) Glycogen (C) Wax (D) Glucose
4.
To get quick energy one should use (A) carbohydrates (B) fats (C) vitamins (D) proteins
5.
Circular and double stranded DNA occurs in (A) golgi body (B) mitochondria (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) Cellulose (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) Phosphorous 23. Proteins are the polymers of (A) amino acids (C) enzymes
(B) natural protein (D) nucleic acids
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
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. 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 (D) glycogen & fructose
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
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 : Ex 6. (i) Find the missing term(s). (ii) Find the wrong term(s). Sol.
NUMBER SERIES
8, 12, 21, 46, 95, ? (A) 188 (B) 214 (C) 148 (D) 216 (D) The pattern is + 22, + 32, + 52, + 72, ....... missing number = 95 + 112 = 216 3, 9, 36, 180, ? (A) 1080 (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) Some Important Patterns : (b) Multiple Series : (i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression)
A multiple series is a mixture of more than one series :
(ii) a, ak, ak2, ak3, ................(Geometric Progression)
a a a (iii) a, , 2 , 3 , .............(Geometric Progression) k k k
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 (D) As per series, a, ak, ak2, 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
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) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............] (ii) 1, 8, __, 64 mixed combination 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.
5, 8, 10, 12, 15, 18, 20, 23 (A) 8 (B) 12 (C) 15 (D) 18 (B) Sol.
12 28 64 140 37 (P) (Q) (R) Which number will come (A) 1412 (C) 696 (A)
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
(Q)
(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
(P)
×2+4
42 23
(S) (T) in place of (T) ? (B) 164 (D) 78
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
2 9 57 3 (P) (Q) Which number will come (A) 113 (C) 3912
(T)
(A) Similarly, (P)
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.
337 (R) (S) in place of (Q) ? (B) 17 (D) 8065
(Q)
×8–7
×7–6
(S)
673
113
17
3
(R)
3361 ×5–4
×6–5
Therefore, the number 113 will come in place of (Q).
ALPHABET SERIES (a) Pattern of Alphabets Show Variation Based on : (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, 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, W, U, ? in which each letter is moved two steps backward to obtain the next term. So, the missing term would be S. AG, LR, WC, 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
A B C 6 D E F 15 G H I 24 , 4 5 6 , 1 2 3 7 8 9 Alphabetical positions Difference in Alphabetical positions
Therefore, the next group of letter would be SY. H
N
S
10 11 12
So, the missing term would be J K L 33 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) WC (C) WD (D) VE (C) The first letter of subsequent groups have a difference of 4, 5 and 6 places respectively, 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 by 2 letters. All the letters differ in the forward direction. Hence, the next choice would be PMR. XYQ, ZAR, BCS, 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, 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, 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, J20R, M43P, P90N (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 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. So, the (A) baa (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' so that we have two a's followed by two b's.
2.
Second blank space should be filled in by 'a' so that we have three a's followed by three b' s.
3. Ex 29.
The last blank space must be filled in by 'a' to keep the series in sequence. _ bca _ ca _ c _ b _ Ex 36. (A) aabbc (B) abbbc (C) aabcc (D) abbac
Sol.
(D)
Sol.
Sol.
Series is abc/ abc/ abc/ abc. So, pattern abc is repeated. Ex. 30
a_h_ _c_ ne_ h_ eac_ _ _ _ _ 21_4 3_5 __2 54 ____ _ _ _ _ The last five terms in the series are (A) 32524 (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. _ m y e _ _ y l x _ y l m _ _ l _ _ _ _ 4 6 _ 5 8 6 _ __ 5 7_ 6 5 8 __ _ _ _ The last five terms of the number series are (A) 46758 (B) 74658 (C) 76485 (D) 46785 (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 that first digit of the first group (i.e. 4) is the third digit of the second group and the last two digits have interchanged their positions. The same rule applies in others groups also.
a _ abb _ aa _ ba _ a _ b (A) ababa (B) aabba (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 way. In the given question, 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, 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.
9
4 (A) 11 (C) 3 Sol.
(B) 6 (D) 2
(C) Clearly, in the column,
6 4 8 3
18 3 27 We take x in place of ? 2 95 15 x Similarly in the column, 3 9 ,x 15 5
Sol.
