Vedanta Excel in Mathematics Book 9 Final (2078)

Example 3: Factorise: a) x2 + 7x + 12 b) 2x2 – 5x + 2 Algebraic Expressions
c) 2x2 – x – 6

Solution: 12 × 1 = 12
4 × 3 = 12
x2 x2 x x x x x + 3
a) x2 + 7x + 12 = x2 + (4 + 3) x + 12 x x
x
= x2 + 4x + 3x + 12 2x xx 12 34
x–2x2 x2 x 56 78
= x (x + 4) + 3 (x + 4) xx 9 101112

= (x + 4) (x + 3) x+4

2×2=4 2x – 1
4×1=4
x2 x2 x
b) 2x2 – 5x + 2 = 2x2 – (4 + 1) x + 2 2 xx xx 21
= 2x2 – 4x – x + 2
= 2x (x – 2) – 1 (x – 2) 2x – 1
= (x – 2) (2x – 1) x–2

2 × 6 = 12 2x + 3 2x + 3
4 × 3 = 12 x2 x2 x x x
x x(x–2) x(x–2)
c) 2x2 – x – 6 = 2x2 – (4 – 3) x – 6 x–2 2 xx xx
1 23
x–2 4 56
x–2
= 2x2 – 4x + 3x – 6 x–2
xx3

= 2x (x – 2) +3 (x – 2) 2x + 3

= (x – 2) (2x + 3) x(x–2) x(x–2) 3x–6

Example 4: Resolve into factors:

Solution: a) a2 – 3 + 2b2 b) 9 (x + y)2 + x + y – 8
b2 a2

a) a2 – 3 + 2b2 = a2 – 3. a . b + 2b2
b2 a2 b2 b a a2

= a2 – a . b – 2 ab . ab + 2b2
b2 b a a2

= a (ab – ab ) – 2b (ab – b )
b a a

=( – b ) ( a – 2b )
a b a

b) Let x + y = a

Then, 9 (x + y)2 + x + y – 8 = 9a2 + a – 8

= 9a2 + 9a – 8a – 8

= 9a (a + 1) – 8 (a + 1)

= (a + 1) (9a – 8)

= (x + y + 1) [9 (x + y) – 8]

= (x + y + 1) (9x + 9y – 8)

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Algebraic Expressions

Example 5: Factorise 2a6 – 19a3 + 24
Solution:
2a6 – 19a3 + 24 = 2a6 – 16a3 – 3a3 + 24

= 2a3 (a3 – 8) – 3 (a3 – 8)
= (a3 – 8) (2a3 – 3)
= (a3 – 23) (2a3 – 3)
= (a – 2) (a2 + 2a + 4) (2a3 – 3)
= (a – 2)(2a3 – 3) (a2 + 2a + 4)

EXERCISE 6.3
General section

1. In each of the following figures write the polynomial as the product of its factors.

a) x 3 b) x 3 c) x

x
xx

22 3

d) x x 1 e) x x 4 x
f) x

x x2 x2 x x
3 2
2 1
3 3

Creative section - A

2. Resolve into factors.

a) 8x3 + y3 b) 1 + 27a3 c) 128t3 – 2t d) x3y – 64y4

e) a6 + b3 f) 64x6y3 – 125 g) a6 – 64 h) 64x6 – y6
j) (x + 2)3 – 27 k) (x – y)3 – 8 (x + y)3
i) (a + b)3 + 1 l) p3 + 1
p3

m) a 3– b3 n) 27x3 – 30x2y + 40xy2 – 64y3 o) 8 – 6a – 9a2 + 27a3
b a

p) x4 + 1 q) p7 + 1
x2 p5

3. Factorise.

a) x2 + 4x + 3 b) x2 – 7x – 8 c) a2 – 27a + 180

d) 2x2 + 7x + 6 e) 3p2 – 7p – 6 f) 2x2 + 3xy – 5y2
h) 9a3bx + 12a2b2x – 5ab3x
g) 3a2 – 16ab + 13b2 k) 2 (x + y)2 + 9 (x + y) + 7 i) 12 a2 + a – 20
n) 2x6 + 17x3 + 8 b2 b
x2 3y2
j) y2 – 2 – x2 l) 3 (x – y)2 – 10 (x – y) + 8

m) 8a4 – 14a2 – 9 o) 3x6 – 79x3 – 54

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Algebraic Expressions

Creative section - B

4. Simplify.

a) a2 – 4 b) x2 x2 – 9 15 c) 4a2 – 1
a2 – a – 6 + 2x – 2a2 – 3a – 2

d) a2 – 9 e) x3 + 3x2 – 4x f) 2x2 + 3x – 2
a3 + 27 x4 – x 2x2 – 5x + 2

g) a3 – a2b + ab2 – b3 h) (x2 – x – 6) (x2 + 4x + 3) i) (x2 – 16) (x + 2)
a4 – b4 (x2 – 9) (x + 1) (x2 – 2x – 8) (x2 + 5x + 4)

5. a) The area of a rectangular field is (x2 + 8x + 15) sq. m.
(i) Find the length and breadth of the field.
(ii) Find the perimeter of the field.

b) The area of a rectangular plot of land is (x2 + 13x + 40) sq. m.

(i) Find the length and breadth of the land.

(ii) If the length and breadth of the land are reduced by 2/2 m respectively, find the
new area of the land.

c) A rectangular ground has area (2x2 + 11x + 12) sq. m. If the length of the ground
is decreased by 2 m and the breadth is increased by 2 m, find the new area of the
ground. (Take a longer side of the rectangles as length)

Project work

6. a) Write three different expressions of your own each in the form of a3 + b3 and a3 – b3.
Then, factorise your expressions.

b) Write an expression in each of the following forms, then, factorise your expressions

(i) x2 + ax + b (ii) x2 – ax – b (iii) x2 + ax – b (iv) x2 – ax + b

c) Write an expression in each of the following forms, factorise your expressions.

(i) ax2 + bx + c (ii) ax2 – bx – c (iii) ax2 + bx – c (iv) ax2 – bx + c

6.2 Highest Common Factor (H.C.F.) of algebraic expressions

To find the H.C.F. of monomial expressions, at first should find the H.C.F. of the
numerical coefficients. Then, the common variable with the least power is taken as
the H.C.F. of the monomial expressions.

To find the H.C.F. of the polynomial expressions, they are to be factorised and a
common factor or the product of common factors is obtained as their H.C.F.

Worked-out examples

Example 1: Find the H.C.F. of a2 – 9 and a2 + 4a – 21.

Solution:

1st expression = a2 – 9 a2 – 9 a2 + 4a – 21
2nd expression = a2 – 32 = (a + 3) (a – 3)
= a2 + 4a – 21 (a + 3) (a – 3) (a + 7)
? H.C.F. = a2 + 7a – 3a – 21
= a (a + 7) – 3 (a + 7)
= (a + 7) (a – 3)
= (a – 3)

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Algebraic Expressions

6.3 Lowest Common Multiple (L.C.M.) of algebraic expressions

The L.C.M. of monomial expressions is the common variable with the highest power.
To find the L.C.M. of the given polynomial expressions, they are to be factorised and
the product of common factors and the factors which are not common is taken as their
L.C.M.

Example 2: Find the L.C.M. of x2 – 25, x3 – 125 and x2 – 8x + 15

Solution:

1st expression = x2 – 25 = a2 – 52 = (a + 5) (a – 5)

2nd expression = x3 – 125 = x3 – 53 = (x – 5) (x2 + 5x + 25)

3rd expression = x2 – 8x + 15

= x2 – 5x – 3x + 15

= x (x – 5) – 3 (x – 5) = (x – 5) (x – 3)

? L.C.M. = (x – 5) (x + 5) (x – 3) (x2 + 5x + 25)

Example 3: Find the L.C.M. of a2 – 3a + 2, a2 – 5a + 6 and a2 – 4a + 3

Solution:

1st expression = a2 – 3a + 2 = a2 – 2a – a + 2 = a (a – 2) – 1 (a – 2) = (a – 2) (a – 1)

2nd expression = a2 – 5a + 6 = a2 – 3a – 2a + 6 = a (a – 3) – 2 (a – 3) = (a – 3) (a – 2)

3rd expression = a2 – 4 + 3 = a2 – 3a – a + 3 = a (a – 3) –1 (a – 3) = (a – 3) (a – 1)

? L.C.M. = (a – 1) (a – 2) (a – 3)

6.4 Simplification of rational expressions 3x , 25y , a + b,
7
Rational expressions can be expressed in the form of , where q ≠ 0.

x+y , etc. are a few examples of rational expressions.
x–y

When we multiply or divide rational expressions, the numerators and denominators
of each expression are factorised (if necessary). Then we simplify the expressions to
the simplest forms.

Worked-out examples

Example 1: Simplify: a) a2 + 3a +2 × a2 – 9 6 b) 4x2 – 81y2 ÷ 2x – 9y
Solution: a2 + a –6 a2 – a – 1 – 4a2 a – 2a2

a) a2 + 3a + 2 × a2 – 9 6 = a2 + 2a + a + 2 × a2 – a2 – 32 – 6
a2 + a – 6 a2 – a – a2 + 3a – 2a – 6 3a + 2a

= a (a + 2) + 1 (a + 2) × (a + 3) (a – 3)
a (a + 3) – 2 (a + 3) a (a – 3) + 2 (a – 3)

= (a + 2) (a + 1) × (a + 3) (a – 3)
(a + 3) (a – 2) (a – 3) (a + 2)

= a+1
a–2

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Algebraic Expressions

b) 4x2 – 81y2 ÷ 2x – 9y = 4x2 – 81y2 × a – 2a2 = (2x)2 – (9y)2 × a (1 – 2a)
1 – 4a2 a – 2a2 1 – 4a2 2x – 9y 12 – (2a)2 2x – 9y

= (2x + 9y) (2x – 9y) × a (1 – 2a)
(1 + 2a) (1 – 2a) 2x – 9y

= a (2x + 9y)
1 + 2a

When we add or subtract rational expressions with unlike denominators, we should find the
L.C.M. of denominators. Then, the simplification is performed as like the simplification of
fractions.

Example 2: Simplify: a) x x 2 – x x 2 b) x2 + ax + a2 + x2 – ax + a2
Solution: + – x+a x–a

a) x x 2 – x = x (x – 2) – x (x + 2) = x2 – 2x – x2 – 2x = –4x
+ x–2 (x + 2) (x – 2) x2 – 4 x2 – 4

b) x2 + ax + a2 + x2 – ax + a2 = (x – a) (x2 + ax + a2) + (x + a) (x2 – ax + a2)
x+a x–a (x + a) (x – a)

= x3 – a3 + x3 + a3 = 2x3
x2 – a2 x2 – a2

Example 3: Simplify: x–1 + (x + 3 – 1) – (1 – 1 – 2x)
Solution: (2x – 1) (x + 2) 2) (x x) (1

x–1 + 3 1) – (1 – 1 – 2x)
(2x – 1) (x + 2) (x + 2) (x – x) (1

= x–1 + 3 – 1
(2x – 1) (x + 2) (x + 2) (x – 1) [– (x – 1)] [– (2x – 1)]

= (x – 1) (x – 1) + 3 (2x – 1) – (x + 2)
(2x – 1) (x + 2) (x – 1)

= x2 – 2x + 1 + 6x – 3 – x – 2
(2x – 1) (x + 2) (x – 1)

= x2 + 3x – 4
(2x – 1) (x + 2) (x – 1)

= (2x x2 + 4x – x – 4 1) = x (x + 4) – 1 (x + 4) = (2x (x + 4) (x – 1) 1) = (2x x+4 2)
– 1) (x + 2) (x – (2x – 1) (x + 2) (x – 1) – 1) (x + 2) (x – – 1) (x +

Example 4: Simplify: p–1 + p2 p–2 6 + p2 p– 5 15
Solution: p2 – 3p + 2 – 5p + – 8p +

p–1 + p2 p–2 6 + p2 p–5 15
p2 – 3p + 2 – 5p + – 8p +

= p2 – p–1 + 2 + p2 – p – 2 + 6 + p2 – p–5 + 15
2p – p 3p – 2p 5p – 3p

=p (p – p – 1 (p – 2) + p (p – p – 2 (p – 3) + p (p – p – 5 (p – 5)
2) – p 3) – 2 5) – 3

= (p p–1 + p–2 + (p – p–5 – 3)
– 2) (p – 1) (p – 3) (p – 2) 5) (p

= p 1 2 + p 1 3 + p 1 3 = p – 3+ p–2 +p – 2 = (p 3p –7 3)
– – – (p – 2) (p – 3) – 2) (p –

