NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions. Furthermore, here we’ve provided you with the latest solution for Class 8 Maths Chapter 9 – Arithmetic Expressions. As a result here you’ll find solutions to all the exercises. This NCERT Class 8 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 8 Maths Chapter 9 – Arithmetic Expressions. The Chapter 9 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise Wise Solution

Exercise 9.1 – Page 140 of NCERT
Exercise 9.2 – Page 143 of NCERT
Exercise 9.3 – Page 146 of NCERT
Exercise 9.4 – Page 148 of NCERT
Exercise 9.5 – Page 151 of NCERT

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise 9.1 Solution

Here you’ll find NCERT Chapter 9 – Arithmetic Expressions Exercise 9.1 Solution.
Exercise 9.1: Solutions of Questions on Page Number: 140

Q1: Identify the terms, their coefficients for each of the following expressions.

1. 5xyz² – 3zy

2. 1 + x + x²

3. 4x²y² – 4x²y²z² + z²

4. 3 – pq + qr – rp

5. 
   
6. 0.3a – 0.6ab + 0.5b

Answer:

The terms and the respective coefficients of the given expressions are as follows.

–	Terms	Coefficients
(i)	5xyz2 – 3zy	5 – 3
(ii)	1x x2	1 1 1
(iii)	4x2y2 – 4x2y2z2 z2	4 – 4 1
(iv)	3 – pq qr – rp	3 -1 1 -1
(v)	– xy	1/2 1/2 -1
(vi)	0.3a – 0.6ab 0.5b	0.3 – 0.6 0.5

Q2: Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd +

da, pqr, p²q + pq², 2p + 2q

Answer:

The given expressions are classified as Monomials: 1000, pqr

Binomials: x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q Trinomials: 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy Polynomials that do not fit in any of these categories are x+ x²+ x³+ x⁴, ab + bc + cd + da

Q3: Add the following.

1. ab – bc, bc – ca, ca – ab

2. a – b + ab, b – c + bc, c – a + ac
   
3. 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
   
4. l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Answer:

The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.

(i)

Thus, the sum of the given expressions is 0.

(ii)

Thus, the sum of the given expressions is ab + bc + ac.

(iii)

Thus, the sum of the given expressions is – p²q² + 4pq + 9.

(iv)

Thus, the sum of the given expressions is 2(l² + m² + n² + lm + mn + nl).

Q4:

1. Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

2. Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

3. Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q

Answer:

The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.

(a)

(b)

(c)

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise 9.2 Solution

Here you’ll find NCERT Chapter 9 – Arithmetic Expressions Exercise 9.2 Solution.
Exercise 9.2: Solutions of Questions on Page Number: 143

Q1: Find the product of the following pairs of monomials.
(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq

(iv) 4p³, – 3p (v) 4p, 0

Answer:

The product will be as follows.

(i) 4 x 7p = 4 x 7 x p = 28p

(ii) – 4p x 7p = – 4 x p x 7 x p = (- 4 x 7) x (p x p) = – 28 p²

3. – 4p x 7pq = – 4 x p x 7 x p x q = (- 4 x 7) x (p x p x q) = – 28p²q

4. 4p³ x – 3p = 4 x (- 3) x p x p x p x p = – 12 p⁴

5. 4p x 0 = 4 x p x 0 = 0

Q2: Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)

Answer:

We know that,

Area of rectangle = Length x Breadth Area of 1^st rectangle = p x q = pq

Area of 2^nd rectangle = 10m x 5n = 10 x 5 x m x n = 50 mn Area of 3^rd rectangle = 20x² x 5y² = 20 x 5 x x² x y² = 100 x²y² Area of 4^th rectangle = 4x x 3x² = 4 x 3 x x x x² = 12x³

Area of 5^th rectangle = 3mn x 4np = 3 x 4 x m x n x n x p = 12mn²p

Q3: Complete the table of products.

	2x	– 5y	3x2	– 4xy	7x2y	– 9x2y2
2x	4x2	…	…	…	…	…
– 5y	…	…	– 15x2y	…	…	…
3x2	…	…	…	…	…	…
– 4

Answer:

The table can be completed as follows.