In the column,
Ex 40.
Sol.
3C
27D
9E
7I
21K
3M
4D
?
7J
Ex 41.
Sol.
Ex 42.
2
Sol.
(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 × 3 + 4 = 22
81 18
Sol.
88 ? 11
9 (B) 21 (D) 81
In second figure, 9
6 ? 3 4
3
8 29 3 5
7
6
174
5
3
2 5 ?
336
3 2 9 2 7 9 6 4 5 (A) 140 (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
9
15
7
21
25
14
2 18 2
Ex 47.
= 84. = 81.
= 88 or x =
88 2 11
= 16.
Sol.
176
?
(A) 184 (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
3
Let the missing number In third figure be x.
x 2
5
8
40
(A) In first figure, 12
Then, 11
2
5
Ex 46.
(A) 16 (C) 61 Sol.
5
(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
11
(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
84 14 12
?
5 26 4 6
1
7
6
5
41 ? 5
6
30 3
Ex 44.
(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.
4
5
5
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.
EXERCISE-1 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) 104 (C) 116
(B) 114 (D) 118
3, 12, 27, 48, 75, 108, ? (A) 192 (C) 162
(B) 183 (D) 147
1, 4, 12, 30, ? (A) 60 (C) 64
(B) 62 (D) 68
94, 166, 258, ?, 4912 (A) 3610 (C) 1026
(B) 1644 (D) 516
22. (B) 180 (D) 452
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) 582 (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) 121 (C) 126
1, 11, ?, 11, 11, 11, 16, 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) 106 (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 7 (P) (Q) Which number will come (A) 172 (C) 270
28.
112 (R) (S) in place of (S) ? (B) 276 (D) 376
(T)
15 159 259 323 7 (P) (Q) (R) (S) (T) Which of the following numbers will come in place of (R) ? (A) 251 (B) 315 (C) 176 (D) 151 PAGE # 70
Directions : (29 to 35) Find the wrong term(s) — 29.
30.
31.
32.
33.
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
9, 15, 24, 34, 51, 69, 90 (A) 15 (B) 24 (C*) 34 (A) 15 (C) 34
8.
9.
10.
11.
35.
(C) 69
(D) 132
105, 85, 60, 30, 0, – 45, – 90 (A) 85 (B) – 45 (C) 105 (D) 0
4.
5.
6.
7.
ZGL, XHN, VIQ, TJU, ? (A) RKX (C) RLZ
(B) RKY (D) RKZ
RML, VIJ, ZFH, 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) 22L (C) 22K
(B) 24L (D) 2M
15.
16.
Directions : (1 to 24) Find the missing term(s) :
3.
(B) OET (D) PEV
(B) ZVX (D) VZX
17.
2.
CYD, FTH, IOL, LJP, ? (A) PET (C) OEY
ZSD, YTC, XUB, WVA, ? (A) VZZ (C) VWZ
EXERCISE-2
1.
(B) ELS (D) DLS
13.
9, 13, 21, 37, 69, 132, 261 (B) 37
JXG, HTJ, FPN, ?, BHY (A) EKS (C) DLR
MTH, QRK, UPN, YNQ, ? (A) CKT (B) ELT (C) CLT (D) EKT
(D) 51 (B) 24 (D) 51
(A) 21
(B) KBO (D) LBN
12.
14. 34.
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 KI, ? (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 12 E (B) R 11F (C) T 11E (D) R 13 D
Q, T, V, 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) 17L21 (B) 17K21 (C) 17M21 (D) 17L23
Z, L, X, J, V, H, T, 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, OMK, TQP, YWU (A) ECA (B) JHF (C) TQP (D) YWU
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, FJW, HIT, JHS, LGQ (A) FJW (B) LGQ (C) JHJ (D) HIT
11.
DVG, FSI, HPK, JNM, LJO (A) DVG (B) JNM (C) HPK (D) LJO
a _ bc _ a _ bcda _ ccd _ bcd _ (A) abddbd (B) acbdbb (C) adbbad (D) bbbddd
12.