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Algebraic Expressions

Example 5: Simplify: m+n + m2 m–n n2 – m4 + 2n3 + n4
Solution: m2 + mn + n2 – mn + m2n2

m2 m+n n2 + m2 m–n n2 – m4 + 2n3 + n4
+ mn + – mn + m2n2

= (m + n) (m2 – mn + n2) + (m – n) (m2 + mn + n2) – (m2 + n2)2 2n3 + m2n2
(m2 + mn + n2) (m2 – mn + n2) – 2m2n2

= m3 + n3 + m3 – n3 – 2n3
(m2 + mn + n2) (m2 – mn + n2) (m2 + n2)2 – (mn)2

= (m2 + mn + 2m3 – mn + n2) – (m2 + mn + 2n3 – mn + n2)
n2) (m2 n2) (m2

= (m2 + mn 2m3 – 2n3 mn + n2)
+ n2) (m2 –

= 2 (m – n) (m2 + mn + n2)
(m2 + mn + n2) (m2 – mn + n2)

= 2( m– n)
m2 – mn + n2

Example 6: Simplify: (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
Solution: a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

(a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

= (a – b + c) (a – b – ) + (b – c + a) (b – c – a) + (c – a + b) (c – a – b)
(a + b + c) (a – b – c) (b + c + a) (b – c – a) (c + a + b) (c – a – b)

= a–b+c + b–c+a + c–a+b
a+b+c a+b+c a+b+c

= a – b + c +b–c+a+ c – a + b
a+b+c

= a + b + c = 1
a + b + c

EXERCISE 6.4
General section

1. Find the H.C.F. and the L.C.M. of the following factors:
a) (x + 2) (x – 1) and (x + 1) (x + 2) b) (a – 5) (a – 4) and (a + 4) (a – 5)
c) (x + y) (2x – y) and (2x – y) (x – y) d) (a + b) (a2 – ab + b2) and (a – b) (a + b)

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Algebraic Expressions

Creative section - A
2. Find the H.C.F.

a) ax2 + 2ax, ax3 – 4ax b) x2 – 1, x3 – 1

c) a3 – 8, a2 + 2a + 4 d) (x – 1)2, x2 – 3x + 2

e) x2 + 5x + 6, x2 – 4, x3 + 8 f) 2x2 + 5x + 2, 3x2 + 8x + 4, 2x2 + 3x – 2

g) m3 – 1, m4 + m2 + 1, m2 + m + 1 h) 2x3 – x2 – x, 4x3 – x, 8x4 + x

i) x3 – y3, x6 – y6, x4 + x2y2 + y4 j) a3 + b3, a4 + a2b2 + b4, a3 – a2b + ab2

3. Find the L.C.M.

a) (x + 1) (x + 2), (x + 2) (x – 2) b) (x – 1) (x – 2), (x – 2) (x – 3), (x – 3) (x – 1)

c) x2 – 1, x3 – 1 d) x2 – 9, 3x + 9

e) 4x2 – 25, 2x2 – x – 15 f) x4 + x2y2 + y4, x3 – y3

g) 2x2 – 3x – 9, 4x2 – 5x – 21, x3 – 9x h) 8x3 + y3, 8x3 – y3, 16x4 + 4x2y2 + y4

i) x2 – 5x + 6, x2 – 4x + 3, x2 – 3x + 2 j) x3 + 27, 2x3 – 6x2 + 18x, x2 – 3x + 9

4. Simplify:

a) x2 – a2 × 7a b) 4y2 – 9z2 × y– 2 c) 4a2 – 9b2 × x2y + xy2
ax + a2 x–a y2 – 4 2y – 3z x2 – y2 4a – 6b

d) a2 – b2 ÷ a2b – ab2 e) 2x + 6 ÷ 3x2 + 9x f) a2 – 4b2 ÷ a + 2b
a2b + ab2 a2b2 x2 – 9 2x2 – 6x a2 – 9x2 a – 3x

5. Simplify:

a) a2 b + b2 b) p 4 + 4 2 c) x2 x y2 + y2 y x2
a– b–a p2 – – p2 – –

d) 1 1 + 1 1 e) 1 – 1 f) y 1 3 – 2 1
2a + 2a – x–2 x–1 – 2y –

g) a 1 b – a+b h) p+q + q 1 p i) x3 + y3
– a2 – b2 p2 – q2 – x–y y–x

j) a–b + b–c + c–a k) x+2 – x–2 l) x2y – xy2
ab bc ca x–2 x+2 x–y x+y

m) y2 1 y) – x2 1 y) n) a2 + 2a – 1 o) 1 + 3
(x – (x – a–1 1–a y2 – 4 y2 + 5y + 6

p) a2 + ab + b2 + a2 – ab + b2 q) x2 + 2x + 4 + x2 – 2x + 4 r) (a 1 – a2 1 b2
a+b a–b x+2 x–2 – b)2 –

s) x+2 2 + 3 1 t) x–2 – x+ 1 u) x–2 – x2 x+1
x2 + x – x2 – x2 + 4x + 4 x2 – 4 x2 – 1 – 2x + 1

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Algebraic Expressions

Creative section - B
6. Simplify:

a) 1 – x 2 1 + x 1 2 b) x 1 1 – x 1 1 + 3 1
x + + + – x2 –

c) x x 2 + x – 4x d) 2xy – x–y + x+y
+ x–2 x2 – 4 x2 – y2 x+y x–y

e) x+1 + x–1 – 1 4 f) 2a 1 2x – 2a 1 2x – a2 x x2
2x3 – 4x2 2x3 + 4x2 x2 – – + +

7. Simplify:

a) 1 + 1 + 1
(x – 3) (x – 4) (x – 4) (x – 5) (x – 5) (x – 3)

b) ( 2 (a – 3) 5) + (3 – a–1 – 4) + (5 – a–2 – 3)
– 4) (a – a) (a a) (a

c) x2 – 2 + 6 – x2 – 2 + 3 + x2 – 1 + 2
5x 4x 3x

d) x2 x–1 2 + x2 x–2 6 + x2 x– 5 15
– 3x + – 5x + – 8x +

e) 2y + 5 + 11 – 16y
y2 + 6y + 9 y2 – 9 8y2 – 24y

8. Simplify:

a) x–y + x2 x + y y2 + 2y3
x2 – xy + y2 + xy + x4 + x2y2 + y4

b) a2 a–2 4 + a2 a +2 4 – a4 + 16 + 16
– 2a + + 2a + 4a2

c) x+3 9 + x–3 9 – 54 81
x2 + 3x + x2 – 3x + x4 + 9x2 +

d) 1 a +2 a2 – a–2 – 1 2a2 a4
+ a+ 1 – a + a2 + a2 +

9. Simplify:

a) (x – y)2 – z2 + (y – z)2 – x2 + (z – x)2 – y2
x2 – (y + z)2 y2 – (z + x)2 z2 – (x + y)2

b) a2 – (b – c)2 + b2 – (c – a)2 + c2 – (a – b)2
(c + a)2 – b2 (a + b)2 – c2 (b + c)2 – a2

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Unit Indices

7

7.1 Indices – review

An index is a number that tells how many times a base is multiplied by itself. An
index is also called an exponent.

Let's take a few examples of repetitive multiplication of the same base.

a = a1, a × a = a2, a × a × a = a3, a × a × a × a = a4, etc.

Here, 1 is the index of a1, 2 is the index of a2, 3 is the index of a3, 4 is rthe index of a4,
and so on.

Thus, the index refers to the power to which a number is raised. For example, in
23 the base 2 is raised to the power 3. Indices is the plural of index.

7.2 Laws of Indices

While performing the various operations of indices, we apply different proven rules
like product rule, quotient rule, power rule, etc. These rules are well known as laws
of indices.

The table given below shows the laws of indices at a glance.

Name of laws Rules Examples
Product law am × an = am + n 23 × 25 = 23+5 = 28

Quotient law am ÷ an = am – n when m > n 37 ÷ 33 = 37 – 3 = 34 and
am ÷ an = 1 when m < n 11

an – m 33 ÷ 37 = 37 – 3 = 34

Power law (am)n = am × n, (ab)m = ambm, (24)2 = 24 × 2 = 28,
a am (2x)5 = 25 × x5, …
Negative index law b m = bm
Zero index law
Root laws of indices 1 1 5–3 = 1 or 53 = 1
a– m = am or am = a–m 53 5–3

a0 = 1, (ab)0 = 1, (a + b)0 = 1 20 = 1, 30 = 1, 990 = 1

1 = 2 31 or 3,

m n am or n am m 32

an = = an 2 = 3 32

33

(i) Product law of indices

If am and an are the two algebraic terms, where m and n are the positive integers,
then am × an = am + n

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Indices

Proof

We know that,

a2 = a × a (two factors)

a3 = a × a × a (three factors)

am = a × a × a × ... ('m' factors)

an = a × a × a × ... ('n' factors)

? am × an = (a × a × a × ... 'm' factors) × (a ×a × a × ... 'n' factors)
= a × a × a × ... ('m + n' factors)
= am + n

Thus, am × an = am + n
Similarly, if m, n, p, q, r, ... are the positive integers, then
am × an × ap × aq × ar × ... = am + n + p + q + r + ...

(ii) Quotient law of indices

If am and an are the two algebraic terms, where m and n are the positive integers,
then am ÷ an = am – n when m > n

am ÷ an = 1 m when n > m
an –
Proof
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors

= a × a × a × ... ('m – n' factors), where m > n

= am – n

Thus, am ÷ an = am – n, when m > n

But, in the case of n > m,
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors

= a ×a × a × 1 – m' factors
... 'n

= 1 m
an –

Thus, am ÷ an = 1 m, when n > m
an –

(iii) Power law of indices

If am be an algebraic term, then (am)n = am × n = amn, where m and n are the

positive integers.

Proof

(am)n = am × am × am × ... (n) factors

= am + m + m + ... 'n' times 'm'

= amn

Thus, (am)n = amn

cor 1. amn = (am)n cor 2. amn = (an)m

cor 3. am = am cor 4. a m = am
bm b b bm

cor 5. (ab)m = ambm cor 6. ambm = (ab)m

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Indices

(iv) Law of negative index

If a–m is an algebraic term, where m is a negative integer,

a–m = 1 or, 1 = a–m or, am = 1
am am a–m

Proof

Here, a–m = am – 2m = am ÷ a2m = am = am = 1
a2m am × am am

Thus, a–m = 1
am

Similarly, 1 = a–m and am = 1
am a–m
(v) Law of zero index

If a0 is an algebraic term, where a ≠ 0, then a0 = 1.

Proof am
am
Here, a0 = am – m = am × a–m = =1

Thus, a0 = 1, where a ≠ 0

(vi) Root law of indices

If amn is an algebraic term, where m and n are the positive integers, then

amn = n am

Proof n

nth order of root in represented as 2

In this way the 2nd order of root is represented as or only
3
The 3rd order of root is represented as and so on.