	2x	– 5y	3x2	– 4xy	7x2y	– 9x2y2
2x	4x2	– 10xy	6x3	– 8x2y	14x3y	– 18x3y2
– 5y	– 10xy	25 y2	– 15x2y	20xy2	– 35x2y2	45x2y3
3x2	6x3	– 15x2y	9x4	– 12x3y	21x4y	– 27x4y2
– 4xy	– 8x2y	20xy2	– 12x3y	16x2y2	– 28x3y2	36x3y3
7x2y	14x3y	– 35x2y2	21x4y	– 28x3y2	49x4y2	– 63x4y3
– 9x2y2	– 18x3y2	45 x2y3	– 27x4y2	36x3y3	– 63x4y3	81x4y4

Q4: Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a², 7a⁴ (ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c

Answer:

We know that,

Volume = Length x Breadth x Height

1. Volume = 5a x 3a² x 7a⁴ = 5 x 3 x 7 x a x a² x a⁴ = 105 a⁷

2. Volume = 2p x 4q x 8r = 2 x 4 x 8 x p x q x r = 64pqr

3. Volume = xy x 2x²y x 2xy² = 2 x 2 x xy x x²y x xy² = 4x⁴y⁴

4. Volume = a x 2b x 3c = 2 x 3 x a x b x c = 6abc

Q5:

Obtain the product of

(i) xy, yz, zx (ii) a, – a², a³ (iii) 2, 4y, 8y², 16y³

(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp

Answer:

1. xy x yz x zx = x²y²z²

2. a x (- a²) x a³ = – a⁶

3. 2 x 4y x 8y² x 16y³ = 2 x 4 x 8 x 16 x y x y² x y³ = 1024 y⁶

4. a x 2b x 3c x 6abc = 2 x 3 x 6 x a x b x c x abc = 36a²b²c²

5. m x (- mn) x mnp = – m³n²p

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise 9.3 Solution

Here you’ll find NCERT Chapter 9 – Arithmetic Expressions Exercise 9.3 Solution.
Exercise 9.3: Solutions of Questions on Page Number: 146

Q1: Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a²b²

(iv) a² – 9, 4a (v) pq + qr + rp, 0

Answer:

(i) (4p) x (q + r) = (4p x q) + (4p x r) = 4pq + 4pr

(ii) (ab) x (a – b) = (ab x a) + [ab x (- b)] = a²b – ab²

(iii) (a + b) x (7a² b²) = (a x 7a²b²) + (b x 7a²b²) = 7a³b² + 7a²b³ (iv) (a² – 9) x (4a) = (a² x 4a) + (- 9) x (4a) = 4a³ – 36a

(v) (pq + qr + rp) x 0 = (pq x 0) + (qr x 0) + (rp x 0) = 0

Q2: Complete the table

—	First expression	Second Expression	Product
(i)	a	b + c + d	–
(ii)	x + y – 5	5 xy	–
(iii)	p	6p2 – 7p + 5	–
(iv)	4p2q2	p2 – q2	–
(v)	a + b + c	abc	–

Answer:

The table can be completed as follows.

–	First expression	Second Expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y – 5	5 xy	5x2y + 5xy2 – 25xy
(iii)	p	6p2 – 7p + 5	6p3 – 7p2 + 5p
(iv)	4p2q2	p2 – q2	4p4q2 – 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

Q3: Find the product.

(i) (a²) × (2a²²) × (4a²⁶)

(ii)

(iii)

(iv) x × x² × x³ × x⁴

Answer:

(i) (a²) × (2a²²) × (4a²⁶) = 2 × 4 ×a² × a²² × a²⁶ = 8a⁵⁰

(ii)

(iii)

(iv) x × x² × x³ × x⁴ = x¹⁰

Q4:

1. Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3, (ii) .
2. a (a² + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = – 1.

Answer:

(a) 3x (4x – 5) + 3 = 12x² – 15x + 3

(i) For x = 3, 12x² – 15x + 3 = 12 (3)² – 15(3) + 3

= 108 – 45 + 3

= 66

2. For
   

(b)a (a² + a + 1) + 5 = a³ + a² + a + 5

(i) For a = 0, a³ + a² + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a³ + a² + a + 5 = (1)³ + (1)² + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = – 1, a³ + a² + a + 5 = ( – 1)³ + ( – 1)² + ( – 1) + 5

= – 1 + 1 – 1 + 5 = 4

Q5:

1. Add: p (p – q), q (q – r) and r (r – p)

2. Add: 2x (z – x – y) and 2y (z – y – x)

3. Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)

4. Subtract: 3a (a + b + c) – 2b (a – b + c) from 4c (- a + b + c)

Answer:

1. First expression = p (p – q) = p² – pq Second expression = q (q – r) = q² – qr Third expression = r (r – p) = r² – pr Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p² + q² + r² – pq – qr – rp.