CDF, DEG, EFH, FHI (A) CDF (C) FHI
cc _ ccdd _ d _ cc _ ccdd _ dd (A) dcdcc (B) dcddc (C) dccdd (D) None of these
13.
a_baa_baa _ba (A) a a b (C) b b a
(B) b a b (D) b b b
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) lmmm (C) lmml
(B) lmlm (D) mllm
(B) DEG (D) EFH
ZLA, BMY, CNW, FOU, HPS (A) ZLA (B) BMY (C) FOU (D) CNW
14.
G4T, J10R, M20P, P43N, S90L (A) G4T
(B) J10R
(C) M20P
(D) P43N
15.
EXERCISE-3
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 alternative. 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
_aabb_a_ab_b (A) b b a a (C) b a a b
(B) b a b a (D) a b a b
aab_aaa_bba_ (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.
_bc__bb_aabc (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 _ _ _ _ _ (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.
ab_cd_a_abd_dba_ 1_3_32_1___4____ The last four terms in the series are (A) 1234 (B) 3112 (C) 3211 (D) 4312 PAGE # 72
21.
22.
23.
_bnt _ _nam _nab_ _a__ _ _ 13_25 3__ 5 24_32 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 _ _ma _ _ _ _ _ _ 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 _ CD 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) 235 (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 _ B A 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 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
EXERCISE-4 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 12 15
11 49 8 54 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
5
27 80
39 45
29
33
42
20.
43
43
(A) 69 (C) 50
30 70
59
31 44
35
43
48
15
56
(A) 1 (C) 3
(B) 142 (D) 198
7 3 4 74
12
3
6
4
1
140 4 10 7
3 (B) 14 (D) 22
4 9
3 24
9 ?
(A) 117 (C) 32
(B) 36 (D) 26
(A) 26 (C) 27
(B) 25 (D) 30
4 13
21.
34 184
12 18 30
16 32 40
36 18 34
30
44
?
22.
(A) 48 (C) 44
23.
17.
(B) 145 (D) 18
8 11
20
6 ? 8 104 (B) 2 (D) 4
5 20
(B) 9 (D) 64
5
21
16 109 2
22 53 19
6
15
(A) 25 (C) 7
(A) 33 (C) 135
8
?
2 6
10
?
(A) 127 (C) 158
1 3 5 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
?
J90
4
(A) 1 (C) 4
16.
6
6 4
8
H70
5
14.
19.
4
C26
(B) 36 (D) 49
6
(A) 2 (C) 4
13.
18.
(B) 7 (D) 0
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
?
34.
3
7
(A) 14 (C) 11
3
8
10
2
?
1
6
56
90
2
20
0
(A) 0 (C) 5
(B) 18 (D) 26
(B) 3 (D) 7
2
15 35.
(A) 9 (C) 10 27.
6
HCA - 138 FED - 456 E?H - 87?
BIG - 792
36 49 26 64 25
9 81 21 16
2
25
64
25 ?
144
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) 127 (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
6
4
36.
(B) I, 9 (D) I, 5
(A) 19 (C) 25
29.
?
(B) 11 (D) 12
(A) G, 6 (C) G, 5
28.
16
13
65
5 26.
7
9
80
(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.
(B) 12 (D) 6
4
9
9 6
16
16 12
? 20
31. (A) 0 (C) 3
(B) 2 (D) 1
(A) 60 (C) 21 40.
(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
8 32
Find the missing letters from left to right.
(A) JSN (C) JRS
2
(A) 3 (C) 8
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 (zoology, physics, botany, 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.
4.
(B) Dr. Davar is from Osmania University.
5.
(D) Dr. Natranjan - Gujarat University is the correct 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) Ramesh (B) Karan (C) Vinay (D) Cannot be determined (D) In this question ranking of Karan is not defined. Consequently, either Ram or Karan occupies the top position with regard to height. Hence, 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 and 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 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 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 (B) The male advocates are A and B, lady doctors be teaching either in Delhi or Gujarat University...(2) Sol. are P, Q and S ; teachers are E, T and U. Form V, we get Dr Joshi teaches history in Delhi Now, A and B will be selected. 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 in C, D, P, Q, S ; teachers are E, T and U. The team the same game. (iv) One day they played four games as follows. consists of 3 teachers i.e. E, T, U. 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 teachers, 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 ; lady 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
Whom is E married to ? (A) A (C) C
(B) B (D) D
Whom is F married to ? (A) A (C) C
(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
will be chosen. C and T have to be together. Now, the second teacher to be selected 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
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, GB, ED does not contradict any of the statements.
