Concept Maps Class VIII Rational Numbers Linear Equation in one variable Compound Interest Square and Square Roots Cube and Cube Roots Mensuration Direct and Inverse Variation Percentage & It's Application Alzebric Expression Exponents Quadrilaterals Algebraic Identities Statistics Probability Solid Shapes
Rational Numbers
Operation on rational Number 1. Addition 2 Ex. Add 3 and –
5 7 3 2 & # Sol. + $' ! 7 5 % "
Stondard form a
Rational No. Between two Rational No. Q Insert 5 rational number between 3 and 5 8 6
A rational number is said to be in standard form ( b ) if a and b are integers having no common factor other then 1, and b is positive. Ex. Standard form of 24 is –8 -15
5
Sol. LCM of 8, 6 is 24 3 = 3 × 3 = 9 , 5 = 5 × 4 = 20 8 8 3 24 6 6 4 24 10 11 12 13 5 , 14 , 5 rational no.be between 3 and 6 are 24 ,24 ,24 24 24 8
=
3 2 15 ' 14 1 ' ( ( 7 5 35 35
2. Subtraction 3 1 Ex. Subtract – from
5 2 1 & 3 # 1 3 2 ) 15 17 Sol. ' $ ' ! ( ) ( ( 5 % 2" 5 2 10 10 3. Multiplication3 1 Ex. Multiply – from
5 2 3 2 6 Sol. ' * ( ' 5 7 35 4. Division 3 5 Ex. Divide – from –
Rational Numbers
2 4 3 5 3 4 6 Sol. + ' ( * (' 2 4 2 '5 5
a b
Ex.
5 ,1 , -3 7 8
Graphical Representation 27 4 A'
Q. Represent O O
27 4
an number line
Important Points
A
Multiplicative inverse of a is
27
Multiplicative identity is 1 Additive identify is 0.
Mark point A and divide OA into four equal part then OA' will represents 27 4
×
×
×
×
×
×
in above table a, b, c are rational number
1 a
Additive inverse of a is-a Absolute value of a number is it's numerical value (value without sign)
Linear Equation in one variable
RULES FOR SOLVING We can add or Subtract same number on both side x+7=8 x + 7- 7 = 8 - 7 x=1
We can multiply or divide both side by non zero number 3x = 6 3x 6 3 =3 x=2
SOLUTION
Kept the variable on one side and constant on other side 3x + 7 = 2x + 10 3x - 2x =10 - 7 x=3
Value of variable which satisfy equation x = 3 is Solution of 3x + 1 = 10 because 3 (3) + 1=10 9 + 1 = 10 10 = 10
LINEAR EQUATION IN ONE VARIABLE A linear equation which has only one variable is called linear equation in one variable. For example : x + 3 = 5. SOME PROBLEM 2x +1
7
Solve: 3x - 5 = 3
2x +1 7 = 3x - 5 3 By cross Multiplication 3(2x+1) = 7 (3x-5) 3 × (2x) + 3 × (1) = 7 × (3x) - 7 × (5) 6x + 3 = 21x -35 21x – 6x = 3 +35 15x = 38 x = 38/15
x – 3x + 1 2 5 x – 3x + 1 2 5 5(x) – 2 (3x +1) =6 10 5x – 2 (3x)–2 (1) =6 10 5x – 6x–2 6 = 10
Solve:
English word
Mathematical meaning
More than, exceeds older than, less than, decreaed, younger than times, of, product Divided by, quotient,per, for What, how many,etc.
+ – × ÷ x (or some other variable)
– x – 2 = 60 x = – 62
APPLICATION Fraction Dr. of fraction is 5 more than Nr Nr.=x , Dr.=x +5 Fraction = x x+5
Money No. of 2 Rs. Coin is 3 times the No.of 5 Rs. Coin No. of 5 Rs.Coin = x, No. of 3 Rs. Coin = 3x Total money = 5 × (x) + 2 × (3x)
Geometry Length of Rectangle is 5 less than twice the breadth. b=x l = 2x-5
Age My present age = x yr. after 2 yr. my age will be (x+2) yr. before 3 yr. my age was (x-3) yr.
