  ICSE Class 10 Expansion Solution New Pattern

## ICSE Class Expansion10 Solution New Pattern By Clarify Knowledge

ICSE Class 10 Expansion Solution New Pattern 2022

## Chapter 4 - Expansion Exercise Ex. 4(A)

Question 1

Find the square of:

(i) 2a + b

(ii) 3a + 7b

(iii) 3a - 4b

(iv) Solution 1

Question 2

Use identities to evaluate:

(i) (101)2

(ii) (502)2

(iii) (97)2

(iv) (998)2Solution 2

Question 3

Evalute:

(i)

(ii)

Solution 3

(i)

(ii)

Question 4

Evaluate:

(i)

(ii) (4a +3b)2 - (4a - 3b)2 + 48ab.Solution 4

(i)Consider the given expression:

(ii)Consider the given expression:

Question 5

If a + b = 7 and ab = 10; find a - b.Solution 5

Question 6

If a -b = 7 and ab = 18; find a + b.Solution 6

Question 7

If x + y = and xy = ; find:

(i) x - y

(ii) x2- y2Solution 7

(i)

(ii)

Question 8

If a - b = 0.9 and ab = 0.36; find:

(i) a + b

(ii) a2 - b2.Solution 8

(i)

(ii)

Question 9

If a - b = 4 and a + b = 6; find

(i) a2 + b2

(ii) abSolution 9

(i)

(ii)

Question 10

If a + = 6 and  a ≠ 0 find :

(i)

(ii) Solution 10

(i)

(ii)

Question 11

If a - = 8 and a ≠0, find :

(i)

(ii) Solution 11

(i)

(ii)

Question 12

If a2 - 3a + 1 = 0, and a≠ 0; find:

(i)

(ii) Solution 12

(i)

(ii)

Question 13

If a2 - 5a - 1 = 0 and a ≠ 0; find:

(i)

(ii)

(iii) Solution 13

(i)

(ii)

(iii)

Question 14

If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2.Solution 14

Question 15

The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.Solution 15

Given x is 2 more than y, so x = y + 2

Sum of squares of x and y is 34, so x+ y= 34.

Replace x = y + 2 in the above equation and solve for y.

We get (y + 2)+ y= 34

2y+ 4y - 30 = 0

y+ 2y - 15 = 0

(y + 5)(y - 3) = 0

So y = -5 or 3

For y = -5, x =-3

For y = 3, x = 5

Product of x and y is 15 in both cases.Question 16

The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.Solution 16

Let the two positive numbers be a and b.

Given difference between them is 5 and sum of squares is 73.

So a - b = 5, a+ b= 73

Squaring on both sides gives

(a - b)= 52

a+ b- 2ab = 25

But a+ b= 73

So 2ab = 73 - 25 = 48

ab = 24

So, the product of numbers is 24.

## Chapter 4 - Expansion Exercise Ex. 4(B)

Question 1

Find the cube of :

(i) 3a- 2b

(ii) 5a + 3b

(iii) (iv) Solution 1

(i)

(ii)

(iii)

(iv)

Question 2

If a2 + = 47 and a ≠ 0 find:

(i)

(ii) Solution 2

(i)

(ii)

Question 3

If a2 + = 18; a ≠ 0 find:

(i)

(ii) Solution 3

(i)

(ii)

Question 4

If a + = p and a ≠ 0 ; then show that:

Solution 4

Question 5

If a + 2b = 5; then show that:

a3 + 8b3 + 30ab = 125.Solution 5

Question 6

If  and a ≠ 0 ; then show: a3 + Solution 6

Question 7

If a + 2b + c = 0; then show that:

a3 + 8b3 + c3 = 6abc.Solution 7

Question 8

Use property to evaluate:

