Category: 9th mathematics exercise question with answer

Factorization[class 9] – Full Chapter Exercise solutions

After studying the chapter Factorization[class 9], you learn to find the factors of simple monomial expressions, find the factors of binomial expressions, find the factors of some cubic expressions.

3.2.1 Introduction to Factorization[class 9]:

Different methods of Factorization[class 9]:

Method 1: Factorization[class 9] taking common factors:

When each term of an expression has a common factor we divide each term by this and take it out as shown below:

Example: Factorise 9x2 + 12xy

Solution:

We write 9x2 + 12xy = (3x)(3x) + (3x)(4y)

Here 3x is the common to both the terms. Hence,

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

Thus, 3, x, (3x + 4y) are  the factors  of 9x2 + 12xy

Method 2: Factorization[class 9] by grouping:

Sometimes, in a given expression it is not possible to take out a common factor directly. However the terms of the given expression are grouped in such a manner that we have a common factor

Example: Factorise ab + bc + ax + bx

Solution:

We have ab + bc + ax + cx = (ab + bc) +(ax +cx) = b(a + c) + x(a+ c) = (a + c)(b + x)


Factorization[class 9] Exercise 3.2.1

  1. Factorise:

(i) 5x2 – 20xy

Solution:

= 5x(x – 4y)


(ii) 2p(x + y) – 3q(x + y)

Solution:

= (x + y) (2p – 3q)


(iii) 4(a + b)2 – 6(a + b)

Solution:

= 2 (a + b) [2(a + b) – 3]

= 2 (a + b) (2a + 2b – 3)


(iv) 18x2y – 24xyz

Solution:

= 6xy (3x – 4z)


(v) 7xa – 70 xb

Solution:

= 7x (a – 10b)


(vi) 13m2 + 156n2

Solution:

=13(m2 + 12n2)


(vii) 3x2 + 6x +6

Solution:

= 3 (x2 + 2x + 2)


(viii) 4abx2 + 8abx + 120by

Solution:

= 4ab (x2 + 2x +3y)


2. Factorise

(i) x2 + xy + xz + yz

Solution:

= x (x + y) + z (x+ y)

=(x + y) (x + z)


(ii) a2 – ab + ac – bx

Solution:

= a (a – b) + c (a – b)

= (a – b) (a – c)


(iii) 4x + bx + 4b + b2

Solution:

= x (4 + b) + b (4 + b)

= (4 + b) (x + b)


(iv) 6ax – 6a + 5a – 5

Solution:

= 6a (x – 1) + 5 (x – 1)

= (3x + 5) (x2 + 1)


(v) a3 + a2b + ab + b2

Solution:

= a2 (a +b) + b (a +b)

= (a + b) (a2 + b)


(vi) 3x3 + 5x2 + 3x +5

Solution:

= x2 (3x +5) + 1 (3x + 5)

= (3x + 5) (x2 + 1)


(vii) y4 – 2y3 + y – 2

Solution:

= y3(y – 2) + 1 (y – 2)

= (y – 2) (y3 – 1)

= (y – 2) (y3 – 13)

= (y – 2) (y + 1) (y2 – y +1)


(viii) t4 + t4 – 2t – 2

Solution:

= t3 (t + 1) – 2(t +1)

= (t – 1) (t3 – 2)


3.2.2 Factorization of the difference of two square – Factorization[class 9]:

We know that (a + b)(a – b) = a2 – b2. We can use this identity to factorise many expressions involving the difference of the two squares.

Example 8: Factorise 25x2 – 64y2

Solution:

We write 25x2 – 64y2

25x2 – 64y2 = (5x)2 – (8y)2

This is in the form a2 – b2 which factorises to (a – b)(a + b). Thus,

25x2 – 64y2 = (5x)2 – (8y)2 = (5x – 8y)(5x + 8y)


Factorization[class 9] – Exercise 3.2.1

  1. Factorize the following:

(i) 9x2 – 16y2

Solution:

= (3x)2 – (4y)2

Using a2 – b2 = (a + b) (a – b)

= (3x + 4y) (3x – 4y)


(ii) 5x2 – 7y2 [ Hint : ( 𝟓 )2 = 5 ]

Solution:

( 5 x)2 – ( 7 y)2

Using a2 – b2 = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)


(iii) 100 – 9x2

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a2 – b2 = (a + b) (a – b)]


(iv) (x + 4y)2 – 4z2

Solution:

= (x + 4y)2 – (2z)2 [∵a2 – b2 = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)


(v) x – 64xy4

Solution:

= x (1 – 64y4)

= x (1 + (8y2)2) [∵a2 – b2 = (a + b) (a – b)]

= x (1 + 8y2) (1 – 8y2)


(vi) a2 + 2ab + b2 – 9c2

Solution:

= (a + b)2 – (3c)2 [∵(a + b)2 = a2 + 2ab + b2 ;a2 – b2 = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)


(vii) 4x2 – 9y2 – 2x – 3y

Solution:

= (2x)2 – (3y)2 – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a2 – b2 = (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)


(viii) a2 + b2 – a + b

Solution:

= a2 + b2 – (a – b) [∵a2 – b2 = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)


(ix) x4 – 625

Solution:

= (x2)2 – 252 [∵a2 – b2 = (a + b) (a – b)]

= (x + 25) (x2 – 25)

= (x2 + 25) (x2 – 52)

= (x2 + 25) (x + 5) (x – 5)


(x) 36 (5x + y)2 – 25 (4x – y)2

Solution:

= [6 (5x + y)]2 – [5 (4x – y)]2 [∵a2 – b2 = (a + b) (a – b)]

= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)

= (50x + y) (10x + 11y)


(xi) (a + 𝟏/𝐚 )2 – 4 (x – 𝟏/𝐱 )2

Solution:

= (a + 𝟏/𝐚 )2 – [2 (x – 𝟏/𝐱 )]2

= [a+ 𝟏/𝐚 + 2( x− 𝟏/𝐱)][ a+ 𝟏/𝐚 – 2(x− 𝟏/𝐱 )]


(xii) 12m2 – 75n2

Solution:

= ( √12 m2)2 – (√75 n2)2 [∵a2 – b2 = (a + b) (a – b)]

= (2 √3m2)2 – (5√3 n2)2 [ 12 =2√3 and 75 = 5√3

= (2√3m2 + 5√3 n2) (2√3 m2 – 5√3 n2)

= √3 (2m2 + 5n2) (2m – 5n) (2m + 5n)


  1. Factorize by adding and subtracting appropriate quantity.

(i) x2 + 6x + 3

Solution:

= x2 + 2. x. 3 + 32 – 32 + 3

= (x + 3)2 – 9 + 3

= (x + 3)2 – 6 [∵a2 – b2 = (a + b) (a – b)]

= (x + 3)2 – 62

= (x + 3 + 6 ) (x + 3 – 6 )


(ii) x2 + 10x + 8

Solution:

= x2 + 2. x. 5 + 52 – 52 + 8

= (x + 5)2 – 25 + 8

= (x + 5)2 – 17 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 172

= (x + 5 + 17 ) (x + 5 – 17 )


(iii) x2 + 10x + 20

Solution:

= x2 + 2. x. 5 + 52 – 52 + 20 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 25 + 20

= (x + 5)2 – 5

= (x + 5)2 – 52 [∵a2 – b2 = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )


(iv) x2 + 2x – 1

Solution:

= x2 + 2. x. 1 + 12 – 12 + 1 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 1)2 – 1 + 1

= (x + 1)2 – 2

= (x + 1)2 – 22 [∵a2 – b2 = (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )


  1. prove that (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖 = 1

Solution:

L.H.S = (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

[∵a2 – b2 = (a + b) (a – b)]

= (1 – 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒) + 𝐱𝟖/𝐲𝟖

= (1 – 𝐱𝟒/𝐲𝟒) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

= (1)2 – (𝐱𝟒/𝐲𝟒)2 + 𝐱𝟖/𝐲𝟖

= 1 – 𝐱𝟖/𝐲𝟖 + 𝐱𝟖/𝐲𝟖

= 1 R.H.S


  1. If x = 𝐚+𝐛/𝐚−𝐛 and y = 𝐚−𝐛/𝐚+𝐛 find x2 – y2.

Solution:

x2 – y2 = (x + y) (x – y)

= (a+b/a−b+a−b/a+b)(a+b/a−b a−b/a+b)

= [[(a+b)2+ (a−b)2 ]/(a2− b2)][[(a+b)2− (a−b)2]/(a2− b2)]

= {(a2+ 2ab + b2+ a2+ b2− 2ab)/[(a2− b2)2]} x (a2 + 2ab + b2 – a2 – b2 + 2ab)

= [(2a2+ 2b2)4ab]/[(a2− b2)2]

= [2(a2+ b2)4ab]/(a2− b2)2

= [8ab(a2+ b2)]/(a2− b2)2


  1. If p = x – y, q = y – z and r = z – x, simplify r2 – p + 2pq – q2

Solution:

(z – x)2 – (x – y)2 + 2 (x – y) (y – z) – (y – z)2

= z2 – 2xy + x2 – (x2 – 2xy + y2) + 2 (xy – y2 – xz + yz) – (y2 + z2 – 2yz)

= z2 + x2 – 2xz – x2 + 2xy – y2 + 2xy – 2y2 – 2xz + 2yz – y2 – z2 + 2yz

= 4xy – 4xz + 4yz – 4y2

= 4[xy – xz + yz – y2]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq


  1. The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.

[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord]

Solution:

Factorization[class 9]

Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory

OA2 = OB2 + AB2

132 = x2 + 52

x2 = 132 – 52

= (13 + 5) (13 – 5)

= 18 (8)

x2 = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm


  1. Is it possible to factorize x2 + 4x + 20? Give reason

Solution:

x2 + 4x + 20

= x2 + 2. x. 2 + 22 – 22 + 20

= x2 + 4x + 4 – 4 + 20

= (x + 4)2 + 16

= (x + 4)2 + 42

We cannot factorize their further since this is nit of the form a2 – b2


3.2.3 More on Factorization of Trinomials – Factorization[class 9]:

Example: Factorize x2 + 4√2x + 8

Solution:

We write this in the form

x2 + 2(x)(2√2) + (2√2)2

This is in the form a2 + 2ab + b2 , where a = x, b = 2√2

But we know that a2 + 2ab + b2 = (a + b)2

Thus,

x2 + 4√2x + 8 = (x + 2√2)2

Example: Factorise x2 + 11x + 30

Solution:

Note that 11 = 5 + 6 and 30 = 5 x 6

This helps us to split the trinomial as,

x2 + 11x + 30 = x2 + 5x + 6x + 30

= x(x + 5)+6(x + 5)

= (x + 6)(x + 5)

Thus (x+ 5) and (x + 6) are the factors of x2 + 11x + 30

Example: Factorise (x2 – 2x)2 – 23(x2 – 2x) + 120

Solution:

We introduce a = x2 – 2x . Then the expression is a2 – 23a + 120. We observe that – 23 = (-15) + (-8) and 120 = (-15)(-8). Hence we can write a2 – 23a + 120 = a2 – 15a – 8a + 120 = a(a – 15) – 8(a – 15) = (a – 15)(a – 8)

Hence, (x2 – 2x)2 – 23(x2 – 2x) + 120 = (x2 – 2x – 15)( x2 – 2x – 8)

Further we see,

(x2 – 2x – 15) = x2 – 5x + 3x – 15 = x(x – 5)+3(x -5) = (x + 3)(x – 5)

( x2 – 2x – 8) = x2 – 4x + 2x – 8 = x(x – 4)+2(x – 4) = (x – 4)(x + 2)

We hence obtain

(x2 – 2x)2 – 23(x2 – 2x) + 120 = (x + 3)(x – 5)(x + 2)(x – 4)


Factorization[class 9] – Exercise 3.2.3

  1. Factorize

(i) x2 + 9x + 18

Solution:

= x2 + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

 

(ii) y2 + 5y – 24

Solution:

= y2 + 8y – 3y – 24

= y (y + 8) – 3 (y + 8)

= (y + 8) (y – 3)

 

(iii) 7y2 + 49y +84

Solution:

= 7 (y2 + 7y + 12)

= 7 (y2 + 4y + 3y + 12)

=7 2017

=7 (y + 4) (y + 3)

 

(iv) 40 + 3x – x2

Solution:

= – [x2 – 3x – 40]

= 3 – [x2 – 8x + 5x – 40]

= – [x (x – 8) + 5 (x – 8)]

= [(x – 8) (x + 5)]

= (8 – x) (x + 5)

 

(v) m2 + 17mn – 84n2

Solution:

= m2 + 21mn – 4mn – 84n2

= m (m + 21n) – 4n (m + 21n)

= (m + 21m) (m – 4n)

 

(vi) 117p2 + 2pq – 24q2

Solution:

= 117p2 + 54pq – 52pq – 24q2

= 9p (13p + 6q) – 3q (13p + 6q)

= (9p – 3q) (13p + 6q)

 

(vii) 15x2 – r – 28

Solution:

= 15x2 – 21x + 20x – 28

= 3x (5x – 7) + 4 (5x – 7)

= (3x + 4) (5x – 7)

 

(viii) 2x2 – x – 21

Solution:

= 2x2 – 7x + 6x – 21

= x (2x – 7) + 3 (2x – 7)

= (2x – 7) (x + 3)

 

(ix) 8k2 – 22k – 21

Solution:

= 8k2 – 28k + 6k – 21

= 4k (2k – 7) + 3(2k – 7)

= (2k – 7) (4k + 3)

 

(x) 𝟏/𝟑 x2 – 2x – 9

Solution:

= (x2+ 6x−27)/3

= 1/3 [x2 – 6x – 27]

= 1/3 [x2 – 9x + 3x – 27]

= 1/3 [x (x – 9) + 3 (x – 9)]

=1/3 (x – 9) (x + 3)


  1. Factorize

(i) √𝟓 x2 + 2x – 3√𝟓

Solution:

= √5 x2 + 5x – 3x – 3√5

= √5 x (x + √5 ) – 3 ((x +√ 5 )

= (x + √5 ) (√5 x – 3)

 

(ii) √𝟑 a2 + 2a – 5√𝟑

Solution:

= √3a2 + 5a – 3a – 5√3 – 5√3 – √3

= a (√3 a + 5) – √3 (√3 a + 5)

= (√3 a + 5) (a – √3 )

 

(iii) 7√2 y2 – 10y – 4√2

Solution:

= 7√2 y2 – 14y + 4y – 4√2

= 7√2 y (y – √2) + 4 (y – 2)

= (y – √2) (7√2 y + 4)

 

(iv) 6√𝟑 z2 – 47z – 5√𝟑

Solution:

= 6√3 z2 – 45z – 2z – 5√3

= 3√3 z (2z –5√3) – (2z – 5√3)

= (2z – 5√3) (3√3z –1)

 

(v) 4√𝟑x2 + 5x – 2√𝟑

Solution:

= 4√3 x2 + 8x – 3x – 2√3

= 4x (√3x + 2) – √3(√3x + 2)

= (√3x + 2) (4x –√3)


  1. Factorize

(i) 2 (x + y)2 – 9 (x + y) – 5

Solution:

Let x + y = p

= 2p2 – 9p – 5

= 2p2 – 10p + p – 5

= 2p(p – 5) + 1 (p – 5)

= (p – 5) (2p + 1)

= (x + y – 5) [2(x + y) + 1]

= (x + y – 5) (2x + 2y + 1)

 

(ii) 2(a – 2b)2 – 25(a – 2b) + 12

Solution:

Put a – 2b = x

= 2x2 – 25x + 12

= 2x2 – 24x – x + 12

= 2x (x – 12) – 1 (x – 12)

= (x – 12) (2x – 1)

= (a – 2b – 12) [2(a – 2b) – 1]

= (a – 2b – 12) [2a – 4b – 1]

 

(iii) 12(z + 1)2 – 25(z + 1) (x + 2) + 12 (x + 2)2

Solution:

Let z + 1 = a x + 2 = b

= 12a2 – 25ab + 12b2

= 12a2 – 16ab – 9ab + 12b2

= 4a (3a – 4b) – 3b (3a – 4b)

= (3a – 4b) (4a – 3b)

= [3 (z + 1) – 4 (x + 2)] [4 (z + 1) – 3 (x + 2)]

= (3z + 3 – 4x – 8) (4z + 4 3x – 2)

= (3z – 4x – 5) (4z – 3x – 2)

 

(iv) 9(2x – y) – 4(2x – y) – 13

Solution:

9(2x – y) – 4(2x – y) – 13

Put 2x – y = a

= 9a + 9a – 13a – 13

= 9a (a + 1) – 13 (a +1)

= (a + 1) (9a – 13)

= (2x – y + 1) [9 (2x – y) – 13]

= (2x – y + 1) (18x – 9y – 13)


  1. Factorize

(i) x4 – 3x2 + 2

Solution:

Put x2 = a

a2 – 3a + 2

= a2 – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 2) (a – 1)

= (x2 – 2) (x2 – 1)

= (x – 2 ) (x + 2 ) (x – 1) (x + 1) [∵a2 – b2 = (a + b) (a – b)]


(ii) 4x4 + 7x2 – 2

Solution:

Put x2 = a

= 4a2 + 7a – 2

= 4a2 + 8a – a – 2

= 4a (a + 2) – 1 (a + 2)

= (a + 2) (4a – 1) [∵a2 – b2 = (a + b) (a – b)]

= (x2 + 2) (4x2 – 1)

= (x2 + 2) (2x + 1) (2x – 1)


(iii) 3x3 – x2 – 10x

Solution:

= x (3x2 – x – 10)

= x (3x2 – 6x + 5x – 10)

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5)


(iv) 8x3 + 2x2y – 15xy2

Solution:

= x (8x2 – 2xy – 15y2)

= x (8x2 – 12xy + 10xy – 15y2)

= x [4x (2x – 3y) + 5y (2x – 3y)]

= x (2x – 3y) (4x + 5y)


(v) x6 – 7x3 – 8

Solution:

Put x3 = a

a3 – 7a – 8 [∵a3 + b3 = (a + b) (a2 – ab + b2)]

= a3 – 8a + a – 8 [∵a3 – b3 = (a – b) (a2 – ab + b2)]

= a (a – 8) + 1 (a – 8)

= (a – 8) (a + 1)

= (x3 – 3) (x3 + 1)

= (x3 – 3) (x3 + 1)

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


3.2.4 Some Miscellaneous Factorization – Factorization[class 9]:

Consider the expression a4 + a2b2 + b4 .

a4 + a2b2 + b4 = (a2 + b2 + ab)(a2 + b2 – ab)

Example: Factorize 4x4 + 9y4 + 6x2y2

Solution:

a = √2x , b = √3y we obtain,

a4 + a2b2 + b4 = 4x4 + 6x2y2 + 9y4

We can use the factorization a4 + a2b2 + b4 = (a2 + b2 + ab)(a2 + b2 – ab)

4x4 + 6x2y2 + 9y4 = (2x2 – √6xy + 3y2)(2x2 + √6xy + 3y2)


Factorization[class 9] – Exercise 3.2.4

  1. Resolve into factors

(i) x4 + y4 – 7x2 y2

Solution:

= x4 + y4 – 2x2 y2 – 2x2 y2 – 7x2 y2

= x4 + y4 + 2x2 y2 – 9x2 y2

= (x2 + y2)2 – (3xy)2

= (x2 + y2 + 3xy) (x2 + y2 – 3xy)

 

(ii) 4x4 + 25y4 + 10x2y2

Solution:

Take a = √2x ;b = √5y

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

4x4 + 25y4 + 10x2y2 = (√2 x)4 + (√5 y)4 + 2 x2 5 y2

= [( 2 x2) + ( 5 y2) + (√10 xy] [( 2 x2) + ( 5 y2) – (√10 xy]

= (2x2 +5y2 + 10 xy) (2x2 +5y2 – 10 xy)

(iii)9a4+100b4+30a2b2

Solution:

Take a = √3a ;b = √10b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

9a4+100b4+30a2b2 = (√3a)4+(√10b)4+3a2.10b2

= (3a2+10b2+30a2b2) (3a2+10b2– 30a2b2)

(iv)81a4 +9a2b2+b4

Solution:

Take a = 3a ;b = b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

81a4 +9a2b2+b4 = (3a)4+(3a)2(b)2+b4

= (9a2+b2+3ab)(9a2+b2-3ab)

 

(v) x4 – 6x2y2 + y4

= x4 + y4 + 2x2y2 – 2x2y2 – 6x2y2

= (x2 + y2)2 – ( 8 xy)2

= (x2 + y2 + 8 xy) (x2 + y2 – 8 xy)

 

  1. vi) m4 + n4 – 18m2n2

= m4 + n4 + 2m2n2 – 2m2n2 – 18m2n2

= (m2 + n2)2 – 20 m2n2

= (m2 + n2 + √20 mn) (m2 + n2 – √20 mn)

 

(vii) 4m4 + 9n4 – 24m2n2

= 4m4 + 9n4 + 12 m2n2 – 12 m2n2 – 24m2n2

= (2m2 + 3n2)2 – 36mn

= (2m2 + 3n2 + 6mn) (2m2 + 3n2 – 6mn)

 

(viii) 9x4 + 4y4 + 11x2y2

= 9x4 + 4y4 + 12x2y2 – 12x2y2 + 11x2y2

= (3x2)2 + (2y2)2 + 2 (3x)2 (2y)2 – x2y2

= (3x2 + 2y2)2 – x2y2

= (3x2 + 2y2 + xy) (3x2 + 2y2 – xy)


  1. Find the factors of the following

(i) x4 + 9x2+81

Solution:

= x4+9x2+81+9x2-9x2

= (x2)2+18x2+81-9a2

=(x2+9)2-(3a)2

=( x2+9 +3a)( x2+9 – 3a)

 

(ii) a4 + 4a2 + 16

Solution:

= a4 + 4a2 + 16 + 4a2 – 4a2

= (a2)2 + 8a2 + (4)2 – 4a2

= (a2 + 4)2 – (2a)2

= (a2 + 4 – 2a) (a2 + 4 + 2a)


  1. Factorise the following

(i) 64a4 + 1

Solution:

Adding and subtracting 16a2 we get

64a4 + 1 + 16a2 – 16a2

= (8a2)2 + 1 + 16a2 – 16a2

= (8a2 + 1)2 – (4a)2

= (8a2 + 1 + 4a) (8a2 + 1 – 4a)

 

(ii) 3x4 + 12y4

Solution:

3(x4 + 4y4)

3[(x2)2 + (2y2)2]

By adding and subtracting 2ab i. e.

2 x x2 x 2y = 4x2y2

= 3 [(x2)2 + 2(y2)2 + 4x2y2 – 4x2y2]

= 3 [(x2 + 2y2)2 – (2xy)2]

= 3 [(x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy)]

 

(iii) 4x4 + 81y4

Solution:

(2x2)2 + (9y2)2

By adding and subtracting 2ab i. e.

2 x x2 x 9y2 = 36x2y2

= (2x2)2 + (9y2)2 + 36x2y2 – 36x2y2

= (2x2 + 9y2)2 – (6xy)2

= (2x2 + 9y2 + 6xy) (2x2 + 9y2 – 6xy)

 

(iv) a8 – 16b8

Solution:

(a4)2 – (4b4)2

Using a2 – b2 = (a + b) (a – b)

= (a4 + 4b4) (a4 – 4b4)

= (a4 + 4b4) [(a2)2 – (2b2)2]

= (a4 + 4b4) (a2 + 2b2) (a2 – 2b2)


3.2.5 Some More Identities – Factorization[class 9]:

Identity: a3 + b3 = (a + b)(a2 – ab + b2)

We know that, (a + b)3 = a3 + b3 + 3ab(a + b)

Hence, a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b)[(a+b)2 – 3ab]

However,

(a + b)2 – 3ab = a2 + 2ab + b2 – 3ab = a2 – ab + b2

Using this we obtain,

a3 + b3 = (a+b)(a2 – ab + b2)

Identity 2: a3 – b3 = (a – b)(a2 + ab + b2)

Here we start with (a – b)3 = a3 – b3 – 3ab(a – b) . This gives, as earlier,

a3 – b3  = (a – b)3 + 3ab(a – b) = (a – b)[(a – b)2 + 3ab] = (a – b)(a2 + ab + b2)

Thus,

a3 – b3  = (a – b)(a2 + ab + b2)

Example: Factorize x3 + 27

Solution:

We do this in several steps

Step 1: Write x3 + 27 in the form x3 + 33 . This is in the form a3 + b3.