The square root of 32 = 3 = 322 = 2 32 or 32

The cube root of 53 = 5 = 533 = 3 53

Thus, 322 = 2 32

3 = 3 53

53

In general, amn = n am

Worked-out examples

Example 1: Find the value of: 243 – 25
32
Solution: a) (16)43 b) (9–3)61 c) d) (160.5)32 e) 3 729 f) 3 64–1

a) (16)43 = (24) 34 = (2)4 × 34 = 23 = 8

b) (9–3)16 = (32)–3 × 61 = 3–6 × 16 = 3–1= 1
3

c) 243 – 25 = 35 – 52 = 25 52 = 2 =5 × 25 2 2= 4
32 25 35 3 3 9

d) (160.5)32 = (24)0.5 × 23 = 24 × 0.5 × 23 = 23 = 8

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Indices

e) 3 729 = 3 93 = 933 = 9 = 32 = 3

1 1 2= 3 1 3 1 3 1
64 8 8 2 2
f) 3 64–1 = 3 = 3 = =

Example 2: Evaluate: a) 25 – 12 125 31 ÷ 16 – 14 b) 127 × 286
16 64 81 217 × 166
1 1 1
Solution: c) (1 – 6–9)–1 + (1 – 69)–1 d) (a – b)–1 + (b – c)–1 + (c – a)–1

a) 25 – 21 125 31 ÷ 16 – 41
16 64 81

= 16 21 125 13 ÷ 81 14
25 64 16

= 4 2 × 12 5 3× 13 ÷ 3 4× 14
5 4 2

= 4 5 ÷ 3 = 4 5 × 2 = 4 × 5 = 2
5 4 2 5 4 3 5 6 3

b) 127 × 286 = (22 × 3)7 × (22 × 7)6 = 214 × 37 × 212 × 76 = 214 + 12 × 76 = 226 – 24 = 4
217 × 166 (3 × 7)7 × (24)6 37 × 77 × 224 224 × 77 27 – 6 7

c) (1 – 6–9)–1 + (1 – 69)–1

= 1 – 1 –1 + (1 – 69)–1
69

= 69 – 1 –1 + (1 – 69)–1 = 69 + 1 = 69 – 1 = 69 – 1 = 1
69 69 – – 69 69 – 69 – 69 – 1
1 1 1 1

d) (a 1 + (b 1 + (c 1 = (a – b) + (b – c) + (c – a) = 0
– b)–1 – c)–1 – a)–1

Example 3: Prove that a) 7m + 2 + 4 × 7m =1 b) 5x – 5x – 1 =1
7m + 1 × 8 – 3 × 7m 4 × 5x – 1

c) 273n + 1 × (243)–45n = 1 d) 3–p × 92p – 2 = 2
9n + 5 × 33n – 7 33p – 2 × (3 × 2)–1 3

Solution: 7m + 2 + 4 × 7m 7m × 72 + 4 × 7m 7m (49 + 4) 53
7m + 1 × 8 – 3 × 7m 7m × 71 × 8 – 3 × 7m 7m (7 × 8 – 3) 53
a) LHS = = = = = 1 = RHS

5x – 5x – 1 5x – 5x 5x 1 – 1 4
4 × 5x – 1 4× 5 5 5
b) LHS = = 5x = 5x = 4 = 1= RHS
c) LHS 5 4 × 5
5

= 273n + 1 × (243)–45n = (33)3n + 1 × (35)–45n = 39n + 3 × 3–4n
9n + 5 × 33n – 7 (32)n + 5 × 33n – 7 32n + 10 × 33n – 7

= 39n + 3 – 4n = 35n + 3 = 35n + 3 – (5n + 3) = 30 = 1 = RHS
32n + 10 + 3n – 7 35n + 3

3–p × 92p – 2
d) LHS = 33p – 2 × (3 × 2)–1

3–p × (32)2p – 2 3–p + 4p – 4 33p – 4 33p – 4 – 3p + 3 3–1 2
= 33p – 2 × 3–1 × 2–1 = 33p – 2 –1 × 2–1 = 33p – 3 × 2–1 = 2–1 = 2–1 = 3 = RHS

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Example 4: Simplify:

Solution: a) 3 50ba5c–2 × 3 20ab5c8 b) 3 27x3y6 ÷ 4 81x4y8 c) 3 (a + b)–8 × (a + b)23

a) 3 50ba5c–2 × 3 20ab5c8 = 3 1000a5+1 b1+ 5 c–2 + 8 = 3 1000a6 b6 c6 = (103a6b6c6)31 = 10a2b2c2

b) 3 27x3y6 ÷ 4 81x4y8 = (33x3y6)31 ÷ (34x4y8)14 = 3xy2 ÷ 3xy2 = 1.

c) 3 (a + b)–8 × (a + b)32 = (a + b)– 38 × (a + b)23 = (a + b)– 83 + 32 = (a + b) – 83+ 2
1
= (a + b)– 63 = (a + b)–2 = (a + b)2

Example 5: Simplify: xa a2 + ab + b2 × b b2 + bc + c2 c c2 + ca + a2
Solution: xb a
xx × xxc

xa a2 + ab + b2 × xb b2 + bc + c2 × xc c2 + ca + a2 = (xa – b)a2 + ab + b2 × (xb – c)b2 + bc + c2 × (xc )– a c2 + ca + a2
xb xc xa

= xa3 – b3 × xb3 – c3 × xc3 – a3

= xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1

Example 6: Simplify: a+b xa2 × b+c xb2 × c+a xc2
xb2 xc2 xa2
Solution:
a2 – b2 b2 – c2 c2 – a2
xa2 xb2 xc2 = x a+b × x b+c × x c+a
a+b xb2 × b+c xc2 × c+a xa2

= xa – b × xb – c × xc – a = xa – b + b – c + c – a = x0 = 1

111
Example 7 : Simplify: 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a
Solution:

111
1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a

= 1 + 1 + 1

1+ xa + xc 1+ xb + xa 1+ xc + xb
xb xb xc xc xa xa
1 1 1
= xb + xa + xc + xc + xb + xa + xa + xc + xb

xb xc xa

= xb + xc + xa = xb + xc + xa =1
xa + xb + xc xa + xb + xc xa + xb + xc xa + xb + xc

Example 8 : Simplify m + (mn2)31 + (m2n)31 × 1 – n31
m–n m13
Solution:

m + (mn2)31 + (m2n)31 × 1 – n13 = m +m31 n23 + m23 n13 × m31 – n31
m– n m31 m–n m13

= m31 (m23 + n23 + m31 n31 ) × m13 – n31
m– n m31

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Indices

= (m31 – n13) [(m31)2 + m13.n31+ (n31)2] = (m31)3 – (n31)3 = m–n =1
m–n m – n m–n

x2 – 1 x× x – 1 y–x
y2 y
Example 9 : Show that = x x+y
Solution: 1 y× 1 x–y y
y2 – x2 y + x

x2 – 1 x× x – 1 y–x x + 1 x x – 1 x x – 1 y–x
y2 y y y y
Here, LHS = =
1 y× 1 x–y 1 y 1 y 1 x–y
y2 – x2 y + x y + x y – x y+ x

x + 1 x x – 1 x+y–x x + 1 x x – 1 y
y y y y
= =
1 y+x–y 1 y 1 x 1 y
y + x y – x y + x y – x

x + 1 x x – 1 y xy + 1 x xy – 1 y
y y yy
= ×
y + 1 y – 1 = xy + 1 × xy – 1
x x xx

= xy + 1 × xy x 1 x× xy – 1 × x 1 y= x x× x y= x x + y = RHS
y + y xy – y y y

111
Example 10 : If pqr = 1, prove that 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1 = 1.
Solution:
1
Here, pqr = 1, then qr = p = p–1

111
Now, LHS = 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1
qr r 1
= qr(1 + p + q–1) + r(1 + q + r–1) + 1 + r + p–1

qr r 1
= qr + pqr + r + r + qr + 1 + 1 + r + qr

qr r 1 qr + 1 + r
= qr + 1 + r + qr + r + 1 + qr + r + 1 = qr + 1 + r = 1 = RHS

Example 11: If x – 1 = 223 + 231 , show that x3 – 3x2 – 3x = 1.

Solution:
Here, x – 1 = 223 + 231
21

or, (x – 1)3 = (23 + 23 )3 [Cubing on both sides]
3 231 3 + 3. 223 . 231
or, x3 – 3x2 + 3x – 1 = 223 223 + 231
+

or, x3 – 3x2 + 3x – 1 = 4 + 2 + 6 (x – 1)

or, x3 – 3x2 + 3x – 1 = 6 + 6x – 6

or, x3 – 3x2 + 3x – 6x = 1

or, x3 – 3x2 – 3x = 1

Hence, x3 – 3x2 – 3x = 1 proved.

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EXERCISE 7.1

General section

1. a) Express am × an as a single base.

b) What is the value of (5a)0, a ≠ 0?

c) Find the value of am – n × an – m.

d) What is the value of (p + q)0 + 1p + q?

2. Evaluate:

a) 29 × 2-6 b) 5–7 × 58 c) 11–5 ÷ 11–3 d) (25) 3
2

e) (64)– 2 f) 1 1 g) 8 – 4 h) 169 – 1
3 64 6 3 2
27 196

i) (32–1)5–1 j) a0 – 2 k) (70.5)2 l) 9 0.5 × 32 0.2
125 3 25 243

m) 3 64–1 729 –1 3 1
64 4 100
n) 3 o) 100 × 4 p) 3 9 3 9 9

Creative section - A

3. Find the value of:

a) 8 – 1 ÷ 4 – 1 b) 125 – 2 ÷ 625 – 1
27 3 9 2 64 3 256 2

c) 27 – 13 81 41 ÷ 4 – 21 d) 25 – 12 125 31 ÷ 8 – 31
8 16 25 16 64 27

4. Simplify: 1 2
3 3
a) (8a3 ÷ 27x–3)– b) (125p3 ÷ 64q–3)– c) 146 × 155 d) 409 × 498
356 × 65 569 × 358

5. Simplify.

a) (xa)b – c × (xb)c – a × (xc)a – b b) (ax + y)x – y × (ay+z)y – z × (az – x)z + x

c) x2a + 3b × x3a – 4b d) 1 1 + 1 1
xa + 2b × x4a – 3b + ax – y + ay – x

e) 1 – 1 n + 1 – 1 m f) (1 – 3–5)–1 + (1 – 35)–1
xm – xn –
y–1 x–1 –1
g) (a + b)–1.(a–1 + b–1) h) x–1 + y–1

6. Show:

a) 3x + 1 + 3x = 1 b) 5n + 2 – 5n =1 c) 72p +1 –3× 49p = 1
4 × 3x 24 × 5n × 49p
4

d) 6m + 2 – 6m = 7 e) 7n + 2 + 4 × 7n = 1 f) 5a + 3 – 55 × 5a – 1 = 1
6m+1 – 6m 7n + 1 × 8 – 3 × 7n 5a + 2 + 89 × 5a

7. Simplify:

a) 3p – 3p –1 b) 5x – 5x – 1 c) 3 5 × 2m – 4 × 2m – 2 1
3p + 1 + 4× 5x – 1 × 2m + 2 – 5 × 2m +
3p

d) 2n +2 × (2n – 1)n + 1 ÷ 4n e) 5–n × 625n –1 f) 9x × 3x – 1 – 3x
2n(n – 1) 53n – 2 × (5 × 32x + 1 × 3x – 2 – 3x
2)–1

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8. Simplify: b) a6b–2c4 ÷ 4 a4b–4c8
a) 25a2b2 × 3 27a3
d) 3 56p7q4
c) 4 16x8y4 ÷ 3 8x6y3 3 7p4q7

e) 4 216m7n5 ÷ 4 6–1m–1n f) 3 (a + b)–7 1

× (a + b) 3

g) 3 (2x – y)–8 ÷ (2x – y)– 2 h) (a + b)–1 × (a – b) (a2 – b2)
3

9. Simplify:

a) xa a+b × xb b+c × xc c + a b) ax x–y × ay y–z × az z – x
xb xc xa a–y a–z a–x

c) xa ×a2 + ab + b2 xb ×b2 + bc + c2 xc d)c2 + ca + a2 xl2+m2 l – m× xm2+n2 m–n× xn2+l2 n–l
xb xc xa x–lm x–mn x–nl

e) xa + b c–a × xb + c a–b × xc + a b – c f) xm + n m–n× xn + p n–p × xp + m p – m
xa – b xb – c xc – a xp xm xn

11 1 1 1 1
p + (pq2) 3 + (p2q) 3 q 3 a–c
g) p– q × 1 – h) b b–a × c c–b × a
p 1
3 xb–c xc–a xa–b

x+ 1 a× 1 – x a a2 – 1 a× a – 1 b–a
y y b2 b
i) j)
1 a× 1 a 1 b× 1 a–b
y+ x x – y b2 – a2 b + a

10. Simplify:

a) yz ay × zx az × xy ax b) pq xr – p × pr xq – r × qr xp – q
az ax ay xr – q xq – p xp – r
1 1 1
c) x+y ax2 × y+z ay2 × z+x az2 d) xb xc xa ab
ay2 az2 ax2 bc × ca ×
xc xa xb
111
e) 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a

111
f) 1 + ax – y + ax – z + 1 + ay – z + ay – x + 1 + az – x + az – y

Creative section - B
11. a) If a3 + b3 + c3 = 0, prove that (xa + b)a2 – ab + b2 × (xb + c)b2 – bc + c2 × (xc + a)c2 – ca + a2 = 1

b) If a = xq + r.yp, b = xr + p.yq and c = xp + q.yr, prove that aq – r × br – p × cp – q = 1

c) If xyz = 1, prove that 1 + 1 + 1 = 1.
1 + x + y–1 1 + y + z–1 1 + z + x–1
1 1 1
d) If a + b + c = 0, prove that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1.