(i) 133 + (-8)3 + (-5)3

(ii)73 + 33 + (-10)3

(iii) 93 - 53 - 43

(iv) 383 + (-26)3 + (-12)3Solution 8

Property is if a + b + c = 0 then a+ b+ c= 3abc

(i) a = 13, b = -8 and c = -5

133 + (-8)3 + (-5)= 3(13)(-8)(-5) = 1560

(ii) a = 7, b = 3, c = -10

73 + 33 + (-10)= 3(7)(3)(-10) = -630

(iii)a = 9, b = -5, c = -4

93 - 53 - 4= 93 + (-5)3 + (-4)= 3(9)(-5)(-4) = 540

(iv) a = 38, b = -26, c = -12

383 + (-26)3 + (-12)= 3(38)(-26)(-12) = 35568Question 9

Solution 9

(i)

(ii)

Question 10

If and a - = 4; find:

(i)

(ii)

(iii) Solution 10

(i)

(ii)

(iii)

Question 11

If and  x + = 2; then show that:

Solution 11

Thus from equations (1), (2) and (3), we have

Question 12

If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.Solution 12

Given that 2x 3y = 10, xy = 16

Question 13

Expand :

(i)  (3x + 5y + 2z) (3x - 5y + 2z)

(ii)  (3x - 5y - 2z) (3x - 5y + 2z)Solution 13

(i)

(3x + 5y + 2z) (3x - 5y + 2z)

= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}

= (3x + 2z)2 - (5y)2

{since (a + b) (a - b) = a2 - b2}

= 9x2 + 4z2 + 2 × 3x × 2z - 25y2

= 9x2 + 4z2 + 12xz - 25y2

= 9x2 + 4z- 25y2 + 12xz

(ii)

(3x - 5y - 2z) (3x - 5y + 2z)

= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}

= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}

= 9x2 + 25y2 - 2 × 3x × 5y - 4z2

= 9x2 + 25y2- 30xy - 4z2

= 9x2 +25y2 - 4z2 - 30xyQuestion 14

The sum of two numbers is 9 and their product is 20. Find the sum of their

(i) Squares (ii) CubesSolution 14

Given sum of two numbers is 9 and their product is 20.

Let the numbers be a and b.

a + b = 9

ab = 20

Squaring on both sides gives

(a+b)= 92

a+ b+ 2ab = 81

a+ b+ 40 = 81

So sum of squares is 81 - 40 = 41

Cubing on both sides gives

(a + b)= 93

a+ b+ 3ab(a + b) = 729

a+ b+ 60(9) = 729

a+ b= 729 - 540 = 189

So the sum of cubes is 189.Question 15

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:

(i) Sum of these numbers

(ii) Difference of their cubes

(iii) Sum of their cubes.Solution 15

Given x - y = 5 and xy = 24 (x>y)

(x + y)= (x - y)+ 4xy = 25 + 96 = 121

So, x + y = 11; sum of these numbers is 11.

Cubing on both sides gives

(x - y)= 53

x- y- 3xy(x - y) = 125

x- y- 72(5) = 125

x- y3= 125 + 360 = 485

So, difference of their cubes is 485.

Cubing both sides, we get

(x + y)= 113

x+ y+ 3xy(x + y) = 1331

x+ y= 1331 - 72(11) = 1331 - 792 = 539

So, sum of their cubes is 539.Question 16

If 4x+ y= a and xy = b, find the value of 2x + y.Solution 16

xy = b ….(i)

4x+ y= a ….(ii)

Now, (2x + y)2 = (2x)2 + 4xy + y2

= 4x2 + y2 + 4xy

= a + 4b ….[From (i) and (ii)]

## Chapter 4 - Expansion Exercise Ex. 4(C)

Question 1

Expand:

(i) (x + 8) (x + 10)

(ii) (x + 8) (x - 10)

(iii) (x - 8) (x + 10)

(iv) (x - 8) (x - 10) Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.Solution 4

Question 5

If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.Solution 5

Question 6

If a + b + c = p and ab + bc + ca = q; find a2 + b2 + c2.Solution 6

Question 7

If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.Solution 7

Question 8

If x+ y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.Solution 8

## Chapter 4 - Expansion Exercise Ex. 4(D)

Question 1

If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz; evaluate: (3z + x)^ Solution 1