Step 2: We substitute a = x and b = 3 in the factorization a3 + b3 = (a + b)(a2 – ab + b2)

We get,

x3 + 33 = (x + 3)(x2 – 3x + 32)

Step 3: Simplify the expression. This gives x3 + 27 = (x + 3)(x2 – 3x + 9)


Factorization[class 9] – Exercise 3.2.5

  1. Factorise:

(i) 8y3 – 1

Solution:

= (2y)3 – 13 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (2y – 1) [(2y)2 + 2y .1 + 12]

= (2y – 1) (4y2 + 2y + 1)

 

(ii) 27x3 – 8

Solution:

= (3x)3 – 23 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (3x – 2) [(3x)2 + 3x .2 + 22]

= (3x – 2) (9x2 + 6x + 4)

 

(iii) x3 + 8y3

Solution:

= x3 + (2y)3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (x + 2y) [x2 – x.2y + (2y)2]

= (x + 2y) (x2 – 2xy + 4y)2

 

(iv) 1 – x3

Solution:

= 13 – x3 [∵a3 – b3 = (a – b) (a2 + ab + b2)]

= (1 – x) (12 + 1.x + x2)

= (1 – x) (1 + x + x2)

 

(v) a3 b3 + c3

Solution:

= (ab)3 + c3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (ab + c) [(ab)2 – ab .c + c2]

= (ab + c) (a2b2 – abc + c2)

 

(vi) a3b – 𝐛/𝟔𝟒

Solution:

= b (a31/64)

= b [a3 – (1/4 )3]

= b (a – 1/4) [a2 + a. 1/4 + ( 1/4 )2]

= b (a – 14) (a2 + a/4 + 1/16 )

 

(vii) 𝐚3/𝟖 + 1

Solution:

= ( a/2 )3 + 13 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= ( a/2 + 1) [( a/2 )2a/2 .1 + 12]

= ( a/2 + 1) [a2/4a/3+ 1]

 

(viii) 3a6 𝐛𝟔/𝟗

Solution:

= 3(a6− b6/27)

= 3[(a2)3− (b2/3)3]

= 3[(a2− b2/3) (a2)3− a2.b2/3 (b2/3)2

= 3(a2− b2/3)( a4+a2 . b2/3+ b4/3 )

 

(ix) 2a3 + 𝟏/𝟒

Solution:

= 2 (a3 + 𝟏/𝟒 )

= 2 (a3 + 𝟏/2 )3

= 2 (a + 𝟏/2 ) [ a2 – a. 𝟏/2 + (𝟏/𝟒 )2]

= 2 (a + 𝟏/2 ) (a2a/2 + 𝟏/𝟒 )

 

(x) x3 – 512

Solution:

= x3 – 83 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (x – 8) (x2 + 8x + 64)

 

(xi) 32x3 – 500

= 4 (8x3 – 125)

= 4 [(2x)3 – 53]

= 4 (2x – 5) [(2x)2 + 2x .5 – 52]

= 4 (2x – 5) (4x2 + 10 x + 25)

 

(xii) x7 + xy6

= x (x6 + y6)

= x [(x2)3 + (y2)3]

= x [(x2 + y2) { (x2)2 – x2 y2 + (y2)2}]

= x (x2 + y2) (x4 – x2 y2 + y2)

 

(xiii) 2a4 – 128a

= 2a (a3 – 64)

= 2a (a3 – 43)

= 2a (a – 4) (a2 + 4a + 42)

= 2a (a – 4) (a2 + 4a + 16)


  1. Factorise

(i) (1 – a)3 + (3a)3

Solution:

Using x3 + y3 = (x + y) (x2 – xy + y2)

(1 – a)3 + (3a)3

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

= (1 + 2a) [(1 – a)2 – 3a + 3a2 + 9a2]

= (1 + 2a) (1 + a2 – 2a – 3a + 12a2)

= (1 + 2a) (13a2 – 5a + 1)

 

(ii) 8x3 – 27y3

Solution:

= (2x)3 – ( 3y)3

= ( 2x – 3y) [(2x)2 + 2x .3y + (3y)2]

= (2x – 3y) (4x2 + 6xy + 9y2)

 

(iii) z4 x3 + 8y3 z4

Solution:

= z4 (x3 + 8y3)

= z4 [x3 + (2y)3]

= z4 (x + 2y) [x2 – x.2y + (2y)2]

= z4 (x + 2y) [x2 – 2xy + 4y2]

 

(iv) 3(x + y)3 + 𝟏/9 (xy)3

Solution:

=  (x + y)3 + 1/27 (xy)3

= (x + y)3 + ( xy/3 )3

= (x + y + xy/3 ) (x + y)2 – [ (x + y) ( xy/3 )+ ( xy/3)2]

= (x + y + xy/3 ) (x2 + y+ 2xy – (x + y)( xy/3) + x2y2/9 )

 

(v) x6 + y6

Solution:

= (x2)3 – (y2)3

= (x2 – y2) [(x2)3 + x2 y2– (y2)3]

= (x2 – y2) (x4 + x2 y2 + y4)

= (x + y) (x – y) (x2 + x y + y2) (x2 – xy + y2)

 

(vi) a3 – 2√𝟐 b3

Solution:

= a3 – (√2 b)3

= (a –√2b) [a2 + a.√2 b + √2 b2]

= (a –√𝟐 b) (a2 + √𝟐 ab + √𝟐b2)


  1. Factorize the following

(i) x6 – 26x3 – 27

Solution:

put x3 = a

a3 – 26a – 27

= a3 – 27a + a – 27

= a ( a – 27) + 1 (a – 27)

= (a – 27) ( a + 1)

= (x3 – 27) (x3 + 1)

= (x – 3) (x2 + 3x + 9) (x – 1) (x2 – x + 1)

 

(ii) z6 – 63z3 – 64

Solution:

put z3 = a

a3 – 63a – 64

= a3 – 64a + a – 64

= a (a – 64) + 1 (a – 64)

= (a – 64) (a + 1)

= (z3 – 64) (z3 + 1)

= (z – 4) (z2 + 4z + 16) (z + 1) (z2 – z + 1)

 

(iii) a3 – b3 – a + b

Solution:

= (a – b) (a2 + ab + b2) – (a – b)

= (a – b) [ a2 + ab + b2 – 1]

 

(iv) x6 + 7x3 – 8

Solution:

put x3 = a

a3 + 7a – 8

= a3 + 8a – a – 8

= a (a + 8) – 1 (a + 8)

= (a – 1) (a + 8)

= (x3 – 1) (x3 + 8)

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

 

(v) a3 𝟏/𝐚𝟑 – 2a + 𝟐𝐚

Solution:

= (a – 𝟏/𝐚) (a2 + a. 𝟏/𝐚+ 𝟏/𝐚2 ) – 2 (a – 𝟏/𝐚 )

= (a – 𝟏/𝐚 ) [a2 + 1 + 𝟏/𝐚2 – 2]

= (a – 𝟏/𝐚 ) (a2 + 𝟏/𝐚𝟐 – 1 )


3.2.6 Factorization of some Cubic Expressions – Factorization[class 9]

  1. a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

We have,

a3 + b3 + c3 – 3abc = (a3 + b3) + c3 – 3abc

[use a3 + b3 = (a + b)3 – 3ab(a + b)]

= (a + b)3 + c3 – 3ab(a + b + c)

[use x3 + y3 = (x + y)(x2 – xy + y2)]

= (a + b + c)[(a + b)2 – (a + b)c + c2] – 3ab(a + b + c)

= (a + b + c)[a2 + b2 + 2ab – ac – bc + c2 – 3ab]

=(a + b + c)(a2 + b2 + c2 –  ab – bc – ca)

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 –  ab – bc – ca)


Corollary: If a + b + c = 0 then a3 + b3 + c3 = 3abc

Proof :

We use the identity

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 –  ab – bc – ca)

Substituting a + b + c = 0, we obtain ,

a3 + b3 + c3 – 3abc = 0

Hence, a3 + b3 + c3 = 3abc

Such an identity is called a conditional identity. note that this is not valid for all choices of a, b, c it is valid only for those a, b, c satisfying a + b + c = 0

Alternatively, we can also do this as follows, Since a + b + c = 0 we get a + b = -c

Hence

(-c)3 = (a+ b)3 = a3 + b3 + 3ab(a+b) = a3 + b3 + 3ab(-c)

This simplifies to

a3 + b3 + c3 – 3abc = 0

or

a3 + b3 + c3 = 3abc


Factorization[class 9] – Exercise 3.2.6

  1. Factorise

(i) a3 – b3 – c3 – 3abc

Solution:

= a3 + (– b3)+(– c3) – 3a(–b) ( –c)

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = a y = –b c = –c

∴ a3 + (– b3)+(– c3) – 3a(–b) (c)

= [a + (–b) +(c)] [a2 + (–b)2 + (–c) – a(–b) (–b)(–c) – (–c)a]

= (a – b – c) (a2 + b2 + c2 + ab – bc + ca)

 

(ii) a3 – b3 + 8c3 + 6abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = a y = –b c = 2c

a3 + (–b)3 + (2c)3 – 3a (–b) (2c)

= [a + (– b) + 2c] [a2 + (-b)2 + (2c)2 – a(– b) – (–b) (2c) – 2c.a]

= (a – b + 2c) (a2 + b2 + 4c2 + ab + 2bc – ca)

 

(iii) 125a3 + b3 + 64c3 – 60abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = 5a y = b c = 4c

(5a)3 + b3 + (4c)3 – 3(5a) b (4c)

= (5a + b + 4c) [(5a)2 + b2 + 16c2 – 5a(b) – b(4c) – 4(c) (5a)]

= (5a + b + 4c) (25a2 + b2+ 16c2 – 5ab – 4bc – 20ca)

 

(iv) 1 + b3 + 8c3 – 6bc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = 1 y = b c = 2c

13 + b3 + (2c)3 – 3 . 1 . b . 2c

= (1 + b + 2c) [12 + b2 + (2c)2 – 1 .b – b.2c – 2c.1]

= (1 + b + 2c) (1 + b2 + 4c2 – b – 2bc – 2c)

 

(v) 8a3 + 125b3 – 64c3 + 120abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = 2a y =5b c = –4c

(2a)3 + (5b)3 + (–4c)3 – 3(2a) (5b) (-4c)

= [2a + 5b + (–4c)] [(2a)2 + (5b)2 + (4c)2 – (2a)(5b) – (5b)(-4c) –

(–4c) (2a)]

= (2a + 5b – 4c) (4a2 + 25b2 + 16c2 – 1ab + 20bc + 8ca)

 

(vi) 2√𝟐 a3 + 16√𝟐 b3 + c3 – 12abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = √2a ;y =2√2 b;  c = c

(√2 a)3 + (2 √2 b)3 + c3 – 3(√2 a) (2√2 b)c

= (√2 a + 2√2 b + c) [(√2 a)2 + (2√2 b)2 + c2 -√2 a. 2√2 b –

2√2 b . c – c. √2 a]

= (√𝟐 a + 2√𝟐 b + c) (2a2 + 8b2 + c2 – 4ab – 2√𝟐 bc – √𝟐 ac)

 

(vii) (x – y)3 + (y – z)3 + (z – x)3

[hint: apply corollary]

Solution:

Using a + b + c =0 then a3 + b3 + c3 = 3abc using this

a = x – y b = y – z c = z – c

we have a + b + c = x – y + y – z + z – c = 0

∴ (x – y)3 + (y – z)3 + (z – x)3

= 3 (x – y) (y – z) (z – x)

 

(viii) p3 (q – r)3 + q3 (r – p)3 + (p – q)3

[Hint: apply corollary]

Solution:

Using identify if a + b + c = 0 then a3 + b3 + c3 = 3abc we get

a = p(q – r) b = q( r – p) c = r(p – q)

a + b + c = p(q – r) + q( r – p) + r(p – q)

= pq – pr – qr – pq + pr – qr

= 0

∴ p3 (q – r)3 + q3 (r – p)3 + (p – q)3

= 3p (q – r) q(r – p) r(p – q)

= 3pqr (q – r) (r – p) (p – q)


  1. Find the product using appropriate identity

(i) (a – b – c) (a2 + b2 + c2 + ab – bc – ca)

Solution:

Using (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = x3 + y3 + x3

We get x = a y = –b z = –c

(a – b – c) (a2 + b2 + c2 + ab – bc – ca)

= a3 + (-b)3 + (-c)3

= a3 – b3 – c3 – 3abc

 

(ii) (x – 2y – z) (x2 + 4y2 + z2 + 2xy – 2yz – zx)

Solution:

Using (a + b + c) (a2 + b2 + c2 – ab – bc ca) = a3 + b3 + c3 – 3abc

We get a = x b = –2y c = –z

[x + (–2y) + (–z)] [ x2 +(–2y)2 +(-z)2 – x (-2y) – (-2y) (-z) + (-z)(x)]

= x3 + (–2y)3 + (-z)3 – 3x (–2y) (–8)

= x3 – 8y3 – z3 – 6xyz

 

(iii) (x + y – z) (x2 + y2 + z2 – xy + yz + zx)

Solution:

Using (a + b + c) (a2 + b2 + c2 – ab – bc ca) = a3 + b3 + c3 – 3abc

We get a = x b = y c = –z

[x + y + (–z)] [x2 + y2 + (–z)2 – xy + y(–z) + (–z) x]

= x3 + y3 + (-z)3 – 3(x) (y) (-z)

= x3 + y3 – z3 + 3xyz


  1. Get the factorization

(a + b + c)3 – a3 – b3 – c3 = 3 (a+ b) (b + c) (c + a) writing the expression

(a + b + c)3 – a3 – b3 – c3 = {(a + b + c)3 – a3} – {b3 + c3}

Solution:

We have to prove {(a + b + c)3 – a3} – {b3 + c3}

= 3(a + b) (b+ c) (b + c)

L.H.S = {(a + b + c)3 – a3} – {b3 + c3}

Using identity a3 – b3 = (a – b) (a2 + ab + b2)

= (a + b + c – a) {(a + b + c)2 + (a + b + c)a + a2} – (b – c) (b2 – bc + c2)

= (b + c) [(a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab + ac + a2) – (b2 – bc + c2)

= (b + c) [(3a2 + b2 + c2 + 3ab + 2bc + 3ca] – (b + c) (b2 – bc + c2)

= [3a2 + b2 + c2 + 3ab + 2bc + 3ca – b2 + bc – c2]

= (b + c) (3a2 + 3ab + 3bc + 3ca)

= (b + c) 3[a (a + b) + c (a + b)]

= 3 (b + c) (a + b) (a + c)

= 3 (a + b) (a + c) (b + c)

= R.H.S


  1. If x + y + 4 = 0, find the value of x3 + y3 – 12xy + 64.

Solution:

x + y = – 4

Cubing on both sides

(x + y)3 = (– 4)3

x3 + y3 + 3xy (x + y) = (– 4)3

x3 + y3 + 3xy (– 4)3 = – 64

x3 + y3 – 12xy + 64 = 0


  1. If x = 2y + 6, find the value of x – 8y – 36xy – 216.

Solution:

x – 2y = 6

Cubing on both sides

(x – 2y)3 = (6)3

= x3 – (2y)3 – 3x (2y) (x – 2y) = 216

= x3 – 8y3 – 6xy (6) – 216 = 0

= x3 – 8y3 – 36xy – 216 = 0


  1. Without actually calculating the cubes, find the values of the following:

(i) (–12)3 + 73 + 53

Solution:

We have – 12 + 7 + 5 = 0

By the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–12)3 + 73 + 53 = 3 (–12) (7) (5)

= -1260

 

(ii) (28)3 + (–315)3 + (–13)3

Solution:

We find that 28 – 15 – 13 = 0

By the identify if a + b + c = 0 then a3 + b3 + c33abc we get

(28)3 + (–15)3 + (–13)3 = 3 x 28 x (–15) (–13)

= 16380

 

(iii) (–15)3 + 73 + 83

Solution:

Since – 15 + 7 + 8 = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–15)3 + 73 + 83 = 3 (–15) 7 x 8

= – 2520

 

(iv) (–10)3 + 33 + 73

Solution:

Since – 10 + 3 + 7 = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–10)3 + 33 + 73 = 3(-10) (3) (7)

= – 630


  1. Factorise the following using the identity

a3 + b3 + c3 – 3abc = 𝟏/𝟐 (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]

(i) a3 + 8b3 + c3 – 6abc

Solution:

= 1/2 (a + 2b + c) [(a – 2b)2 + (2b – c)2 + (c – a)2]

= 1/2 (a + 2b + c) (a2 + 4b2 – 4ab + 4b2 + c2 – 4bc + c2 +

a2 – 2ac)

= 1/2 (a + 2b + c) (2a2 + 8b2 + 2c2 – 4ab – 4bc – 4ca)

 

(ii) 125a3 + b3 + 64c3 – 60abc

Solution:

= 1/2  (5a + b + 4c) [(5a – b)2 + (b – 4c)2 + (4c – 5a)2]

= 1/2  (5a + b + 4c) (25a2 + b2 – 10ab + b2 + 16c2 – 8bc + 16c2

+ 25a2 – 40ac)

= 1/2 (5a + b + 4c) (50a2 + 2b2 + 32c2 – 10ab – 8bc – 40ac)

 

(iii) 13 + b3 + 8c3 – 6bc

Solution:

= 1/2 (1 + b + 2c) [(1 – b)2 + (b – 2c)2 + (2c – 1)2]

= 1/2 (1 + b +2c) (1 + b2 – 2b + b2 – 4bc + 4c2 + 4c2 + 1 – 4c)

= 1/2 (1 + b + 2c) (2 + 2b2 + 8c2 – 2b – 4bc – 4c)

 

(iv) 125 – 8x3 – 27y3 – 90xy

Solution:

= 1/2 (5 – 2x – 3y) [(5 – 2x)2 + (–2x – 3y)2 + (–3y – 5)2]

= 1/2 (5 – 2x – 3y) (25 + 4x2 – 20x + 4x2 + 9y2 – 12xy + 9y2 +

25 – 30y)

= 1/2 (5 –2x – 3y) (50 + 8x2 + 18y2 – 20x – 12xy – 30y)


  1. Factorise the following:

(i) (x – 2y)3 + (2y – 3z)3 + (3z – x)3

Solution:

We find that x – 2y + 2y – 3z + 3z – x = 0

Hence using identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(x – 2y)3 + (2y – 3z)3 + (3z – x)3

= 3(x – 2y) (2y – 3z) (3z – x)

 

(ii) [(𝐱𝟐− 𝐲𝟐)𝟑+ (𝐲𝟐− 𝐳𝟐)𝟑+(𝐳𝟐− 𝐱𝟐)𝟑(𝐱 − 𝐲)𝟑+ (𝐲 − 𝐳)𝟑+(𝐳 − 𝐱)𝟑]/[(x – y)3 (y – z)3 (z – x)3]

Solution:

We find that x2 – y2 + y2 – z2 + z2 – y2 = 0

and x – y + y – z + z – x = 0, using the identify if

a + b + c = 0 then a3 + b3 + c33abc we get

[(x2− y2)3+ (y2− z2)3+(z2− x2)3] /[(x − y)3+ (y − z)3+(z − x)3]

= [3(x2− y2)+ (y2− z2)+(z2− x2)3]/[(x – y) + (y – z)+(z – x)]

= [(x – y)(x + y)(y – z)(y + z)(z – x)(z + x)] /[(x – y)(y – z)(z – x)]

[∵ a2 – b2 = (a – b) (a + b)]

= (x + y) (y + z) (z + x)

 

(iii) (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

Solution:

We find that 2x – 3y + 4z – 2x + 3y – 4z = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

= 3 (2x – 3y) (4z – 2x) (3y – 4z)

= 6 (2x – 3y) (4z – 2x) (3y – 4z)

 

(iv) (a – 3b)3 + (3b – c)3 + (c – a)3

Solution:

We find that a – 3b + 3b – c + c – a = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(a – 3b)3 + (3b – c)3 + (c – a)3

= 3 (a – 3b) (3b – c) (c – a)


  1. Factorise the following:

(i) (x – y – z)3 – x3 + y3 + z3

Solution:

= [x + (–y) + (–z)]3 – x3 – (–y)3 – (–z)3

Using the identify if (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = x; b = –y; c = – z

= 3[x + (–y)] [(–y) + (–z)] [(–z) + x]

= 3(x – y) (–y –z) (–z + x)

= – 3(x – y) (y + z) (x – z)

= 3(x – y) (y + z) (z – x)

 

(ii) (a – b – 1)3 – a3 + b3 + 1

Solution:

= [a + (–b)3 + (–1)3 – a3 – (–b)3 – (–1)3

Using the identify if (x + y + z)3 – x3 – y3 – z3 = 3(x + y) (y +z) (z +x) we get

x = a y = –b z = – 1

= 3 [a + (–b)] [(–b) + (–1)] [(–1) + a]

= 3(a – b) (–b – 1) (–1 + a)

= 3 (a – b) (1 + b) (a – 1)

 

(iii) (2x + y – z)3 – 8x3 – y3 + z3

Solution:

= [2x + y + (–z)]3 – (2x)3 – y3 – (–z)3

Using the identify if (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 2x b = y c = – z

= 3(2x + y) 2017 [(–z) + 2x]

= 3 (2x + y) (y – z) (-z + 2x)

= 3 (2x + y) (y – z) (2x – z)

 

(iv) ( a – 𝟏/𝟐 b + 𝟐/𝟑 c )3 – a3 𝟏/𝟖 b3 𝟏/𝟐𝟕 c3

Solution:

= (a – (𝟏/𝟐 b) + 𝟐/𝟑 c )3 – a3 – ( 1/8 b)3 – 𝟏/𝟐𝟕 c3

Using the identify if (x + y + z)3 – x3 – y3 – z3 = 3(x + y) (y +z) (z +x) we get

x = a ;y = – b/𝟐; z = – c/𝟑

= 3 (a – b/𝟐 ) ( −b2 + c/3 ) ( c/3 – a)

= 3 (a – b/𝟐 ) ( c/3b/2) (a + c/3 )

 

(v) (x + y + z)3 – (2x – y)3 – (2y – z)3 – (2z – x)3

Solution:

Let us consider 2x – y + 2y – z + 2z – x = (x + y + z)

∴ The given problem can be written as

(2x – y + 2y – z + 2z – x)3 – (2x – y)3 – (2y – z)3 – (2z – x)3

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 2x – y; b = 2y – z; c = 2z – x

= 3{(2x – y) + (2y – z)}{(2y – z) + (2z – x)}{(2z – x) + (2x – y)}

= 3 (2x – y + 2y – z) (2y – z + 2z – x) (2z – x + 2x – y)

= 3 (2x + y – z) (2y + z – x) (2z + x – y)

 

(vi) (x + y + z – 3)3 – (x – 1)3 – (y – 1)3 – (z – 1)3

Solution:

= [(x – 1) + (y – 1) + (z – 1)]3 – (x – 1)3 – (y – 1)3 – (z – 1)3

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = x – 1; b = y – 1; c = z – 1

= 3 {x – 1 (y – 1)} + {y – 1 + (z – 1)} {z – 1 + (x – 1)}

= 3 (x – 1 + y – 1) (y – 1+ z – 1) (z – 1 + x – 1)

= 3 (x + y – z) (y + z – 2) (z + x – 2)


  1. Find the factors of the following numbers

(i) 303 – 123 – 103 – 83

Solution:

We find that 12 + 10 + 8 = 30

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 12; b = 10 ;c = 8

= 3 (12 + 10) (10 + 8) (8 + 12)

= 3 x 22 x 18 x 20

= 24 x 33. 5.11

 

(ii) 853 – 683 + 53 – 223

Solution:

We find that 68 – 5 + 22 = 85

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 68 b = –5 c = 22

= 3 {68 + (–5)} {–5 + 22} {22 + 68} = 3 x 63 x 17 x 90

= 2.35.5.7.17

 

(iii) 1003 – 493 + 103 – 613

Solution:

We find that 49 – 10 + 61 = 100

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 49; b = –10; c = 61

= 3 [49 + (10)] (–10 + 61) (61 + 49)

= 3 x 39 x 51 x 110

= 2.33.5.11.13.17

 

  1. Prove that

(x + y + z)3 – (x + y – z)3 – (y + z – x)3 – (z + x – y)3 = 24xyz

[Hint: find a relation between x + y + z and x + y – z, z + x – y]

Solution:

L.H.S. (x + y + z)3 – (x + y – z)3 – (y + z – x)3 – (z + x – y)3

= (x + y + z)3 – (x + y – z)3 – [(y + z – x)3 + (z + x – y)3]

= [x + y + z – (x + y – z)] [(x + y + z)2 + (x + y + z) (x + y – z)] –

(x + y – z)2] – (y + z – x + z + x – y)] [(y + z – x)2 – (y + z – x)

(z + x – z) + (z + x –y)2]

= [2z {y2 + z2 + x2 + 2yz – 2zx – 2xy – (yz + z2 – zx + xy + zx – x2 – y2

– yz + xy) + z + x + y + 2xy – 2xy – 2yz}]

= [2z (3x2 + 3y2 + z2 + 6xy) – 2z (y2 + z2 + x2 + 2yz – 2zx – 2xy – yz – z2

+ zx – xy – zx + x2 + y2 + yz – xy + z2 + x2 + y2 + 2xy – 2xy – 2yz)

= (x2z + 6y2z + 2z3 + (x + y) – 2z (3x2 + 3y2 – 6xy)

= 6x2z + 6y2z + 12xyz – 6x2z – 6y2z – 2z3 + 12xyz

= 24xyz

= R.H.S


 

Multiplication of Polynomials – Full chapter – Class IX

After studying chapter Multiplication of Polynomials you learn to find the product of a monomial and a binomial; find the product of two polynomials; find the product of binomials using some special identities; prove some of the important identities; expand the binomial terms using the identities; solve problems under identities; prove problems under identities; prove some of the important conditional identities.