2. First expression = 2x (z – x – y) = 2xz – 2x² – 2xy Second expression = 2y (z – y – x) = 2yz – 2y² – 2yx Adding the two expressions, we obtain

Therefore, the sum of the given expressions is – 2x² – 2y² – 4xy + 2yz + 2zx. (c) 3l (l – 4m + 5n) = 3l² – 12lm + 15ln

4l (10n – 3m + 2l) = 40ln – 12lm + 8l² Subtracting these expressions, we obtain

Therefore, the result is 5l² + 25ln.

(d) 3a (a + b + c) – 2b (a – b + c) = 3a² +3ab + 3ac – 2ba + 2b² – 2bc

= 3a² + 2b² + ab + 3ac – 2bc

4c ( – a + b + c) = – 4ac + 4bc + 4c² Subtracting these expressions, we obtain

Therefore, the result is – 3a² – 2b² + 4c² – ab + 6bc – 7ac.

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise 9.4 Solution

Here you’ll find NCERT Chapter 9 – Arithmetic Expressions Exercise 9.4 Solution.
Exercise 9.4: Solutions of Questions on Page Number: 148

Q1: Multiply the binomials.

(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)

(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)

(v) (2pq + 3q²) and (3pq – 2q²)

(vi)

Answer:

(i) (2x + 5) × (4x – 3) = 2x × (4x – 3) + 5 × (4x – 3)

= 8x² – 6x + 20x – 15

= 8x² + 14x – 15 (By adding like terms)

(ii) (y – 8) × (3y – 4) = y × (3y – 4) – 8 × (3y – 4)

= 3y² – 4y – 24y + 32

= 3y² – 28y + 32 (By adding like terms)

(iii) (2.5l – 0.5m) × (2.5l + 0.5m) = 2.5l × (2.5l + 0.5m) – 0.5m (2.5l + 0.5m)

= 6.25l² + 1.25lm – 1.25lm – 0.25m²

= 6.25l² – 0.25m²

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq – 2q²) = 2pq × (3pq – 2q²) + 3q² × (3pq – 2q²)

= 6p²q² – 4pq³ + 9pq³ – 6q⁴

= 6p²q² + 5pq³ – 6q⁴

(vi)

Q2: Find the product.

(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y)

(iii) (a² + b) (a + b²) (iv) (p² – q²) (2p + q)

Answer:

(i) (5 – 2x) (3 + x) = 5 (3 + x) – 2x (3 + x)

= 15 + 5x – 6x – 2x²

= 15 – x – 2x²

(ii) (x + 7y) (7x – y) = x (7x – y) + 7y (7x – y)

= 7x² – xy + 49xy – 7y²

= 7x² + 48xy – 7y²

(iii) (a² + b) (a + b²) = a² (a + b²) + b (a + b²)

= a³ + a²b² + ab + b³

(iv) (p² – q²) (2p + q) = p² (2p + q) – q² (2p + q)

= 2p³ + p²q – 2pq² – q³

Q3: Simplify.

(i) (x² – 5) (x + 5) + 25

(ii) (a² + 5) (b³ + 3) + 5

3. (t + s²) (t² – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd) (v) (x + y) (2x + y) + (x + 2y) (x – y)

(vi) (x + y) (x² – xy + y²)

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c) (a + b – c)

Answer:

(i) (x² – 5) (x + 5) + 25

= x² (x + 5) – 5 (x + 5) + 25

= x³ + 5x² – 5x – 25 + 25

= x³ + 5x² – 5x

(ii) (a² + 5) (b³ + 3) + 5

= a² (b³ + 3) + 5 (b³ + 3) + 5

= a²b³ + 3a² + 5b³ + 15 + 5

=a²b³ + 3a² + 5b³ + 20

(iii) (t + s²) (t² – s)

= t (t² – s) + s² (t² – s)

= t³ – st + s²t² – s³

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

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

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= (ac + ac + 2ac) + (ad – ad) + (bc – bc) + (2bd – bd – bd)

= 4ac

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

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

= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²

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

= 3x² – y² + 4xy

(vi) (x + y) (x² – xy + y²)

= x (x² – xy + y²) + y (x² – xy + y²)

= x³ – x²y + xy² + x²y – xy² + y³

= x³ + y³ + (xy² – xy²) + (x²y – x²y)

= x³ + y³

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y

= 1.5x (1.5x + 4y + 3) – 4y (1.5x + 4y + 3) – 4.5x + 12y

= 2.25 x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y

= 2.25 x² + (6xy – 6xy) + (4.5x – 4.5x) – 16y² + (12y – 12y)

= 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)