Compound Interest In SI. the interest is calculated on principal, for all years, so interest is same for all years. When interest is calculated QLY R/4# & A ( P $1 ) ! 100 " %
T*4
In CI, the interest is calculated an amount of the previous year so interest is different for all years.
Depreciation As the time passes the valve of object depreciate is called depreciation.
When interest in calculdted HLY
T
R # & A ( P$1 ' ! 100 % "
T* 2
& R/2# A ( P$1 ) ! % 100 " When Time Period is in fraction Ex. T ( 2
If the present population of a town is P. and it is growing at R% P.A.
3 yr 4
Compound Interest
3 # *R! 4 $1 ) ! 100 ! $ $ ! % " &
2$
R # & A ( P$1 ) ! % 100 "
T
R # & A ( P$1 ) ! % 100 "
T
R # & Then population of town after T year is P$1 ) ! % 100 " P Then population of town before T year in T R # & $1 ) ! % 100 "
CI = A – P
When Rate of interest is different for diff. year T
T2 R1 # 1& & R ! $1 ) 2 # A ( P$ 1 ) ! $ 100 ! % 100 " ........ % "
Q. In what time a sum fo 1000 become 1331 at 10% P.A at CI. Sol. A = 1331, P =1000, R = 10% P. A. T
R # & A ( P$1 ) ! 100 % "
Q. If CI -SI = 50 for 2 yr at R= 10% P.A. find P
T
Sol. SI (
PRT P * 10 * 2 P ( ( 100 100 5 T
2
2
R # & & 10 # &11 # 121 A ( P$1) P ! ( P$1) ! ( P$ ! ( 100 100 % " % " %10 " 100
121P 21 P CI ( A ' P ( 'P ( 100 100
CI - SI = 50 2 1P P ' ( 50 100 5
P ( 50 - P ( 5000 100
Q. If a money become double in 5 yr. In what time it will become 8 times. T
R # & , A ( P $1 ) ! % 100 " 5
R # & 2P ( P $1 ) ! % 100 " R # & 1/ 5 $1 ) !( 2 % 100 "
T
R # & A ( P$1) ! % 100 "
3
T
R # & 8P ( P$1 ) ! % 100 " 8 ( (21/ 5 )T 23 ( 2T / 5 - 3 (
T
1331 & 11 # ($ ! 1000 %10 "
A ( 8P, T ( ?
Sol. A ( 2P, T ( 5 yr.
10 # & 1331 ( 1000$1 ) ! 100 % "
T - T ( 15 yr. 5
T
& 11 # & 11 # $ ! ( $ ! - T ( 3 yr. %10 " %10 "
Square and Square Roots Squre Root
Perfect Square
Squre root of no x, in that no whose square is x
A natural no. is perfect sq. if it is sq. some natural no. 36 is perfect sq because it is the me sq. of 6
2
Ex. Sq. root of 64 is 8 become 8 = 64. Þ 64 = 8
Prime factorization Q.
Properties of perfect square
Find sq. root of 36
Sol.
No. ending with 2,37,8 never be a perfect square No. ending with odd no of zero never be a perfect sq.
36 = 2 × 2 ×3 × 3 =2×3=6
diff of square of two consecutive natural no. is equal 2 2 to their sum. (n+1) – n = (n+1) + (n) 2 2 2 Pythogorean triplet (x,y,z) if z = x + y
2 36 2 18 3 9 3 3 1
Long Division method. Q. find sq. root 58081
Square and Square Roots
Square by diagonal method. 2 25 2 5 2
0
5
1
1 0
4 2
5
0 0
6
241 58081 4 180 176 481 481 × 58081 = 241 2 2 44 +4 481
When a number is multiplied with it self Square of 7 is 7× 7 = 49.
2
5
2
\ 25 = 625
Successive subtraction Square by column method 2 to find 25 , take a = 2, b = 5 2
a
2
2ab
Squre root of fraction
2
b
a×b= a × b
2
2
2 ´ 2´ 5
5
4 +2 6
20 +2 22
25
a a = b b Ex.
529 529 23 = = 841 841 29
2
25 = 625
Ex.