3.1.1 Introduction to Multiplication of Polynomials

Definition: An algebraic expression of the form,

f(x) = a0 + a1x + a2x2 + a3x3 + … + anxn

where, a0, a1, a2, … an are real numbers and n a non-negative integer, is called a polynomial with real coefficients or (real polynomial)

Here a0, a1, a2, … an are called the coefficients of the polynomial f(x) and a0 ,a1x , a2x2 ,a3x3, … ,anxn are called the terms of f(x). If an ≠ 0, we say that n is the degree of the polynomial f(x).

3.1.2 Some Products – Multiplication of Polynomials

  1. Product of a monomial by a binomial:

Consider a monomial ab and a binomial 2a + 1. We multiply them as follows:

ab(2a + 1)  = ab(2a) + ab = 2a2b + ab

Example: Multiply  caa and b2+2bc

Solution: Using distributive property

c2a(b2 + 2bc) = c2ab2 + c2a(2bc) = c2ab2 + 2abc3

  1. Product of a binomial by a binomial:

Example: Find the product (x + 2) and (x + 3)

Solution:

We use our rule: multiply term by term. Thus,

(x + 2)(x + 3) = x(x + 2) + 3(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6

  1. Some special products of binomials:

Some products occur so frequently In algebra that it is advantageous to recognize them by sight and write their product by memory. This will be particularly useful when we talk about factoring.

(a) Difference of two squares:

(a+b)(a – b) = a(a – b) + b(a – b)= a2 – ab + ab – b2 = a2 – b2

(b) Square of a binomial:

(a + b)2  = (a+b)(a+b) = a(a+b)+b(a+b) = a2 + ab  + ab + b2 =a2 + 2ab  + b2

(a – b)2  = (a-b)(a-b) = a(a-b)-b(a-b) = a2 – ab  – ab + b2 =a2 – 2ab  + b2

Example : Find the product (3 + √2)(3 + √2)

Solution:

(3 + √2)(3 + √2) = 3(3 + √2)+ √2(3 + √2) = 9 +3√2 + 3√2 + 2 = 9 + 6√2 + 2 = 11 + 6√2


Multiplication of Polynomial – Exercise 3.1.2

  1. Evaluate the following products:

(i) ax2 ( bx + c)

= ax2 (bx) + ax2 (c)

= abx3 + acx2


(ii) ab (a+b)

= ab (a) + ab (b)

= a2b + ab2


(iii) a2b2 (ab2+a2b)

= a2b2 (ab2) + a2b2 (a2b)

= a3b4 + a4b3


(iv) b4(b6 + b8)

= b4 (b6) + b4 (b8)

= b4 (b6) + b4 (b8)

= b10 + b12


  1. Evaluate the following products:

(i) (x+3) (x+2)

= (x+3) x + (x+2) x

= x2 + 3x + 2x + 6

= x2 + 5x + 6


(ii) (x+5) (x–2)

= (x+5) x + (x+5) (–2)

= x2 +5x – 2x –10

= x2 + 3x –10


(iii) ( y – 4 ) ( y + 6 )

= (y – 4) y +(y – 4)6

= y2 – 4y + 6y – 24

= y2 + 2y – 24


(iv) (a–5) (a–6)

= (a–5) a + (a–6) (–6)

= a2 – 5a – 6a + 30

= a2 – 11a +30


(v) (2x+1) (2x–3)

= (2x+1)2x + (2x+1) (–3)

= 4x2 + 2x – 6x – 3x

= 4x2 – 4x –3


(vi) ( a + b ) ( c + d )

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

= ac + bc + ad + bd


(vii) ( 2x – 3y ) ( x – y )

= (2x – 3y) x + (2x – 3y) (–y)

= 2x2 – 3xy – 2xy +3y2

= 2x2 – 5xy + 3y2


(viii) ( √𝟕𝐱 + √𝟓 ) (√𝟓𝐱 + √𝟕 )

= (√𝟕𝐱 + √𝟓 ) (√𝟓𝐱) + (√𝟕𝐱 + √𝟓 ) (√𝟕)

= √𝟑𝟓 x2 + 5x + 7x + √𝟑𝟓

= √𝟑𝟓x2 + 12x + √𝟑𝟓


(xi) (2a+3b) (2a–3b)

= (2a+3b) 2a + (2a+3b) (–3b)

= 4a2+ 6ab – 6ab – 9b2

= 4a2 – 9b2


(xii) (6xy–5) (6xy+5)

= (6xy – 5) (6xy) + (6xy – 5) 5

= 36x2y2 – 30xy + 30xy – 25

= 36x2y2 – 25


(xiii)(2/x+3) (2/x –7)

= (2/x+3)2/x +(2/x+3)(-7)

= 4/x2 + 6/x14/x – 21

= 4/x28/x –21


  1. Expand the following using appropriate identity:

(i) (a +5)2

Using (a + b)2 = a2 +2ab +b2 we get

a = a b = 5

(a +5)2 = a2 +2.a.5 +b2

= a2 +10a +25


(ii) (2a +3)2

Using (a + b)2 = a2 +2ab +b2 we get

a = 2a; b = 3

(2a +3)2 = (2a)2 +2.2a.3 +b2

= 4a2 + 12a + 9


(iii) ( x + 𝟏/𝐱 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 2a; b = 1/x

(X + 𝟏/ )2 = x2 + 2.x. 1x + (𝟏/𝐱 )2

= x2 + 2 + (𝟏/x )2


(iv) ( √(12a) + √(6b) )2

Using (a + b)2 = a2 +2ab +b2 we get

a = √(12a) and b = √(6b)

(√(12a) + √ (6b))2= (√12a)2 + 2.√12 a + √(6b) + (√(6b))2

= 12a2 +2√72 ab +6b2

= 12a2 +2 √ (36 ×2) ab +6b2

= 12a2 + 12 √𝟐ab +6b2


(v) (𝛑 + 𝟐𝟐/7 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 𝛑 and b = 22/7

(𝛑 + 22/7 )2 = 𝛑2 + 2. 𝛑 22/7 + (22/7)2

= 𝛑2 + 44π/7 + ( 22/7)2

= 𝛑2 + 𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/49


(vi) (y – 3)2

Using (a – b)2 = a2 – 2ab + b2

a = y and b = –3

(y – 3)2 = y2 – 2. y.3 + 32

= y2 – 6y + 9


(vii) (3a – 2b)2

Using (a – b)2 = a2 – 2ab + b2

a = 3a and b = –2b

(3a – 2b)2 = (3a)2 – 2.3a.2b + (2b)2

= 9a2 – 12ab + 4b2


(viii) ( y – 𝟏/y )2

Using (a – b)2 = a2 – 2ab + b2

a = y and b = 𝟏/y

(y – 𝟏/y)2 = y2 – 2.y. 𝟏/y + (𝟏/y )2

= y2 – 2 + 𝟏/


(ix) ( √(10x) – √(5y))2

Using (a – b)2 = a2 – 2ab + b2

a = √(10x) and b = √(5y)

(√(10x) – √(5y))2

= (√(10x))2 – 2. √(10x).√(5y)+ √(5y))2

= 10x2 – 2√50 xy + 5y2

= 10x2 – 2.5√2 xy + 5y2

= 10x2 – 10√2 xy + 5y2


(x) (𝛑 – 𝟐𝟐/7 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 𝛑 and b = 𝟐𝟐/𝟕

(𝛑 – 𝟐𝟐/ )2 = 𝛑2 – 2. 𝛑 𝟐𝟐/𝟕 + (𝟐𝟐/𝟕)2

= 𝛑244/𝟕 π + (𝟐𝟐/𝟕)2

= 𝛑2𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/𝟒𝟗


(xi) (2x+3) (2x+5)

Using (x + a) (x + b) = x2 + x (a + b) ab we get

x = 2x, a = 3 and b = 5

(2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5

= 4x2 +16x +15


(xii) (3x – 3) (3x + 4)

Using (x + a) (x + b) = x2 + x (a + b) ab we get

x = 3x, a = –3 and b = 4

(3x – 3) (3x + 4) = (3x)2 + 3x [(–3)+(4)] + (-3)4

= 9x2 + 3x –12

= 9x2 + 3x –12


  1. Expand : 

(i) (x + 3 ) (x – 3)

Using (a + b) (a – b) = a2 – b2 we get

a = x, b = 3

(x + 3) (x – 3) = x2 – 32

= x2 – 9


(ii) (3x – 5y) (3x + 5y)

Using (a + b) (a – b) = a2 – b2 we get

a = 3x, b = 5y

(3x – 5y) (3x + 5y) = (3x)2 – (5y)2

= 9x2 – 25y2


(iii) (x/3+y/2)( x/3y/2)

Using (a + b) (a – b) = a2 – b2 we get

a = x/3, b = y/2

(x/3+ y/2)( x/3y/2)= (x/3 )2 – (x/3)2

= 𝐱𝟐/9-𝐲2/𝟒


(iv) (x2 + y2) (x2 – y2)

Using (a + b) (a – b) = a2 – b2 we get

a = x2, b = y2

(x2 + y2) (x2 – y2) = (x2)2 – (y2)2

= x4 – y4


(v) (a2 + 4b2) (a + 2b) (a – 2b)

Using (a + b) (a – b) = a2 – b2 for 2nd and 3rd term we get

(a2 + 4b2) (a + 2b) (a – 2b) = (a2 + 4b2) [a2 – (2b)2]

= (a2 + 4b2) (a2 – 4b2)

Using the above identity once again we get

= (a2)2 – (4b2)2

= a4 – 16b4


(vi) (x – 4) (x + 4) (x – 3) (x + 4)

Using (a + b) (a – b) = a2 – b2 we get

(x – 4) (x + 4) (x – 3) (x + 4) = (x2 – 42) (x2 – 32)

= (x2 – 16) (x2 – 9)

Using (a + b) (a – b) = x2 – x (a + b) + ab

= (x2)2 – x (16+9) +16×9

= x4 – 25x2 +144


(vii) (x – a) (x + a)(𝟏/𝐱𝟏/𝐚)( 𝟏/𝐱 +𝟏/𝐚 )

(x2 – a2) )(( /𝐱 )2– (1/a )2)

x2 x 𝟏/𝐱 2 – x2 x 𝟏/𝐚 2 – a2 x 𝟏/𝐱 2 + a2 x 𝟏/𝐚 2

1 – x2/a2 − a2/ x2 + 1

2 – 𝐱𝟐/a𝟐 − 𝐚𝟐 /x𝟐


  1. Simplify the following:

(i) (2x – 3y)2 + 12xy

= (2x)2 + (3y)2 – 2.2x.3y + 12xy

= 4x2 + 9y2 -12xy +12xy

= 4x2 + 9y2

(ii) (3m + 5n)2 – (2n)2

= (3m)2 + (5n)2 + 2.3m.5n – 4n2

= 9m2 + 25n2 + 30mn – 4n2

= 9m2 + 30 mn +21n2

(iii) (4a – 7b)2 – (3a)2

= (4a)2 – 2.4a.7b + (7b)2 – (3a)2

= 16a2 – 56ab + 49b2 – 9a2

= 7a2 -56ab + 49b2

(iv) (x + 𝟏/x)2 (m + 𝟏/𝐦 )2

= (x2 + 2. x. 𝟏/x + 𝟏/x2)2 – (m2 + 2. m. 𝟏/𝐦 + 𝟏/𝐦 2)2

= x2 + 2 + 𝟏/x 2 – (m2 + 2 + 𝟏/𝐦 2)

= x2 + 2 + 𝟏/x 2 – m2 + 2𝟏/𝐦 2

= x2 – m2+ 𝟏/x 𝟐𝟏/𝐦 𝟐 + 4

(v) (m2 + 2n2)2 – 4m2n2

= m4 +2m4.2n2 +4n4 – 4m2 n2

= m4+ 4m2n2 + 4n2 – 4m2n2

= m4 + 4n2

(vi) (3a – 2)2 – (2a -3)2

= (9a2 – 2.3a.2 + 22) – (4a2 – 2.3a.2 + 92)

= 9a2 –12a + 4 – 4a2 + 12a – 9

= 5a2 – 5

=5(a2 – 1)


3.1.3 Identities – Multiplication of Polynomials

Consider 5x + 10 = 15 and (2x + 3)2 = 4x2 + 12x + 9. Only x = 1 satisfies the first relation, whereas any value of x satisfies the second relation. We say 5x + 10 = 15 is an equation and (2x + 3)2 = 4x2 + 12x + 9 is an identity. Thus an identity is an equation which is true for all values of the variables in it.

Multiplication of Polynomials

Proof of identities:

Identity: (x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc

Let us expand LHS, (x + a)(x + b)(x + c) = [x2 + xb + ax + ab](x + c)

= x3 + x2b +  ax2 + abx + x2c + xbc + axc + abc

= x3 + x2(a + b + c) + x(ab + bc + ac) + abc

This proves the identity.

Example 7: Find the product of the binomials (x+2), (x + 3) and (x + 4)

Solution:

(x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc

Here a = 2; b = 3; c = 4 then,

(x + 2)(x + 3)(x + 4) = x3 + x2(2+3+4)+x(2×3 + 3×4 + 4×2) + (2x3x4)

= x3 + x2(9) + x(6 + 12 + 8) + (24)

= x3 + 9x2 + 26x + 24

Identity: (a+b)3 = a3 + 3a2b + 3ab2 + b3

If a = b = c then, the product (x+a)(x+b)(x+c) simply reduces to (x+a)3. In this case a+ b+ c = a+a+a = 3a, ab+bc+ca = a2 + a2 + a2 = 3a2 and abc = a3 thus we obtain,

(x + a)3 = x3 + 3x2a + 3xa2 + a3

Taking x = b in this, we obtain,

(a+b)3 = a3 + 3a2b + 3ab2 + b3

This can also be written as,

(a + b)3 = a3 + 3ab(a+ b) + b3

Example: Expand (2x + 3y)3

Solution: We use the above identity: (a + b)3 = a3 + 3a2b + 3ab2 + b3 . Take a = 2x and b = 3y and we get,

(2x + 3y)3 = (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3

= 8x3 + 36x2y + 54xy2 + 27y3


Multiplication of Polynomials – Exercise 3.1.3

  1. Find the following products:

(i)(x + 4) (x + 5) (x + 2)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

a = 4, b = 5 and c =2

(x + 4) (x + 5) (x + 2) = x3 + x2(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2

= x3 + 11x2 + x (20 + 10 + 8) +40

= x3 + 11x2 + 38x +40

 

(ii) (y + 3) (y + 2) ( y – 1)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get

x = y, a = 3, b = 2 and c = -1

(y + 3) (y + 2) (y – 1) = y3 + y2 (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)

= y3 + 4y2 + y (6 – 2 – 3) – 6

= y3 + 4y2 + y – 6

 

(iii) (a + 2) (a – 3) (a + 4)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = a, a = 2, b = –3 and c = 4

(a + 2) (a – 3) (a – 4) = a3 + a2 (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4

= a3 + 3a2 + a (–6 – 12 + 8) – 24

= a3 + 3a2 – 10a – 24

 

(iv) (m – 1) (m – 2) (m – 3)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = m, a = –1, b = –2 and c = –3

(m – 1) (m – 2) (m – 3) = m3 + m2 (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +

(– 3)(–1)] + (–1) (–2) (–3)

= m3 + m2 (–6) + m [2 + 6 + 3] – 6

= m3 – 6m2 + 11m – 6

 

(v) ( √𝟐 + √𝟑) (√𝟐+ √𝟓) (√𝟐+ √𝟕 )

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = 2, a = 3, b = 5 and c = 7

(√2 + √3) (√2+ √5) (√2+ √7 ) = (√2)3 + (√2)2 [√3 + √5+ √7]+ √2 [√3. √5 + √5. √7 + √7. √3] + √3. √5. √7

= 2√2 + 2(√3 + √5 + √7) + √2 (15+ √35 + √21) + √105

 

(vi) 105 x 101 x 102

We can write this as

(100 + 5) (100 + 1) (100 + 2)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = 5, b = 1 and c = 2

(100 + 5) (100 + 1) (100 + 2) = 1003 + 1002 (5 + 1 + 2) + 100 (5.1 + 1.2

+ 2.5) + 5.1.2

= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10

= 1000000 + 80000 + 1700 +10

= 1081710

 

(vii) 95 x 98 x 103

We can write this as

(100 – 5) (100 – 2) (100 + 3)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = -5, b = -2 and c = 3

(100 – 5) (100 – 2) (100 + 3) = 1003 + 1002 (–5 – 2 +3) + 100(–5) (–2) + (–2) 3

+ 3 (-5) + (-5) (-2) 3

= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30

= 1000000 – 40000 – 1100 +30

= 958930

 

(viii) 1.01 x 1.02 x 1.03

We can write this as

(1 + 0.01) (1 + 0.02) (1 + 0.03)

Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc

we get

x = 1, a = 0.01, b = 0.02 and c = 0.03

(1 + 0.01) (1 + 0.02) (1 + 0.03) = 13 + 12 (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +

(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)

= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006

= 1.061106


  1. Find the coefficients of x2 and x in the following:

(i) (x + 4) (x + 1) (x + 2)

= x3 + x2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2

= x3 + 7x2 + 14x + 8

Coefficient of x2 is 7

Coefficient of x is 14

 

(ii) (x – 5) (x – 6) (x – 1)

= x3 + x2 (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)

= x3– 12x2 + x (30 + 6 + 5) – 30

= x3– 12x2 + 41x – 30

Coefficient of x2 is –12 and x is 41

 

(iii) (2x + 1) (2x – 2) (2x – 5)

= (2x)3 + (2x)2 [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)

= 8x3 + 4x2 (–6) + 2x [–2 +10–5] + 10

= 8x3 – 24x2 + 6x + 10

Coefficient of x2 is –24 and x is 6

 

(iv) ( 𝐱/𝟐 + 1) ( 𝐱/𝟐 + 2) ( 𝐱/𝟐 + 3)

= (𝐱/ )3 + (𝐱/𝟐)2 [1 + 2 + 3] + 𝐱/𝟐 [1.2 + 2.3 + 3.1] + 1.2.3

= x3/8 + (x2/4 )(6) + 𝐱/𝟐 (2 + 6 + 3) + 6

= x3/8 + 3/2 x2 + 11/2 x + 6

Coefficient of x2 is 𝟑𝟐 and x is 𝟏𝟏𝟐


  1. The length, breadth and height of a cuboids are (x +3), (x – 2) and (x -1) respectively. Find its volume.

Solution:

Volume of a cuboids = length x breadth x height

V = (x +3) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get

V = x3 + x2 (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)

= x3 – 0x2 + X (–6 + 2 – 3) + 6

= x3 – 7x2 + 6


  1. The length, breadth and height of a metal box are cuboid are (x +5), (x – 2) and (x – 1) respectively. What is its volume?

Solution:

Volume of the metal box = length x breadth x height

V = (x +5) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get

V = x3 + x2 + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)

= x3 + 2x2 + x [–10 + 2 – 5] + 10

= x3 + 2x2 – 13x + 10


  1. Prove that

(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc

[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d]

Solution:

x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get

  1. H.S. = (a + b + c)3 + (a + b + c)2 (–c – a – b) + (a + b + c) [(–c) (–a) +

(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)

= (a + b + c)3-(a + b + c)2[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc

= (a + b + c)3– (a + b + c)3 + (a + b + c) (ac + ab + bc) – abc

= (a + b + c) (ac + ab + bc) – abc

= R. H. S.


  1. Find the cubes of the following:

(i) (2x +y)3

Solution:

Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get

a = 2x, b = y

(2x +y)3 = (2x)3 + 3(2x)2 y + 3 (2x) y2 +y3

= 8x3 + 12 x2y + 6 xy2 +y3

 

(ii) (2x + 3y)3

Solution:

Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get

a = 2x, b = 3y

(2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3 (2x) (3y)2 + (3y)3

= 8x3 + 36 x2y + 54 xy2 + 27y3

 

(viii) 1013

Solution:

We write 101 as (100 + 1)3

Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get

a = 100, b = 1

(100 + 1)3 = 1003 + 3. 1002 + 3. 100.12 + 13

= 1000000 + 30000 + 300 + 1

= 1030301

 

(viii) 2.13

Solution:

We write 2.1 (2 + 0.1)3

Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get

a = 2, b = 0.1

(2 + 0.1)3 = 23 + 3 x 22(0.1) + 3 x 2 x (0.1)2 + (0.1)3

= 8 + 1.2 + 0.06 + 0.001

= 9.261


  1. Find the cubes of the following:

(i) (2a – 3b)3

Solution:

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 2a, b = 3b

(2a – 3b)3 = (2a)3 – 3 (2a)2(3b) + 3 (2a)(3b)2 – (3b)3

= 8a3 – 36a2b + 54ab2 -27b³

 

(ii) ( x – 𝟏/𝐱 )3

Solution:

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = x, b = 𝟏/𝐱

(x – 𝟏/ )3 = x3 – 3x2 𝟏/𝐱 + 3x(𝟏/𝐱)3 – (𝟏/𝐱)3 = x3 – 3x + 3x/ x21/x3

= x3 – 3x + 3/x1/x3

 

(iii) (√3 x – 2)2

Solution:

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = √3 x, b = 2

(√3 x – 2)2 = (√3x)3 – 3 (√3x)2 .2 + 3. √3x x 22 – 23

= 3√3 x3 – 6. 3 x2 + 12√3 x – 8

=3√3 x3 – 18x2 + 12√3 x – 8

 

(iv) (2x – √5)3

Solution:

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 2x, b = √5

(2x – √5)3 = (2x)3 – 3(2x)2 √5 + 3. 2x. √5)2 – (√5)3

= 8x3 – 12√5x2 + 30x – 5√5

 

(v) 493

Solution:

We can write 49 = 50 – 1

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 50, b = 1

(50 – 1)3 = 503 – 3.502.1 + 3.50.12 – 13

= 125000 – 3 x 2500 + 150 – 1

= 125000 – 7500 +149

= 117649

 

(vi) 183

Solution:

Let us write 18 = 20 – 2

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 20, b = 2

(20 – 2)3 = 203 – 3.202.2 + 3.20.22 – 23

= 8000 – 6×400 + 60×4 – 8

= 8000 – 2400 +240 – 8

= 5832

 

(vii) 953

Solution:

We write 95 = 100 – 5

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 100, b = 5

(100 – 5)3 = 1003 – 3.1002.5 + 3.100.52 – 53

= 1000000 – 150000 + 7500 – 125

= 857375

 

(viii) 1083

Solution:

We write 1083 = (110 – 2)

Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get

a = 110, b = -2

(110 – 2)3 = 1103 – 3. (110)2.2 + 3.110×22 – 23

= 1331000 – 72600 + 1320 -8

= 1259712


  1. If x + 𝟏/𝐱 = 3, prove that x3 + 𝟏/𝐱𝟑 = 18.

Solution:

Given x + 1/x = 3

Cubing both sides we get

(x + 1/x )3 = 33

Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get

a = x b = 1/x

(x + 1/x )3 = (x)3 + ( 1/x )3 + 3x. 1/x (x + 1/x )

27 = x3 + 1/x3 + 3 (3)

x3 + 1/x3 = 27 – 9

= x3 + 1/x3 = 18


  1. If p + q = 5 and pq = 6, find p3 + q3

Solution:

(p + q)3 = p3 + 3pq (p + q) + q3

53 = p3 + 3.6 (5) + q3

125 = p3 + 90 + q3

p3 + q3 = 125 – 90

p3 + q3 = 35


  1. If a – b = 3 and ab = 10, find a3 – b3

Solution:

Given a – b = 3 and ab = 10

(a – b)3 = a3 – b3 – 3ab (a – b)

33 = a3 – b3 – 3.10 (3)

27 = a3 – b3 – 90

a3 – b3 = 27 + 90

a3 – b3 = 117


  1. If a2 + 𝟏/𝐚𝟐 = 20 and a3 + 𝟏/𝐚𝟑 = 30, find a + 𝟏/𝐚

Solution:

a3 + 1/a3 = (a + 1/a) (a2 + 1/a2 – a x 1/a )

30 = (a + 1/a ) = (20 – 1)

30 = (a +1/a ) x 19

30/19 = a + 1/a

a + 1/a = 𝟑𝟎/𝟏𝟗


3.1.4 Square of a trinomial – Multiplication of Polynomials

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Example: Expand (x + 2y + 3z)2

Solution:

We take a = x; b = 2y ; c = 3z

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(x + 2y + 3z)2 = x2 + (2y)2 + (3z)2 + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)

= x2 + 4y2 + 9z2 + 4xy + 12yz + 6xz

Multiplication of Polynomials – Exercise 3.1.4

  1. Expand the following:

(i) (a + b + 2c)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get

a = a, b = b and c = 2c

(a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2(2c)a

= a2 + b2 + 4c2 + 2ab + 4bc + 4ca

 