Given that x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0

Therefore, x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

Now

Question 2

If a + = m and a ≠ 0 ; find in terms of 'm'; the value of :

(i)

(ii) Solution 2

(i)

(ii)

Question 3

In the expansion of (2x2 - 8) (x - 4)2; find the value of

(i) coefficient of x3

(ii) coefficient of x2

(iii) constant term.Solution 3

Question 4

If x > 0 and find: .Solution 4

Given that

Question 5

If 2(x2 + 1) = 5x, find :

(i) (ii) Solution 5

(i)

2(x2 + 1} = 5x

Dividing by x, we have

(ii)

Question 6

If a2 + b2 = 34 and ab = 12; find:

(i) 3(a + b)2 + 5(a - b)2

(ii) 7(a - b)2 - 2(a + b)2Solution 6

a2 + b2 = 34, ab= 12

(a + b)2 = a2 + b2 + 2ab

= 34 + 2 x 12 = 34 + 24 = 58

(a - b)2 = a2 + b2 - 2ab

= 34 - 2 x 12 = 34- 24 = 10

(i) 3(a + b)+ 5(a - b)2

= 3 x 58 + 5 x 10 = 174 + 50

= 224

(ii) 7(a - b)2 - 2(a + b)2

= 7 x 10 - 2 x 58 = 70 - 116 = -46Question 7

If 3x -  and x ≠ 0 find : .Solution 7

Given 3x -

We need to find

Question 8

If x2 + = 7 and  x ≠ 0; find the value of:

.Solution 8

Given that

We need to find the value of

Consider the given equation:

Question 9

If x = and x ≠ 5 find .Solution 9

By cross multiplication,

=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x

Dividing both sides by x,

Question 10

If x =  and x ≠ 5 find .Solution 10

By cross multiplication,

=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x

Dividing both sides by x,

Question 11

If 3a + 5b + 4c = 0, show that:

27a3 + 125b3 + 64c3 = 180 abcSolution 11

Given that 3a + 5b + 4c = 0

3a + 5b = -4c

Cubing both sides,

(3a + 5b)3 = (-4c)3

=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3

=>27a3 + 125b3 + 45ab x (-4c) = -64c3

=>27a3 + 125b3 - 180abc = -64c3

=>27a3 + 125b3 + 64c3 = 180abc

Hence proved.Question 12

The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.Solution 12

Let a, b be the two numbers

.'. a + b = 7 and a3 + b3 = 133

(a + b)3 = a3 + b3 + 3ab (a + b)

=> (7)3 = 133 + 3ab (7)

=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210

=> 21ab = 210 => ab= 2I

Now a2 + b2 = (a + b)2 - 2ab

=72 - 2 x 10 = 49 - 20 = 29Question 13

In each of the following, find the value of 'a':

(i) 4x2 + ax + 9 = (2x + 3)2

(ii) 4x2 + ax + 9 = (2x - 3)2

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2Solution 13

(i) 4x2 + ax + 9 = (2x + 3)2

Comparing coefficients of x terms, we get

ax = 12x

so, a = 12

(ii) 4x2 + ax + 9 = (2x - 3)2

Comparing coefficients of x terms, we get

ax = -12x

so, a = -12

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Comparing coefficients of x terms, we get

(7a - 5)x = 30x

7a - 5 = 30

7a = 35

a = 5Question 14

If

(i)  (ii) Solution 14

Given

Question 15

The difference between two positive numbers is 4 and the difference between their cubes is 316.

Find:

(i) Their product

(ii) The sum of their squaresSolution 15

Given difference between two positive numbers is 4 and difference between their cubes is 316.