36 × 49 = 36 × 49 = 6 × 7 = 42
We successive subtract odd no from the given no. till we get zero. The number of time we subtract is the square root of the no. Ex.
16 16 – 1 = 15 15 – 3 = 2 12 – 5 = 7 7 – 7 = 0 so 16 = 4
Cube and Cube Roots
Cube roots
Perfect cube A natural no. is said to be a perfect cube if it is the cube of same natural no.
The cube root of a no. is x that no whose cube gives x. Ex. Cube root of 8 is 2 because 23 = 8 3 8 =2
Properties of perfect cube (i) Cube of even no is even. (ii) Cube of odd no. is odd (iii) Cube of negative no is negative. (iv) The sum of the cube of first n natural no. is equal to the squre of their sum. 13 + 23 + 33 + .... + n3 = (1 + 2 + 3 + .... n)2
Cube root by prime factorization 3 216
Cube & Cube roots The cube of no. is obtained when no. is multiplied by itself 3 times. Cube of x is x ´ x ´ x
(v) Cubes of the numbers ending in digits 1, 4, 5, 6 and 9 are the number ending in the same digit. Cubes of numbers ending in digit 2 ends in 8, and cube of numbers ending in digit 8 ends in 2. The cubes of the numbers ending in digits 3 and 7 ends in 7 and 3 respectively.
2 2 2 3 3 3
Ex.
216 108 54 27 3 3
Cube root by pattern 3
3
2
8 +7 15 3
2
3a b
2
3ab
2
3´ 2 ´5 3´ 2´52 60 +16 76
25 = 15625
3
3
' 512 3
125
' 8 * '8 * '8 3
5* 5* 5 8 =– 5 3 216 = 3 2 ´ 2 ´ 2 ´ 3 ´ 3 ´ 3 = 2 ´ 3 = 6
Cube by column method 3 to find 25 , take a = 2, b = 5 a
' 512 = 3 125
150 +12 162
3
b
3
5
125
We have to successively subtract 1,7,19,37,61,91 ............... from number till we get zero. The no of time we subtract give the cube root. Ex. 3 64 64 – 1 = 63 63 – 7 = 56 56 – 19 = 37 37 – 37 = 0 So 3 64 = 4
Basic Geometrical figures Triangle Rectangle 1 Area = 2 ´ base ´ Height A = l ´ b, P = 2(l + b) Eq. D = 3 side2 Square
Mensuration Rhombus a
Area = a2, P = 4a
a
Parallelogram Area = Base ´ height
a
Area =
T.S.A. = 2(lb+bh+hl) L.S.A. = 2h(l + b) V= lbh
4
a
1 2
Cuboid
Circle Area = pr2 circumference = 2pr
b l
Cube L.S.A. = 4a2 T.S.A. = 6a2 V= a2
d1d2
Perimeter = 4a d12
h
a a
a
d2
a2 = 4 + 42
Mensuration Cylinder
Trapezium a
Plane Figures
h b
Area =
1 2
Figure having two dimensions are called plane figures. Ex. Square, Rectangle, Circle, Triangle etc.
(a+b)h
C.S.A. = 2prh T.S.A. = 2pr (h+r) V=pr2h
Solid Figures Figure having three dimensions are called solid figures. Ex. Cube, Cuboid, Cylinder, etc.
h r
Quadrilateral C
D
Problem of Plane Figures
h1 h2 B
A 1
Area = 2 (AE)(h1+h2)
Q. Find the area of trapezium D 13 A
15 25
C 13 B
Sol.Draw CE ||AD \ AECD is || cm D C 15 EC = AD = 13 13 13 h 13 AE = DC = 15 E B A 25 F \ BE = AB – AE = 25 – 15 1 = 10 EF = FP = 2 EB = 5 In DCFB h = 132 – 52 = 12 1 ar trap. ABCD = 2 (25 + 15)12 = 240 sq. unit
Problem of solid Figures Q. A cube of 9 cm edge is immersed completely in a rectangular vessel contaning water. If the dimension of the base are 15 cm and 12 cm find rise is water level in the vessel. Sol. After emerging cube
9 cm
Initial level
x 15
12
Volume of cube = Volume of cuboid of height x 9 ´ 9 ´ 9 = 15 ´ 20 ´ x x=
9* 9* 9 15 * 20
= 4.05 cm
Direct and Inverse Variation 1 m
1 8
1 6
N1 cos t 1 ( N2 cos t 2
24 Days. 7
5 60 7 * 60 ( - cos t 2 ( ( 84 R s. 7 cos t 2 5
Direct and Inverse Variation
Speed (
m1 d2 10 d2 10 * 6 ( ( - d2 ( ( 20 days. m2 d1 3 6 3
1 8
total dis tan ce total time
5 m / sec . 18
1 10 (
5' 4 1 ( 40 40
dista nce time
Avg. speed (
km / h. (
,
1 1 3)4 7 ) ( ( 8 6 24 24
1 1 ' 8 10
dis tan ce speed
L ) 125 5
Percentage & It's Application
Profit & Loss
Ex.