(ii) (x + y + 3z)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = x, b = y and c = 3z

(x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x

= x2 + y2 + 9z2 + 2xy + 6yz + 6zx

 

(iii) (p + q – 2r)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = p, b = q and c = -2r

(p + q – 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p

= p2 + q2 + 4r2 + 2pq – 4pr – 4pr

 

(iv) ( a/2+b/2+c/2 )2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = a/2, b = b/2 and c = c/2

(a/2+ b/2+ c/2  )2 =( a/2 )2 + (b/2 )2 + (c/2  )2 + 2 (a/2 ) (b/2) + 2 (b/2 ) (c/2) +

2 (c/2  ) (a/2 )

= a2/4+b2/4+c2/4 + ab/2+bc/2+ca/2

 

(v) (x2 + y2 + z)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = x2, b = y2 and c = z

(x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2

= x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2

 

(vi) (m – 3 – 1/m )2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = m, b = -3 and c = – 1/m

(m – 3 – 1/m )2 = m2 + (-3)2 + (1/m )2 + 2.m(-3) + 2(-3)(- 1/m ) + 2 (-1/m) m

= m2 + 9 + 1/m 2 – 6m + 6m – 2

= m2 + 1/m2 + 6/m – 6m + 7

 

(vii) (-a + b – c)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = –a b = b c = –c

(-a + b – c)2 = (–a)2 + b2 + (–c)2 + 2(–a)b + 2b(–c) +2(–c)a

= a2 + b2 + c2 – 2ab – 2bc +2ca

 

(viii) (x + 5 + 1/2x )2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

a = x; b = 5; c = 1/2x

(x + 5 + 1/2x  )2 = x2 + 52 + (1/2x)2 + 2.x.5 + 2.5. 1/2x  + 2(1/2x)x

= x2 + 25 + 1/4x2 +10x + 5/x + 1

= x2 + 1/4x2 +10x + 5/x + 26


  1. Simplify the following:

(i) (a – b + c)2 – (a – b – c)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca (a – b + c)2 – (a – b – c)2

= [ a2 + (-b)2 + c2 + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)

+ 2(–b)(–c) + 2(–c)a]

= a2 + b2 + c2 – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]

= a2 + b2 + c2 – 2ab – 2bc +2ca – a2 – b2 – c2 + 2ab – 2bc +2ca

= 4ac – 4bc

= 4c (a – b)

 

(ii) (3x + 4y + 5)2 – (x + 5y – 4)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

(3x + 4y + 5)2 – (x + 5y – 4)2

= [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x2 + (5y)2 + (-4)2 +

  1. X.5y + 2.5y (-4) + 2 (-4). x]

= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – [x2 + 25y2 + 16 + 10xy – 40y-8x

= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – x2 – 25y2 – 16 – 10xy + 40y + 8x

= 8x2 – 9y2 + 14xy + 80y + 38x + 9

 

(iii) (2m – n – 3p)2 + 4mn – 6np + 12pm

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get

(2m – n – 3p)2 + 4mn – 6np + 12pm

= (2m)2 + (–n)2 + (-3p)2 + 2.2m(–n) + 2(n)(–3p) + 2 (-3p)(2m) + 4mn –

6np + 12pm

= 4m2 + n2 + 9p2 – 4mn – 6np – 12pm + 4mn – 6np + 12pm

= 4m2 + n2 + 9p2

 

(iv) (x + 2y + 3z + r)2 + (x + 2y + 3z – r)2

Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get

a = (x + 2y) b = 3z c = r

(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2

= (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 +

(3z)2 – (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y)

= 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +

6 (x + 2y) z – 6zr – 2r (x + 2y)

= 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z

= 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz

= 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz


  1. If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca.

Given a + b + c = 12 squaring both sides

(a + b + c)2 = 122

a2 + b2 + c2 + 2ab + 2bc + 2ca = 144

Given a2 + b2 + c2 = 50

50 + 2ab + 2bc + 2ca = 144

2(ab + bc + ca) = 144 – 50

2(ab + bc + ca) = 94

ab + bc + ca = 94/2

ab + bc + ca = 47


  1. If a2 + b2 + c2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c.

Given a2 + b2 + c2 = 35 and ab + bc + ca = 23,

(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

= a2 + b2 + c2 + 2(ab + bc + ca)

= 35 + 2 (23)

= 35 + 46

(a + b + c)2 = 81

(a + b +c) = ± √81 = ±9


  1. Express 4x + 9y + 16z + 12xy – 24yz – 16zx as the square of a trinomial

Using and comparing the coefficient of

(a +b+c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get

4x + 9y + 16z + 12xy – 24yz – 16zx

= (2x)2 + (3y)2 + (–4z2) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x

= (2x + 3y – 4z)2


  1. If x and y are real numbers and satisfy the equation

(2x + 3y – 4z)2 + (5x – y – 4)2 = 0 find x, y.

[Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.]

Given if a2 + b2 = 0, then a = b = 0

(2x + 3y – 4z)2 = 0 and (5x – y – 4)2 = 0

(2x + 3y = 5)x1 i.e., 2x + 3y = 5————-(1)

(5x – y = 4)x3 i.e., 15x – 3y = 12—————-(2)

Therefore,

2x + 3y = 5

15x – 3y = 12

—————

17x = 17

—————

x = 1

substitute the value of x in (1)

2(1)+3y = 5

2+ 3y = 5

3y = 5 – 2 = 3

y = 3/3 = 1

Thus, x = 1 and y = 1


3.1.5 Conditional Identity – Multiplication of Polynomials

Example :

If x + y = 3, prove that x3 + y3 + 3ab(a + b)

Solution:

We know that (x + y)3 = x3 + y3 + 3xy(x + y) . Substitute x + y = 3 in this identity. We get,

27 = x3 + y3 + 3xy(3) = x3 + y3 + 9xy


Multiplication of Polynomials – Exercise 3.1.5

  1. If a + b + c = 0, prove the following:

(i) (b + c) (b – c) + a (a + 2b) = 0

Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have

L.H.S = (b + c) (b – c) + a (a + 2b)

= (-a) (b – c) + a (a + b + b)

= -ab + ac + a (-c + b)

= -ab + ab – ac + ac

= 0= R.H.S

 

(ii) a (a2 – bc) + b (b2 – c) + c (c2 – ab) = 0

L.H.S = a (a2 – bc) + b (b2 – c) + c (c2 – ab)

= a3 – abc + b3 – abc + c3 – abc

= a3 + b3 + c3 – 3abc

We know that if a + b + c = 0 then

a3 + b3 + c3 = 3abc

Hence we have

= 3abc – 3abc

= 0 = R.H.S

 

(iii) a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = –3abc

L.H.S = a (b2 + c2) + b (c2 + a2) + c (a2 + b2)

= ab2 + ac2+ bc2 + ba2 + ca2 +cb2

= ab2 + ba2 + b2c + bc2 + ac2 + a2c

= ab (a + b) + bc (b + c) + ac (a + c)

= ab (–c) + bc (–a) + ac (–b)

= –abc – abc – abc

= –3abc = R.H.S

[a + b + c =0, a + b = -c, b + c = -a, c + a = -b]

 

(iv) (ab + bc + ca)2 = a2b2 + b2c2 + c2a2

L.H.S = (ab + bc + ca)2

= (ab)2 + (bc)2 + (ca)2 +2ab.bc + 2bc. ca + 2ca.ab

= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b

= a2b2 + b2c2 + c2a2 + 2abc + (0)

= a2b2 + b2c2 + c2a2 = R.H.S

 

(v) a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)

a.a – bc = a(-b-c)-bc = -ab-ac-ba = -(ab+bc+ca)——(1)

b2 – ca = b.b – ca = b(-c-a)-ca = -bc-ab-ca = -(ab+bc+ca) ———-(2)

c2 – ab = c.c – ab = c(-a-b) – ab = -ac-bc-ab = -(ab+bc+ca) ———-(3)

From equation (1) (2) and (3)

a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)

 

(vi) 2a2 + bc = (a – b) (a –c)

L.H.S = 2a2 + bc

= a2 + a2 + bc

= a2 + a x a + bc

= a2 + a (- b – c) + bc

= a2 – ab – ac + bc

= a (a – b) – c (a – b)

= (a – b) (a –c) = R.H.S

 

(vii) (a + b) (a – b) + ca – cb = 0

We have a + b + c = 0

a + b = c

L.H.S = (a + b) (a – b) + ac – cb

= –c (a – b) + ac – cd

= – ca + bc + ac – cb

= 0 = R.H.S

 

(viii) a2 + b2 + c2 = -2(ab + bc + ca)

We have a + b + c = 0

Squaring we get

(a + b + c)2 = 0

a2 + b2 + c2 + 2ab + 2bc +2ca = 0

a2 + b2 + c2 = – 2ab – 2bc – 2ca

a2 + b2 + c2 = – 2(ab + bc + ca)

Hence the proof


  1. Suppose a, b, c are non-zero real numbers such that a + b + c = 0,

Prove the following:

(i) 𝐚𝟐/𝐛𝐜 + 𝐛𝟐/𝐜𝐚 + 𝐜𝟐/𝐚𝐛 = 3

L.H.S = a2/bc + b2/ca + c2/ab

= (a2.a+b2.b+c2.c)/abc

= (a3+b3+c3)/abc ………. (1)

We have a + b + c = 0, a + b = –c

Cubing we get

(a + b)3 = (–c)3

a3 + b3 + 3ab + (a + b) = –c3

a3 + b3 – 3ab = –c3

a3 + b3 + c3 = 3abc ………. (2)

Substituting (2) in (1)

L.H.S = 3abc/abc = 3

(ii) ( +𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/𝐛 ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )

Whenever b + c ≠ 0, c + a ≠ 0, a + b ≠ 0

We have a + b + c =0

a + b = –c

b + c = –a

c + a = –b

L.H.S = (𝐚+𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/ ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )

= ( −c/c + −a/a + −a/b ) ( b/−b + c/−c + a/−a )

= (-1-1-1) (-1-1-1)

= (-3) (-3)

= 9 = R.H.S

 

(iii) 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛 = 1, provided the denominators do not become 0.

L.H.S = 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛

= 𝐚𝟐/(a-b)(a-c) + 𝐛𝟐/(b-c)(b-a) + 𝐜𝟐/(c-a)(c-b)

= 𝐚𝟐/(a-b)(a-c) – 𝐛𝟐/(a-b)(a-c + 𝐜𝟐/(a-c)(b-c)

= [a2(b−c)– b2(a−c) + c2(a−b)] /(a−b)(b−c)(a−c)

= [a2b− a2c − b2a + b2c + c2a− c2b] /(ab− b2− ac +bc) (a−c)

= [a2b− a2c − b2a + b2c + c2a− c2b]/[a2b − ab2− a2c + abc − abc + b2c + ac2]

= 1 = R.H.S


  1. If a + b + c = 0, prove that b2 – 4ac is a square.

We have a + b + c = 0

b = – (a + c)

Squaring on both sides

b2 = [- (a + c)]2

= (a + c)2

b2 = a2 + c2 + 2ac

Subtracting 4ac on both sides

b2 – 4ac = a2 + c2 + 2ac – 4ac

= a2 – 2ac + c2

b2 – 4ac = (a – c)2

We find that b2 – 4ac is the square of (a – c)


  1. If a, b, c are real numbers such that a + b + c = 2s, prove the following:

(i) s (s – a) + s (s – b) + s (s – c) = s2

L.H.S. = s (s – a) + s (s – b) + s (s – c)

= s2 – as + s2 – bs + s2 – cs

= 3s2 – as – bs – cs

= 3s2 – s (a + b + c)

= 3s2 – s (2s) (a + b + c = 2s)

= 3s2 – 2s2

= s2 = R.H.S.

 

(ii) s2 (s – a)2 + s (s – b)2 + s (s – c)2 = a2 + b2 + c2

L.H.S. = s2 (s – a)2 + s (s – b)2 + s (s – c)2

= s2 + s2 + a22sa + s2 + b2 + s2 + c2 – 2as – 2bs – 2cs

= 4s2 + a2 + b2 + c2 – 2s – 2bs – 2cs

= 4s2 + a2 + b2 + c2 – 2s (a + b + c)

= 4s2 + a2 + b2 + c2 – 2as (2s) (a + b + c = 2s)

= 4s2 + a2 + b2 + c2 – 4s2

= a2 + b2 + c2

= R.H.S.

 

(iii) (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2 = ab + bc + ca

L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2

= s2 – as – bs + ab + s2 – bs – cs + bc + s2 – cs – as + ac + s2

= 4s2 – 2 as – 2bs – 2cs + ab + bc +ca

= 4s2 – 2s (a + b + c) + ab + bc +ca

= 4s2 – 2s (2s) + ab + bc +ca

= 4s2 – 4s2 + ab + bc +ca (a + b + c = 2s)

= ab + bc +ca

= R.H.S.

 

(iv) a2 – b2 – c2 + 2bc = 4 (s – b) ( s – c)

L.H.S = a2 – b2 – c2 + 2bc

= a2 – (b2 + c2 – 2bc)

= a2 – (b – c)2

= (a + b – c) [a – (b – c)]

= (a + b – c) (a + b – c)

= (2s – c – c) (2s – b – b)

= (2s – 2c) (2s – 2b)

= 2(s – c) (2) (s – b)

= 4 (s – c) (s – b)

= R.H.S.


  1. If a, b, c are real numbers, a + b + c =2s and s – a ≠ 0, s – b ≠ 0, s – c ≠ 0,

Prove that 𝐚/(𝐬𝐚) + 𝐛/(𝐬𝐛) + 𝐜/(𝐬𝐜) + 2 = 𝐚𝐛𝐜/(𝐬𝐜)( 𝐬𝐛) (𝐬𝐜)

L.H.S = a/(s−a) + b/(s−b) + c/(s−c) + 2

=a(s−b)( s−c) + b(s−a)(s−c) + c(s−a)( s−b) + 2(s−a)(s−b)(s−c)/ (s−a)(s−b)(s−c)

= [a(s-b)(s-c)+b(s-a)(s-c) +c(s-a)(s-b)+2(s-a)(s-b)(s-c)] /(s−a )(s−b)(S−c)

=[a(s2−bs−cs+bc) +b(s2−as−cs+ac) +c(s2−as−bs+ab) +2(s2−as−bs+ab)(s−c)] /(s−a)(s−b)(S−c)

= [as2−abs−acs+abc+bs2−abc−bcs+abc) +cs2−acs−bcs+abc+2(s3−as2−bs2+abs −cs2+acs+bcs−abc] /(s−a)(s−b)(s−c)

= [s2(a+b+c) − 2abc −2acs − 2bcs − 3abc + 2s3−2as2− 2bs2 + 2abs−2cs2+ 2acs+2bcs−2abc] /(s−a)(s−b)(s−c)

=  [s2(2s) + abc+2s3−2s2 (a+b+c)]/(s−a)(s−b)(S−c)

=[2s3+ abc+2s3−2s2(2s)]/(s−a) s−b (S−c)

=[4s3+ abc−4s3]/(s−a)(s−b)(S−c)

= abc/(s−a)(s−b)(S−c)

= R.H.S.


  1. If a + b + c = 0, prove that a2 – bc = b2 – ca = c2 – ab = (𝐚𝟐 + 𝐛𝟐+𝐜𝟐)/𝟐

We have a + b + c = 0

Squaring will get

(a + b + c)2 = 0

a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

a2 + b2 + c2 + 2b (a + c) + 2ca = 0

a2 + b2 + c2 + 2b (-b) + 2ca = 0

a2 + b2 + c2 = 2b2 – 2ca                 (hint: a + c = -b)

a2 + b2 + c2 = 2 (b2 – ca)

a2 + b2 + c2 = b2 – ca ……… (1)

IIIly (a + b + c)2 = 0

a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

a2 + b2 + c2 + 2ab + 2c (b + a) = 0

a2 + b2 + c2 + 2ab + 2c (-c) = 0 (a + c = -c)

a2 + b2 + c2 = 2 (c2 – ab)

[a2 + b2 + c2]/2 = (c2 – ab)……………(2)

Also a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

a2 + b2 + c2 + 2a (b + c) + 2bc = 0

a2 + b2 + c2 + 2a (-a) + 2bc = 0

a2 + b2 + c2 = 2 (a2 – bc) ……..(3)

From (1), (2) and (3) we get,

a2 – bc = b2 – ca = c2 – ab = [a2 + b2 + c2]/2


  1. If 2(a2 + b2) = (a + b)2, prove that a = b.

2a2 + 2b2 = a2 + b2 + 2ab

2a2 + 2b2 – a2 – b2 – 2ab = 0

a2 + b2 – 2ab = 0

(a – b)2 = 0

a – b =0

a = b


  1. If x2 – 3x + 1 = 0, prove that x2 + 𝟏/𝐱𝟐 = 7.

Given x2 – 3x + 1 = 0

x2 + 1 = 3x

x + 1/x = 3 (dividing both sides by x)

Squaring both sides we get

(x + 1/x )2 = 32 = x2 + 1/x 2 + 2x. 1/x = 9

x2 + 1/x 2 + 2 = 9

x2 + 1/x 2 = 9 – 2 = 7


IX – Table of Contents


 

Ratio and Proportion – Full Chapter – Class IX

After studying the chapter Ratio and Proportion and their general form; to understand and differentiate between different types of proportion; to acquire skills of writing proportion; to solve problems on time and work involving proportions; to apply proportion in day to day life situations.

2.4.1 Introduction to Ratio and Proportion

In a ratio a : b, the first term a is called the antecedent and the second term b is called the consequent. Ratio is an abstract quantity and has no unit. Ratio tells how many times the first term is there in the second term.

Ratio and ProportionExample 1: In the adjacent figure, find the ratio of the shortest side of the triangle to the longest side. 

Solution:

We see that the shortest side is of length 5 cm and the longest side is of length 13 cm. Hence the ratio is 5:13

Example : Suppose the ratio of boys to girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

Solution:

For every 7 boys there are 5 girls. Thus out of 12 students, 7 are boys and 5 are girls. Hence the number of boys is 7/12 x 720 = 420.

The number of girls is 720 – 420 = 300. Now we want the ratio of boys to girls to be 1:1. This means the number of boys and girls must be same. Since the deficiency of  girls is 420 – 300 = 120, the school must admit 120 girls to make the ratio 1:1.

Example 5: Consider the ratio 12:5 If this ratio has to be reduced by 20% which common number should be added to both the numerator and denominator?

Solution:

Consider 12/5. This has to be reduced by 20%. This means we have to consider 80% of this number. Thus, we must get,

12/5 x 80/100 = 48/25

We have to find a such that,

12+a/5+a = 48/25

Cross multiplying

25(12 + a) = 48(5 + a)

48a – 25a = (25 x 12) – (48 x 5) =

23a = 60

a = 60/23

If we add 60/23 to both terms of 12 : 5 we get a ratio which is 20% less than the original ratio.

Ratio and Proportion – Exercise 2.4.1

1.Write each of these ratios in the simplest form.

(i) 2:6

(ii)24:4

(iii) 14:21

(iv) 20: 100

(v) 18:24

(vi) 22:77

Solution:

(i) 2:6 = 1:3 (dividing both by 2)

(ii) 24:4 = 6:1 (dividing both by 2)

(iii) 14:21 = 2:3 (dividing both by 7)

(iv) 20:100 = 1:5 (dividing both by 2)

(v) 18:24 = 3:4 (dividing both by 6)

(vi) 22:77 = 2:7 (dividing both by 11)


 2. A shop-keeper mixes 600 ml of orange juice with 900 ml of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in the fruit drink in its simplest form.

Solution:

Ratio of volumes of

Orange juice and apple juice O:A

= 600:900

= 6:9

= 2:3


3. a builder mixes 10 shovels of cement with 25 shovels of sand. Write the ratio of cement to sand.

Solution:

Ratio of cement to sand = 10 shovels :25 shovels


4.In a school there are 850 pupils and 40 teachers. Write the ratio of teachers to pupils.

Solution:

Number of teachers : Number of pupils

= 40 : 850 = 4:85


5. On a map, a distance of 5cm represent an actual distance of 15km. Write the ratio of the scale of the map.

Solution:

Let x be the number to be added them

(49 + x) = (68 + x) = 3:4

4(49+X) = (68 + X)3

196+4X = 204 + 3X

4X – 3X = 204 – 196

x = 8


2.4.2 Proportion – Ratio and Proportion

Ratio and Proportion – Exercise 2.4.2

1. In the adjacent figure, two triangles are similar find the length of the missing side

Ratio and Proportions

Solution:

Let the triangles be ABC and PQR

BC/QR = AC/PR

5/X  = 13/39

13X = 5 X 39

X = 5X39/13  = 5 X3 = 15


  1. What number is to 12 is 5 is 30?

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = 12×5/30   = 2


  1. Solve the following properties:

(i). x : 5 = 3 : 6

(ii) 4 : y = 16 : 20

(iii) 2 : 3 = y : 9

(iv) 13 : 2 = 6.5 : x

(v) 2 : π = x : 22/7

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = 15/6

 

(ii) 4 : y = 16 : 20

4×20 = 16y

y = 4×20/16

y = 5

 

(iii) 2 : 3 = y : 9

2×9 = 3y

y = 2×9/3 = 2×3 = 6

 

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = 13/13 = 1

 

(v) 2 : π = x : 22/7

2x22/7 = πx

x = (2x22/7) /π =(2x22/7) /(22/7)

x = 2


  1. find the mean proportion to :

(i) 8, 16

(ii) 0.3, 2.7

(ii)162/3 , 6

(iv)  1.25, 0.45

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then 8/x = x/16

x2 = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

 

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then 0.3/x = x/2.7

x2 = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

 

(ii)162/3 , 6

Let x be the mean proportion to 162/3  and 6

Then (162/3) /x = x/6

x2 = 162/3  x 6 = 50/3 x 6 = 100

x = √100 = 10

 

(iv)  1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then 1.25/x = x/0.45

x2 = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x√(100×100)/ √(100×100)

= √(125×45)/√(100×100) = √(25x5x5x9)/√(100×100)  = 5x5x3/10×10  = 3/4


  1. Find the fourth proportion for the following:

(i) 2.8,  14, 3.5

(ii) 31/3, 12/3, 21/2

(iii)15/7, 23/4, 33/5

Solution:

(i) 2.8,  14, 3.5

Let x be the  fourth proportion

Then, 2.8 : 14  :: 3.5  : x

2.8x = 14×3.5

x  = 14×3.5/2.8 = 17.5

(ii) 31/3, 12/3, 21/2

Let x be the  fourth proportion

Then, 31/3: 12/3  :: 21/2: x

10/3 :5/3 : : 5/2 : x

10/3x = 5/3 x 5/2

10/3x = 25/6

x  = 25/6 x 3/10 = 75/60  = 15/12 = 5/4

(iii)15/7, 23/14, 33/5

Let x be the  fourth proportion

Then, 15/7: 23/14:: 33/5: x

12/7 :31/14 : : 18/5 : x

12/7 x = 31/14 x 18/5

12/7 x = 31×18/14×5

x  = 31×18/14×5 x 7/12 = 31×3/5×4 = 93/20 = 413/20


  1. Find the third proportion to:

(i) 12, 16

(ii) 4.5, 6

(iii) 51/2 , 161/2

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = 16×16/12 = 64/3  = 211/3

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = 6×6/4.5 = 36/4.5  = 360/45  = 8

(iii) 51/2 , 161/2

Let x be the third proportion

Then 161/2: 51/2:: x : 161/2

11/2 x = 33/2 x 33/2

x = 33/2 x 33/2 x 2/11= 33×3/2  = 99/2  = 491/2


  1. In a map 1/4 cm represents 25km, if two cities are 21/2c apart on the map, what is the actual distance between them?

Solution:

Let 21/2 cm represemts x km

1/4cm: 25km :: 21/2cm : x km

1/4 x x = 25 x 21/2

x/4 = 25 x5 /2

x = 25×5/2 x 4 = 25x5x2 = 250km


  1. Suppose 30 out of 500 components for a computer were found defective. At this rate how many defective components would he found in 1600 components?

Solution:

Number of defective components in 500 components = 30

Let x be the number of defective components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = 30×1600/500 =96


2.4.3 Time and Work – Ratio and Proportion

Ratio and Proportion – Exercise 2.4.3

  1. Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?

Solution:

Number of days in which A and B together can finish the work = 12 days

Number of days in which B alone can finish the work = 30

1/T = 1/m + 1/n

1/12 = 1/m + 1/30

1/m = 1/301/12 = 5-2/60 = 3/60 = 1/20

A can finish the work in 20 days.

Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?

Solution:

A is twice as good a workman as B

i.e if B can finish a work in t days A can finish it in 1/2 days

1/T = 1/m + 1/n

1/24 = 1/t/2 + 1/t = 2/t + 1/t = 3/t

1/24 = 3/t

t = 24 x 3 = 72

i.e, B takes 72days to finish the job

A takes 72/2 = 36 days to finish it


  1. Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?

Solution:

A can finish a work in 15 days.