12. a) If x = 1 + 2– 1 , prove that 2x3 – 6x = 5.
3
23

b) If a = 1 – p– 1 , prove that a3 + 3a = p – 1 .
3 p
p3

12
c) If x – 2 = 3 3 + 3 3 , show that x(x2 – 6x + 3) = 2.

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Indices

7.3 Exponential equation

Let's take any two equations: 2x = 4 and 2x = 4.

In the equation 2x = 4, variable x is a base. It is a linear equation. However, in 2x = 4,

variable x is an exponent of the base 2. Such an equation in which variable appears

as an exponent of a base is known as exponential equation.

To solve an exponential equation, we need to have equations with the same base on

either side of the 'equal' sign. Then, we compare the powers of equal base and solve

the equation.

Let's study, the following axioms which are used in solving the exponential equations.

(i) If ax = ab, then x = b (ii) If ax = 1, then ax = a0 and x = 0

In this way, while solving an exponential equation, we should simplify the equation
till the equation is obtained in the form ax = ab or ax = 1.

Worked-out examples

Example 1: Solve: a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
Solution: 25

a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
or, 2x = 25 25 or, 7x – 2 = 70
? x=5 1 or, x – 2 = 0
or, 52x = 52 ? x=2

or, 52x = 5–2

or, 2x = – 2

or, x = – 1

Example 2: Solve: a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 =10

Solution:

a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 = 10

x–3 x+1 or, 3x + 3x × 32 = 10
or, 2 2 = 2 4

or, x – 3 = x + 1 or, 3x (1 + 9) = 10
2 4
or, 4x – 12 = 2x + 2 or, 3x = 10 =1
10
or, 2x = 14 or, 3x = 30

or, x = 7 or, x = 0

Example 3: Solve: a) 2x + 1 × 3x – 2 = 48 b) 9y + 1 = 32y + 1 + 54

Solution: b) 9y + 1 = 32y + 1 + 54
a) 2x + 1 × 3x – 2 = 48
or, 32y + 2 – 32y + 1 = 54
or, 2x × 21 × 3x × 3–2 = 48 or, 32y × 32 – 32y × 3 = 54
or, 32y (32 – 3) = 54
or, 2x × 3x × 2 = 48 or, 32y =9
32 or, 32y = 32
or, (2 × 3)x = 216 or, 2y =2

or, 6x = 63 ? y =1

? x =3

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Indices

Example 4: If ax = b, by = c and cz = a, prove that xyz = 1.

Solution: Another process:
by = c
Here, ax = b, by = c and cz = a
Now, ax = b or, axy = c [putting b = ax in by = c]
or, cxz = b [By putting a = cz in ax = b] or, cxyz = c [putting a = cz in axy = c]
or, bxyz = b [Putting c = by in cxz = b] or, xyz = 1
or, xyz = 1 proved.

Example 5: If 4p = 5q = 20–r, show that 1 = 1 + 1 = 0.
Solution: p q r

Another process:

Let, 4p = 5q = 20–r = k 4p = 5q, i.e. 4 = 5pq

Then, 4p = k ? 4 = kp1 20–r = 5q, i.e. 20 = 5– q
r

5q = k ? 5 = kq1 Now, 4 × 5 = 20
and 20–r = k 1 or, q q
or, = 5– r
? 20 = k r 5p × 5

Now, 4 × 5 = 20 5q +1 = 5– q
p r
11 1
or, = k– r q q
kp × kq p r
or, +1 = –
k1 + 1 k– 1
or, p q = r q q
p r
1 1 –1 or, +1 + =0
p q r
+ = or, q( 1 + 1 + 1 ) =0
p q r
? 1 + 1 + 1 = 0 Proved 1 1 1
p q r or, p + q + r =0

Example 6: Solve: 2x + 1 = 881 .
2x
Solution:
1 = 818
Here, 2x + 2x

or, 2x + 1 = 65
2x 8
1 65
Let 2x = a, then equation becomes a + a = 8

Now, a + 1 = 65
a 8
a2 + 1 65
or, a = 8

or, 8a2 + 8 = 65a

or, 8a2 – 65a + 8 = 0

or, 8a2 – 64a – a + 8 = 0

or, 8a(a – 8) –1(a – 8) = 0

or, (a – 8) (8a – 1) = 0

Either a – 8 = 0 i.e. a = 8 or, 2x = 23 ?x=3
1 1
or, 8a – 1 = 0 i.e. a = 8 or, 2x = 23 = 2–3 ? x = –3
Hence, x = ±3.

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Indices

EXERCISE 7.2

General section
1. a) If (ap × aq) ÷ ar = ax then express x in terms of p, q and r.

b) If (bm ÷ bn) × bp = by then express y in terms of m, n and p.
c) If (xm)n = xm × xn, express m in terms of n.
d) If ax = a , find the value of x.
e) If am = 1, is the value of m?
f) If xx = 4, what is the value of x?

2. Solve. b) 23x = 8 c) 32x = 1 d) 53x = 1
a) 2x = 4 9 125

e) mx – 2 = m2 f) 52x + 3 = 1 g) 1 = 64 h) 1 = 1
43x 7– 2x 49

i) 2 x 8 j) 4 2x= 16 –2 k) 22x + 1 = 23x – 1 l) 32x + 1 = 92x – 1
3 27 5 25
2=

m) 3 = 3x n) 2x – 2 = 2 × 82 o) 33x – 2 = 27 × 92 p) 53x–1 = 25 × 5x+1
3x 3

q) 42x – 1 – 2x + 1 = 0 r) 9x +1 = 1 s) 252x – 3 = 1 t) 49x +2 = 1
27x 625 343–2x
3. a) If px ÷ p4 = 1, find the value of x. b) If 3m ÷ 34 = 27, find the value of m.

c) If 5t = 0.04, find the value of t. d) If 103r = 1 , find the value of r.
0.001

Creative section - A

4. Solve.

a) 2x+5 = 16 b) (35)x – 1 = 25 c) ( 2)3x – 1 = ( 4)x – 2 d) ( 9)x – 3 = ( 3)x +2
f) (0.3)35x = 0.027
e) (0.5) x = 0.25 g) 2 1 2x = 2x h) 9 1 = 27–x
2 × × 32x

5. Solve.

a) 2x + 1 – 2x = 8 b) 3x + 1 – 3x = 54 c) 2x + 2x + 2 = 5

d) 7x + 7x + 1 = 56 e) 11x + 1 + 11x = 12 f) 3x + 2 + 3x + 1 = 1 1
3

g) 2x + 2x – 1 = 3 h) 3x – 3x – 2 = 8 i) 3x + 5 = 3x + 3 + 8
3

j) 2x + 2 + 2x + 3 = 1 k) 3x + 3 + 3x + 4 = 162 l) 5x + 5x + 1 + 5x + 2 = 155
2 3

m) 32x + 3 – 2.9x + 1 = 1 n) 9x – 2 + 2 × 32x – 3 = 63 o) 23x – 1 + 3 × 8x – 1 = 56
9

6. Solve.

a) 2x + 3 × 3x + 4 = 18 b) 2x + 3 × 3x + 2 = 432
c) 2x – 5 × 5x – 4 = 5
e) 23x – 5 × ax – 2 = 2x – 2 × a1 – x d) 72x + 1 × 52x – 1 = 7
5

f) 75x – 4 × a4x – 3 = 72x – 3 × ax – 2

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Indices

7. a) If a = bc, b = ca and c = ab, prove that abc = 1.

b) If xa = y, yb = z and zc = x, prove that abc = 1.
c) If a1x = b13 and ab = 1, prove that x + 3 = 0.

d) If a = 7x, b = 7y and aybx = 49, show that xy = 1.

e) If ax = by and ay = bx, show that x = y.

f) ax = by = cz and b2 = ac, show that y = 2xz
x+z

Creative section - B

8. a) If (a–1 + b–1) (a + b)–1 = ambn, prove that am – n = 1.

b) If m–1n2 7÷ m3n–5 –5 = mxny, prove that mx – 2y = 1.
m2n–4 m–2n3

9. Solve:

a) 3x + 1 = 9 1 b) 4x + 1 = 16 1
3x 9 4x 16

c) 4x – 6 × 2x + 1 + 32 = 0 d) 25x – 6 × 5x + 1 + 125 = 0

e) 2x + 21–x = 3 f) 3x – 1 + 32–x = 4

Project work

10. a) Take any base number and index number greater than 1 and less than 6. Then
verify the following laws of indices.

(i) Product law (ii) Quotient law

(iii) Power law (iv) Law of negative index

(v) Root law of indices (vi) Law of zero index

b) Write any five exponential equations of your own in the form of ax = b, where a is
positive integer greater than 1 and b is the integer of power of a. Then, solve your
equations.

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Unit Simultaneous Linear Equations

8

8.1 Simultaneous equations - review

Let’s take a linear equation, y = x + 4.
It is a first degree equation with two variables x and y because each variable is raised
to the first power, i.e. 1.
The standard form of linear equations in two variables is ax + by = c, where a, b, and
c are constants, and a or b both are not zero.
The equation, y = x + 4 has as many pairs of solutions as we wish to find. The table
given below shows a few pairs of solutions.

x 0 1 2 3 –1 –2 –4
y4567320

Thus, (0, 4), (1, 5) (2, 6), (3, 7), (–1, 3), (–2, 2), (–4, 0), … are a few pairs of solutions
that satisfy the equation y = x + 4.
Again, let’s consider another linear equation y = 2x + 1
A few pairs of solutions of this equations are shown in the table below:

x 0 1 3 4 –1 –2
y 1 3 7 9 –1 –3

Thus, (0, 1), (1, 3), (3, 7), (4, 9), (–1, –1), (–2, –3) are a few pairs of solutions that satisfy
y = 2x + 1.
Here, (3, 7) is the common pair of solution that satisfies both the equations
simultaneously. Such pair of equations that have only one pair of solution which
satisfies both the equations simultaneously at a time are called simultaneous
equations.

8.2 Method of solving simultaneous equations

There are various methods of solving simultaneous equations. Here, we discuss only
three methods:
(i) Graphical method (ii) Elimination method (iii) Substitution method

(i) Graphical method

In this method, we find a few pairs of solutions of each of two given equations.
The pairs of solutions of each equation are plotted in a graph and by joining
them two separate straight lines are obtained. The coordinates of the point of
intersection of two straight lines are the solutions to the given equations.

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Worked-out examples

Example 1: From the given graph, find the solutions of the two equations represented
Solution: by the two straight lines.

In the graph, the coordinates of the point of intersection of the (3, 2)
two straight line l1 and l2 is (3, 2).

? (3, 2) is the common solution to the equations
represented by l1 and l2.

? x = 3 and y = 2 are the required solutions represented by
the two straight lines.

Example 2: Solve graphically x + y = –3 and 3x + y = 1.
Solution:

Here, x + y = –3 x0 1 2 (0, 1)
or, y = –x – 3 y –3 –4 –5

Again, 3x + y = 1 x0 1 2 (1, –2)
or, y = 1 – 3x y 1 –2 –5

The coordinates of the point of intersection of the two (0, –3) (2, –5)
(1, –4)
straight lines represented by the equations is (2, –5).
? (2, –5) is the common solution to the equations.
? x = 2 and y = –5.

EXERCISE 8.1
General section

1. Write down the solution to each pair of simultaneous equations represented by

straight-line graphs.

a) b) c)

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d) e) Simultaneous Linear Equations
f)

2. Copy and complete the table of values for x and y. Plot the coordinates separately

from each table in graph. Find the common solution to the equations from the graphs.

a) y = 5 – x y=x–1 b) y = 2x + 2 y = –x – 4

x0 6 x4 x1 –2 x2

y3 y0 –3 y8 y –4 –8

x–8 y = –2x – 4 x–1 –4 – x
c) y = 2 d) y = 3 y= 2

x2 4 x 3 –3 x –2 7 x2
y –1 y –6
y1 y –2 –1

Creative section

3. Solve the following simultaneous equations graphically:

a) y = 8 – x b) y = 2x + 2 c) y = x + 4 d) y = x + 3
y=x–2 y=x–1 y=2–x y = 3x – 1

e) y = x + 4 f) y = 2x – 4 g) x + y = 8 h) x – y = 1
y = 2x – 1 y=x–3 x–y=2 2x + 3y = 12

i) x + 2y = 14 j) x + 3y = 13 k) 5x – 3y = 11 l) 2x – 3y = 4
x–y=5 x+y=7 2x – 3y = –1 3x + y + 5 = 0

m) x + 3y = 7 n) x –2 = 2x – 6 o) x–8 = 3x – 3 p) x + y = 3
3x – y = 11 2 3 4 5 5 4 2

x – y = – 1
10 2 2

(ii) Elimination method

In this method, we add or subtract the given equations to eliminate one of the
two variables by making their coefficients equal. Then, a single equation with
one variable so obtained is solved to find the value of the variable. The value of
the variable is substituted to any one equation to find the value of the eliminated
variable.