Profit = SP – CP
Problem on percentage
Loss = CP – SP
Ex. In 800 student 25% are girls, find the number of boys Sol. Boys percentage = (100–25)% = 75% No. of boys = 75 of total student 75 ( * 800 ( 600 100 Ex. Ram salary is decreased by 20% and then increased by 20% find % change in his salary Sol. Let has salary is Rs. 100 His salary after 20% decrease = 100 – 20% of 100 = 100 – 20 = 80 Now when his salary increased by 20% it become = 80 + 20% of 80 = 80 +16 = 96 So Ram income is decreased by (100 – 96) = 4%
SP ' C P * 100 CP
P r ofit % (
Loss% (
CP ' SP * 100 CP
Profit & Loss are Calculated on CP. &100 ) gain% # SP ( $ !CP 100 % "
& 100 ' Loss % SP ( $ 100 %
# !CP "
Sol.
A man sold an article at 450 and having a loss of 10% in order to gain 20% at what price should be sold. Initially SP = 450 loss = 10%
&100 ' loss% # SP ( $ !CP 100 % " &100 ' 10 # 450 ( $ !CP % 100 " CP ( Now
450 * 100 ( 500 90 CP = 500 gain = 20%
& 100 ) gain% # So the New SP ( $ ! CP 100 % " & 100 ) 20 # ( $ ! 500 % 100 " ( Rs. 600
Percentage and its Application Percentage means per hundred or for every hundred
x x% ( 100
25 1 Ex 25 % ( ( 100 4
Discount Discount = MP – SP D is c o u n t (
MP ' SP * 100 MP
Discount always given on MP
Value added Tax
% increase and Decrease % increase (
% decrease (
Tax is always calculated on the price at which article is sold.
increase * 100 original value decrease * 100 original value
& 100 ) tax % # SP W ith tax ( $ ! SP 100 % "
The cost of article in shop is Rs. 60 The sales tax was 5% find bill amount Sol. SP = 60 , tax% = 5
SP (
100 ' Discount% * MP 100
Ex. An article marks Rs 600 and a discount of 20% is given find selling price of it. Sol. MP = 600 Discount % = 20
Q.
&100 ) 5 # SP With tax ( $ ! * 60 ( 63 % 100 "
&100 ' discount% # SP ( $ !MP 100 % " 100 ' 20 & # ( $ ! * 600 ( 480 Rs. % 100 "
Alzebric Expression
Term
2
9x
–3yx
9 x x Factors
–3 x y Factor
Degree
Like & Unlike terms
Algebraic Exp. : 9x2 – 3xy + 5
Like term having same algebraic factor.. Ex. 3xy, 5xy 3xy = 3 × p × y 5xy = 5 × x × y Unlike term having diff. algebraic factor. Ex. 4x2y = 4 × x × x × y 2xy = 2 × x × y.
5
Highest power of the variable in a term. 2 Ex 3x + x + 1 deg 2 2 ex. x'y'z' + x + 1 deg 3 * if term contain more than one variable we have to add the power of all variable.
Numerical coefficient is -3 2
In – 3x y
Coefficient of y is -3x
Division
2
Ex. Divide. 3 2 15x + 12x + 21x by 3x
2
Coefficient of x is -3y
Sol.