Work done A in 1 day = 1/15

B is 60% more efficient

Work done by B in 1 day

1/15 + 1/15 x 60/100

= 1/15 (1 + 60/100)

= 1/15 ( 8/5)

= 8/75

Number of days in which B alone can finish the work = 1/(8/75)  = 75/8 = 93/8 days


  1. Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?

Solution:

M = 14 days

N = 21 days

Part of work done in 6 days

= (1/14 + 1/21)6

= 6(3+2/42) = 5×6/42 = 5/7

Remaining part of the work = 1-5/7 = 2/7

Days taken by B to finish

2/7 part of the work = (2/7)/(1/21) = 2/7 x 21/7 = 6 days

Total number of days in which the work is completed = 6+6 = 12 days


  1. Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?

Solution:

If A alone takes to t1 days to do the work , B finishes it in t1/2  and C is t1/3 days

1/T = 1/t1 +1/t2 + 1/t3

= 1/t1 +1/(t1/2) + 1/(t1/3)

= 1/t1 +2/t1 + 3/t1

= 6/t1

1/T = 1/2

1/2 = 6/T1

i.e. t1 = 12 dyas

B takes 12/2 = 6days

C takes 12/3 = 4 days

Part of work done by B

In one day = 1/6

Part of work done by C in one day = 1/4

If B and C together takes t days to finish the work 1/T = 1/6 + 1/4 = 2+3/12 = 5/12

T = 12/5 = 2.4 days


 

 

Hire Purchase and Installment Buying – Class IX

After studying the chapter Hire Purchase and Installment Buying you learn the meaning of hire purchase and installment buying; to differentiate hire purchase and installment buying; to calculate interest in installment buying; to find out the equated monthly installment hire purchase.

2.3.1 Introduction to Hire Purchase and Installment Buying – Hire Purchase and Installment Buying

You might have seen advertisements like pay initially some amount and take home your dream car, television etc., and pay the remaining amount in easy monthly installments  at 0%  interest, 12% interest   and the like. What these kinds of advertisements mean? Do these refer to a scheme of business? What benefits do the giver and taker get? In this unit Hire Purchase and Installment Buying we study we study some concepts related with these kinds of transactions.

2.3.2 Hire Purchase and Installment Buying:

Hire purchase system is system of purchase and sale of articles. The purchaser pays the price of the article in installments. The articles are delivered to the purchaser at the time of agreement before the payment of installments, but the ownership or the title remains with the vendor till the purchaser pays off all the installments.

Installment buying – Hire Purchase and Installment Buying

It is another method of selling goods. Here the buyer takes home an article by paying some initial amount out of its total cost. the balance amount is to be paid periodically in installments.

Installment is concerned with the periodicity of amount to be paid. The cost of an article inclusive of interest and such other charges is divided by the number of months of the loan period. The amount so obtained is the installment amount.

2.3.3 Difference between hire purchase and installment – Hire Purchase and Installment Buying

Hire Purchase schemeInstalment scheme
1. On hire purchase, buyer gets the article but ownership lies with the vendor till the full payment.1. Buyer gets the article with ownership in instalment scheme.
2. The hirer cannot resell, pledge or cause any damage before full payment.2. Buyer has the liberty to resell or pledge as he is the owner.
3. If the buyer fails to pay off all the instalments, vendor can repossess the article3. If the buyer fails to pay the instalments, seller cannot repossess but can go to court of law.
4. This is bound with the provisions of Hire Purchase Act.4. This is bound with the provisions o Sale of Goods Act

2.3.4 Some terminologies associated with hire purchase and instalment buying:

Cash price – Hire Purchase and Installment Buying

It is the price on the article delivered to the customer on immediate payment of cash. Cash price is always lower than hire purchase price.

Down payment or initial payment – Hire Purchase and Installment Buying

A certain amount out of the total cost of the article is to be paid for the possession of the article. This is known as initial payment or cash down payment.

Installment – Hire Purchase and Installment Buying

The total cost of an article inclusive of interest and such other charges is divided by the number of months of the loan period. The amount so got is installment.

0% interest – Hire Purchase and Installment Buying

It is the total cost of the article paid as loan to the buyer. The interest collect 3 – 5 months installments in advance.

100% finance – Hire Purchase and Installment Buying

It is the total cost of the article paid as loan to the buyer. The interest and processing charges are also included in this scheme.

Equated monthly Installments – Hire Purchase and Installment Buying

The amount to be paid as installment decreases with period. For the article bought under hire purchase, the borrower has to repay the cost of the article, interest and other charges. The total amount to be paid is divided by the number of months of repayment. This amount is called equated monthly installment. EMIs are used to pay off both interest and principal each month so that the loan amount is paid off fully over a period of time.

EMI is obtained by dividing the principal and interest together by the number of months.

EMI = Principal + Interest/Number of months.

2.3.5 Calculation of interest in installment buying:

When the installment amount (I), number of installments (n) and the excess amount paid (E) are known, we can find the rate of interest directly using the formula,

Hire Purchase and Installment Buying

Example 1: The cost of a cell phone is Rs. 8000 and the down payment is Rs. 1000. The balance amount is to be paid in 8 equal instalments of Rs. 1000 each. Find the rate of interest.

Solution:

Given: Cost price = Rs. 8000, down payment = Rs. 1000

Balance, P = 8000 – 1000 = Rs. 7000

Number of installments n = 8

Installment amount = Rs.  1000

Thus, excess amount paid , E = nI – P = (8 x 1000) – 7000 = Rs. 1000

Hire Purchase and Installment Buying


Hire Purchase and Installment Purchase Exercise 2.3.5

1.A wash machine costs 10,200 cash down. It was bought by paying a down payment of 2,000 and the balance was agreed to be paid in 6 equal monthly installments of 1,500 each find the rate of interest.

Solution:

Cost price = 10,200.00
Down payment = 2,000.00
Balance = 10,200.00 – 2,000.00 = 8,200.00

Number of installments = 6
Amount of each installment (I) = 1,500.00
Amount paid in 6 installments (n)= 1,500 x 6 = 9,000.00
Excess amount paid = 9,000 – 8,200 = 800.00

Rate of interest = 2400E /n[(n + 1 )I – 2E]

= (2400×800)/6[(6 + 1) 1500 – 2×800]

= 2400×800/6(7×1500 – 1600)

= (2400×800) /6(10500 – 1600)

= 2400 x 800/6×8900 = 35.95%


  1. The cost of an android mobile phone is 8,990. Joseph bought it by paying 500 Cash down and the balance he agreed to pay in 10 monthly installments of 900 each. Nizam bought the same phone by initially paying 900 and the Remaining balance is 8 installments of 1,200 each. Who has paid more rate of interest?

Solution:

Cost price of the phone = 8,990.00

(i) cash down payment

Paid by Joseph = 500

Balance = 8,990.00 – 500.00 = 8,490.00

Number of installments (n) =10

Amount of each installments (i) = 900.00

Amount paid in installments = 900×10 = 9,000.00

Extra amount paid = 9000 – 8490 = 510

Rate of interest = 2400E/N[(n+1)l – 2E]

= 2400 x 510/10[(10+1)900 – 2x 510]

= 2400 x 510/10[11 x 900 – 2 x 510]

= 2400 x 510/10 x 880

= 13. 78%

 

(ii) Cash down payment

made by Nizam = 900.00

Balance = 8,990.00 – 900.00 = 8090.00

Number of installments(n) = 8

Amount of each installments(i) = 1,200.00

Amount paid in installments = 1,200 x 8 = 9,600.00

Extra amount paid(E) = 9,600.00 – 8,090.00 = 1,510.00

Rate of interest = 2400E/N[(n+1)I – 2E]

= 2400×1510/8(10,800 – 3,020)

= 2400×1510/ 8×7780

= 58.226%

Nizam is paying a higher rate of interest.


  1. The cost of a motor bike is 48,000.The company offers it in 30 months of Equal Installments at 10% rate of interest. Find the equated monthly installment.

Solution:

 R = 48,000.00 R = 10%

N= 30(number of installments)

Monthly installment I = P(2nR+2400) /N[2,400+(n+1)R]

= 48,000(2x30x10+2,400)/30[2,400+(30-1)10]

= 48,000(600+2,400)/30(2,400+290)

= 48,000×3,000/30×2690 = 48,00,000/2690 = 4,80,000/2690

= 480000/269

= 1784.38


  1. The cost of a set of home appliances is 36,000. Siri wants to buy them under a scheme of 0% interest and by paying 3 EMI in advance. The firm charges 3% as processing charges. Find the EMI and the installment for a period of 24 months.

Solution:

Cost of the set of home appliances (P) = 36,000.00

Number of installments (n) = 24

Amount of each installment = P/n = 36,000/24 = 1,500.00

Amount paid in advance = SEMI = 1500 x 3 = 4,500.00

Processing charge at the rate of 3% = 36,000 x 3/100  = 1,080.00

The total amount paid = 1,500 x 24+1,080

= 36,000+1,080

= 37080


 

Compound Interest – Full Chapter – Class IX

After studying the chapter Compound Interest you will learn to define compound interest; to distinguish simple interest and compound interest;  to derive the formula for amount when interest is compounded; to calculate compound interest using ready rockers; to solve problems using the formula of compound interest.

2.2.1 Introduction – Compound Interest:

In banks, insurance corporations and the other financial institutions, when they issue loans or accept deposits, the interest is calculated not on the original principal throughout the period. They calculate interest at regular intervals like quarterly, half yearly or yearly. The interest is added to the principal and further interest is calculated on the principal and the accrued interest. In these cases, the principal increases. We say that the interest is compounded in all these situations. The concept of compounding is very useful in predicting the population of a country,, growth in the food production, decay of a radioactive substance etc.

In the unit compound interest we study the calculation of compound interest and derive the formula to find the compound interest.

2.2.2 Compound Interest – Compound Interest:

Simple interest is calculated using the formula,

I = P x T  x R/100

Where I = simple interest;

P = principal amount;

T = Time;

R = Rate of interest.

Example: If the principal Rs. 2200 and rate 10%, then the interst after 1 year, I = P x T  x R/100

= 2200 x 1 x 10/100 = 200.

2.2.3 Difference between simple interest and compound interest

Let us consider some examples for calculating compound interest and compare it with simple interest.

Example 1:

Rs. 20000 is lent for 2 years at 5% compound interest. Calculate the compound interest and the amount after 2years if the interest is compounded annually. What would be have been the interest earned if the same amount was lent for simple interest at the same rate for the same duration?

Solution:

Principal for the first year = 20,000 x 1 x 5/100 = 1000(Using simple interest formula)

Principal for the second year = 20,000 + 1000 = 21000

Interest for the second year = 21000 x 5 /100 = 1050.

Amount after 2 years = 21000 + 1050 = 22050

Compound interest = amount – principal = 22050 – 20000 = 2050

Hence compound interest ater 2 years is 2050 Rs.

Simple interest on 20000 for 2 years at the rate of 5% is, I = P x T  x R/100 = 20,000 x 2 x 5/100 = 2,000.

Major differences between Simple interest and Compound interest:

SIMPLE INTERESTCOMPOUND INTEREST
1. It is the interest calculated on the principal alone.1. It is the interest calculated on the principal and accrued interest.
2. Principal remains the same throughout the period.2. Principal goes on increasing for every period of time.

2.2.4 Derivation of Compound Interest – Compound Interest:

Let P be the principal R be the rate of interest per annum and n be the period or time or number of years. We derive the formula in three steps:

Compound Interest


Example: Find the compound interest on Rs. 12000 for 2 years at 4% per annum

Solution:

Given P = Rs. 12000;

R = 4%

n = 2years

Using the formula,

Compound Interest

Therefore, Compound interest = A – P = 12979.2 – 12000 = Rs. 979.2


Example: Find the amount on Rs. 2000 after 2years if the rates  of interest are 2% and 3% for successive years.

Solution:

We have P = 20000 , R1 = 2% and R2 = 3% and n = 2 years

Here, we can directly use the formula,

Compound Interest

= 20000 (1.02)(1.03)

= 21, 012

Hence the amount after 2years is Rs. 21,012.


2.2.5 Calculation of Compound Interest Using the ready Reckoners – Compound Interest:

Compound interest can be easily calculated using ready reckoners. A ready reckoner is a pre-calculated table for interest for different periods and different times.

Compound Interest


Compound Interest – Exercise 2.2.5

  1. Calculate the amount and compound interest for the following:

(a) Rs12,000 for 2 years at 10% compounded annually.

(b) Rs 20,000 for 3 years at 8% compounded annually.

(c) Rs 5,000 for 1 year at 4% compounded semi-annually.

(d) Rs 10,000 for 11/2 years at 5% compounded half – yearly.

(e) Rs 500 for 1year at 2% compounded quarterly.

Solution:

(a) Rs12,000 for 2 years at 10% compounded annually.

P = 12,000;

R = 10%

n = 2

A = P [1 + R/100]n

=12,000[1 + 10/100]2

=12,000[1+0.1]2

=12,000[1.1]2

= 12,000[1.21]

=14,520.

Compound interest = Amount – Principal

= 14,520 – 12,000

= 2520

 

(b) Rs 20,000 for 3 years at 8% compounded annually.

P = 20,000;

R = 8%

n = 3

A = P [1 + R/100]n

= 20000[1 + 8/100]3

= 20,000 [1.08]3

= 20,000[1.259712]

=25,194.24

Compound interest = Amount – Principal

= 25,194.24 – 20,000

= 5194.24

 

(c) Rs 5,000 for 1 year at 4% compounded semi-annually.

P = 5,000

n = 1

R = 4

A = P(1 + R/2×100)2n

A = 5,000x(1 + 4/2×100)2×1

=5000x(1.02)2

= 5,202

Compound interest = Amount – Principal

=5,202 – 5,000

= 202

 

(d) Rs 10,000 for 11/2 years at 5% compounded half – yearly.

P = 10,000

n = 11/2

R = 5

A = P(1 + R/2×100)2n

= 10,000(1 + 5/2×100)2×1.5

= 10,000(41/40)3

= 10,000(1.02)3

= 10,768.91

Compound interest = Amount – Principal

=10,768.91 – 10,000

= 768.91

=769.00

 

(e) Rs 500 for 1year at 2% compounded quarterly.

P = 500

n = 1

R = 2

A = P(1 + R/4×100)4n

=500(1 + 2/4×100)4×1

= 500(1 + 1/200)4

= 500(201/200)4

=510.07

Compound interest = Amount – Principal

= 510.07 – 500

= 10.07


  1. A man invests Rs 5000 for 2 years at compound interest. After one year his money amount to Rs 5150. Find the interest for second year.

Solution:

Principal for the first year = P1 = 5000.00

Amount at the end of first year = A1 = 5150.00

Interest = A1 – P1 = 5150 – 5000 =150

Rate of interest = R = 100xI/PT = 100×150/5000 = 15000/5000 = 3

P = 5000 + 150 = 5150

Interest in the second year = PTR/100 = 5150x1x3/100 = 154.5


  1. Find the amount on 10000 after 2 years if the rate of interest are 3% and 4% doe successive years.

Solution:

P1 = 10000.00

R1 = 3%

T = 1

I1 = PTR/100 = 10000x1x3/100 = 300

Principal for the second year P2 = 10000 + 300 = 10300

R2 = 4%

I2 = PTR/100 = 10300x1x4/100 = 412

Amount at the end of second year P2 = 10300 + 412 = 10712


  1. Pralhad invests a sum of money in a bank and gets Rs 3307.5 Rs 3472.87 in 2nd and 3rd year respectively. Find the sum he invested.

Solution:

Amount at the end of second year A2 = RS 3307.50

Amount at the end of third year A3 = Rs 3472.87

Interest on Rs 3307.50 for a year = 3472.87 – 3307.50 = 165.37

Rate of interest = 100xI/PT = 100*165.37/3307.50*1 = 5%

Let P be the initial investment

Amount after 2 years A = P(1+ R/100)n

3307.50 = P(1+5/100)2

3307.50 = P(21/20)2

P = 3020.00


  1. On what sum of money will be difference between the simple interest and compound interest for 2 years at 4% per annum will be equal to RS 100? (Hint : Assuming the principal to be RS 100 first calculate the SI and CI, then proceed)

Solution:

Let the principal be Rs. 100.00

R = 4

T = 2 years

S.I = PTR/100 = 100*2*4/100 = Rs. 8.00

Amount at the end of 2 years if C.

I is calculated = 100(1 + 4/100)2

= 100*26/25*26/25=108.16

C.I = 108.16 – 100.00 = 8.16

Difference between SI and CI = 8.16 – 8.00 = 0.16

When the difference is Rs. 0.16

The amount invested = Rs. 100.00

The amount invested when the difference between SI and CI is Rs. 100.00 = 100*100/0.16 = Rs. 62500.00


  1. A sum of money is invested at compound interest payable annually. The interest in two successive year are RS 275.00 and Rs 300.00. Find the rate of interest.

Solution:

Interest at the end of 1 year = 275

Interest at the end of II year = 300

Interest per year = 300 -275 = RS. 25

Interest on RS 100 per year = 25

I = PTR/100

25 = 275*1*R/100 = 25*100/275 = 2500/275 = 9.09

Rate of interest = 9.09


  1. The difference between compound interest and simple interest on a certain sum for 2 year at 7 ½% per annum is RS. 360. Find the sum and verify answer.

Solution:

To find the simple interest and compound interest let us assume that the principal be Rs. 100

Time = 2 years

Rate =7 ½ % = 15/2%

SI = PTR/100 = 100*2*15/100*2 = 15

CI = P[(1 + R/100)n – 1]

= 100[(1 + 15/2×100)2 – 1]

= 100[(215/200)2 – 1]

= 100[46225/40000 – 1 ]

= 100[46225-4000040000]

= 100*6225/40000 = 6225/400

= 15.5625

CI – SI = 15.5625 – 15 = 0.5625

When P = 100 ; Difference  in CI and SI = 0.5625

If the difference is Rs. 360 then principal = 100/0.5625 x 360 = 64000

Thus, the principal is Rs. 64000

Verification:

S.I = PTR/100 = 64000x2x15/2×100  =  9600

CI = P[(1 + R/100)n – 1] = 64000x[(1+15/2×100)2 – 1] = 9960

SI – CI = 9960 – 9600 = 360


  1. Teju invests Rs. 12,000 at 5% interest compounded annually. IF he receives an amount Rs. 13320 at the end find the period.

Solution:

P = 12,000

R = 5%

A = 13230

A = P(1+R/100)n

13230 = 12000(1+ 5/100)n

13230/12000 = (1+5/100)n

21/20 = (21/20)n

n = 2


  1. The present population of a village is Rs . 18000. It is estimated that the population of the village grows by 3% per year . Find the population of the village after 4 years.

Solution:

P = 18000

R = 3%

n = 4

A = P(1+R/100)n

= 18000(1 + 3/100)4

= 18000x103x103x103x103/100x100x100x100

= 20256.15858

= 20259

Population after 4 years = 20259


  1. Jeshu purchased a bike by paying Rs. 52000. If the value depreciates by 2% every year. Find the value of the bike after 3 years .

Solution:

Original price of the bike =rs.52,000

Rate of depreciation =2%

Price after 3 years = P(1 – R/100) n

= 52000 x (1 – 2/100)3

= 52000 x (98/100)3

= 48941.98 ≈ 48942


  1. Using the ready reckoner find the compound interest in the following:

(a). Principal Rs 15000 for 4 years at 6.5% p.a.

(b). Principal Rs 22000 for 5 years at 12% p.a.

Solution:

(a) Principal Rs 15000 for 4 years at 6.5% p.a.

P = 15000

T = 4

R = 6.5%

From the table the interest on Rs. 1 t 6.5% for 4 years is Rs. 0.2865

CI on Rs. 15000 = 0.2865*15000 = Rs. 4297.5

(b)P = Rs. 22000

R = 12%

n = 5years

From the table the interest on Rs. 1 t 12% for 5 years is Rs. 0.7623

CI on Rs. 22000 for 5 years

= 0.7623*22000 = Rs. 16770.60 ≈ 16771


  1. Using the ready reckoner find the period of interest in the following :

(a) Compound interest Rs 4026 at 7% p.a. principal Rs 10000

(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.

Solution:

(a) P = 30000 ; R = 7% ; CI = Rs. 4026

Thus, Rs. 4026 is the compound interest for the principal of Rs. 10000 at 7%.

Now, we have to find compound interest for Re. 1. For Rs. 10000, th compound interest is Rs. 4026

For the principal Re. 1, the compound interest is = 4026/10000 = 0.4026

From the ready recknors 0.4026 corresponds to 5 years.

Therefore, n = 5.

(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.

We have A = Rs. 16939.2

P = 12000

R = 9

Compound interest = A – P = 16939.2 – 12000 = 4939.2

Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.

Now, we have to find compound interest for Re. 1. For Rs. 12000, the compound interest is Rs. 4939.2

For the principal Re. 1, the compound interest is = 4939.2/12000 =0.4116

From the ready recknors 0.4116 corresponds to 4 years.

Therefore, n = 4.


  1. Using the ready recknors ,Find the rate of interest in the following :

(a) Compound interest RS 1733.6 for 3 years on principal of RS 11000

(b) Amount RS 35246 principal SR 20000 for 5 years .

Solution :

(a) Compound interest = Rs. 1733.6

n = 3

P = Rs. 11000

Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.

Now, we have to find compound interest for Re. 1. For Rs. 11000, the compound interest is Rs. 1733.6

For the principal Re. 1, the compound interest is = 1733.6 /11000 = 0.1576

From the ready recknors 0.1576 corresponds to 5%

Therefore the rate of interest is 5%.

(b)

Amount = Rs. 35246

P = Rs. 20000

n = 5

Compound interest = A – P = 35246 – 20000 = 15246

Now, we have to find compound interest for Re. 1. For Rs. 20000, the compound interest is Rs. 15246

For the principal Re. 1, the compound interest is = 15246 /20000 = 0.7623

From the ready recknors 0.7623 corresponds to 12%

Therefore the rate of interest is 12%.


IX – Table of Contents


 

Statistics[Class 9] – Full Chapter with Exercise

After studying the chapter Statistics[Class 9] you will be  able to calculate range and coefficient of range for given data; to find out quartile deviation for regrouped and grouped data; to calculate mean deviation for regrouped data and grouped data; to draw histogram of varying width for inclusive and exclusive class intervals; to draw commutative frequency curve and identity quartile and median n it; to construct frequency polygons for inclusive and exclusive class intervals; to identify random experiment and types of probability.

1.5.1 Introduction – Statistics[Class 9]

The collection of numerical facts with particular information during experiment is called data. These data can be grouped into a table that displays frequencies of scores corresponding to various class interval. While grouping the data, if the end points o the groups do not overlap we call it inclusive method if grouping the data and if the end points of consecutive groups overlap, we calling it exclusive method.

Range – Statistics[Class 9]:

The difference between the highest and the lowest scores in a given distribution is called range.

Mean – Statistics[Class 9]:

It is the average of the scores, which s equal to the one of the scores divided by the number of scores.

For ungrouped data, the mean is calculated using the formula,

Ẍ = x/N

For grouped data, the mean is given by Ẍ = fx/N

Median – Statistics[Class 9]:

Median is the middle most score in a given set of scores. In ungrouped data, median is the middle score( when the scores are odd) or the average of two middle scores (when the scores are even), after the scores being arranged in ascending or descending order.

Median for grouped data is calculated using the formula,

Statistics[Class 9]

Mode – Statistics[Class 9]:

Mode is the score that occurs frequently in a given set of scores. Most repeated score in a ungrouped data is the mode. Mode of the value around which the other scores cluster around densely. In a grouped data, the scores corresponding to the maximum frequency is the mode.

A collection of data can have more than one mode. If the data has only one mode, we say it has Uni mode, if it has 2 modes, we say it has bi mode and it has more than 3 modes, we say it has multi-mode.

1.5.2 Measures of dispersion – Statistics[Class 9]:

There are four measures of dispersion viz.

  1. range(R)
  2. Quartile Deviation(QD)
  3. Mean Deviation(MD)
  4. Standard deviation(SD)

(a) Range – Statistics[Class 9]:

To understand the range, consider the following set of data:

24, 52, 35, 28, 49, 21

Find out the highest and the lowest scores. Have you observed the highest scores is 52 and the lowest is 21? Take the differenece of these two scores. It is 52 – 21 = 31. What is the difference called? This difference os called range.

Example : Calculate the range from the following data:

Marks263854657288
No. of students51015202530

Solution:

We observe that, highest scores H = 88 and lowest scores L = 26

Therefore, range H – L = 88 – 26 = 62 marks


Range is the simplest measure of dispersion. The difference between the highest and the lowest scores of distribution is called  range.

Range = Highest Score (H) – Lowest Score(L)


(b) Coefficient of range[Class 9]:

Consider the following example:

The wages of  six workers of a factory in  rupees are:

1600, 1500, 1750, 1800, 1250, 1400

What is the highest  and the lowest wages? The highest is 1800 and the lowest is 1250. Let us calculate the ratio of the difference of the highest and the lowest wages to its sum. It is

H L/H + L = 1800 1250/1800 + 1250 = 550/3050 = 0.18 (approximately)

Coefficient of range is a relative measure of dispersion and it is based on the value of range. It is also called the range coefficient of dispersion.