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Worked-out examples

Example 1: Solve: x + 2y = 11 and x + y = 7.
Solution:

x + 2y = 11 ....................... (i)

x + y = 7 .......................... (ii) The coefficients of x are the
Subtracting equation (ii) from (i) same in both equations. So,
we subtract one equation
x + 2y = 11 from the other to eliminate x.
+– x +– y =+– 7

y =4

Now, substituting the value of y in equation (ii), we get,

x+4 =7 Checking:
or, x =3 When x = 3 and y = 4,
? x = 3 and y = 4 From equation (i): LHS = x + 2y = 3 + 2 × 4 = 11 = RHS
From equation (ii): LHS = 3 + 4 = 7 = RHS

Example 2: Solve: 2x + 5y = 9 and x – y = 1

Solution:

2x + 5y = 9 ....................... (i)

x – y = 1 .......................... (ii)

Multiplying equation (ii) by 5 and adding to (i), Here we are eliminating y.
2x + 5y = 9 So, to make the coefficient
5x – 5y = 5 of y same, equation (ii) is
7 = 14 multiplied by 5.

or, x = 2

Now, substituting the value of x in equation (ii), we get,

2–y =1 Checking:

or, y = 1 When x = 2 and y = 1,
? x = 2 and y = 1 From equation (i): LHS = 2x+5y = 2×2 + 5×1 = 9 = RHS
From equation (ii): LHS =x – y = 2 – 1 = 1 = RHS

Example 3: Solve: 3 + 2 = 1 and 4 + 3 = 17
Solution: x y x y 12

3 + 2 = 1 ........................ (i)
x y

4 + 3 = 17 ....................... (ii)
x y 12

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Multiplying equation (i) by 3 and (ii) by 2 and subtracting (ii) from (i),

9 + 6 = 3 ........................ (i) Checking:
x y When x = 6 and y = 4,
From equation (i);
± 8 ± 6 = – 34 ....................... (ii)
x y 12
3x+2y 3 2
1 = 1 LHS = = 6 + 4
x 6
= 21+21 = 1 = RHS
or, x = 6

Now, substituting the value of x in equation (i), we get, From equation (ii);

3 + 2 =1 LHS =4x+3y = 4 + 3
6 y 6 4

or, y = 4 = 23+34 = 17 = RHS
? x = 6 and y = 4 12

Example 4: Solve x 10 y + 9 = 5 and x 15 y – 3 = 2.
Solution: + x–y + x–y

10 y + 9 = 5 ... (i) and x 15 y – 3 = 2 ... (ii)
x+ x–y + x–y

Let x + y = a and x – y = b. Then, equation (i) and (ii) reduce to

10 + 9 = 5 ... (iii) and 15 – 3 = 2 ... (iv) . Then, multiply equation (iv) by 3 and
a b a b adding (iii) and (iv),

10 + 9 + 45 – 9 =5+6
a b a b

or, 55 = 11 ? a=5
a
Now, substituting the value of a in equation (iii), we get,

10 + 9 =5 or, 9 = 3 ?b=3 Checking:
5 b b When x = 4 and y = 1,
From equation (i);
Again x + y = a, i.e. x + y = 5 ... (v)

and x – y = b, i.e. x – y = 3 ... (vi) LHS = 10 + 9 = 10 + 9
x+y x–y 4+1 4–1
Adding equations (v) and (vi), we get,

x+y+x–y=5+3 = 10 + 9 = 5 = RHS
or, x = 4 5 3
Substituting the value of x in equation (v), we get,
From equation (ii);

4+y=5 ?y=1 LHS 15 – 3 = 15 – 3
x+y x–y 4+1 4–1
15 3
Hence, x = 4 and y = 1. = 5 – 3 = 2 = RHS

(iii) Substitution method

In this method, a variable is expressed in terms of another variable from one
equation and it is substituted in the remaining equations.

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Example 5: Solve x + 3y = 1700 and 7x – y = 900

Solution:
x + 3y = 1700 ... (i) and 7x – y = 900 ... (ii)

From equation (i),
x = 1700 – 3y ... (iii)

Substituting for x from equation (iii) in equation (ii), we get,
7(1700 – 3y) – y = 900

or, 11900 – 21y – y = 900
or, 11000 = 22y
? y = 500
Now, substituting the value of y in equation (iii), we get,

x = 1700 – 3 × 500 = 1700 – 1500 = 200
? x = 200 and y = 500.

Example 6: Solve the given system of equation by substitution method.

x–1 = 3 and x+2 = 4
y+1 4 y–2 3
Solution:
x–1 3
y+1 = 4

or, 4x – 4 = 3y + 3

or, 4x – 3y = 7

or, 4x = 7 + 3y
? 7 + 3y
x = 4 ... (i)
Also,
x+2 = 4
y–2 3

or, 3x + 6 = 4y – 8

or, 3x – 4y = –14 ... (ii) Checking:
Now, substituting the value of x in equation (ii), we get, When x = 10 and y = 11,

3 7 + 3y – 4y = –14 From equation (i);
4

or, 21 + 9y – 16y = –14 LHS = yx+–11=1110+–11=192= 3 RHS
4 4

or, 21 – 7y = –56 From equation (ii);

or, 77 = 7y ? y = 11 LHS = xy+–22=1101+–22=192= 4 RHS
3

Again, substituting the value of y in equation (i), we get,

x= 7 + 3× 11 = 10
4

Hence, x = 10 and y = 11.

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Example 7: The sum of two numbers is 20. If three times the smaller number is equal
Solution: to two times the bigger one, find the numbers.

Let the bigger number be x and the smaller be y.

From the first condition, From the second condition,

x + y = 20 3y = 2x .................... (ii)

or, y = 20 – x ............... (i)

Substituting the value of y from equation (i) in equation (ii), we get,

3 (20 – x) = 2x Checking:
or, 60 – 3x = 2x x = 12 and y = 8
or, x = 12 x + y = 12 + 8 = 20
Now, substituting the value of x in equation (i), we get,
2x = 2 × 12 = 24
y = 20 – 12 = 8 3y = 3 × 8 = 24
Hence, the required numbers are 12 and 8.

Example 8: The cost of four pencils and three pens is Rs 68. If two pencils and a pen
Solution: cost Rs 26, find the cost of each pencil and each pen.

Let, the cost of each pencil be Rs x and each pen be Rs y.

From the first condition, From the second condition,

4x + 3y = 68 ............. (i) 2x + y = 26

or, y = 26 – 2x ................. (ii)

Substituting the value of y from equation (ii) in equation (i), we get

4x + 3 (26 – 2x) = 68

or, 4x + 78 – 6x = 68

or, x = 5

Now, substituting the value of x in equation (ii), we get, y = 26 – 2 × 5 = 16

Hence, each pencil costs Rs 5 and each pen costs Rs 16.

Example 9: A father is three times as old as his daughter. Six years ago, he was four
Solution: times as old as his daughter was. Find their present age.

Let the present age of the father be x years and that of the daughter be y years.

From the first condition, From the second condition,

x = 3y .................. (i) x – 6 = 4 (y – 6) ............. (ii)

Substituting the value of x from equation (i) in equation (ii),

3y – 6 = 4 – 24

or, y = 18

Substituting the value of y in equation (i), we get, x = 3 × 18 = 54

Hence, the present age of the father is 54 years and that of the daughter is 18 years.

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Example 10: The sum of the digits of a two-digit number is 7. When the places of the
digits are interchanged, the new number so formed will be 27 more than
the previous one. Find the number.

Solution:

Let the digit at tens place be x and at ones place be y.

Then, the number is 10x + y

When the places of the digits are interchanged the new number is 10y + x.

From the first condition, From the second condition,

x+y =7 10y + x = 10x + y + 27
or, y = 7 – x ................... (i) or, y – x = 3 ...................(ii)

Substituting the value of y from equation (i) in equation (ii),

7–x–x =3

or, x = 2

Substituting the value of x in equation (i), we get,

y =7–2=5

Hence, the required number is 10x + y = 10 × 2 + 5 = 25.

EXERCISE 8.2

General section

1. Find the value of the variables as indicated.

a) If x = 2 in 2x + y = 5, find the value of y.

b) If x = –1 in x – y = 3, find the value of y.

c) If x = –y in x + 2y = –2, find the values of y and x.

d) If y = 3 in 2x – y = 5, find the value of x.
x
e) If y = 2 in x + y = 9, find the value of x.

2. Select the correct pair of solutions of the given pair of equations.

a) x + y = 7 and x – y = 3 (i) (4, 3) (ii) (6, 3) (iii) (5, 2)

b) 2x – y = –2 and x + 2y = 9 (i) (1, 4) (ii) (4, 6) (iii) (3, 3)

c) y – 3x = 10 and 3y – x = 6 (i) (7, –1) (ii) (–3, 1) (iii) (–1, 1)

d) y = x + 2 and y = 3y – x = 6 (i) (5, 3) (ii) (4, 6) (iii) (3, 5)

e) y = x – 1 and y = x + 1 (i) (5, 2) (ii) (2, 5) (iii) (2, 1)
2 3

Creative section - A

3. Solve each pair of simultaneous equations by elimination method.

a) x + y = 5 b) 3x + y = 13 c) x + 2y = 7
x–y=1 x–y=3 x+y=4

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d) 3x + 5y = 11 e) 2x + 3y = 3 f) 6x + 7y = 5

4x – y = 7 x + 2y = 1 7x + 8y = 6
g) 2x + 3y – 1 = 0 h) 2x – 3y – 12 = 0 i) 3x + 2y – 15 = 0

5x + 2y + 3 = 0 3x – 2y – 13 = 0 5x – 3y – 25 = 0
j) 4(x – 1) + 5(y + 2) = 10 k) 2(x + 2) – 3(y – 1) = 6 l) 2x + 5 = 4(y + 1) – 1

5(x + 1) – 2(y – 2) = 14 7(x – 1) + 4(y + 2) = 12 3x + 4 = 5(y + 1) – 3

m) 2 + 6 = 3 n) 9 – 4 = 5 o) x + 6 = 4
x y x y 5 y
10 9 3 8 x 9
x – y = 2 x + y =– 3 2 – y = 2

4. Solve each pair of simultaneous equations by substitution method.

a) y = 2x b) y = 5x c) y = –4x d) 3x + 2y = 26

x+y=9 x + y = 12 x + 3y = – 11 x=y+2

e) y = x – 3 f) y = 3x + 1 g) x = 3 – y h) x + 2y = 9
2x – y = 3 x = 2y + 1
3x + 4y = 2 2x + y = 6

i) 2x + y = 8 k) x + 3y = 7 j) 2x + y = 4 l) 2x – 3y = 6
2x – y = 8 x + 3y = 3
x–y=1 3x – y = 11

m) x–1 = 1 n) x–2 = 1 o) x+ y–2 =6
y+1 2 y–2 2 3

x–2 = 1 x + 2 = 3 y + x + 1 = 8
y+2 3 y + 2 4 2

Creative section - B
5. a) The sum of two numbers is 30. If three times the smaller number is equal to two

times the bigger one, find the numbers.
b) The total cost of a cap and a sunglasses is Rs 1,250. If the cost of the sunglasses is

Rs 250 more than the cost of the cap, find the cost of each of these two items.
c) The total cost of a bag and an umbrella is Rs 2,000. If the cost of the bag is Rs 1,000

less five times the cost of the umbrella, find the cost of each of these two items.

6. a) The cost of a pen is three times and Rs 4 more than the cost of a pencil. If three
pencils and two pens cost Rs 71, find the cost of each item.

b) The cost of tickets of a comedy show of 'Gaijatra' is Rs 700 for an adult and
Rs 500 for a child. If a family paid Rs 3,100 for 5 tickets, how many tickets were
purchased in each category?

c) The cost of 4 kg of chicken and 5 kg of mutton is Rs 7,200. If the cost of 4 kg of
chicken is the same as the cost of 1 kg of mutton, find the rate of cost of chicken
and mutton.

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7. a) Mother is three times as old as her daughter. Three years ago she was four times as
old as her daughter was. Find their present ages.

b) The sum of the present ages of a father and his son is 40 years. Four years hence
father will be three times as old as his son will be. Find their present ages.

c) The differences of the present ages of a father and his daughter is 30 years. Four
years ago father was seven times as old as his daughter was. Find their present
ages.

d) Three years hence a father will be four times as old as his son will be. Before two
years he was seven times as old as his son was. Find their present ages.