Algebraic Expression Multiplication 2
15 x 3 ) 12 x 2 ) 21x 3x 15 x 3 12 x 2 21x ( ) ) 3x 3x 3x =
2
5x + 4x + 7
2
Ex. Multiply (x + y) (x + y – xy) 2 2 2 2 Sol. x(x +y – xy) +y (x +y – xy) 3 2 2 3 2 x + xy – x y + y – xy 3 3 =x –y
*
When the remainder is zero the divisor is called a factor of the dividend..
Ex. Find the value of a if 2x – 3 is a 4 3 2 factor of 2x – x – 3x – 2x + a. 4 3 2 Sol. First we divide 2x – x – 3x – 2x + a by 2x – 3.
Types of Algebraic Exp Acording to the No. of terms . Monomial Single term Binomial
Two term
Addition & Subtraction 2xy
Ex1 x + y
Trinomial
Three term Ex x + y + z
Multinomial
More than 3 term Ex. x3 + x2 + 7x + 1
We can add & Subtract only the like terms. Ex. Add 5ab , 4ab = 5ab + 4ab = (5+4) ab = 9ab Ex Subtract of 7xy from 2xy Sol. 2xy – 7xy = (2– 7) xy = – 5xy.
4
3
2
2x – 3 is a factor of 2x – x – 3x – 2x + a if, a – 3 = 0 Hence, a = 3.
Exponents Laws of exponent m
(i) Scientific notation or standard form ex. speed of light = 300000000 m/s is written in standard form as 8 = 3 × 10 m/s
No. in expanded form with the help of exponents 123.45=1×102+2×101+3×100 –1 –2 + 4 × 10 + 5 × 10
n
m
m) n
&a #
&a # &a # &a # $ ! x$ ! ($ ! % b " %b " %b "
(ii) $% b !"
n
n
m' n
&a # &a # +$ ! ($ ! %b " %b " n
m mxn 4 &a # .& a # 1 . (iii) 3$ b ! 0 ( $ b ! % " . 2% " . /
n
n
4 (a / b) 1 ( a / b )n 0 ( (v) 3 (c / d)n 2 (c / d) /
a
'n
b
n
& # & # (vi) $ b ! ( $ a ! % " % "
0
&a # %b "
(vii) $ ! ( 1
Exponents (–1) even =1 (–1)odd = –1
a × a × a = a3, read as a raised to power 3 where a is base 3 is exponent
Q. find x 2x–1 If 24 = 625 Sol. (52)2x–1 = 54 54x–2 = 54 \ 4x –2 = 4 4x = 6 3 x= 6 = 4
2
Q. find x –6
–6
–6
&x #
& # & # If $% 7 !" * $% 9 !" ( $$ y !! % " 2
14
–6
–6
3–5 * 10 –5 * 125 5 –7 * 6–5 –5 –5 3 * 10 * 125
Q. Simplify Sol.
5–7 * 6–5
–6
&x # &2 # &14 # Sol. $ 7 ! * $ 9 ! ( $$ y !! % " % " % " 14 # &2 * $ ! 9 " %7
& 4 # $ ! % 9 "
–6
–6
&x ( $$ %y
& x # ! ( $$ ! % y "
# ! ! "
3 –5 * ( 2 * 5 ) –5 * 5 3 5 – 7 * (2 * 3 ) – 5
–6
(
3 –5 * 2 –5 * 5 –5 * 5 3 5 –7 * 3 –5 * 2 –5
–6
–5+5
=2
0
,
x y
(
4 9
n
&a c # &a # &c # (iv) $ b x d ! ( $ b ! x$ d ! % " % " % "
–5+5
×3 0
–5+3+7
×5 5
= 2 × 3 × 5 = 3125
gm
Quadrilaterals
Property of parallelogram Opposite angle are equal Opposite sides are equal Diagonal bisect each other Diagonal divide parallelogram in to two congruent triangle.