Coefficient of range is given  by = H L/H + L 

Example : Calculate the coefficient of range for the following data:

No. of wards12345678
No. of houses32572896138906658

Solution:

Here, H = 8 ; L = 1

Hence, coefficient of range = H L/H + L = 8-1/8+1 = 7/8


Merits and demerits of range[Class 9]:

Merits:

  1. It is the simplest measure of dispersion and easy to calculate.
  2. It does not require special knowledge to understand.
  3. Its calculation takes less time.

Demerits:

  1. It does not take into account  all the scores/items of distribution.
  2. It is affected by extreme scores.
  3. It does not indicate the direction of variability.

(c) Quartile Deviation – Statistics[Class 9]:

The points that divide the distribution in to four equal parts are called quartiles. If we take the difference between the third quartile and the first quartile, it gives us a value called inter quartile range. It is equal to

Q3 – Q1. Half of this is the Semi-Inter Quartile Range or quartile deviation which is (Q3 – Q1)/2 . Quartile deviation is also called the semi-inter quartile range.

(d) Quartile Deviation for ungrouped data:

Example: The runs scored by a batsman in five innings are 28, 60, 85, 58, 74, 20, 90. Find Q1, Q2, Q3 and quartile deviation.

Solution:

Arranging the scores in ascending we get, 28, 60, 85, 58, 74, 20, 90.

There are 7 scores and n = 7

  1. First quartile(Q1) = n+1/4 th score = 7+1/4 = 2nd score = 28
  2. Median(Q2) = n+1/2 th score = 7+1/2 = 4th score = 60
  3. Third quartile(Q3) = 3(n+1)/4 = 3(7+1)/4 th score = 6th score = 85
  4. Quartile deviation = Q3 Q1/2 = 85 28/2 = 28.5 ≅ 29

(e) Quartile deviation for grouped data – Statistics[Class 9]:

Example: Find the median and quartile deviation for the following data:

x34567
f1235524118

Solution:

Let us find the commutative frequency for the data given:

xfCommutative frequencyRemarks
311212, the first 12 scores correspond to x = 3
43547(12 + 35) from 13th to 47th scores correspond to x = 5
55299(47+52) from 48th to 99th scores correspond to x = 5
641140(99+41) from 100th to 140th scores correspond to x = 6
718158(140+18) from 141th to 158th scores correspond to x = 7

Observe that, n = 158, the last commutative frequency.

Median = n+1/2 th score = 79.5th score ≅ 80th score

From the column fc, from 48th to 99th score corresponds to x = 5. Hence 80th position = 5. Therefore, median = 5

To find the Quartile Deviation:

Q1 = n/4 th score = 158/4 th score = 39.5 i.e., 40th score

From column fc, 13th to 47th scores correspond to x = 4. Hence the 40th position is 4. Therefore,

Q1 = 4

Similarly, 3n/4 = 3×158/4 = 118.5

Hence, Q3 is 119th score. The column for fc shows that 100th to 140th scores correspond to x = 6. Therefore, Q3 = 6.

Quartile Deviation = Q3 Q1/2 = 6 4/2 = 1 mark.

(e) Quartile deviation for grouped data with class intervals:

Example: The heights of 100 students in 9th standard are given below:

Height(cm)100 – 110110 – 120120 – 130130 – 140140 – 150150 – 160
No. of Students(f)101216301220

Find quartile deviation.

Solution:

Step 1: First let us find the commutative frequency corresponding to the frequencies given.

Height(cm)f1fc
100 – 1101010
110 – 1201222
120 – 1301638(Q1 Class)
130 – 1403068
140 – 1501280
150 – 16020100(Q3 Class)
N = 100

Step 2: To find Q1 : Recall the formula to find the median

Statistics[Class 9]

where LRL = lower classs limit, fc = commutative frequency just above the median class fm = frequency corresponding the median class and I = size of the class interval. Now, to find Q1 replace N/2 by N/4 in the formula for median. Thus we get,

Now, find out, N/4 : N/4 = 100/4 = 25

Locate 25 in the commutative frequency column. This corresponds to the CI 120 – 130. This is Q1­ class.

From this class, LRL = 120, fc = 22 , fm = 16 and I = 10. Substituting these values in Q1 , we get,

Statistics[Class 9]

= 120 + (0.1875  x 10)

= 120 + 1.875

= 121.88

Thus, Q1 ≅ 122

Step 3: To find Q3:

In the median formula, replace, N/2 by 3N/4 and follow the same steps as in step 2.

Statistics[Class 9]

Here, 3N/4 = 3×100/4 = 75. In the commutative frequency column 75 corresponds to class interval 140 – 150. Therefore,

LRL = 140, fc = 68 , fm = 12 and I = 10

Statistics[Class 9]

= 140 + (0.58 x 10)

= 120 + 5.8

= 145.8

Thus, Q3 ≅ 146

Step 4: Now, we can find quartile deviation using the formula:

Quartile deviation = Q3 – Q1/2 = 146 – 122/2 = 12.


Statistics Exercise 1.5.3

  1. Calculate the range and coefficient of range from the following data.

a) The heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155, 133, 160, 140

Solution:

Heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155,133,  160,140.

Range: H – L = 168 – 101 = 67

Coefficient of Range: H – L/H + L

= 67/168 + 101 = 67/269 = 0.249


b) Marks scored by 12 students in a test: 31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.

Solution:

 Marks scored by 12 students

31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.

H = 49; L = 13

Range: H – L = 49 – 13 = 36

Coefficient of Range: H – L/H + L = 36/49 + 13 = 36/62 = 0.58.


c) Number of trees planted in 6 months: 186, 234, 465, 361, 290, 142.

Solution:

No .of trees planted in 6 months:

186, 234, 465, 361, 290, 142.

H = 465; L = 142

Range: H – L = 465 – 142 = 323

Coefficient of Range: H – L/H + L = 323/465+142 = 323/607 = 0.532


  1. State quartile deviation for the following data:

a) 30, 18, 23, 15, 11, 29, 37, 42, 10, 21.

Solution:

 Arrange the scores in ascending order:

n=10,

10, 11, 15, 18, 21, 23, 29, 30, 37, 42.

(a) First Q1 = n+1/4

= 10+1/4 = 11/4 = 2.75 = 3rd score = 15

(ii) Third Quartile

Q3 = 3(n+1)/4 = 3×11/4 = 33/4 = 8.25 = 9th = 37

(iii) Quartile Deviation:

= (Q3−Q1)/2 = 37−15/2 = 22/2 = 11


b) 3, 5, 8, 10, 12, 7, 5.

Solution:

3,5,5,7,8,10,12

n = 7

(i) Quartile Q1 = n+1/4 = 7+1/4 = 8/4 = 2nd score

Q1 = 5

(ii) Quartile Q3 = 3(n + 1)/4 = 3[(8)/4] = 6th score

Q3 = 10

(iii) Quartile Deviation:

= [Q3 − Q1]/2 = 10 – 5/2 = 5/2 = 2.5


(c)

Age3691215
No. of children4811712

Solution:

xfcommutative frequency
344
6812
91123
12730
151242

\ n = 42

Q1 = n/4 ; score = 42/4 = 10.5;11th score

From fc  ∴Q1 = 6

Q3 = 3n/4 = 3 ×42/4 = 31.5; 32nd score ∴Q3 = 15

Q.D = [Q3−Q1]/2 = 15−6/2 = 9/2 = 4.5


d)

Marks scored102030405060
No. of students12716081822

Solution:

xffc
101212
200719
301635
400843
501861
602283

n = 83

Q1= n/4 = 83/4 = 20.75; 21st score

∴Q1 = 30

Q3 = 3n/4 = 3 ×83/4 = 3X20.75 = 62.25; 62nd score

∴Q3 = 60

Q.D = [Q3−Q1]/2 = 60 – 30/2 = 30/2 = 15

∴QD = 15


  1. Compute quartile deviation for each of the following tables.

a)

Class intervalFrequencyfc
5 – 151111
15 – 25516
25 – 351531
35  – 45940
45 – 552262
55 – 65870
65 – 751787

Solution:

n = 87

Q1= n/4 = 87/4 = 21.75

22nd score CI = 25 – 35

∴LRC = 25

fc = 31; i = 10

Q2 = LRL +( [N/4−fc]/fm)i

= 25 + [(87/4−16)/15] 10

= 25 + [21.75 −16/15] ×10

Q2 = 28.83

Q3 = LRL + [(3N/4 – fc)/fm]*i

3N/4 = 3 × 87/4 = 65.25  class interval 55 – 65

L = 55, fc = 62, fm = 8, CI = 10

LRL = 55 +(65.25 – 62)/8 ×10

= 55 + 4.06

LRL = 59.06

Quartile Deviation = [Q3 − Q1]/2

= 59.06 – 28.83/2

= 30.23/2

Q.D = 15.11


(b)

class intervalfrequencyfc
1 – 944
10 – 1937
20 – 292027
30 – 391239
40 – 49544
50 – 59852
60 – 691466
70 – 792793
80 – 89295
90 – 995100

H = 100

Solution:

n = 100

100/4 = 25th Score 20 – 29; LRL = 19.5

Fc = 7; fm = 20

Q1 = LRL+[ (N/4−fc)/fm]*i

= 19.5 + [25 – 7/20]× 10

= 19.5 + 18/20*10

Q1 = 28.5

3N/4 = 3 × 100/4 = 3 × 25 = 75th score cl 70 – 79

LRL = 69.5; fc = 66; fm = 14

Q3 = 69.5 + [(75 – 66)/14] × 10

= 69.5 + 3.33

Q3 = 72.83

Quartile Deviation = (Q3 − Q1)/2

= 72.83 – 28.5 /2

= 44.33/2

Q.D = 22.16


1.5.3 Mean Deviation – Statistics[Class 9]:

Calculation of mean deviation for ungrouped data about median[Class 9]:

Consider the following set of scores:

15, 11, 13,20, 26, 18, 21

arranging scores in ascending order:

11, 13, 15, 18, 20, 21, 26.

Here  we have 7 scores which is odd.

Therefore, median =  (N+1/2) th score = (7+1/2)th score = (8/2) th score = 4th score = 18

Now let us find the deviation of each score from the median. the deviation D = score(X) – median.

Let us take only positive value of D. The positive value of D is called absolute value and denoted by |D|.

Scores(X)Deviations from median

D = X – median

|D|
1111 – 18 = -77
1313 – 18 = -55
1515 – 18 = -22
1818 – 18 = 00
2020 – 18 = 22
2121 – 18 = 33
2626 – 18 = 88
N = 7⅀|D| = 28

Add all |D| and divide it by the total number of scores. It gives you mean deviation.

i.e.,

|D|/N = 28/7 = 4

Therefore, mean deviation = 4

Calculation f mean deviation for ungrouped data about mean:

Example 10: Calculate the mean deviation from the mean for the scores given below:

15, 11, 13, 20, 26, 18, 21

Solution:

Arrange the scores in an order:

11, 13, 15, 18, 20, 21, 26

We known mean = sum of all the scores/number of scores = x/N

= 11+13+15+18+20+21+26/7 = 124/7 ≅17.7

Now calculate the deviation D of each score from mean and find out ⅀|D|.

Scores(X)Deviations from mean

D = X – mean

|D|
1111 -17.7 = -6.76.7
1313 -17.7 = -4.74.7
1515 -17.7 = -2.72.7
1818 -17.7 = 0.30.3
2020 -17.7 = 2.32.3
2121 -17.7 = 3.33.3
2626 -17.7 = 8.38.3
N = 7⅀|D| = 28.3

Now mean deviation from mean is = |D|/N = 28.3/7 = 4.04

Calculation of mean deviation for grouped data about median:

Example : Calculate the mean deviation for the following data about median.

Class interval0 – 45 – 910 – 1415 – 1920 – 2425 – 29
Frequency111217122028

Solution:

First let us find the median

CIffcmidpoint of CI(X)Deviation

D = X – median

|D|fx|D|
0 – 4111122 – 18.7 = -16.716.7183.7
5 – 9122377 – 18.7 = -11.711.7140.4
10 – 1417401212 – 18.7 = -6.76.7113.9
15 – 191251717 – 18.7 = -1.71.720.4
20 – 2420722222 – 18.7 = 3.33.366.0
25 – 29281002727 – 18.7 = 8.38.3232.4
N = 100⅀f|D|= 756.8

Statistics[Class 9]

= 14.5 + (50 – 40/12) x 5

≅ 14.5 + (0.83 x 5)

= 14.5 + 4.15

= 18.65

≅ 18.7

After finding the median , the deviation of median from midpoint of the class intervals are calculated. This gives D. Frequency (f) is multiplied with |D| to get f|D|. By adding all F|D|, we get⅀f|D|. these are shown above.

Mean deviation = ⅀f|D|/N = 756.8/100 = 7.57

Calculation of mean deviation for grouped data about mean:

Example 12: Calclate the mean deviation for the data given below:

Class interval0 – 1010 – 2020 – 3030 – 4040 – 50
Frequency539126

Solution:

Class intervalFrequencymidpoint xfxDeviation

D = x – X

|D|f|D|
0 – 1055255 – 28.1 = – 23.123.1115.5
10 – 203154515 – 28.1 = – 13.113.139.3
20 – 3092522525 – 28.1 = – 3.13.127.9
30 – 40123542035 – 28.1 = 6.96.982.8
40 – 5064527045 – 28.1 = 16.916.9101.4
⅀fx⅀f|D| = 366.9

Mean x = ⅀fx/N = 985/35 = 28.1

Mean deviation = ⅀f|D|/N = 366.9/35 = 10.4


Statistics[Class 9] Exercise 1.5.3

  1. Find the mean deviation about mean for the following date:

a) 14, 21, 28, 21, 18

Solution:

Mean = 14+21+28+21+18/5 = 102/5 = 20.4

ScoreDeviation from mean|D|
1414 – 10.4 = -6.46.4
1818 – 20.4 = -2.42.4
2121 – 20.4 = 0.60.6
2121 – 20.4 = 0.60.6
2828 – 20.4 = 7.67.6

Statistics - Exercise 1.5.3 – Class IX


(b)

Score(x)62081816121410
Frequency(f)2711271813175

Solution:

xffxD= x – x|D|f|D|
6212 – 8.588.5817.16
207140 + 5.425.4237.94
81188 – 6.586.5872.38
1827486 – 3.423.4292.34
1618288 – 1.421.4225.56
1213156 – 2.582.5833.56
1417238 – 0.580.589.86
10550 – 4.584.5822.9
N = 1001458311.68

Statistics - Exercise 1.5.3 – Class IX


  1. Find the mean deviation about mean for the following data:

a) 15, 18, 13, 16, 12, 24, 10, 20

Solution:

15+ 18+ 13+ 16+ 12+ 24+ 10+ 20 =128

N = 8

⅀x = 1258

Mean= x/N =128/8 = 16

xD = x – x|D|
1010-16=-66
1212-16 = -44
1313 – 16 = -33
1515 – 16 = -11
1616 – 16 = 00
1818 – 16 = 22
2020 – 16 = 44
2424 – 16 = 88
⅀28

MD =|D|/N = 28/8 = 3.5


(b)

CIf
10-196
20-294
30-3910
40-499
50-5911
60-698
70-792

Solution:

CIfxfxD = x – x|D|
10-19614.587-29.5177
20-29424.598-19.578
30-391034.5345-9.595
40-49944.5400.50.54.5
50-591154.5599.510.5115.5
60-69864.551620.5164.0
70-79274.514930.561
N  = 50⅀fx=2195⅀|D|=695

Statistics[Class 9]


(c)

Class intervalFrequency
0-59
5-1013
10-156
15-2012
20-259
25-306
30-3510
35-4015
40-456
45-504

Solution:

Class intervalFrequencyxfxD= x – x|D|
0-592.522.5-20.5184.5
5-10137.597.5-15.5201.5
10-15612.575.0-10.563.0
15-201217.5210.0-5.566.0
20-25922.5202.5-0.54.5
25-30627.5165.0-4.527.0
30-351032.5325.09.595.0
35-401537.5562.514.5217.5
40-45642.5255.019.5117.0
45-50447.5190.024.598.0
N = 90⅀fx=2105⅀|D|=1074.0

Statistics[Class 9]


  1. Find the mean deviation about median for the following data:

a) 18, 23, 9, 11, 26, 4, 14, 21

Solution:

4, 9, 1, 14, 18, 21, 23, 26

Median = N+1/2 = 8+1/2 = 9/2 =  4.5th

14+18/2 = 32/2= 16

median = 16

xD =  x – median|D|
44 – 16 = -1212
99 – 16 = -77
1111 – 16 = -25
1414  – 16 = -22
1818 – 16 =  22
2121 – 16 = 055
2323 – 16 = 077
2626 – 16 = 1010
50

MD = 50/8 = 6.25


(b)

Class intervalFrequency
8 – 1214
13   – 178
18 –  2220
23 – 277
28 – 3211
33 – 3710
38 – 4224
43  – 476

Solution:

Class intervalFrequencyxfxD = x –  medianf|D|
8 – 1214101410 – 23 = -13182
13   – 178152215 – 23 =  -864
18 –  2220204220 – 23 = -360
23 – 277252925 – 23 = 214
28 – 3211306030 – 23 = 777
33 – 3710357035  –  23 = 12120
38 – 4224409440  – 23 = 17408
43  – 4764510045 – 23 = 22132
N = 100⅀f|D|=1057

Statistics[Class 9]


(c)                                                       

Class intervalfrequency
20 – 309
30 – 4018
40 – 507
50 – 6021
60 – 7011
70 – 804

Solution:

Class intervalfrequencyxfcD = x – medianf|D|
20 – 309251925-42=-17153
30 – 4018352735-42=-7126
40 – 507453445-42=321
50 – 6021555555-42=13273
60 – 7011656665-42 = 23253
70 – 804757075-42 =33132
N = 70⅀f|D|=958

Statistics[Class 9]


  1. Find the mean deviation about mean and median for the following data:

a)

Cl1-56-1011-1516-2021-25
f295410

Solution:

ClffcxfxD = x – xf|D|
1-5223063 – 15 = -1224
6-109118728 – 15 = -763
11-15516136513 – 15 = -210
16-20420187218-15 = 312
21-2510302323023 – 15 = 880
N=30⅀fx=445189

Mean = 445/30 = 14.83 = 15

MD from Mean = 189/30 = 6.3

CIffcxDfx
1-5223-11.523
6-109118-6.558.5
11-1551613-1.57.5
16-2042018±3.514.0
21-25103023±8.585.0
N=30⅀fx = 188

Statistics[Class 9]


(b)

CI5 – 1010-1515-2020-2525-30
f5123119

Solution:

CIffcxfxD=x-xf|D|x-medianf|D|
5-10557.537.57.5-18=10.552.5-7.562.5
10-15121712.5150-5.566-2.590
15-2032017.552.5-0.51.52.57.5
20-25113122.5247.54.549.57.529.5
25-3094027.5247.59.585.512.50.5
N= 40⅀fx=735235255

Mean = 735/40 = 18.375 ≈18

Statistics[Class 9]


1.5.4 Graphical Representation – Statistics[Class 9]:

1. Construction and Interpretation of Histogram- Statistics[Class 9]:

  • Histogram is the most properly and widely used methods of graphical representations.
  • Histogram is a two dimensional graphical representation of a continuous frequency distribution.
  • In a histogram the area of rectangular are proportional to the frequencies.

Class intervals are marked on the x – axis and frequencies on the Y – axis. Class intervals must be exclusive. IF the class intervals are in inclusive form, they  are to be converted into exclusive form. Rectangles of width equal to class interval and length equal to frequencies re drawn.

(a) Histograms of varying width:

The width of each class interval is calculated by the corresponding frequencies. This is done by using a concept Frequency Density.

Frequency density is the ratio of frequency and its class width. when we consider the frequency density the length of the rectangle is to be modified accordingly. Length of the rectangle is the product of frequency density and the minimum class width of given data.

Length of the rectangle = frequency/class width X C

Here C is the minimum class width of the given data.

  1. Commutative Frequency Curve – Statistics[Class 9]:

Commutative frequency curve is a graph drawn with commutative frequency against the upper limit of class interval. The points are joined by a smooth curve and the curve is joined to the lower limit of the first class interval.

  1. Frequency Polygon – Statistics[Class 9]:

Frequency polygon is also a graphical representation of data, where the frequency is plotted against midpoint of the class interval. The frequencies corresponding to the mid points of class intervals are joined by line segments to get the frequency polygon.

A frequency polygon is drawn by drawing a histogram for the given data and joining the midpoints of the top of the rectangles. It can be drawn by marking the midpoints of the class intervals corresponding to their respective frequencies and joining them by line segments.


Statistics[Class 9] Exercise 1.5.4

  1. Construct histogram of variable width for the following data:

a)

CI      25-2930-3536-4041-5051-5657-60
f102415201216

Solution:

CI24.5-2929.5-35.535.5 – 40.540.5-50.550.5-56.556.5-60.5
f102415201216
class width5651064
length of  the rectangles102015101020

Statistics[Class 9]


(b)

CI0 – 1010- 1515 – 2020 – 3030 – 4040 – 6060-70
f2015102553050

Solution:

CI0 – 1010- 1515 – 2020 – 3030 – 4040 – 6060-70
f2015102553050
class width105510102010
Length of the rectangles10151012.52.57.525

Statistics[Class 9]


  1. Draw given (cumulative frequency curve) for the data given below:
Class intervalfrequency
1000 – 110052
1100 – 120035
1200 – 130025
1300 – 140014
1400 – 150041
1500 – 160033

Solution:

Class intervalfrequencyCumulative frequency
1000 – 11005252
1100 – 12003587
1200 – 130025112
1300 – 140014126
1400 – 150041167
1500 – 160033200

Statistics[Class 9]


(b)

Class intervalFrequency
5 – 144
15 – 248
25 – 3412
35 – 4414
45 – 546
55 – 644
65 – 7418
75 – 8424

Solution:

Class intervalcorrective factorFrequencycumulative frequency
5 – 144.5 – 14.544
15 – 2414.5 – 24.5812
25 – 3424.5 – 34.51224
35 – 4434.5 – 44.51438
45 – 5444.5 – 54.5644
55 – 6454.5 – 64.5448
65 – 7464.5 – 74.51866
75 – 8474.5 – 84.52490

Statistics[Class 9]


  1. Construct frequency polygon for the following data :

a)

CI5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 45
f257619148

Solution:

Statistics[Class 9]

(b)

CI30-3535-4040-4545-5050-55
f122016814

Solution:

Statistics[Class 9]


1.5.5 Random Experiment and the concept of Probability – Statistics[Class 9]:

Based on some assumptions ,  uncertainity can be measured mathematically by what is called probability. Probability has wide applications in the feild of physical science, commerce, biological sciences, weather forecasting, insurance, economics, sociology, investments and in various such other areas.

(a) Trial – Statistics[Class 9]:

A trial is an action which results in one or more outcomes.

Consider the following examples:

  1. rolling an unbiased die.
  2. Picking up a red card from a deck of playing cards.
  3. Drawing a marble from a bag of different colored marbles.

    (b) Random Experiment – Statistics[Class 9]:

A random experiment is one which exact outcomes are not possible to predict. For example, in tossing a coin we cannot predict outcomes head or tail.

(c) Sample Space – Statistics[Class 9]:

Consider the event of throwing an unbiased die. The possible outcomes are 1, 2, 3, 4,5 6. The set of all these possible outcomes is called Sample space.

S = {1, 2, 3,4, 5, 6}

(d) Empirical probability – Statistics[Class 9]:

It is the probability based on actual experiment leading to the possibility of outcomes.

Consider an experiment of tossing a coin 10 times. Let the frequency of head appearing would be 6 and that tail would be 4. Then the empirical probability of appearing head would be 6/10 = 0/6 and that of tall would be 4/10 = 0.4 .

The empirical probability of a certain event of an experiment is based on the outcomes of actual experiment. For this reason, empirical probability is also known as experimental probability.

If the number of tosses increases, the empirical probability of a head (or also tail)seems to approach the number 1/2 or 0.5. This is actually known as the theoretical probability of getting a head (or a tail)

IF n is the number of trails of an event E, then the empirical probability p€ is given by,

P(E) = Number of outcomes favourable to E/Number of possible outcomes to the experiment.

 


Statistics[Class 9] – Exercise 1.5.5

  1. Two unbiased 6 – faced die are thrown. What is the total number out comes?

Solution:

total number of outcomes = 6 × 6 = 36


  1. A die has the faces numbered 2, 4, 6, 8, 10 and 12. It is thrown once. What is the probability that an even numbered face shows up?

Solution:

Probability that even number farm shows up = 6/6 = 1


  1. In a pack of 52 playing cards, a card was selected at random. What is the probability that the card selected was both red and black?

Solution:

Zero: A card cannot be both red and black.


  1. Weather forecast made for 30 days in a month was recorded and found that it was correct for 21 days. What is the probability that on a randomly selected day, the forecast is (i) Correct and (ii) Not correct?