8. a) The sum of the digits of a two-digit number is 7. When the places of the digits are
interchanged the new number so formed will be 9 more than the previous one.
Find the number.

b) The sum of the digits of a two-digit number is 10. If 18 is subtracted from the
number, the places of the digits are reversed. Find the number.

c) A certain number of two-digit is three times the sum of its digits. If 45 is added to
it, the digits are reversed. Find the number.

Project work!

9. a) Make a pair of simultaneous equations and solve them to get the following values

of the variables.

(i) x = 3 (ii) x = 5 (iii) x = 4 (iv) x = –2 (v) x = –3

y=2 y=7 y = –1 y=5 y = –2

b) Think the values of x and y of your own choice. Then, complete the each pair of

equations and solve them. x y
2 3
(i) x + y = ....... (ii) 2x+y = ....... (iii) 3x–2y = ....... (iv) + = .......
2x – y = .......
x – y = ....... x +2y = ....... x + y = .......
3 2

c) Make word problems reflecting to the real life situations as far as possible from the

given pairs of simultaneous equations, then solve them.

(i) x + y = 30 (ii) x+y = 36 (iii) 3x = y (iv) x + 2y = 70

x – y = 10 x = 2y y – x = 20 x:y=3:2

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k) a = a2 + b2 l) xa – yc = a2 + c2
b c2 + d2 xb – yd b2 + d2

m) a2 – c2 = a4 – c4 n) ac = a3 + c3 o) a2 + c2 = a4 + c4
b2 – d2 b4 – d4 bd 3 b3 + d3 b2 + d2 4 b4 + d4

4. If a = c = e , prove that
b d f

a) a+c+e = c+e b) 2a – 3c + 5e = e c) a2 + c2 + e2 = ac
b+d+f d+f 2b – 3d + 5f f b2 + d2 + f2 bd

d) ac + ce + ae = a2 e) a3 + c3 + e3 = ace f) (a + c – e)3 = a(a – c + e)2
bd + df + bf b2 b3 + d3 + f3 bdf (b + d – f)3 b(b – d + f)2

g) la + mc + ne = ace h) a3c3 + c3e3 + a3e3 = ace
lb + md + nf 3 bdf b3d3 + d3f3 + b3f3 bdf

5. If x = y = z , prove that
a b c

a) (x + y + z)3 = x3 + y3 + z3 b) x3 + y3 + z3 = xyz
(a + b + c)2 a2 b2 c2 a3 + b3 + c3 abc

c) y–z 3 x3 – y3 d) x2 – yz = a2 – bc
b–c a3 – b3 y2 – zx b2 – ac
=

6. If a = b prove that
b c
a a–b a+b 2 a2 + b2 (a – b)2 (b – c)2
a) a+b = a–c b) b+c b2 + c2 c) a = c
=

d) a2 + ab = b2 + bc e) (a + b)2 = (b + c)2 f) a+b+c = (a + b + c)2
b2 c2 b2 c2 a–b+c a2 + b2 + c2

g) a2 + ab + b2 = a h) a3 + b3 = a (a – b) i) a+b = a2 (b – c)
b2 + bc + c2 c b3 + c3 c (b – c) b+c b2 (a – b)

7. a) If a, b, c are in continuous proportion, show that (a2 + b2) (b2 + c2) = b2 (a + c)2.

[Hint: a = b , i.e. b2 = ac]
b c
b) If a, b, c are in continuous proportion, show that (ab + bc + ca)3 = abc (a + b + c)3

[Hint: a = b , i.e. b2 = ac. Then, L.H.S. = (ab + bc + b2)3]
b c

c) If x, y, z are in continuous proportion, prove that

x2y2z2 1 + 1 + 1 = x3 + y3 + z3
x3 y3 z3
p q
d) If q = r , show that p3q3 + q3r3 + r3p3 = pqr (p3 + q3 + r3)

[Hint: p = q , i.e. q2 = pr, then L.H.S. = p3.pqr + r3.pqr + q3.pqr]
q r
e) If b is the mean proportional between a and c, prove that

(a + b – c) (a + b + c) = a2 + b2 + c2.

f) If y is the mean proportional between x and z, prove that (x – y)2 = (y – z)2.
x z

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Ratio and Proportion

8. If a = b = c , prove that
b c d

a) a2 + b2+ ca = a+b b) a–b = a2 + b2 + c2
b2 + c2 + bd c+d c–d b2 + c2 + d2

c) ab – cd = b2 – d2 d) a2 – ab = b2 – bc = c2 – cd
ab + cd b2 + d2 b2 c2 d2

e) a3 + b3+ c3 = a f) a3 + b3+ abc = a
b3 + c3 + d3 d b3 + c3 + bcd d

g) a3 – b2c+ c3 = a3 h) (ax – by – cz)3 = bc
abc – bd2 + d3 b3 (bx – cy + dz)3 d2

i) (a + b + c) (b + c + d) = ab + bc + cd

j) ab – bc + cd = (a – b + c) (b – c + d)

Creative section B
9. If a : b : : b : c : : c : d, prove that

a) (b + c) (b + d) = (c + a) (c + d)
b) (a + d) (b + c) – (a + c) (b + d) = (b – c)2
c) (a – c)2 + (b – c)2 + (b – d)2 = (a – d)2

d) (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2
10. a) If a, b, c, and d are in continues proportion, prove that (a2 – b2) (c2 – d2) = (b2 – c2)

b) If p, q, r and s are in continuous proportion show:

(p2 + q2 + r2) : (q2 + r2 + s2) = (p2 + qr) : (q2 + rs)

c) If a, b, c, d, and e are in continuous proportion, prove that a : e = a4 : b4.

Project work

11. a) Take any two ratios of positive integers and find the compounded ratio.

b) Take a ratio of any two positive integers and find:

(i) duplicate and sub-duplicate ratios (ii) triplicate and sub-triplicate ratios

c) How many students are there in your class? Write the ratio of the number of boys
and girls. Then, make an equation to find the number of boys and girls.

d) How many teachers are there in your school? Write the ratio of the number of
male and female teachers. Then, make an equation to find the number of male and
female teachers.

12. a) Take any four positive integers which are in proportion. Then, verify the following
properties of proportion.

(i) Invertendo (ii) Alternendo (iii) Componendo

(iv) Dividendo (v) Componendo and dividendo (vi) Addendo

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Objective Questions

1. Which one is the equal expression to a2 + b2?

(A) (a + b)2 – 2ab (B) (a – b)2 + 2ab

(C) (a + b) (a – b) (D) Both (A) ands (B)

2. The factorization of x4 + x2y2 + y4 is

(A) (x2 – xy + y2) (x2 + xy + y2) (B) (x2 – xy + y2) (x2– xy + y2)

(C) (x2 + xy – y2) (x2 + xy + y2) (D) (x2 + xy – y2) (x2 + xy – y2)

3. What is the HCF of p2 – 1 and p3 + 1?

(A) p + 1 (B) p – 1 (C) (p – 1) (p3 +1) (D) (p + 1) (p3 – 1)

4. If m + n = m, what is the value of mn?

(A) 0 (B) 1 (C) m (D) n

5. What is the value of a0 – 2 ? (A) 4 (B) 16 (C) 64 (D) 512
64 3

6. The value of 5m+2 – 5m is
5m+1 – 5m
(A) 2 (B) 4 (C) 6 (D) 8

7. For what value of x, will the equation 2x + 2x + 1 = 12 be satisfied?

(A) 0 (B) 1 (C) 2 (D) 3

8. If 23x – 5 × ax – 2 = 2x – 2 × a1 – x, what is the value of 4x?

(A) 1 (B) 1.5 (C) 4 (D) 64

9. The sum of two numbers is 30. If twice the bigger number is thrice the smaller, the
smaller is

(A) 6 (B) 8 (C) 12 (D) 18

10. According to marriage ritual, a groom takes the bride to his house. If Ram’s daughter gets
married to Sita’s son, both the family will have equal members. But, if Ram’s son gets
married with Sita’s daughter, Ram’s family will be twice as large as Sita’s family. How
many members are there in Ram and Sita’s family?

(A) 4, 3 (B) 5, 3 (C) 7, 5 (D) 9, 5

11. The equation ax2 + bx + c = 0, a ≠ 0 becomes pure quadratic if -

(A) b < 0 (B) b > 0 (C) b = 0 (D) c = 0

12. What are the roots of the quadratic equation ax2 + bx + c = 0, a ≠ 0?

(A) – b ± b2 – 4ac (B) – a ± b2 – 4ac (C) – b ± b2 + 4ac (D) – b ± b2 – ac
2a 2b 2a a
13. What are the roots of the equation x2 + 1 = 5 (x – 1) ?

(A) 1, 2 (B) 2, 3 (C) 1, 5 (D) 1, 6

14. The duplicate ratio of 4:9 is

(A) 2:3 (B) 4:9 (C) 16:81 (D) 64:729

15. The number of girls and boys in a school are in the ratio 8:9. If 24 new girls join and
7 boys leave the school; the ratio becomes 9:8. How many students are there in the
school?

(A) 120 (B) 135 (C) 255 (D) 272

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Unit Geometry - Triangle

11

11.1 Various types of angles B
A
In the given figure, straight line segments AO and Bo meet at the O
point O to from ‘AOB. Here, O is the vertex, AO and BO are the
arms of ‘AOB.

Types of angles on the basis of their sizes
According to the size of angles, they are classified in the following ways.

Acute angle Right angle Obtuse angle Straight angle Reflex angle Complete turn

B B B O A OA
OA B Exactly 360°
OA OA QO P
Less than 90° Exactly 90° Greater than 90°
but less than 180° Exactly 180° Greater than
180° but less
than 360°

Types of pairs of angles

Following are the different types of pairs of angles according to their construction and
properties.

Adjacent angles Vertically opposite Complementary Supplementary
angles angles angles

DC C B
B
C B O B 180°
B A OA

O AC O A ‘AOB + ‘BOC = 90° CO A
‘AOB and ‘BOC
‘AOB and ‘BOC are a ‘AOB and ‘COD are complementary ‘AOB + ‘BOC = 180°
pair of adjacent angles. ‘AOD and ‘BOC are angles. ‘AOB and ‘BOC are
two pairs of vertically supplementary angles.
opposite angles.

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Geometry - Triangle

Pairs of angles between two line segments intersected by a transversal
When a transversal intersects two straight line segments at two distinct points, following
pairs of angles are formed.

Co-interior angles Alternate angles Corresponding angles

ac ac ac ac ac ae
bd bd db db gc
b
bf
d hd
Fig. 1
Fig. 2

Fig. 1 Fig. 2 Fig. 1 Fig. 2 ‘a and ‘b, ‘c and ‘d are

‘a and ‘b, ‘c and ‘d are two pairs ‘a and ‘b, ‘c and ‘d are two pairs of corresponding

of co-interior angles. two pairs of alternate angles. angles.
The pair of corresponding
The sum of a pair of co-interior The pair of alternate angles angles between parallel line
segments are equal.
angles between parallel line between parallel lines

segments is 180°. segments are equal.

In figure 2, In figure 2, In figure 2

‘a + ‘b = 180° and ‘c + ‘d = 180° ‘a = ‘b and ‘c = ‘d ‘a = ‘b and ‘c = ‘d

‘e = ‘f and ‘g = ‘h

11.2 Axioms and postulates

Axioms

Axioms and postulates are the most important building blocks of geometric proofs.

An axiom is a statement that is taken to be true, to serve as a premise or starting point

for further reasoning and arguments. In other words, an axiom is a self-evident truth.

Following are a few examples of axioms.

1. The reflective axiom: It states that any quantity is equal to itself. For example,

lines segments, angles, and polygons are always equal to themselves.

2. The transitive axiom: It states that if two quantities are both equal to a third

quantity, then they are equal to each other. For example,

If ‘A = ‘B and ‘B = ‘C, then, ‘A = ‘B = ‘C

3. The substitution axiom: It states that if two quantities are equal, then one can

be replaced by the other in any expression, and the result won’t be changed. For

example, If ‘A = ‘B and ‘B + ‘C = 180° , then, ‘A + ‘C = 180°.