Q. ABCD is || find all angle of || Sol. ÐC = ÐA = 50° D \ ÐC = 50° as AB||CD 50° \ ÐA + ÐD = 180° A 50°+ÐD = 180° ÐD=130° ÐB = ÐD = 130° \ ÐB = 130°
gm
Angle sum property sum of all interior angle of quadrilateral is 360° Q. Angle of Quadrilateral are in the ratio 1 : 2 : 3 : 4 find angle
C B
Sol. Let angle are x, 2x, 3x, 4x x + 2x + 3x + 4x = 360 10x = 360 x = 36 So, angle are 36, 2´ 36 3 ´ 36, 4 ´ 36 i.e. 36°, 72°, 108°, 144°
gm
Q. If ABCD is || find x,y Sol. In ||gm diagonal bisect each other C D \ OA = OC x–5=7 x =12 and OB = OD 15 = x + y B 15 = 12 + y A y=3
Quadrilaterals A quadrilateral is four sided closed figure.
Polygon Sum of interior angle = (n –2)180 Sum of exterior angle = 360° For Regular polygon (n ! 2)180 Each Interior angle =
Types of Quadrilateral
D Trapezium Quadrilateral with one pair of opposite side is parallel. D
A
Each Interior angle =
360 n
n
C
Parallelogram A B Quadrilateral in which both pair of opposite side is parallel.
Kite Quadrilateral in which adjacent sides equal but unequal opp. side
C
B
If AD = BC, it is known as isosceles trapezium
Rectangle
||gm
with all angle 90° *Length of diagonal are equal
Rhombus
||gm
with all side equal *diagonal bisect each other at 90°
Square
||gm
with all side equal and all angle 90° *Length of diagonal are equal *Diagonal bisect at 90°
Algebraic Identities
Identities 2
2
2
1. (a + b) = a + 2ab + b 2 2 2 2. (a –b) = a – 2ab + b 3. (a + b) (a – b)2 = a2 – b2
Factorization The process of finding two or more expression whose product is the given expression is called factorization.
Problem Based on Identity
I. Factorization by taking out the common facto r. 3 2 2 Ex. 8x y – 4yx = 4xy(2x y – 1) Ex. 2(x + 3) + 2(x + 3) = (x + 3) = (x + 3) (x + 2)
2
Q. Expand (2x – 5y) Sol.(2x – 5y)2 2 2 = (2x) – 2(2x)(5y) + (5y) 2 2 = 4x – 20xy + 25y 107 2 ! 103 2 210 2 2 (107 " 103 )(107 ! 103 ) 107 ! 103 Sol. = 210 210 210 # 4 = =4 210 Q. Find the value of
Q. If x + Sol. x +
1 1 = 3 find x2 + 2 x x
1 =3 x 2
1& ) 2 'x " $ x %= 3 (
1 1 =9 + x x2 1 2 x +2+ 2 =9 x 2
x + 2x ´
2
x +
1 x2
=9–2=7
Algebraic Identities An identity is an equality, which is true for all values of the variables.
II. Factorization by grouping. Ex. ax + by + ay + bx = ax + ay + bx + by = a(x +y) + b(x + y) = (x + y)(a + b) III. Factorization the difference of two squares. 2 2 a – b = (a + b) (a – b) Ex. 9x2 – 16y2 = (3x)2 – (4y)2 = (3x – 4y) (3x + 4y) IV. Factorization of quadratic trinomial 2 Ex. Factorize x + 9x + 18 2 2 Sol. x + 9x + 18 = x + 6x+ 3x + 18 = x(x + 6) + 3(x + 6) = (x + 6) (x + 3) Ex. Factorize 9x2 – 22x + 8 2 2 Sol. 9x – 22x + 8 = 9x – 18x – 4x + 8 = 9x(x – 2)– 4(x –2) = (x – 2) (9x – 4)
Draw the bar graph for the given table.
In a survey of 20 families, each family is found to have the following number of children : 1, 2, 2, 3, 2, 3, 3, 4, 1, 1, 4, 4, 2, 2, 3, 1, 5, 1, 1, 2 Make a frequency distribution table. Sol.Arrange in ascending order. 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5. Number of children 1 2
Class No. of Students
Tally Marks
No. of families
I I
6
60 50 40
6
30
IIII
4
4
III
3
5
I
1
Total
20
3
Histogram
Bar-Chart
Frequency distribution table
Draw a histogram of the following frequency distribution.