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
12.
(D) E is married to D.
13.
(A) F is married to A.
(C) The advocates are A, B and R ; the doctors are 14. C, D, P, Q and S ; male teacher is E. Clearly, E will 15. be selected. E and Q have to be together. C and Q cannot be together. So, C will not be selected. P Ex 16. also cannot be selected because U is not selected. 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.
Ex 11.
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 :
Sol.
(A) A B C P T U
(B) A C P R T U
(C) A E P Q R T
(D) B C E Q R T
(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 supply last ? (A) 5.25 hr (B) 6 hr (C) 4.5 hr (D) 3 hr (B) He has got =
49 7 cigarettes. 7
The duration of time he will take to smoke these 3 7 cigarettes = 7 hr = 5.25 hr (i.e. 5 hr and 15 4
each of these combinations consists of four ladies.
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
B C E Q R T is incorrect because B and R cannot
cigarette. This will take him another
(A) A C P R T U and A E P Q R T are wrong because
go together.
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.
Sol.
17. 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) Business (B) Engineer (C) Lawyer (D) Doctor Agewise who among the following lies between A and C ? (A) Lawyer (B) Doctor (C) Cloth merchant (D) Engineer What is B’s occupation ? (A) Business (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) Business questions can be solved very easily. Directions : (6 to 10) Study the given information carefully Male (40) Female (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 . They are wearing caps of different colours - green, 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. unmarried 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 EXERCISE seated at one end.
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, B is a Shikh while A is a Hindu and runs a factory.
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 of the cap worn by the Japanese? (A) Red (B) Green (C) Yellow (D) White
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 F like reading. (ii) D is taller than E who is not as tall as B. III. D does not like badminton, but likes music. (iii) The oldest is the shortest. IV. Both B and E like painting and music. (iv) The youngest would be fourth if the children V. A and D do not like painting, but they like cricket. stood in a line according to their height and one VI. All student except one like badminton. started counting from the tallest. VII. Two students like tennis. (v) D is younger than F but older than E who is VIII. F does not like cricket, music 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.
16.
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 W hose 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, 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, 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. 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 th (C) 12 (D) None The organisation would end with which play ? (A) P (B) Q (C) S (D) None
PAGE # 79
27.
Which day is play Q organised ? (A) Tuesday (B) Wednesday (C) Thursday (D) None
28.
Which of the following is the correct sequence of organising plays ? 32. (A) PTRUQS (B) QSTURP (C) SUTRQP (D) None
31.
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 following statements must be true ? (i) Henna cannot go (ii) Shamma cannot go (iii) Reena cannot go (iv) Rama cannot go (A) (i) and (ii) (B) (iii) and (iv) (C) (i), (iii) and (iv) (D) (i) and (iv)
Directions : (29 to 30) Read the following information carefully and answer the questions given below it. I. Seven books are placed one above the other in a particular way . 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 open a new office in another city using three males 29. To find the number of books between the civics and two females of the present staff. To do so they and the science books, which other extra piece of plan to separate certain individuals who do not information is required, from the following ? function well together. The following guidelines (A) There are two books between the geography were established and the science books. I. Females A and C are not to be together (B) There are two books between the mathematics II. C and E should be separated and the geography books . III. D and G should be separated (C) There is one book between the English and IV. D and F should not be part of a team. the science books. (D) The civics book is placed before two books 33. above the economics book. 30.
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. (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.
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. 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 A is chosen to be moved, which of the following cannot be a team ? (A) ABDEH (B) ABDGH (C) ABEFH (D) ABEGH If C and F are to be moved to the new office, how many combinations are possible ? (A) 1 (B) 2 (C) 3 (D) 4 If C is chosen to the new office, which number of the staff cannot be chosen to go with C ? (A) B (B) D (C) F (D) G 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 chosen II. A cannot be chosen III. H must be chosen. (B) II only (A) I 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, 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 daughter - in applicable to all the questions below : law of L. (vi) The businessman is married to the chartered 38. If two of the members have to be boys, the team 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.