4. The partition axiom: It states that a quantity is equal to the B P
sum of its parts. It is well known as ‘whole part axiom’ in

geometry. OA
For example, in the given figure, ‘AOB = ‘AOP + ‘POB

5. The addition, subtraction, multiplication, and division axioms:

(i) Axiom of equality of addition
When the equal quantities are added to both sides of equal quantities, the sum
is also equal.
If a = b, then a + c = b + c

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Geometry - Triangle

(ii) Axiom of equality of subtraction

When the equal quantities are subtracted from both sides of equal quantities,

the difference is also equal.

If a = b, then a – c = b – c

(iii) Axiom of equality of multiplication

When both sides of the equal quantities are multiplied by equal quantities, the

product is also equal.

If a = b, then a × c = b × c

(iv) Axiom of equality of division

When both sides of the equal quantities are divided by equal quantities, the

quotient is also equal.

If a = b, then a = b
c c
Postulates

A statement that is accepted as true without any proof is called a postulate. A postulate
forms the basis of a theory. A few geometrical postulates which are used as the basis
to prove different theorems are given below.

(i) Only one straight line can be drawn by joining any two points.

(ii) Infinite number of straight lines can be drawn from a point.

(iii) Two straight lines can intersect only at a point.

(iv) A straight line can be extended infinitely to both directions.

(v) A line segment can have only one mid-point.

(vi) Only one perpendicular can be drawn from a point to a line.

(vii) Only one line can be drawn from a point parallel to the given straight line.

(viii) The distance between a point and a straight line is the length of the perpendicular

drawn from the point to the line.

(ix) Only one straight line can be the bisector of an angle.

EXERCISE 11.1

General section

1. Identify the types of these angles on the basis of their sizes. Then, list them separately:

a) b) R c) d)

B A Y X
G

QP

C Z FE

e) f) A BX R

QO P g) h)
Q
Y O W O

CD Z P

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Geometry - Parallelogram

Theorem 15

The opposite angles and sides of a parallelogram are equal.

Given: ABCD is a parallelogram in which AB // DC and
AD // BC.

To prove: (i) ‘ABC = ‘ADC, ‘BAD = ‘BCD
(ii) AB = DC, AD = BC

Construction: Diagonal AC is drawn.

Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) AB // DC and alternate angles

(i) ‘BAC = ‘ACD (A) (ii) Common side

(ii) AC = AC (S) (iii) AD // BC and alternate angles

(iii) ‘ACB = ‘CAD (A) (iv) A. S. A. axiom
2. Corresponding angles of congruent triangles
(iv) ? ∆ABC # ∆ACD 3. Drawing the diagonal BD and same as above

2. ‘ABC = ‘ADC in ∆s ABD and BCD
4. Corresponding sides of congruent triangles
3. ‘BAD = ‘BCD
Proved.
4. AB = DC and AD = BC

Converse (I) Theorem 15

If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.
Given: ABCD is a quadrilateral in which AB = DC and

AD =BC.
To prove: ABCD is a parallelogram, i.e. AB // DC and AD // BC.
Construction: Diagonal AC is drawn.
Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) Given
(ii) Given
(i) AB = DC (S) (iii) Common side
(iv) S. S. S. axiom
(ii) BC = AD (S) 2. Corresponding angles of congruent triangles

(iii) AC = AC (S) 3. From statement (2), alternate angles being
equal
(iv) ? ∆ABC # ∆ACD
4. Opposite sides are parallel
2. ‘BAC = ‘ACD and Proved

‘ACB = ‘CAD

3. AB // DC and AD // BC

4. ABCD is a parallelogram

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Geometry - Parallelogram

Converse (II) of Theorem 15

If the opposite angles of a quadrilateral are equal, the quadrilateral is a parallelogram.

Given: ABCD is a quadrilateral in which ‘ABC = ‘ADC and
‘BCD = ‘BAD.

To prove: ABCD is a parallelogram, i.e., AB // DC and AD // BC.

Proof

Statements Reasons

1. ‘A + ‘B + ‘C + ‘D = 360° 1. Sum of the angles of a quadrilateral
2. ‘A + ‘B + ‘A + ‘B = 360q 2. Given, ‘A = ‘C and ‘B = ‘D.

or, 2 (‘A + ‘B) = 360°

or, ‘A + ‘B = 180° 3. Being the sum of co-interior angles 180°
3. ? AD // BC 4. Same as above
4. Similarly, ‘A + ‘D = 180° 5. Same as reason (3)
5. ? AB // DC 6. Opposite sides are parallel
6. ABCD is a parallelogram

Theorem 16 Proved

The diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram. Diagonals AC and BD intersect
at O.

To prove: AC and BD bisect each other at O, i.e., AO = OC and
BO = OD.

Proof

Statements Reasons
1.
1. In ∆s AOB and COD (i) AB // DC and alternate angles
(ii) Opposite sides of a parallelogram
(i) ‘OAB = ‘OCD (A) (iii) AB // DC and alternate angles
(iv) A. S. A. axiom
(ii) AB = DC (S) 2. Corresponding sides of congruent triangle.
3. From statement (2)
(iii) ‘OBA = ‘ODC (A)
Proved
(iv) ? ∆AOB # ∆COD
2. AO = OC and BO = OD
3. AC and BD bisect each other at O.

Converse of Theorem 16

If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

Given: ABCD is a quadrilateral in which diagonals AC and
BD bisect each other at O.
? AO = OC and BO = OD

To prove: ABCD is a parallelogram, i.e., AB // DC, AD // BC,
AB = DC, AD = BC.

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Geometry - Parallelogram

Proof

Statements Reasons
1. In ∆s AOB and COD
1.
(i) AO = OC (S) (i) Given
(ii) Vertically opposite angles
(ii) ‘AOB = ‘COD (A) (iii) Given
(iv) S. A. S. axiom
(iii) BO = OD (S) 2. Corresponding sides of congruent triangle
3. Corresponding angles of congruent triangles
(iv) ? ∆ AOB # ∆ COD 4. From statements (3), alternate angles are equal
2. AB = DC 5. AD and BC join the ends of two equal and parallel
3. ‘OAB = ‘OCD
4. AB // DC lines towards the same side.
5. AD = BC and AD // BC 6. Opposite sides are equal and parallel.

6. ABCD is a parallelogram Proved

Worked-out examples

Example 1: In the given figure, PQRS is a parallelogram. P S
Solution: If PT = SR, ‘ PSR = 4x° and ‘ QPT = x°, find the x° 4x°
value of ‘ QRS.
Q R
T

(i) ‘PQR = ‘PSR = 4x° [Opposite angles of parallelograms are equal]

(ii) PQ = PT [ PQ = SR and PT = SR]

(iii) ‘PQT + ‘PTQ = 4x° [ PQ = PT]
(iv) ‘PQT + ‘PTQ + ‘QPT = 180° [Sum of angles of 'PQT]

or, 4x° + 4x° + x° = 180°

or, 9x° = 180°

? x = 20°

(v) ‘QRS = 180° – ‘PQR [ PQ // SR and co-interior angles]

= 180° – 4 × 20° = 100° A D
35° C
Example 2: In the figure alongside, ABCD is a rhombus.
If ‘DAC = 35°, find the measure of ‘ABC. B

Solution:

(i) ‘DAC = ‘ACD = 35° [AD = CD, sides of rhombus]
(ii) ‘ADC + ‘DAC + ‘ACD = 180° [Sum of angles of 'ADC]
or, ‘ADC + 35° + 35° = 180°

? ‘ADC = 110°
(iii) ‘ABC = ‘ADC [Opposite angles of rhombus]
? ‘ABC = 110°

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Geometry - Parallelogram

Example 3: In the figure alongside, WXYZ is a square. ZY
If ‘ZNX = 115°, find the sizes of ‘WZN and ‘ZMY.
M
Solution: 115°

(i) ‘WZN + ‘ZWN = ‘ZNX [‘ZNX is the exterior angle in 'ZWN] W NX
or, ‘WZN + 90° = 115°

? ‘WZN = 25°

(ii) ‘ZWY = ‘XWY=45° [Diagonal of square bisects its vertical angle]

(iii) ‘ZMY = ‘WZM+‘ZWM [In 'WZM, ‘ZMY is the exterior angle]

? ‘ZMY = 25° + 45° = 70°

Example 4: P, Q, R and S are the mid–points of the sides AB, BC,
CD and DA of the quadrilateral ABCD respectively.
Prove that PQRS is a parallelogram.

Solution:

Given: P, Q, R and S are the mid–points of the sides AB, BC,
CD and DA of the quadrilateral ABCD respectively.

To prove: PQRS is a parallelogram.

Construction: Diagonal AC of the quadrilateral ABCD is drawn.

Proof

Statements Reasons

1. In ∆ABC, PQ // AC and PQ = AC 1. PQ joins the mid–points of the sides AB and
BC.

2. In ∆ACD, SR // AC and SR = AC 2. SR joins the mid–points of the sides AD and
DC.

3. ? PQ // SR and PQ = SR 3. From the statements (1) and (2)
4. SP // RQ and SP = RQ 4. From the statement 3.
5. PQRS is a parallelogram 5. Being opposite sides equal and parallel

Proved

Example 5: In the given figure, PQRS is a parallelogram in which P TS
QT is the bisector of ‘PQR. Prove that PT = SR. Q R
Solution:
Given: In parallelogram PQRS,QT is the bisector of ‘PQR.
To prove: PT = SR.
Proof

Statements Reasons
1. ‘PQT = ‘TQR 1. QT being bisector of ‘PQR.

2. ‘PTQ = ‘TQR 2. PS // QR and alternate angles.
3. ‘PQT = ‘PTQ 3. From the statements (1) and (2)
4. PQ = PT 4. From the statement 3.
5. PQ = SR 5. Opposite sides of parallelogram
6. PT = SR 6. From statements (4) and (5).

Proved

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Geometry - Parallelogram

Example 6: In the figure, ABCD is a parallelogram. The A Q
D
Solution: diagonal BD is produced in either sides to
Given: C
To prove: the points P and Q such that BP = DQ. Prove
Proof
that: B
(i) ‘BCP = ‘DAQ

(ii) AQ // PC. P

In parallelogram ABCD, the diagonal BD is produced to the points P and Q
such that BP = DQ.

(i) ‘BCP = ‘DAQ (ii) AQ // PC.

Statements Reasons
1. AD // BC, alternate angles
1. ‘DBC = ‘ADB 2. Supplements of equal angles, ‘DBC = ‘ADB
3.
2. ‘PBC = ‘ADQ (i) Given
3. In 'BPC and 'ADQ (ii) From statement(2)
(i) BP = DQ (S) (iii) Opposite sides of parallelogram ABCD
(ii) ‘PBC = ‘ADQ (A) (iv) By S.A.S, axiom
(iii) BC = AD (S) 4. Corresponding angles of congruent triangles
(iv) 'BPC # 'ADQ
4. ‘BCP = ‘DAQ and 6. From statements (4), alternate angles ‘BPC
and ‘AQD are equal.
‘BPC = ‘AQD Proved
5. AQ // PC

Example 7: In the given parallelogram ABCD, M and N are the A M B
mid-points of sides AB and DC respectively. Prove X C
Solution: that (i) ANCM is a parallelogram and Y
Given: D N
To prove: (ii) BX = XY = DY.
Proof
In parallelogram ABCD, M and N are the mid-points of sides AB and DC
respectively.

(i) ANCM is a parallelogram (ii) BX = XY = DY.

Statements Reasons
1. AM = NC and AM // NC 1. AB = DC, AB // DC and given

2. MC = AN and MC // AN 2. From statement (1)
3. ANCM is a parallelogram 3. From statements (1) and (2)
4. BX = XY 4. In 'BAY, BM = MA and MX // AY
5. DY = XY 5. In 'DCX, DN = NC and YN // XC
6. BX = XY = DY 6. From statements (4) and (5)

Proved

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Geometry - Parallelogram

Example 8: In a trapezium ABCD, if AB // DC and AD = BC, A B
prove that (i) ‘ADC = ‘BCD and (ii) AC = BD. D EC

Solution: In trapezium ABCD, AB // DC and AD = BC.
Given:

To prove: (i) ‘ADC = ‘BCD (ii) AC = BD

Construction: BE // AD is drawn to meet DC at E.