VI
VII
VIII
IX
X
40
20
55
50
35
Class (Age in years)
0–5
5 – 10
No. of students
72
103
25
Scale : 1 unit : 10 Students 125 – 100 – 75 –
20
50 – 25 – 0
The marks obtained by 40 students of class VIII in an examination are given below : 18, 8, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16, 4. Present the data in the form of a frequency distribution using the same class size, one such class being 15 –20 (where 20 is not included).
50
Y
10
Group frequency distribution table
10 –1 5 15 – 20
VI VII Class
VIII
IX
5
X
X
10 15 20
Class interval (Age in year)
Statistics It is defined as the science of collection, presentation, analysis and interpretation of numerical data.
Pie-chart Central angle for a variable =
Frequency of the variable ´ 360º Total of frequencie s
The main source of energy is used by each house in a street is listed below : Sol. The frequency distribution is as given below :
Source of Energy Electricity No. of houses 20
Solar 10
Gas 12
Oil 6
Represent the above data by a pie chart 11 7 9 7
For class 10-15, 10 is lower limit,15 is upper limit Class mark = UL " LL = 10 " 15 = 12.5 2 2 Class size = UL – LL = 15 10 = 5
Some definitions
Source of energy
Number of houses
Data is defined as information in numerical facts.
Electricity
20
Solar
10
Gas
12
Oil
6
Range Is defined as the difference between maximum and minimum value of observation. Frequency is defined as the number of times an observation occur.
Sol.
Central angle 20 # 360 * 150 48
10 # 360 * 75 48 12 # 360 * 90 48 6 # 360 * 45 48
Electricity
150° Solar 75° 45° 90° Oil Gas
Probability Sample Space Collection of all possible out comes eg : In a throw of a die S. S. = (1,2,3,4,5,6)
Outcome
Experiment
Different possibilities which can occur eg : In tossing a coin outcome are Head and Tail
Any kind of activity eg : Tossing a coin
Probability
Event A subset of sample space associated with a random experiment is called an event. e.g. : getting six in a throw of a die
Deals with the measurement of uncertainty of the occurrence of some event in terms of percent or ratio
A bag contains 5 red balls, 8 white balls, 4 green balls and 7 blackballs. If one ball is drawn at random, then probability that it is : 7 24 4 1 * (b) Green P(G) = 24 6 (a) Black i.e. P(B) =
(c) P(Not red) = 1–P (Red) * 1 !
Equally likely outcomes
When a coin is tossed
There is equal uncertainty of each outcome of an experiment
Total number of outcomes = 2 ie. T,H 1 Probability of getting head P(H) = 2 1 Probability of getting tail P(T) = 2
Probability of an event A = P (A) *
When P(A) = 0, then A is called as impossible event e.g. Probability of getting a number greater than 7
5 24 ! 5 19 * * 24 24 24
0 <
Number of outcomes in favour of A Total num ber of possible outcom es
P (A) < 1
When P(A) = 1, then A is called as sure event e.g. Probability of getting Tuesday after Monday
Solid Shapes Solid Shapes
Types of Solids (a) Prism : A solid whose base and top are identical polygons and side faces are rectangles, is called prism.
(b) Pyramid : A solid whose base is any polygon and side faces are triangles, all of which meets at the top to form a vertex is called a pyramid. Figures shows a pentagonal pyramid.
(c) Sphere : Sphere is a solid whose every point is equidistant from a fixed point. Figure shows the sphere.
Objects that occupy space and have three dimensions [length, breath and heigh or depth]
2 - D Representation of a 3- D figure Solids
Polyhedron A solid which is made up of polygonal regions called faces is called a polyhedron. (a) Convex polyhedrons :The idea of convex polyhedrons comes from convex polygon. A convex polyhedron is one whose all faces are convex polygons. (b) Regular polyhedron : A polyhedron is regular if all its faces are regular polygons and same number of faces meet at each vertex.
View of 3-D Shapes Top
Euler's formula V+F–E=2 Vertex Face Edges eg : For Triangular pyramid V=4;E=6;F=4 \ 4+4-6=2
Side Front
Object
Vertex Front view Edge
NETS
Side view (from right)
Top view
Nets