40.
(D) C E S P Q
45.
How is P related to Q. (A) Brother (B) Sister (C) Cousin (D) Either brother or sister
46.
Which of the following is the profession of M ? (A) Professor (B) Chartered accountant (C) Businessman (D) Medical representative
If R be one of the members, the other members of the team are : (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 : 47. (A) P Q B C (B) P Q C E (C) P S A B (D) P S C E
41.
If A and C are members, the other members of the 48. team cannot be : (A) B E S (C) E S P
42.
(B) D E S (D) P Q E
If including P at least three members are girls, the members of the team other than P are : (A) Q S A B (C) Q S C E
(B) Q S B D (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
49.
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 (C) Pinku and Sangeeta
(B) Anil and Saroj (D) Gaurav and Rakhi
Which of the following is one of the couples ? (A) QO (B) OM (C) PL (D) None of these 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, 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 is married to P. IV. R, the Jeweller is 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. Sangeeta’s brothers. Saroj is not Ratan’s sister. 43.
Which of the following is the profession of L ? (A) Professor (B) Charted accountant (C) Businessman (D) Engineer
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.
What is the profession of P ? (A) Doctor (B) Lawyer (C) Jeweller (D) Manager
54.
Which of the following is one of the pairs of couples in the family ? (A) PQ (B) PR (C) PS (D) Cannot be determined
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.” 55.
What is Chandra’s age ? (A) 32 years (B) 35 years (C) 29 years (D) 21 years
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.
III. Two persons from the same side of the roads cannot move in consecutive “movements”. IV. If one person crosses the road in a particular movement, he or she cannot immediately move back to the other side. V. Jai and Reema did not take part in first 3 movements. W hat is the minimum possible number of movements that took place in the entire game ? (A) 3 (B) 4 (C) 5 (D) 6 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 ? (A) Nitin, Reema amd Deepti (B) Nitin, Jai and Deepti (C) Deepti, Jai and Ashutosh (D) Ashutosh, Nitin and Deepti You have 12 similar looking 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 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
300 years =
= 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.
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.
15
After counting the odd days, we find the day according to the number of odd days.
A leap year has 366 days. Every year which is divisible by 4 is called a leap year. For example 1200, 1600, 1992, 2004, etc. are all leap years.
Sunday for 0 odd day, Monday for 1 odd day and so on as shown in the following table.
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, 1300, 1400, 1500, 1700, 1800, 1900, 2100 are not leap years.
Ordinary Year
Days
The day on which calendar started (or the very first day ) i.e., 1 Jan, 0001 was Monday.
January
31
3
February
28
0
March
31
3
April
30
2
May
31
3
Table : 1 (Odd days for week days) Days
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Odd Days
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.
0
1
2
3
4
5
6
Table : 2 (Odd days for months in a year) Odd Leap year Days
Days
Odd Days
January
31
3
February
29
1
March
31
3
April
30
2
May
31
3
June
30
2
June
30
2
Total
181 days
6
Total
182 days
0
July
31
3
July
31
3
August
31
3
August
31
3
September
30
2
September
30
2
October
31
3
October
31
3
November
30
2
November
30
2
December
31
3
December
31
3
Total
184 days
1
Total
184 days
2
Table : 3 (Odd days for every quarter)
Odd days of 100 years = 5, Odd days of 200 years = 3, Odd days of 300 years = 1, Odd days of 400 years = 0.
M onths of years
I st three m onths 1 Jan to 31 M arch
Total days
90 / 91 Ord. / Leap
Odd days
6 /0 0 Ord. / Leap O dd 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 =
124 7
= 5 odd days.
IInd three IIIrd three m onths m onths 1 Apr to 1 July to 30 June 30 Sep. 91
92 1 Odd day
Iv th three Total year m onths 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 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 3 + 5 = 8 odd days = 1 odd day. Saturday + 1 odd day = Sunday. PAGE # 83
Ex 2. Sol.
Ans. Ex 3. Sol.
Ans. Ex 4. Sol.
Ans.