Solution:

Probability of correct forecast = 21/30 = 7/10

Probability of correct forecast= 9/30 = 3/10

Sets – Full Chapter with Exercise Solutions – Class IX

After studying the chapter Sets you will learn the concept of a set, to represent sets in roster method and set builder method; operations on sets like union, intersection, difference and symmetric difference; to use the formula for number of elements in a set and to solve simple word problems.

1.4.1 Introduction – Sets

One of the fundamental concepts in modern mathematics is the notion of a set. The concept of set is vital to mathematical thought and is being used in things every branch of mathematics. Understanding set theory helps us to see things in terms of systems, to represent things into sets and begin to understand logic. You can also represent sets and operations using diagrams known as Venn diagrams and apply it to solve problems more effectively.

1.4.2. Concept of a set – Sets:

We often deal with a collection of objects like collection of books , teachers in a school, farmers in a village, state in a country and so on.

A well-defined collection if objects is called a set. The objects in a set are called the elements or members of the set.

Example 1: Consider  A = {3, 5 7, 11}

  1. 3 is a member of A, we can write 3∈A(Read as 3 belongs to A)
  2. 4 is not a member of A, we write 4∉A(Read as 4 does not belong to A)
  3. 11 is an element of A, written as 11∈A

Example 2: Consider A = {1, 3, 5, 7, 9}. Fill in the blank spaces with the appropriate symbol ∈ or ∉.

  1. 1___________ A;
  2. 4___________ A;
  3. 11__________A;
  4. 3___________A;

Solution:

  1. 1∈ A
  2. 4∉A
  3. 11∉A
  4. 3∈ A

1.4.3 Representation of Sets – Sets

  1. Roster method:

For the three sets namely the set A of all odd numbers less than 10, the set B of all states of South India  and the set C of all  even natural numbers less than  or equal to 10, list out all their elements.

The list of elements of A are 1, 3, 5, 7, 9.

The list of elements of B are Karnatka, Andra Pradesh, Tamilnadu and Kerala.

The list of elements of C are 2, 4, 6, 8, 10.

We use the following notation to represent this A = {1, 3, 5, 7, 9}; B = {Karnatka, Andra Pradesh, Tamil Nadu and Kerala} and C = {2, 4, 6, 8, 10}. This method of representing a   set by  writing all its elements is called Roster method or  Tabular method. So, Roster method or Tabular method is a method  in which we write all the elements inside a pair of brackets {}.

2. Set builder method or rule method:

For many sets it may not be possible;e to represent it in Roster form. For example” The set of ll natural numbers. If we try to write this in Roster form we will not be able to list out all the elements since the set is infinite. When our list cannot be completed, we use in such cases what is known as the Set builder method or Rule method for representing a set. Set builder method can be used in some cases where Roster method is also possible.

In this method we write a general element x and a ‘:’ meaning such that and state its property within brackets. (i) The set of all vowels in English alphabet is represented by

V = {x : x is a vowel in English alphabet}


Example 3: List the elements of the following sets in roster form : the set of all positive integers which are multiples of 5.

Solution: The set of positive integers which are multiples of 5 in roster form is {5, 10, 15, 20, …}

Example 4: Write the set A = {x : x is a natural number ≤ 10} in roster form.

Solution:

A = {x : x is a natural number ≤ 10} So, the elements of the set are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Hence the set in roster form is A = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Subset of a set – Sets:

Let A be the set of states in India and let B be the set of all states in South India. We know B = {Karnataka, Andra Pradesh, Kerala, Tamil Nadu}. Karnataka is the element of B, but Karnataka is also a state of India.  So, Karnataka is also an element of A. Similarly, Andra Pradesh, Kerala, Tamil Nadu are also state in India.  Thus every element of B is an element of A. In such situation we say B is subset of A. So, given a set A, as set B is called subset of A if every element of B is also an element of A.

Equality of sets – Sets:

Let A be the set of all even natural numbers less than 10 and B be the set of natural numbers less than 10 which are not odd. Then A = {2, 4, 6, 8} and B = {2, 4, 6, 8}. Note that every element of B is also an element of A and every element of A is also an element of B. In such cases A and B are called equal sets.

Empty set – Sets:

Let A be the set of prime numbers less than 2. What are the elements in A? No prime number is less than 2. So no prime numbers can belong to A. This means there is no element in A.

LetB be the set of all perfect squares which are negative, we know that perfect sqyares are all non-negative. So no perfect square belong to B. This means there is no elemnet in B. Such sets are called Empty set.

A set containing no element is called a null set or  a void set. It is denoted by ∅.

Empty set is a subset of every set. This means ∅ ⊆ A for every set A. 

If the elements of all sets which we are considering tthe integers, then all sets under our consideration are subsets of the set of all integers. Similarly elements of all sets which we are considering are students of your school, then, all sets under our consideration re subsets of set of all students of your school. If all sets under consideration are subsets of a particular set, this particular set is called  universal set. We use U to denote universal set. If we are working with subsets of natural numbers, then the set of natural number N is the universal set.

The fixed set of which the sets we are working are subsets is called universal set.

Power set – Sets:

Given set A, the collection of all subsets of A is called the power set of A.It is usually denoted by P(A) or 2A.

Example: If A = {1, 2, 3} then P(A) = {∅, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}}

Singleton set – Sets:

A set considering of any one element is called a singleton set.

Examples: {1}, {a}, {(a, b)} are singleton set.

A set A is called a finite set if it has finite number of elements. Otherwise it is called infinite set. The number of element in a finite set A is denoted by n(A) or |A|. This number is called the cardinality of the set A


Sets – Exercise 1.4.3 – Solutions:

  1. Define a set. Give examples to illustrate the difference between a collection and a set

Solution:

Definition: A set is a collection of well defined objects. The objects of a set are called elements or members of the set.

Examples:

a) The collection set of all prime numbers between 100 and 200.

b) The collection of all planets in the universe.

c) The collection of all fair people in the city

Here (a) and (b) are examples of sets but (c) is not one cannot define fair.


  1. Which of the following collection are sets?

(a) All the students of your school.

(b) Members of Indian parliament.

(c) The colures of rainbow.

(d) The people of Karnataka having green ration card.

(e) Good teachers in a school

(f) Honest persons of your village.

Solution:

(a), (b) and (c) are sets.

(d), (e) and (f) are not sets.


  1. Represent the following sets in roster method:

(a) Set of all alphabet in English language.

(b) Set of all odd positive integers less than 25.

(c) The set of all odd integers.

(d) The set of all rational numbers divisible by 5.

(e) The set of all colors in the Indian flag.

(f) The set of letters in the word ELEPHANT.

Solution:

 (a) A = {a, b, c……..x, y, z}

(b) Z = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23}

(c) P = {±1, ±3, ±5….}

(d) R = {5, 10, 15……}

(e) Y = {saffron, white, green}

(f) S = {E, L, P, H, A, N, T}


  1. Represent the following sets by using their standard notations.

(a) Set of natural numbers

(b) Set of integers

(c) Set of positive integers

(d) Set of rational numbers

(e) Set of real numbers

Solution:

(a) N = {1, 2, 3……}

(b) Z = {0, ±1, ±2, ±3…….}

(c) Z + = {1, 2, 3…….}

(d) Q = {p/q, p, q  z and q ≠ 0}

(e) R = (Q U Z}


  1. Write the following sets in set builder form:

(a) {1, 4, 9, 16, 25, 36}

(b) {2, 3, 5, 7, 11, 13, 17, 19, 23……}

(c) {4, 8, 12, 16, 20, 24……}

(d) {1, 4, 7, 10, 13, 16……}

Solution:

(a) A = {x: /x=k2 for some k€N, 1 ≤ k ≤6}

(b) P = {x/x is a prime number}

(c) X = {x/x is a multiple of 4}

(d) Z = {x/x=3n – 2 when n = 1, 2, 3…..}


  1. State whether the set is finite or infinite:

(a) The set of all prime numbers.

(b) The set of all sand grains on this earth.

(c) The set of points on a line.

(d) The set of all school in this world.

Solution:

(a) infinite set

(b) finite set

(c) Infinite set

(d) finite set


  1. Check whether the sets A and B are disjoint

(a) A is the set of all even positive integers. B is the set of all prime numbers.

(b) A = {3, 6, 9, 12, 15……}

B = {19, 24, 29, 34, 39……..}

(c) A is the set of all perfect squares; B is the set of all negative integers.

(d) A = {1, 2, 3} and B = {4, 5,{1, 2, 3}}

(e) A is the set of all hydrogen atoms in this universe; B is the set of all water molecules on earth.

Solution:

(a) A and B are not disjoint sets since A ∩ B = {2}

(b) A and B are not disjoint since A ∩ B = {24….}

(c) A and B are disjoint.

(d) A ∩ B = {1, 2, 3}. Hence they are not disjoint.

(e) A and B is disjoint.


1.4.4 Operations on Sets – Sets

Union of sets – Sets :

Suppose D is the set of all players of cricket team and E is the set of all players of hockey team, of your school. some players may be common to both the terms. School wants to be felicitate these players in a function. when all the members are called, you see that the collection consists of all players who are either in cricket team or in hockey team. this is precisely the idea of union of sets.

Suppose A and B are two sets. The union of the sets A and B is the set of all those elements which belong to B. It is denoted by AUB(read as A union B). Thus, AUB = {x |x ∈ A or x ∈ B} Example:

If A = {1, 2, 4, 5} and B = {2, 4, 6, 8, 10} then AUB = {1, 2, 4, 5, 6, 8, 10}

SetsWe can represent the union of two sets using the Venn diagram. In the adjacent figure. A = {x, z} ; b = {y, z }{ and AUB = {x , y, z} the sets A and B are represented by circles. The shaded portion represents AB.

 


Intersection of Sets – Sets

Consider again the example of cricket team and hockey team of your school where three players, say Ram, John and Ismail are playing in both the teams. Suppose the school wants to felicitate players who plays for both the teams. What will be the set of all players who felicitated? It will be the set C = {Ram, John, Ismail}.Suppose we denote the set of all cricket players by A and the set of all hockey players by B. Then C is precisely the set of all players who are both in set A and set B. Thus

C = {x : x ∈A and x ∈B}

Given two sets A and B their intersection is the set of all those elements which are both in A and B. This is denoted by A∩B (read as A intersection B). Thus, A∩B = {x |x ∈A and x ∈B}.

Example : Let A = {1,2, 3, 4} and B = {2, 4, 5, 6} Find A∩B.

Solution:

Note that only common elements of A and B are {2, 4}. Hence A∩B = {2, 4}In the adjecent Venn diagram, the shaded portion represent A∩B.

Observe that A∩B = B∩A. Thus intersection is a commutative operation.

Two sets A and B are disjoint if and only if A∩B = ∅.


Sets – Exercise 1.4.4

  1. Find union of A and B and represent it using Venn diagram.

(i) A = {1, 2, 3, 4, 8, 9}, B = {1, 2, 3, 5}

(ii) A = {1, 2, 3, 4, 5}, B = {4, 5, 7, 9}

(iii) A = {1,2,3}, B = {4, 5, 6}

(iv) A = {1, 2, 3, ,4 ,5}, B = {1, 3, 5}

(v) A = {a, b, c, d}, B = {b, d, e, f}

Solution:

(i) A U B = {1, 2, 3, 4, 5, 8, 9}

Sets

(ii) A U B = {1, 2, 3, 4, 5, 7, 9}

Sets

(iii) A U B = {1, 2, 3, 4, 5, 6}

Sets

(iv) A U B = {1, 2, 3, 4, 5}

Sets

(v) A U B = {a, b, c, d, e, f}

Sets


  1. Find the intersection of A and B, and respect it by Venn diagram:

(i) A = {a, c, d, e}, B = {b, d, e, f}

(ii) A = {1, 2, 4, 5}, B = {2, 5, 7, 9}

(iii) A = {1, 3, 5, 7}, B = {2, 5, 7, 10, 12}

(iv) A = {1, 2, 3}, B = {5, 4, 7}

(v) A = {a, b, c}, B = {1, 2, 9}

Solution:

(i) A ∩ B = {d, e}

Sets

(ii) A ∩ B = {2, 5}

Sets

(iii) A ∩ B = {5, 7}

Sets

(iv) A ∩ B = { }

Sets


  1. Find A B and A B when:

(i) A is the set of all prime numbers and B is the set of all composite natural numbers:

(ii) A is the set of all positive real numbers and B is the set of all negative real numbers:

(iii) A = N and B = Z:

(iv) A = {x /x Z and x is divisible by 6} and

B = {x / x Z and x is divisible by 15}

(v) A is the set of all points in the plane with integer coordinate and B is the set of all points with rational coordinates

Solution:

(i) A = {2, 3, 5, 7….}

B = {1, 4, 6….}

A U B = {1, 2, 3, 4} = N

A ∩ B = { }

 

(ii) A = R+

B = R+

A U B = R – {10}

i.e. A U B ={set of non zero real numbers}

A U B = { }

 

(iii) A = N B = N

A U B = Z

A ∩ B = N

 

(iv) A U B = {x / x € Z and x is divisible by 6 and 15} and

A ∩ B = {x / x € Z and x is divisible by 30}

[LCM of 6 and 15 = 30]

(v) A U B = the set of all points with rational co-ordinates = B.

 

A ∩ B = the set of all points with rational co-ordinates = A.


  1. Give examples to show that

(i) A U A = A and A A = A

(ii) If A B, then A U B = B and A B =A. can you prove these statements formally?

Solution:

(i) If A = {2 4 6 8}

Then A U A = {2, 4, 6, 8……} = A

A ∩ A = {2, 4, 6, 8…….} = A

Hence A U A = A and A ∩ A = A

 

(ii) A = {1, 3, 5, 7, 9……}

B = {1, 2, 3, 4, 5……}

We see that A ⸦ B

A U B = {1 2 3 4…..} = B

A ∩ B = B

A ∩ B = {1, 3, 5……} = A

A ∩ B = A


  1. What is A U Φ and A ∩ Φ for a set A?

Solution:

 AUΦ = A ;  A ∩Φ = Φ


1.4.5 Complement of a set – Sets

Suppose U is a set and A⊆ U. the complement of A in U is the set of all those elements of U which are not members of A. This is denoted by

AC or A’. Thus,

A’ = { x : x ∈ U but x∉ A}

For any subset A of U, we have A∩A’ = ∅ and AUA’ = U.

For any set U, U’ = ∅ and ∅’ = U.

Difference of two sets – Sets:

Given two sets A and B, we define  B  ⃥   A as all those elements o B which are not in A. this is read as B difference A this is also called the complement o B in A. Thus,

B  ⃥  A = {x | x ∈ B and x ∉ A}

B  ⃥  A:

Sets

B difference A

Sets

Symmetric difference of two sets – Sets:

Let us take A = {1, 3, 5, 7} and B = {5, 7, 8}. Then A difference B = {1, 3} and B difference A = {8}

The union of (A difference B) and (B difference A) is {1, 3, 8}. It is the union of all those elements in A which are not in B and those elements in B which are not in A. It is called the symmetric difference of A and B. t is denoted by A∆B.

For any two sets A and B  A∆B = (A difference B)U(B difference A)

Sets


Sets – Exercise 1.4.5

  1. If A’ = {1, 2, 3, 4}, U = {1, 2, 3, 4, 5, 6, 7, 8}, find A in U and draw Venn diagram

Solution:

 A’ = {5, 6, 7, 8}


  1. If U = {x/x € 25, x€N}. A = {x/x € U, x ≤ 15} and B = {x/x € U, 0 < x ≤ 25}, list the elements of the following sets and draw Venn diagram:

(i) A’ in U:

(ii) B’ in U

(iii) A\B;

(iv) A Δ B

Solution:

U = {1, 2, 3, 4 ……….25}

A = {1, 2, 3, 4……….15}

B = {1, 2, 3 ….25}

(i) A’ = {16, 17, 18, 19….25}

SETS - EXERCISE 1.4.5 – Class 9

(ii) B’ = { }

SETS - EXERCISE 1.4.5 – Class 9

(iii) A\B = { }

SETS - EXERCISE 1.4.5 – Class 9

(iv) A Δ B = A \ B U B \ A

= { } U {16, 17, 18… 25}

= {16, 17, 18 …..25}

SETS - EXERCISE 1.4.5 – Class 9


  1. Let A and B subsets of a set U. Identify the wrong statements:

(i) (A’)’ = A

(ii) A \ B = B \ A

(iii) A U A’ = U

(iv) A Δ B = B Δ A

(v) (A \ B)’ = A’ \ B’

Solution:

 If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 9}

(i) (A’)’ = {2, 4, 6, 8}

(A’)’ = {1, 3, 5, 7, 9} = A

(A’)’ = A

 

(ii) A \ B = {1, 3, 5, 7}

B \ A = {2, 4, 6, 8}

We see that A \ B ≠ B \ A

 

(iii) A U A’ = {1, 3, 5, 7, 9} U {2, 4, 6, 8}

= {1, 2, 3, 4, 5, 6, 7, 8, 9}

= U

A U A’ = U

(iv) A Δ B =(A \ B) U (B \ A)

= {1, 3, 5, 7} U {2, 4, 6, 8}

= {1, 2, 3, 4, 5, 6, 7, 8}

B Δ A = (B \ A) U (A \ B)

= {2, 4, 6, 8} U {1, 3, 5, 7}

= {1, 2, 3, 4, 5, 6, 7, 8}

A Δ B = B Δ A

 

(v) A \ B = {1, 3, 5, 7}

(A \ B)’ = {2, 4, 6, 8, 9}

A’ = {2, 4, 6, 8} and B’ = {1, 3, 5, 7}

A’ \ B’ = {2, 4, 6, 8}

Hence (A \ B)’ ≠ A’ \ B’


  1. Suppose U = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. A = {3, 4, 5, 6, 9}, B = {3, 7, 9, 5} and C = {6, 8, 10, 12, 7}. Write down the following sets and draw Venn diagram for each:

(i) A’

(ii) B’

(iii) C’

(iv) (A’)’

(v) (B’)’

(iv) (C’)’

Solution:

(i) A’ = {7, 8, 10, 11, 12, 13}

SETS - EXERCISE 1.4.5 – Class 9

(ii) B’ = {4, 6, 8, 10, 11, 12, 13}

SETS - EXERCISE 1.4.5 – Class 9

(iii) C’ = {3, 4, 5, 9, 11, 13}

SETS - EXERCISE 1.4.5 – Class 9

(iv) A’ = {7, 8, 10, 11, 12, 13}

(A’)’ = {3, 4, 5, 6, 9} = A

SETS - EXERCISE 1.4.5 – Class 9

(v) (B’)’ = B’ = {4, 6, 8, 10, 11, 12, 13}

(B’)’ = {3, 7, 9, 5} = B

SETS - EXERCISE 1.4.5 – Class 9

(vi) (C’)’ = {3, 4, 5, 9, 11, 13}
(C’)’ = {6, 8, 10, 12, 7} = C

SETS - EXERCISE 1.4.5 – Class 9


5. Suppose U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, write down the following sets and draw Venn diagram.
(i) A’

(ii) B’

(iii) A U B

(iv) A ∩ B (v) (A U B)’ (vi) (A ∩B)’
How (A UB)’ is related to A’ and B’? What relation you see between
(A ∩ B)’ and A’ and B’
Solution:

(i) A’ = {5, 6, 7, 8, 9}

SETS - EXERCISE 1.4.5 – Class 9

(ii) B’ = {1, 3, 5, 7, 9}

SETS - EXERCISE 1.4.5 – Class 9

(iii) A U B = {1, 2, 3, 4, 6, 8}

SETS - EXERCISE 1.4.5 – Class 9

(iv) A ∩ B = {2, 4}

SETS - EXERCISE 1.4.5 – Class 9

(v) (A UB)’
(A U B) = {1, 2, 3, 4, 6, 8}
(A U B)’ = {5, 7, 9}

SETS - EXERCISE 1.4.5 – Class 9

vi) (A ∩ B)’
(A ∩ B) = {2, 4}
(A ∩ B)’ = {1, 3, 5, 6, 7, 8}

SETS - EXERCISE 1.4.5 – Class 9

 

We see that (A U B)’ = A’ ∩ B’
(A ∩ B)’ = A’ U B’


6. Find (A \ B) and (B \ A) for the following sets and draw Venn diagram.
(i) A = {a, b, c, d, e, f, g, h} and
B = {a, e, i, o, u}
(ii) A = {1, 2, 3, 4, 5, 6} and
B = {2, 3, 5, 7, 9}
(iii) A = {1, 4, 9, 16, 25} and
B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) A = {x | x is a prime number less than 5} and
B = {x | x is a square number less than 16}
Solution:
(i) A = { a, b, c, d, e, f, g, h}
B = {a, e, i, o, u}
A \ B = {b, c, d, f, g, h}

SETS - EXERCISE 1.4.5 – Class 9

B \ A = {i, o, u}

SETS - EXERCISE 1.4.5 – Class 9

(ii) A = {1, 2, 3, 4, 5, 6} and
B = {2, 3, 5, 7, 9}
A \ B = {1, 4, 6}

SETS - EXERCISE 1.4.5

B \ A = {7, 9}

SETS - EXERCISE 1.4.5 – Class 9

(iii) A = {1, 4, 9, 16, 25} and
B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A \ B = {16, 25}

sets-exercise-1-4-5-class-9

B \ A = {2, 3, 5, 6, 7, 8}

SETS - EXERCISE 1.4.5 – Class 9.png

(iv) A = {x | x is a prime number less than 5}
= {2, 3}
B = {x | x is a square number less than 16}
= {1, 4, 9}
A \ B = {2, 3}

SETS - EXERCISE 1.4.5 – Class 9

B \ A = {1, 4, 9}

SETS - EXERCISE 1.4.5 – Class 9


7. Looking at the Venn diagram list the elements of the following sets:
(i) A \ B
(ii) B \ A
(iii) A \ C
(iv) C \ A
(v) B \ C
(vi) C \ B

SETS - EXERCISE 1.4.5 - Class 9

Solution:
(i) A \ B = {1, 2, 7}
(ii) B \ A = {5, 6}
(iii) A \ C = {1, 2, 3}
(iv) C \ A = {6, 8, 9}
(v) B \ C = {5, 3}
(vi) C \ B = {7, 8, 9}


8. Find A Δ B and draw Venn diagram when:
(i) A = {a, b, c, d} and B = {d, e, f}
(ii) A = {1, 2, 3, 4, 5} and B = {2, 4}
(iii) A ={1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 6}
(iv) A = {1, 4, 7, 8} and B = {4, 8, 6, 9}
(v) A = {a, b, c, d, e} and B = {1, 3, 5, 7}
(vi) A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7}
Ans:
(i) A = {a, b, c, d} B = {d, e, f}
A \ B = {a, b, c}
B \ A = {e, f}
A Δ B = {a, b, c, e, f}

SETS - EXERCISE 1.4.5 – Class 9

(ii) A = {1, 2, 3, 4, 5}                            B = {2, 4}
A \ B = {1, 3, 5}
B \ A = { }
A Δ B = {1, 3, 5}

SETS - EXERCISE 1.4.5 – Class 9.png

(iii) A ={1, 2, 3, 4, 5} ;          B = {1, 2, 3, 4, 5, 6}
A \ B = {.}
B \ A = {6}
A Δ B = {6}

SETS - EXERCISE 1.4.5 – Class 9.png

(iv) A = {1, 4, 7, 8}; B = {4, 8, 6, 9}
A \ B = {1, 7]
B \ A = {6, 9}
A Δ B = {1, 6, 7, 9}

SETS - EXERCISE 1.4.5 – Class 9.png

(v) A = {a, b, c, d, e} and B = {1, 3, 5, 7}
A \ B = {b, d}
B \ A = {g}
A Δ B = {b, d, g}

SETS - EXERCISE 1.4.5 – Class 9.png

(vi) A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7}
A \ B = {2, 4}
B \ A = {7}
A Δ B = {2, 4, 7}

SETS - EXERCISE 1.4.5 – Class 9.png


 

Surds – Full Chapter Surds – Class IX

1.3.1 Introduction – Surds

You have studied about the irrational numbers like √3, ∛7, ∜5 and so on. You have here square root, cube rot and forth root of numbers. There are also irrational numbers which cannot be written in such a form; for example √(3+∛2), π. In the Surds we study more about numbers of the form Surds is a natural number. ‘a’ is a rational number. these are special class of irrational numbers.

1.3.2 Rational Exponent of a Number – Surds

Recall the statement from unit 2: Real Numbers given a natural number n, for every positive real number a there exists a unique positive real number b such that bn  = a. Here b is called the n-th root of a and we write

Surds

We can now define rational power of a positive number. let a be a positive real number. Let r = p/q to be a rational number, where p is an integer and q is a natural number.