Let the positive numbers be a and b

a - b = 4

a- b= 316

Cubing both sides,

(a - b)= 64

a- b- 3ab(a - b) = 64

Given a- b= 316

So 316 - 64 = 3ab(4)

252 = 12ab

So ab = 21; product of numbers is 21

Squaring both sides, we get

(a - b)= 16

a+ b- 2ab = 16

a+ b= 16 + 42 = 58

Sum of their squares is 58.

## Chapter 4 - Expansion Exercise Ex. 4(E)

Question 1

Simplify:

(i) (x + 6)(x + 4)(x - 2)

(ii) (x - 6)(x - 4)(x + 2)

(iii) (x - 6)(x - 4)(x - 2)

(iv) (x + 6)(x - 4)(x - 2) Solution 1

Using identity:

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

(i) (x + 6)(x + 4)(x - 2)

= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)

= x3 + 8x2 + (24 - 8 - 12)x - 48

= x3 + 8x2 + 4x - 48

(ii) (x - 6)(x - 4)(x + 2)

= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2

= x3 - 8x2 + (24 - 8 - 12)x + 48

= x3 - 8x2 + 4x + 48

(iii) (x - 6)(x - 4)(x - 2)

= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)

= x3 - 12x2 + (24 + 8 + 12)x - 48

= x3 - 12x2 + 44x - 48

(iv) (x + 6)(x - 4)(x - 2)

= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)

= x3 - 0x2 + (-24 + 8 - 12)x + 48

= x3 - 28x + 48 Question 2

Solution 2

Question 3

Using suitable identity, evaluate

(i) (104)3

(ii) (97)3 Solution 3

Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)

(i) (104)3 = (100 + 4)3

= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)

= 1000000 + 64 + 1200 × 104

= 1000000 + 64 + 124800

= 1124864

(ii) (97)= (100 - 3)3

= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)

= 1000000 - 27 - 900 × 97

= 1000000 - 27 - 87300

= 912673Question 4

Solution 4

Question 5

Solution 5

Question 6

If a - 2b + 3c = 0; state the value of a- 8b3 + 27c3.Solution 6

a- 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

Since a - 2b + 3c = 0, we have

a- 8b3 + 27c= a3 + (-2b)3 + (3c)3

= 3(a)( -2b)(3c)

= -18abc Question 7

If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.Solution 7

x + 5y = 10

⇒ (x + 5y)3 = 103

⇒ x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000

⇒ x3 + (5y)3 + 3(x)(5y)(10) = 1000

= x3 + (5y)3 + 150xy = 1000

= x3 + (5y)3 + 150xy - 1000 = 0 Question 8

Solution 8

Question 9

If a + b = 11 and a2 + b2 = 65; find a3 + b3.Solution 9

Question 10

Prove that:

x2+ y+ z- xy - yz - zx  is always positive.Solution 10

x+ y+ z- xy - yz - zx

= 2(x+ y+ z- xy - yz - zx)

= 2x+ 2y+ 2z- 2xy - 2yz - 2zx

= x+ x2 + y+ y2 + z2 + z- 2xy - 2yz - 2zx

= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)

= (x - y)2 + (z - x)2 + (y - z)2

Since square of any number is positive, the given equation is always positive.Question 11

Find:

(i) (a + b)(a + b)

(ii) (a + b)(a + b)(a + b)

(iii) (a - b)(a - b)(a - b) by using the result of part (ii)Solution 11

(i) (a + b)(a + b) = (a + b)2

= a × a + a × b + b × a + b × b

= a2 + ab + ab + b2

= a2 + b2 + 2ab

(ii) (a + b)(a + b)(a + b)

= (a × a + a × b + b × a + b × b)(a + b)

= (a2 + ab + ab + b2)(a + b)

= (a2 + b2 + 2ab)(a + b)

= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b

= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2

= a3 + b3 + 3a2b + 3ab2

(iii) (a - b)(a - b)(a - b)

In result (ii), replacing b by -b, we get

(a - b)(a - b)(a - b)

= a3 + (-b)3 + 3a2(-b) + 3a(-b)2

= a3 - b3 - 3a2b + 3ab2

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