Proof

Statements Reasons
1. ABED is a parallelogram 1. By construction and given
2. AD = BC and AD = BE
2. AD = BE = BC 3. Base angles of the isosceles 'BEC
3. ‘BEC = ‘BCE
4. AD // BE, corresponding angles
i.e. ‘BEC = ‘BCD 5. From statements (3) and (4)
4. ‘ADC = ‘BEC 6.
5. ‘ADC = ‘BCD (i) Given
6. In 'ADC and 'BCD (ii) From statement (5)
(i) AD = BC (S) (iii) Common side
(iv) By S.A.S. axiom
(ii) ‘ADC = ‘BCD (A) 7. Corresponding sides of congruent triangles
(iii) DC = DC (S)
(iv) 'ADC # 'BCD Proved
7. AC = BD

Example 9: In the figure alongside, ABCD is a parallelogram.
If 2MO = OD, prove that M is the mid-point of BC.

Solution:

Given: ABCD is a parallelogram in which BC // AD and
To prove: AB // DC, 2 MO = OD
M is the mid-point of BC, i.e. BC = 2 MC

Proof:

Statements Reasons

1. In ∆ AOD and ∆ MOC 1.
(i) (i) AD // BC and alternate angles
(ii) ‘ADO = ‘CMO (ii) Vertically opposite angles
(iii) (iii) AD // BC and alternate angles
(iv) ‘AOD = ‘COM (iv) A.A.A. axiom

2. (i) ‘DAO = ‘MCO 2. (i) Corresponding sides of similar triangles

∆AOD ~ ∆MOC

AD = OD = AO
MC MO CO

(ii) BC = 2MO (ii) AD = BC (opposite sides of parallelogram)
MC MO and OD = 2MO (given)

or, BC = 2 MC (iii) From statement (ii)
(iii) M is the mid-point of BC

Proved

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Geometry - Parallelogram

EXERCISE 13.1 A B
General section C

1. a) In the figure alongside, AB // DC and AB = DC. Write down the Q
relation between AD and BC.

D
P

b) In the given figure, PQ // RS and PQ = RS. What is the relation

between PS and QR? R S
T
B
S
c) In the figure alongside, BEST is a parallelogram. Write down Y
the relation between BT and ES, BE, and TS.

E
Z

d) In the parallelogram WXYZ, diagonals, ZX and WY intersect O X
at O. Write the relation between OW, OY and OZ, OX. W D

A O C
P S
2. a) In the given figure, ABCD is a rectangle. AC and BD are

diagonals. If the length of diagonal AC is 10 cm, what is the

length of diagonal BD? B

b) In the square PQRS given alongside, what is the value of ‘QPR?

Q R
R D

c) In the given figure, READ is a rhombus. If ‘RED = 40°, what is the A
measure of ‘RDE? Z

E Y

d) In the given rectangle WXYZ, O is the point of intersection W

of its diagonal WY and XZ. If ‘XOY = 120°, find the size of O

‘OWZ. X

3. Calculate the size of unknown angles:

a) A yD b) P Q c) L p+q 2p–50° E d) A D
5y x
3x a 60°

2x zw b 4y p+10° r O
B C ES R I K 2x

e) L 115° B CE

x D f) D E F g) P T h) A F
x 110° b x y
y A 100° S
ET
55° 50° E
D

aa 60°
CQ BC
A R
B

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4. a) In the given figure, ABCD is a parallelogram. Geometry - Parallelogram
If ‘BCD = 125° and AB = AE, find the size of
‘BAE. 110°

b) In the figure alongside, DBCE is a parallelogram.
If ‘AFE =110° and ‘FCE = 40°, find the value
of ‘DBC.

c) In the adjoining figure, PQRS is a parallelogram. H ND25°
Find the value of ‘SMR. A
M
d) In the given figure, HEAD is a parallelogram. If E
AM A HE, AN A HD and ‘NAD = 25°, find the
measure of ‘MAE and ‘MAN.

5. a) In the given figure, ABCD is a rhombus. If ‘ADB = 40°,
find the size of ‘CDE.

b) In the given figure, ABCD is a rhombus. If
‘BCD = 114°, find the measure of ‘ABD.

c) In the adjoining figure, ABCD is a rhombus. If CD
‘OAB = 30°, find the measures of ‘ODC and
‘ABC. O
30°
d) In the given figure, PQRS is a rectangle.
If ‘POS = 124°, find the sizes of ‘PQO and BA
‘OQR. PS

124°
O

QR

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Geometry - Parallelogram

e) In the figure alongside, PQRS is a square. If ‘SMR = 70°, SR
find the measures of ‘PST and ‘STQ.
70° Q
f) In the given figure, FIVE is a square. If PQ // FV and M E
‘QPR = 75°, find the measure of ‘FIR and ‘FPI. Q
PT
P FP V

75°
R

M

I
S

g) In the given figure, PQRS is a rhombus and 55° M
SRM is an equilateral triangle. If SN A RM and RN
‘PRS = 55°, find the size of ‘QSN.

Q

6. a) In the given figure, ABCD is a parallelogram. If AB = 3x cm, 3x cm A 19 cm D

BC = (5y – 1) cm, CD = 12 cm, and AD = 19 cm, find the 12 cm

values of x and y. B (5y – 1) C

b) In the figure alongside PQRS is a parallelogram. If (x – y) cmP (x + y) cm S 10 cm
PQ = (x – y) cm, QR = 20 cm, RS = 10 cm, and Q 20 cm R
PS = (x + y) cm, find the value of x and y.

L P
5cm
c) In the adjoining figure, LMNP is a rhombus. If
MN = (2a – 3) cm, OM = (3b + 2) cm, OP = 5 cm, and (3b+2)Ocm
perimeter = 28 cm, find the values of a and b.
M (2a – 3)cm N
d) In the adjoining figure, PQRS is a rectangle. If
OP = (2x – 3) cm and OR = (x + 1) cm, find the length S R
of diagonal QS. (x+1)cm Q

(2x–3)cm O
P

e) RACE is a rectangle in which diagonal RC = 18 cm, R E
OA = (p + q) cm and OE = 3p cm, find the values of p 3p cm C
and q.
(p+q)cmO
A

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Geometry - Parallelogram

D C

7. a) In the adjoining figure ABCD is a square and ABE is an E
equilateral triangle. Find the measure of ‘ADE and ‘DCE.

A A B
B E
b) In the given figure, ABCD is a square and BEC is
an equilateral triangle. Find ‘AEB and ‘DAE.

D C
P
Creative section -A Q
A
8. a) In the adjoining figure, PQ = AB and PQ // AB. B
D
Prove that (i) AP = BQ (ii) AP // BQ.
C
b) In the given quadrilateral, AB = DC and BC = AD. A C
Prove that the quadrilateral ABCD is a parallelogram. B
B
c) In the figure alongside, diagonals AC and BD of the D S
quadrilateral bisect each other at O. Prove that ABCD is a O
parallelogram. R

A

P

d) In the adjoining quadrilateral PQRS, ‘P = ‘R and ‘Q = ‘S.
Prove that PQRS is a parallelogram.

Q

9. a) In the given figure, ABCD is a parallelogram. AP bisects
‘A. Prove that DP = BC.

E P M

b) In the given figure, EXAM is a parallelogram, the
bisector of ‘A meets the mid-point of EM at P. Prove that
AX = 2AM.

XA

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Geometry - Parallelogram AB
DC
c) In the adjoining figure, ABCD is a rhombus in which CD is
produced to E such that CD = DE. Prove that ‘EAC = 90°.

E

d) In the given figure, P and R are the mid–points of the sides
AB and BC of ∆ABC respectively. If PQ // BC, prove that
BP = RQ.

Creative section -B

10. a) In the given quadrilateral PQRS, the mid–points
of the sides PQ, QR, RS and SP are A, B, C, and D
respectively. Prove that ABCD is a parallelogram.

b) In the adjoining figure, P, Q, R and S are the mid-points
of AB, BC, CD and AD respectively. Prove that PQRS is a
parallelogram.

A

c) In the given figure, P, Q, R and S are the mid-points of AB, BC, P S
CD and AD respectively. Prove that PQRS is a parallelogram. C

QR
BD

d) In the given figure, PQRS is a square. A, B, C, and D are the
points on the sides PQ, QR, RS and SP respectively.
If AQ = BR = CS = DP, prove that ABCD is also a square.

11. a) In the given parallelogram PQRS, PA bisects ‘P and RB
bisects ‘R. Prove that PA // BR.

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Geometry - Parallelogram

b) ABCD is a parallelogram. P and Q are two points on the
diagonal BD such that DP = QB. Prove that APCQ is a
parallelogram.

c) ABCD is a parallelogram. DE A AC and BF A AC. Prove
that BEDF is a parallelogram.

P S
M
d) In parallelogram PQRS, the bisectors of ‘PQR and
‘PSR meet the diagonal at M and N respectively. Prove N
that MQNS is a parallelogram.

QR

e) In the given figure, ABCD is a parallelogram. If P and A D
Q are the points of trisection of diagonal BD, prove that Q
PAQC is a parallelogram.
P

BC

12. a) In the figure, ABCD is a parallelogram. M and N are
the mid-points of the sides AB and DC respectively.
Prove that
(i) MBCN is a parallelogram
(ii) DMBN is a parallelogram
(iii) DB and MN bisect each other at O.

b) In the given parallelogram PQRS, M and N are the
mid-points of the sides PQ and SR respectively. Prove
that
(i) PNRM is a parallelogram
(ii) QA = AB = BS

c) ABCD is a square. P and Q are any points on the sides AB
and BC respectively. If AQ = DP, prove that AQ and DP are
perpendicular to each other.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

d) In the figure along side, ABCD and DEFC are parallelograms.
Prove that
e) (i) AE = BF (ii) 'ADE # 'BCF

f) In the given quadrilateral ABCD, AD = BC, P and Q are the DN C

13. a) mid-points of the diagonals AC and BD respectively. If M and PQ
b) N are the mid-points of the sides DC and AB respectively,
c)
d) prove that PMQN is a rhombus. A M B
e) D C
f) (Hint: Apply mid-point theorem in 'BCD, 'ACD, 'ABD and ' ABC)

In the adjoining figure, ABCD is a parallelogram. P S
AS, BS, CQ, and DQ are the bisectors of ‘A, ‘B, ‘C R
and ‘D respectively. Prove that PQRS is a rectangle. Q

Prove that the diagonals of a rectangle are equal. A B

If the diagonals of a rhombus are equal, prove that it is a square.

Prove that the diagonals of a rhombus bisect each other perpendicularly.

Prove that the diagonal of a parallelogram divides it into two congruent triangles.

If P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively of
a quadrilateral ABCD, prove that PQRS is also a parallelogram.

Prove that the line segments joining the mid-points of the diagonals of a trapezium
is parallel to parallel sides and equal to half their difference.

Project work

14. a) Take two rectangular sheets of paper of the same size. Fold one sheet through one
diagonal and another sheet through other diagonal. Then, cut out each sheet of
papers through diagonals.

(i) Are two diagonals equal?

(ii) Do these diagonals bisect each other?

b) Take a square sheet of paper and fold it through both diagonals.

(i) Are two diagonals equal?

(ii) Do these diagonals bisect each other perpendicularly?

Vedanta Excel in Mathematics - Book 9 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Statistics

17

17.1 Statistics - Review

In the age of information technology, statics has a wide range of applications.
Different departments and authorities require various facts and figures to frame
policies and guidelines in order to function smoothly. Statistical information helps to
understand the economic problems and formulation of economic policies. In social
science, statistics is used in the field of demography for studying mortality, fertility,
population growth rate, and so on. In science and technology, there is a regular use of
statistical tools for collecting, presenting and analysing the observed data for various
researches.
The present time is the time of information and communication. In social, economical
and technical area, we frequently require information in the form of numerical figures.
The information collected in the form of numerical figures are called data. Statistics
is a branch of mathematics dealing with data collection, organisation, analysis,
interpretation, and presentation. Statistics in plural form refer to data, whereas in the
singular form it refers a subject.

17.2 Types of data

(i) Primary data

The data collected by the investigator him/herself for definite purposes are
called primary data. These data are highly reliable and relevant.
(ii) Secondary data

The data collected by someone other than the user oneself are called secondary
data.
(iii) Raw data

The data obtained in original form are called raw data.
(iv) Array data

The data arranged in ascending or descending order are called array data.

17.3 Frequency tables

Let’s consider the following marks obtained by 10 students in a unit test in mathematics.

18, 10, 15, 15, 12, 12, 10, 15, 12, 12 o Raw data

10, 10, 12, 12, 12, 12, 15, 15, 15, 18 o Array data

Here, the marks 10 are repeated 2 times. So, 2 is the frequency of 10.

The marks 12 are repeated 4 times. So, 4 is the frequency of 12.

The marks 15 are repeated 3 times. So, 3 is the frequency of 15.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265 Vedanta Excel in Mathematics - Book 9