Surds

Let a and b be positive real numbers. Let r1 and r2 be two rational numbers. Then we have:

Surds

Example 1: Simplify 2^1/2/4^1/6

Solution:

41/6 = (22)1/6  = 22x(1/6) = 21/3

Hence,

2^1/2/4^1/6 = 2^1/2/2^1/3 = 2(1/2)-(1/3) = 21/6


Surds – Exercise 1.3.2

  1. Simplify the following using laws of indices:

(i) (16)-0.75 x (64)4/3

(16)-0.75 x (64)4/3                              [16 = 24 ]

= (24)-0.75 x (24)4/3                         [64 = 26 ]

= 24 x -3/4 x 26 x 4/3

= 2-3 x 28

= 25

= 32


(ii) (0.25)0.5 x (100)-1/2

(0.25)0.5 x (100)-1/2

= (0.25)1/2 x (1/100 )1/2

= (0.25)1/2 x (1/102 )1/2

= (0.5) x ( 1/10 )

= ( 5/10 ) x ( 1/10)

= 5/100

= 𝟏/𝟐𝟎


(iii) (6.25)0.5 x 102 x (100)-1/2 x (0.01)-1

= ( 625/100 )1/2 x 102 x (1/10)1/2 x ( 1/100 )-1

= ( 25/10 )1/2 x 102 x (1/10 )1/2 x (1/100)-1

= ( 252/10 )1/2 x 102 x (1/102 )1/2 x (100)1

= ( 25/10 ) x 100 x 1 10 x 100

= 2500


(iv) (3-1/2 x 2-1/3) ÷ (3-3/4 x 2-5/6)

= 3−1/2× 2−1/3 3−3/4× 2−5/6

= 3-1/2+3/4 x 2-1/3+5/6

= 31/4 x 23/6

= 31/4 x 21/2

= 31/4 x 21/4

=(3 × 22)1/4

= (3 × 4)1/4

= 𝟏𝟐1/4


  1. Find the value of the expression.

[ 31/3 {5-1/2 x 3-1/3 x (2252)1/3}1/2]6

Solution:

= [ 31/3 {5-1/2x-1/2 x 3-1/3x1/2 x (2252)2/3x1/2]6

= [ 31/3 {5-1/4 x 31/6 x (2252)-2/6]6

= [ 31/3 x 6 {51/4 x 6 x 31/6 x 6 x (2252)-2/6 x 6]

= [ 32 x 51/4 x 6 x 31/6 x 6 x 2252-2/6 x 6]

= [32 x 53/2 x 31 x 225-2]

= [32 x 52/3 x 31 x 15-4]

= [32 x 5-2/3 x 31 x 3-4 x 5-4]

= [32 + 1 – 4 x 53/2 – 4]

= 1/31 x 1/55/2

= 1/31  x 1/55/2

= 1/31  x 1/√55

= 1/3 x 1/√3125

= 1/3√3125


  1. Simplify:

[{(35/2 x 53/4) ÷ 2-5/4} ÷ {16 / (52 x 21/4 x 31/2)}]1/5

Solution:

= [{(35/2 × 53/4) ÷ 2−5/4} ÷ {16÷ (52× 21/4× 31/2)}] 1/5

= [{(35/2 × 53/4)/(2−54) ÷ 16/(52× 21/4× 31/2)}] 1/5

= [(35/2 × 53/4)/(2−54) ÷ (52× 21/4× 31/2)/24]1/5

= [(35/2 × 53/4 × 52/1 × 21/4 × 31/2)/(2−5/4 × 24 )]1/5

= [21/4 × 35/2 + 1/2 × 53/4 + 2/1/2−5/4 + 4/1]1/5

= 21/4 × 35/2 + 1/2 × 511/4/211/4]1/5

= [21/4 4/11 × 33 × 511/4 ] 1/5

= [210/4 × 1/5 × 33 × 1/5 × 511/4 + 1/5]

= 21/2 × 31/5 × 511/20


1.3.3 Surds and their properties – Surds:

Consider the following real numbers:

√17, 8 + ∛12 , 3/5 + √(7/11) , ∛(3 + √5))

They are all irrational numbers. Nevertheless, you see that they are all different types.

A surd is real number of the form  , where n is an integer larger than 1 and a is a rational number such that it is not an n-th power of any rational number. For example, 25/36 is the square of 5/6. Thus, √(5/6) is not a surd. On the other hand √(24/17) is a surd.

A simplest form of a surd:

If , where c does not contain any n-th power of a rational number, then it is the simplest form. Here b is the coefficient of the surd.

Pure surd:

A surd which in its simplest form has 1 as coefficient.

Mixed Surd:

A surd which has some rational number not equal to 1 as its coefficient written its simplest form.

Similar surds:

Two surds are called similar surds or like surds if when written in simplest form they have the same order and the same radicand. Otherwise they are called unlike surds.

Reduction of Surds of different orders to the same order:

Example: Reduce the surds √(24/98) and ∛16 to the surds of the same order

Solution:

Here the order are 2 and 3 respectively. Hence their LCM is 6. We write

Surds


Surds – Exercise 1.3.3

  1. Write the following surds in their simplest form:

(i) √𝟕𝟔

= √(19 × 4)

= √(19 ×24)

= 𝟐√𝟏𝟗

 

(ii) (𝟏𝟎𝟖)

= ∛(33 × 22)

= ∛22

= ∛𝟒

 

(iii) 𝟓𝟎𝟎𝟎

= 5000

= ∜(511 × 84)

= 𝟓𝟖

 

(iv) 𝟏𝟖𝟗/𝟐𝟓

= ∛[(33 × 7)/52]

= 3𝟕/𝟐𝟓

 

(v) 𝟒𝟎𝟎/𝟒𝟗

= ∜(24 × 52)/724

= 252/72

= 2𝟐𝟓/𝟒𝟗


  1. Classify the following in to like surds

(i) √𝟐𝟒 , √𝟏𝟐𝟖 , √𝟕𝟓 , √𝟕𝟐 , √𝟓𝟒 , √𝟐𝟒

= 35 , 27 , √(52×3) , √(32×23) , √(2 × 33) , √(23×3)

= 323 , 232 , 53 , 3×22 , 3(2×3), 2(2×3)

= 93, 82, 53, 62, 36, 26

= {82, 62}, {93, 53}, {26, 36,}

={ √𝟏𝟐𝟖, √𝟕𝟐}, { √𝟐𝟒𝟑, √𝟕𝟓}, { √𝟓𝟒, √𝟐𝟒,}


(ii) ∛𝟐𝟎𝟎𝟎, ∛𝟔𝟖𝟔, ∛𝟔𝟒𝟖, ∛𝟑𝟕𝟓, ∛𝟏𝟐𝟖, ∛𝟐𝟒

= ∛(2×52), ∛(73×2), ∛(63× 3), ∛(53×3), ∛(43×2), ∛(23× 3)

= 10∛2, 7∛2, 6∛3, 5∛3, 4∛2, 2∛3

= {4∛𝟐, 7∛𝟐, 10∛𝟐} & {2∛𝟑, 5∛𝟑, 6∛𝟑}

= { ∛𝟐𝟎𝟎𝟎, ∛𝟔𝟖𝟔 , ∛𝟏𝟐𝟖} & { ∛𝟔𝟒𝟖, ∛𝟑𝟕𝟓, ∛𝟐𝟒}


  1. Which of the following are pure surds?

(i) 296 = √(33×37) = 2√(37×2) = 2√𝟕𝟒 – not pure surds

(ii) 729 = 36 = 33 = 27 –  not pure surds

(iii) ∛ 211 Cannot be reduced further hence it is a pure surd. – Yes, a pure surd

(iv) 75 is also a pure surd. – Yes, a pure surds

(v) ∛ 296  = ∛(23×37) = 2 ∛𝟑𝟕 – no, not a pure surd

(vi) 296 = cannot be reduced further, hence it is a pure surd – Yes


  1. Write the following irrational numbers sure from

(i) √(𝟏𝟓√(𝟐𝟕))

= 27 = 33 = 33

= [15(27)1/2]1/2

= 152/4 x 271/4             [1/2 = 4/2]

= ∜(153 × 27)

= ∜(225 × 27)

= ∜𝟔𝟎𝟕𝟓

 

(ii) √(𝟒𝟎(𝟏𝟐))

= [40 × (12)1/2] ½

= (40)1/2 × (12)1/4

= ∜(402 × (12)1)

= ∜(402× 121)

= ∜(1600×12)

= ∜19200

 

(iii) √(𝟓(𝟒𝟖))

= [51/2 ×(48)1/2]½

= [51/2 × 481/4]

= ∜ (52× 48)

= ∜(25×48)

= ∜1200


5. Reduce the following to surds of the same order.

(i) , √𝟐and 𝟓1/5

The orders are 3, 2, and 5

LCM of 2, 3 and 5 is 30

∛2 = 21/3 = 210/30 = (𝟐𝟏𝟎) 1/3𝟎

2 = 21/2 = 215/30 = (𝟐𝟏𝟓 ) 1/3𝟎

51/5 = 51/5 = 56/30 = (𝟓𝟔) 1/3𝟎

(1024)1/3𝟎 , (32768) 1/3𝟎 , (15625) 1/3𝟎

They have the same order 30

 

(ii) √𝟓 , (√𝟏𝟓)1/4 , and (√𝟓𝟎)1/8

Order is 2, 4, 8

Their LCM is 8

5 = 51/2 = 54/8 = (54 )1/8 = (125)1/8

(√15)1/4 = (15)1/4 = (15)2/8 = ((15)2) 1/8 = (225)1/8

(50)1/8 is in its simplest form

(50)1/8  , 𝟐𝟐𝟓𝟖 , 𝟏𝟐𝟓𝟖

∴ Thus they all are of the same order 8.

 

(iii) √𝟐 , √𝟕1/3 , (√𝟏𝟏)1/4 and (√𝟏𝟔𝟕𝟏)𝟏/1𝟐

Order is 2, 3, 4, and 12

Their LCM is 12

2 = 21/2 = 26/12 = (26)1/12 = (64)1/12

71/3 = 74/12 = (74 )1/12 = (2401)1/12

(√11)1/4 = (11)3/12 = ((11)3)1/12 = (1331)1/12

167112 is in its simplest form

𝟔𝟒𝟏/1𝟐 , 𝟐𝟒𝟎𝟏𝟏/1𝟐 , 𝟏𝟑𝟑𝟏𝟏/1𝟐 , 𝟏𝟔𝟕𝟏𝟏/1𝟐

∴ Thus they all are of the same order 12


1.3.4 Comparing Surds and Some Irrational Numbers:

Example 7: Find which surd is larger

Surds


Surds – Exercise 1.3.4

  1. Find which is larger:

(i) 3𝟑 and 4𝟒

∛(3×32)                                                        4×424

= 27 x 3                                                   = 256 x 4

= 81                                                             = 1024

= ((81)4)1/12                                                  = ((1024)4)1/12

= (43046721)1/12                                         = (1073741824)1/12

4𝟒 is greater 3𝟑


  1. Compare the following and decide which is larger.

(i) (∜(30))1/7 and ∛(281/10)

(281/10) 1/3 = (281/10)1/3 = 281/30

(301/4) 1/7 = (301/4)1/7 = 301/28

LCM of 30 and 28 is 420

(281/10)1/3 = 281/30 = (281/14)1/420

(301/4) 1/7 = 301/28 = (3015)1/420

2814 =22 x 714 = 228 x 714

3015 = (5 x 6)15 = 515 x 615

Comparing the 2 numbers we conclude

3015 > 2814

(∜(30))1/7 > ∛(281/10)

 

(ii) √(∜8)  and ∛(∛9)

√(∜8) = (81/4)1/2 = 81/8 = 23/8          [8 = 23]

∛(∛9) = (91/3)1/3 = (32/3)1/3 = 32/9

LCM of 8 and 9 is 72

√(∜8)  = 23/8 = (23/8)9/9 = 227/72 = (227)1/72

∛(∛9) = 32/9 = (32/9)8/8 = 316/72 = (316)1/72

By comparing we find that (227) is larger than 316.

Hence √(∜8)   > ∛(∛9)


  1. Write the following in ascending order:

√𝟐 , 𝟑𝟑 , 𝟔1/6

21/2 , 31/3 , 61/6

LCM of 2, 3 and 6 is 6

21/2 = 23/6 = (23)1/6 = 81/6

31/3 = 32/6 = (33)1/6 = 91/6

61/6 = 63/6 = (6)1/6 = 61/6

Ascending order is 61/6  , 21/2, 31/3


  1. Write the following descending order:

√(𝟔) , (∜𝟏𝟐) , (∜𝟖)

(61/3)1/2 , (121/4)1/3 , (81/4)1/2

61/6, 121/12 , 81/8

LCM of 6, 12, 8 is 24

64/24, 122/24, 83/24

(64)1/24, (122)1/24, (83)1/24

(1296)1/24, (144)1/24, (512)1/24

Descending order is √(𝟔)  , √ (∜𝟖)  , ∛(∜𝟏𝟐)


Real Numbers

Square root


 

Circles – Exercise 4.4.3 – Class IX[state board]

  1. In the figure ∟ACB = 42˚. Find ∟x

Circles – Exercise 4.4.3 – Class IX

Solution:

In the figure ∟AOB = 2∟ACB [angle at the centre = twice the angle at the circumference]

∟AOB = 2 * 42˚ = 84˚

∟AOB = x = 84˚

 


  1. In the figure ∟ABC is a circle with centre O and reflex ∟AOB = 250 ˚ find ∟x.

Circles – Exercise 4.4.3 – Class IX

Solution:

In the figure ∟ACB angle at the circumference ∟AOB reflex angle at the centre.

∟ACB = 1/2 reflex ∟AOB

∟ACB = 1/2 * 250 ˚ = 125 ˚

∟x = 250 ˚


  1. In the fig. AOB is diameter, find ∟C

Circles – Exercise 4.4.3 – Class IX

Solution:

In the given figure AOB is diameter of the circle with centre O.

∟AOB = 180 ˚

∟ACB = 1/2 * ∟AOB

∟ACB = 90 ˚

Angle at the circumference reflex angle at the centre.


  1. Circles – Exercise 4.4.3 – Class IXIn the figure ABCD is a circle with centre O. ∟x = 180 ˚ Find

(i) ∟d

(ii) ∟ y

(iii) ∟a

(iv) ∟b

(v) ∟ b + ∟d

 

Solution:

(i) ∟ADC angle at the circumference ∟AOC angle at the centre.

∟ADC = 1/2 ∟AOC

∟d = ∟ADC = 1/2*108˚

∟d = 54˚

(ii) reflex ∟AOC = 360˚ – ∟AOC = 360˚ – 108˚ = 252˚

∟y = 252˚

(iii) ∟b = ∟ABC = 1/2 reflex ∟AOC

= 1/2 * 252˚ = 126˚

(iv) ∟a + ∟b = 180˚

∟a + 126˚ = 180˚

∟a = 180˚ – 126˚ = 54˚

(v) ∟b + ∟d = 126˚ + 54˚ = 180˚

(vi) ∟e = 1/2 ∟AOC  = 1/2*108˚ = 54˚

∟b + ∟e = 126˚ + 54˚ = 180˚


  1. In the figure diameter AB and chord QR are produced to meet at P. If ∟QPA = 260˚ , ∟QAR = 360˚.Find ∟x and ∟y

Circles – Exercise 4.4.3 – Class IX

Solution:

∟AOB = 180˚

∟AQB =1/2*180˚

∟AQB = 90˚

∟PAR = ∟BQR = x˚

In the ∆ AQP

∟PAQ + ∟AQP + ∟APQ = 180˚

(36˚+ x) + (90˚ + x + 26˚) =180˚

2∟x – 152˚ = 180˚

2∟x = 180˚ – 152˚ = 28˚

∟x = 14˚

In the ∆ARQ,

∟AQR + ∟QAR + ∟ARQ = 180˚

(90˚+ x) + 36˚+ ∟y = 180˚

90˚+14˚+36˚+∟y = 180˚

140˚ + ∟y = 180˚

∟y = 180˚ – 140˚

∟y = 40˚


  1. In the fig. ∟CBD = 110˚. Find ∟AOC.

Circles – Exercise 4.4.3 – Class IX

Solution:

∟ABC + ∟DBC = 180˚

∟ABC + 110˚ = 180˚

∟ABC  = 180˚ – 110˚ = 70˚

∟AOC = 2∟ABC = 2*70˚  = 140˚


  1. In the fig ∟ADC = 84˚ and AB = BC Find ∟BDC.

Circles – Exercise 4.4.3 – Class IX

Solution:

ABCD is a cyclic quadrilateral.

∟ABC + ∟ADC = 180˚

∟ABC + 84˚ = 180˚

∟ABC = 180˚ – 84˚ = 96˚

In ∆ABC , AB = BC

∟BAC = ∟BDC

∟BDC = 42˚


  1. In the fig. AB is diameter ∟BAC = 38˚. Find ∟ADC

Circles – Exercise 4.4.3 – Class IX

Solution:

AOB ia diameter of the circle with centre O.

∟ACB = 1/2 ∟AOB = 1/2*180˚ = 90˚

∟ABC + ∟ACB + ∟BAC = 180˚

∟ABC = 180˚ – 90˚ – 38˚ = 52˚

∟ABC + ∟ADC = 180˚

(ABCD is cyclic quadrilateral)

52˚ + ∟ADC = 180˚

∟ADC = 180˚ – 52˚ = 128˚


  1. In the fig. ∟QXR = 25˚, ∟QRX = 33˚. Find ∟XYZ and ∟PZQ

Circles – Exercise 4.4.3 – Class IX

Solution:

In ∆QXR: RQ is produced to P

Exterior ∟PQX = ∟Q XR + ∟QRX = 25˚ + 33˚ = 58˚

∟PQX = ∟PYX [angles of same segement]

but ∟PQX = 58˚

∟PYX = 58˚

∟XYZ = 58˚

In the ∆XYZ

∟YXZ + ∟XYZ + ∟XZY = 180˚

25˚  + 58˚ + ∟XZY = 180˚

∟XZY = 180˚ – 25˚ – 58˚ = 97˚

Therefore, ∟ABC = 97˚


  1. In the fig AFD and BFE are straight lines. Find ∟ACF.

Circles – Exercise 4.4.3 – Class IX

Solution:

In the fig ABCD is cyclic quadrilateral.

∟BCF = 180˚ – 110˚ = 70˚

∟BCA = 23˚

∟ACF = ∟BCF – ∟BCA = 70˚ – 23˚ = 47˚ ………(1)

∟FCD = ∟FED – 180˚

∟FCD = 180˚ – 115˚ = 65˚

 

∟ACB = ∟AFB but ∟ACB = 23˚

∟AFB = 23˚

 

∟EFD  = ∟ECD [angle in the same segment]

We have ∟ECD = 23˚

∟FCE = ∟FCD – ∟ECD = 65˚ – 23˚ = 42˚

 

∟ACD = ∟ACF + ∟FCE = 47˚ + 42˚ = 89˚

∟ACD = 89˚


 

Circles – Exercise 4.4.2 – Class IX

  1. In a circle whose radius is 8cm., chord is drawn at a point 3cm. from the centre of the circle. The chord is divided into two segments by a point on it. If one segment of the chord is 9cm .What is the length of the other segment?

Solution:

Circles – Exercise 4.4.2 – Class IX

From the data given,

Radius, OB = 8 cm

OC = 3cm

AB is a chord, The chord is divided into two line segments AP and BP , let BP = 9 cm. Now we have to find AP,

To find AP:

Proof:

In Δ OCB ; ∟OCB = 90˚[since OC⊥AB]

BC2 + OC2 = OB2

BQ2 = OB2 – OC2

BQ = √(82 – 32) = √(64-9) = √55

AB = 2√55

AP = AB – BP = (2√55 – 9)cm


  1. Suppose two chords of a circle are equidistant from the center of the circle. Prove that the chords have equal length.

Solution:

Circles – Exercise 4.4.2 – Class IX

Figure is drawn by the data given in the problem. Let O be the centre of the circle. AB and CD are 2 chords which are equidistant from the centre O of the circle, since OB = OD

Now we have to prove OB = OD

Proof:

In ΔMBO and ΔNDO

∟BMO = ∟DNO = 90˚

∟BMO = ∟DNO = 90˚

MO = NO (data given)

OB = OD (radii)

ΔMBO = ΔNDO (RHS)

MB = DN

Hence, AB = CD


  1. Suppose two chords of a circle are unequal in length. Prove that the chord of larger length is nearer to the centre than the chord of smaller length

Solution:

Circles – Exercise 4.4.2 – Class IX

In the figure, O is the center of the circle, where AB and CD are chords. Ab is larger and CD is smaller chord. OP and OQ the distance from the chord AB and CD respectively.

To prove: OP < OQ

Proof:

In ΔAPO;  AQ2 = AP2 + PQ2 …………(1)

In ΔCQO; OC2 = CQ2 + OQ2 ………….(2)

AO = OC radius of the circle.

From (1) and (2),

AP2 + PO2 = CQ2 + OQ2

Let AP = CQ + x                 [AB < CQ; 1/2 AB > 1/2 CD ; AP > CQ]

(CO+X)2 + PO2 = CQ2 + OQ2

CQ2 + 2. CQ. X + PO2 = CQ2 + OQ2

OQ2 = OP2 + 2.CQ.X

or

OQ2 > OP2

OQ > OP

OP < OQ


  1. Let AB and CD be parallel chords of a circle with center O. M is the midpoint of AB and N is the midpoint of CD . Prove that O, N ,M are collinear. If MN = 3cm , AB= 4cm, CD = 10 cm find the radius of the circle.

Solution:

Circles – Exercise 4.4.2 – Class IX

Given: AB and CD are to parallel chords. M and N are midpoint of AB and CD respectively.

To prove: O, N, M are collinear

To find: radius of the circle i.e, OD

Proof:

∟OMB = 90˚ (M is the mid point)

∟ONB = 90˚ and ∟MND = 90˚

∟OND + ∟MND = 180˚

Therefore, O, N , M are collinear.

To find the radius of the circle with centre O.

Let ON = x, in ΔOMB

OB2 = OM2 + MB2

= (3 + x)2 + 22 (AB = 4cm and MB = 2 cm)

= 9 + 6x + x2 + 4

OB2 = x2 + 6x + 13 ……(1)

In ΔOND, OD2 = ON2 + ND2

=  x2 + 52 (CD = 10cm and ND = 5 cm)

OD2 = x2 + 25

OB2 = x2 + 25 …………(2) (OB = OD)[CD = 10cm and ND = 5cm)

x2 + 6x + 13 = x2+ 52

6x = 25 – 13 = 12

x = 12/6 = 2

Thus, x = 2 cm

OD2 = x2 + 52

OD2 = 22 + 52

= 4 + 25 = 29

OD = √29


  1. Circles – Exercise 4.4.2 – Class IXIn the figure ,ABCD is a straight line two concentric circles with centre O. Is it possible to have CD = 6cm, OD = 9cm and OB = 5 cm. Justify.

 

 

 

Solution:

Let us solve the problem using the Pythagoras theorem.

In the figure it is given that OD = 9 cm, OB = 5cm and CD = 6 cm.

Constrction: Draw a line OP from O perpendicular to AD, which cuts the chord AD at P.

Therefore, we have ΔOPD where ∟OPD = 90 ˚, Thus, ΔOPD is a right angled triangle.

By Pythagoras theorem, OD2 = OP2+ PD2

We know OD = 9 thus, 92= OP2+ PD2, where 92 is not a Pythagorian Number.

Therefore,  CD = 6cm, OD = 9cm and OB = 5 cm is not possible.


  1. Circles – Exercise 4.4.2 – Class IX

In the fig A, C, B are collinear centres of three circles and AC =CB. Prove that PQ = RS where the line PQRS passes through the point of intersection Q and R and cuts other two circles at P and Q.

[Hint: draw perpendiculars from A, C, B to PQ]

Solution:

StatementReason
AC = CBGiven

 

Draw AD, CE and BF perpendicular to PQ

 

Construction
AD||CE||BFcorresponding angles are equal i.e., 90˚
DE = EFequal intercepts
DQ + QE = ER + RF

 

⊥ from the centre bisects the chord
1/2PQ + QE = ER + 1/2RS

 

QE = ER
1/2PQ = 1/2RS

 

 
PQ = RS

 

 

  1. Suppose AB and CD are two parallel chords in a circle .prove that the line joining this midpoint passes through the centre of the circle.

Solution:

7.png

Given : O center of the circle and AB ||CD

Construction : Join OM and ON

Proof: M is the midpoint of CD

∟OMB =90 ˚

Similarly N is midpoint of CD

∟OND =90 ˚

∟OMB + ∟BMN = 180 ˚ (∟BMN =90 ˚)

i.e OMN is a straight line

MN passes through O.


  1. Prove that a parallelogram inscribed in a circle is a rectangle

Solution:

Circles – Exercise 4.4.2 – Class IX

Parallelogram ABCD  is cyclic

∟A + ∟C = 180 ˚ [opposite  angle of cyclic quadrilateral]

But ∟A = ∟C = 90 ˚ ……….(1) [opposite  angle of parallelogram are equal]

Similarly, ∟B + ∟D = 180˚

∟B = ∟D = 90˚ ………………..(2) [opposite  angle of parallelogram are equal]

AB = CD and AD = BC ………..(3) [Opposite sides of the  parallelogram]

Opposite sides are equal and each angle is right angle therefore, ABCD is rectangle.