Ex 5. 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. ......................(B) Sol. 68 years have 17 leap years and 51 ordinary years. Thus = (17 × 2 + 51 × 1 ) = 85 odd days ' 01' odd day ...(C) 16 January has = ' 02' odd days..(D) Adding (A) + (B) +(C) +(D), We get, 0 + 01 +01 +02 = 04 odd days Thursday
Ex 7.
Find the day of the week on 18 July, 1776 (leap Sol. year). Here 1600 years have ‘0’ odd day.....................(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.............................(C) Now, the no. of days from 1st January to 18 July, 1776 = 182 + 18 = 200 days = (28 × 7 + 4) days = ‘4’ odd days.....................(D) Adding (A) + (B) +(C) +(D), We get, 0 + 5 + 2 + 4 = 04 odd days Thursday 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.....................(A). 300 years have ‘01’ odd days.....................(B). 74 years have (18 leap years + 56 ordinary years) 2 × 18 + 1 × 56 = 92 odd days = ‘01’ odd days.............(C) Days from 1st January to 1st Oct., 1975 1st Jan – 30 June + 1st 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.
The year next to 1996 having the same Calendar will be 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 year 1996. Prove that last day of a century cannot be Tuesday, Thursday or Saturday. 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 cannot be Tuesday, Thursday or Saturday.
Important Notes :
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.
Minute hand and hour hand are opposite once in
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) and 08 ordinary years (1995,1997,1998,1999, 2001, 2002, 2003,2005) Total odd days = (2 × 3 + 1 × 8) = 14 = 0 odd days (Thus Proved)
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.
Minute hand is 55 min. ahead of hour hand in
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
60 min. Minute 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
MIRROR IMAGE OF CLOCK
to find the mirror image, time is subtracted from 11 : 60.
Minute hand is 15 min. ahead of hour hand in 60 15 180 4 = = 16 min. 55 11 11
4 Hence the right time is 16 minute past 3. 11
HANDS ARE OPPOSITE Ex.9 Sol.
60 40 480 7 = = 43 min. 55 11 11
Hence, the right time is 43
23 : 60. Ex.11
The time in the clock is 4 : 46, what is the mirror image ?
Sol.
(11 : 60) – (4 : 46) = 7 : 14.
7 min. past 2. 11
Case-I
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
Sol.
hand and minute hand ? At 4 O’ clock angle between hour and min. hand is
HANDS ARE PERPENDICULAR
Sol.
If the time is between 11 O’clock to 1 O’clock, then to find the mirror image, time is subtracted from
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
If the time is between 1 O’clock to 11 O’clock, then
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 will also cover 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.
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 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.
EXERCISE 1.
Find the day of the week on 26 January, 1950. (A) Tuesday (B) Friday (C) Wednesday (D) Thursday
2.
W hich two months in a year have the same calendar ? (A) June, October (B) April, November 13. (C) April, July (D) October, December
3.
4.
5.
12.
Are the years 900 and 1000 leap years ? (A) Yes (B) No (C) Can't say (D) None of these If it was Saturday on 17th November, 1962 what will be the day on 22nd November, 1964 ? (A) Monday (B) Tuesday 14. (C) Wednesday (D) Sunday Sangeeta remembers that her father's birthday was certainly after eighth but before thirteenth of 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.
Find the day of the week on 15 August, 1947. (A) Tuesday (B) Friday (C) Wednesday (D) Thursday
7.
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) Wednesday (D) Monday
8.
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) Wednesday (C) Thursday (D) Tuesday
16.
If the day before yesterday was Saturday, what day will fall on the day after tomorrow ? (A) Friday (B) Thursday (C) Wednesday (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 ? (A) Sunday (B) Monday (C) Wednesday (D) Friday Mohini went to the movies nine days ago. She goes to the movies only on Thursday. What day of the week is today ? (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 clock be in the straight line, but not together ? (A) 16 minutes past 9 4 (B) 16 minutes past 9 11 6 (C) 16 minutes past 9 11 9 (D) 16 minutes past 9 11 At what time between 5 & 5 : 30 will the hands of a clock be at right angle ? 10 (A) 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 Ajay left home for the bus stop 15 minutes earlier 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 ? (A) 8.30 a.m. (B) 8.45 a.m. (C) 8.55 a.m. (D) Data inadequate
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. and 2.00 P.M. The second group of ten works between 10.00 27. A.M. and 4.00 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.m.. If he rose at 12 noon, for how many hours did he sleep ? (A) 5 hours 15 min. (B) 16 hours 15 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 following 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.
29.
23.
How many times are the hands of a clocks coincide in a day ? (A) 10 (B) 11 (C) 12 (D) 22
24.
At what time between 2 and 3 O’ clock the hands of a clock will make an angle of 160º ? (A) 20 minutes past 2 (B) 30 minutes past 2 (C) 40 minutes past 2 (D) 50 minutes 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
31.
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 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 : 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) 8 days later (B) 10 days later (C) 6 days later (D) can't be determined 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 am (B) Friday noon (C) Friday 8 pm (D) Tuesday noon PAGE # 87
33.
34.
35.
A clock loses 2 minutes in a hour and another clock 38. gains 2 minutes in every 2 hours. Both these clocks 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, 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 Ramu purchased a second hand Swiss watch which is very costly. In this watch the minute-hand and hour hand coincide after every 65
36
37.
3 minutes. 11
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 will be the time that the clock will show exactly 14 days from the time it was set right ? (A) 1 : 36 : 48 (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. W henever 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 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 the hour hand in that defective clock ? (A) Between 2 and 3 (B) Between 4 and 5 (C) Between 10 and 11 (D) Between 3 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. Here 3
2
2 2
1 1
3
2
3 2 2 3
2 1 1 2
2 1
1 2
3
2 2
3
2 3 1 2 1 2 2 3
3
2 2 3
n=
2 1 1
side of big cube 4 4 side of small cube 1
3
2 1 1 2
2
Ex 1.
How many smaller cubes have three surfaces
2
painted ?
3
(A) 4
(B) 8
(C) 16
(D) 24
2
Sol.
(B) Number of smaller cubes with three surfaces
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.
painted = 8
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 smaller 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
(D) Number of smaller cubes with two surfaces
length of edge of big cube = length of edge of one smaller cube
(D) Number of smaller cubes with no surface
TYPE I
Sol.
How many smaller cubes have two surfaces painted ? (A) 4
(B) 8
(C) 16
(D) 24
painted = (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 painted = (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
painted = (n – 2)3 = (4 – 2)3 = (2)3 = 8
TYPE II If a cube is painted on all of its surfaces 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 64 Total cubes of obtained = 1 1 1
of another opposite surfaces and red on remaining 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 of cubes with two surfaces painted Sol. with black and blue colour = 8. (b) Number of cubes with two surfaces painted with blue and red colour = 8. (c) Number of cubes with two surfaces painted with black and red color = 8. Ex 11.
Ex 7.
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 of cubes 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 with one surface painted with black colour =8. TYPE IV (b) Number of cubes with one surface painted with blue colour = 8. If a cube is painted on its surfaces in such a way (c) Number of cubes with one surface painted that one pair of adjacent surfaces is left unpainted. with red colour = 8. Directions : (12 to 15 )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 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 number 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
Sol.
(A) Number of smaller cubes with no surfaces Step IV. 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. 3 2 = (n – 2) + (n – 2) × 2 + (n – 2) 3 2 Step V. Write remaining number in the remaining block. = (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. )
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)
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.
EXERCISE 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 (D) 48 (C) 8 (D) 10 9.
10.
11.
How many smaller cubes have three surfaces 19. painted ? (A) 4 (B) 8 (C) 16 (D) 24
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 ? (A) 24 (B) 20 14. How many cubes will have at least blue colour on (C) 8 (D) 12 its surfaces ? (A) 20 (C) 24 15.
16.
(B) 8 (D) 32
24.
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
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
35.
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. 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.
(X)
(A)
(B) 51.
(B)
45.
Which symbol will appear on the opposite surface to the symbol x?
(D)
(A)
(B) =
(C)
(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 4
4 2 ?
(i)
(ii)
(iii)
(A) 1 (C) 3
(X) 52.
(A)
(C)
(B)
(B) 6 (D) 5
On the basis of the following figures you have to tell which number will come in place of ‘?’
3 6 1
4 2 6
? 1 5
(i)
(ii)
(iii)
(A) 2 (C) 6
(D)
(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
(A)
(A)
M
X
M (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
1
7
(B)
(C)
7 (D)
8
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)
(A) A (C) 1
J (C)
(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 67. 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.
I C A B 2
68.
Which number/letter is opposite O?
L N M 2 I O (A) L (C) N
(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
ANSWER KEY FORCE AND NEWTON’S LAW OF MOTION(PHYSICS) Que.
1
2
3
4
5
6
7
8
10
11
12
13
14
Ans.
B
C
C
B
A
B
D
A ACD B
9
C
C
C
C
C
Que.
16
17
18
19
20
21
22
23
26
27
28
29
30
24
25
15
Ans.
B
D
A
A
A
C
C
D
B
B
B
B
A
D
C
Que.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans.
B
C
B
A
A
C
B
D
A
C
A
B
D
A
A
Que.
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans. Que. Ans.
D
D
CD
D
A
D
D
C
B
A,C
C
A,B,C
B
B
B
61
62
63
64
65
66
67
68
69
B
A
C
C
D
C
B
B
C
CARBON(CHEMISTRY) 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Que s. 1 Ans. C C B B B A D A C B C D C B B Que s. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. B A C B C B B A C C D C A B C Que s. 31 32 33 34 35 36 37 38 39 40 41 Ans. C B C A B D C B C A A
NUMBER SYSTEM(MATHEMATICS) 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
TRIGONOMETRY(MATHEMATICS) 1
Q.
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
PROTOPLASM (BIOLOGY) Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A. C D C A B A A C A D B A A D D Q. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A. C B D B B C D A D A C A A A B Q. 31 32 33 34 A. A B D A
SERIES COMPLETION(MENTAL ABILITY) 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
17
18
19
20
21
22
23
24
25
26
27
28
29
30
A
B
D
D
A
C
B
B
C
A
12
13
14
15
Ans.
B
C
C
C
B
Que.
31
32
33
34
35
Ans.
C
C
C
D
D
EXERCISE- 2 (Alphabet Series) Que.
1
2
3
4
5
6
7
8
9
10
11
Ans.
D
A
D
C
C
A
D
C
D
B
D
C
C
C
D
Que.
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) Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
D
C
B
Ans.
D
D
A
A
C
B
A
C
D
D
C
Que.
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) Que .
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
Que .
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
A
C
D
B
A
Ans .
B
C
A
B
B
A
C
A
D
D
Que .
31
32
33
34
35
36
37
38
39
40
Ans .
B
B
A
C
A
C
B
C
D
B
PUZZLE-TEST(MENTAL ABILITY) Que. Ans. Que. Ans. Que. Ans. Que. Ans.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
B
D
C
A
D
C
C
C
C
C
D
C
C
B
16
17
18
19
20
21
22
23
24
25
26
C
C
D
D
C
D
D
B
B
C
A
31
32
33
34
35
36
37
38
39
40
41
27 A 42
28 D 43
29 C 44
30 D 45
D
C
B
A
B
A
D
A
D
C
D
A
C
B
D
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
CALENDAR AND CLOCK-TEST(MENTAL ABILITY) Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D 16 B 31 D
C 17 B 32 B
B 18 B 33 D
D 19 B 34 B
D 20 C 35 A
B 21 B 36 A
C 22 C 37 C
B 23 D 38 A
C 24 C 39 D
C 25 B 40 A
A 26 D 41 C
B 27 C
C 28 B
B 29 B
A 30 D
Que. Ans. Que. Ans.
CUBE AND DICE TEST(MENTAL ABILITY) Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
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. Ans. Que. Ans. Que. Ans.
31
32
33
34
35
36
37
38
39
40
41 B 56 C
42 B 57 B
43 A 58 A
44 D 59 C
45 B 60 C
C
D
C
D
C
A
D
C
B
C
46 D 61 A
47 A 62 D
48 B 63 E
49 C 64 D
50 D 65 C
51 A 66 A
52 B 67 B
53 D 68 B
54 B
55 D
PAGE # 100 100