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

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

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

= a² + b² – c² + 2ab

NCERT Solutions for Class 8 Maths Chapter 9 – Arithmetic Expressions Exercise 9.5 Solution

Here you’ll find NCERT Chapter 9 – Arithmetic Expressions Exercise 9.5 Solution.
Exercise 9.5: Solutions of Questions on Page Number: 151

Q1: Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a – 7) (2a – 7) (iv)

(v) (1.1m – 0.4) (1.1 m + 0.4) (vi) (a² + b²) ( – a² + b²)

(vii) (6x – 7) (6x + 7) (viii) ( – a + c) ( – a + c) (ix) (x) (7a – 9b) (7a – 9b)

Answer:

The products will be as follows. (i) (x + 3) (x + 3) = (x + 3)²

= (x)² + 2(x) (3) + (3)² [(a + b)² = a² + 2ab + b²]

= x² + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)²

= (2y)² + 2(2y) (5) + (5)² [(a + b)² = a² + 2ab + b²]

= 4y² + 20y + 25

(iii) (2a – 7) (2a – 7) = (2a – 7)²

= (2a)² – 2(2a) (7) + (7)² [(a – b)² = a² – 2ab + b²]

= 4a² – 28a + 49

(iv)

[(a – b)² = a² – 2ab + b²]

(v) (1.1m – 0.4) (1.1 m + 0.4)

= (1.1m)² – (0.4)² [(a + b) (a – b) = a² – b²]

= 1.21m² – 0.16

(vi) (a² + b²) ( – a² + b²) = (b² + a²) (b² – a²)

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

= b⁴ – a⁴

(vii) (6x – 7) (6x + 7) = (6x)² – (7)² [(a + b) (a – b) = a² – b²]

= 36x² – 49

(viii) ( – a + c) ( – a + c) = ( – a + c)²

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

= a² – 2ac + c²

(ix)

[(a + b)² = a² + 2ab + b²]

(x) (7a – 9b) (7a – 9b) = (7a – 9b)²

= (7a)² – 2(7a)(9b) + (9b)² [(a – b)² = a² – 2ab + b²]

= 49a² – 126ab + 81b²

Q2: Use the identity (x + a) (x + b) = x² + (a + b)x + ab to find the following products.

(i) (x + 3) (x + 7) (ii) (4x +5) (4x + 1)

(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)

(v) (2x +5y) (2x + 3y) (vi) (2a² +9) (2a² + 5)

(vii) (xyz – 4) (xyz – 2)

Answer:

The products will be as follows.

(i) (x + 3) (x + 7) = x² + (3 + 7) x + (3) (7)

= x² + 10x + 21

(ii) (4x + 5) (4x + 1) = (4x)² + (5 + 1) (4x) + (5) (1)

= 16x² + 24x + 5

(iii)

= 16x² – 24x + 5

(iv)

= 16x² + 16x – 5

(v) (2x +5y) (2x + 3y) = (2x)² + (5y + 3y) (2x) + (5y) (3y)

= 4x² + 16xy + 15y²

(vi) (2a² +9) (2a² + 5) = (2a²)² + (9 + 5) (2a²) + (9) (5)

= 4a⁴ + 28a² + 45

(vii) (xyz – 4) (xyz – 2)

=

= x²y²z² – 6xyz + 8

Q3: Find the following squares by suing the identities.
(i) (b – 7)² (ii) (xy + 3z)² (iii) (6x² – 5y)²

(iv) (v) (0.4p – 0.5q)² (vi) (2xy + 5y)²

Answer:

(i) (b – 7)² = (b)² – 2(b) (7) + (7)² [(a – b)² = a² – 2ab + b²]

= b² – 14b + 49

(ii) (xy + 3z)² = (xy)² + 2(xy) (3z) + (3z)² [(a + b)² = a² + 2ab + b²]

= x²y² + 6xyz + 9z²

(iii) (6x² – 5y)² = (6x²)² – 2(6x²) (5y) + (5y)² [(a – b)² = a² – 2ab + b²]

= 36x⁴ – 60x²y + 25y²

(iv) [(a + b)² = a² + 2ab + b²]

(v) (0.4p – 0.5q)² = (0.4p)² – 2 (0.4p) (0.5q) + (0.5q)²

[(a – b)² = a² – 2ab + b²]

= 0.16p² – 0.4pq + 0.25q²

(vi) (2xy + 5y)² = (2xy)² + 2(2xy) (5y) + (5y)²

[(a + b)² = a² + 2ab + b²]

= 4x²y² + 20xy² + 25y²

Q4: Simplify.

(i) (a² – b²)² (ii) (2x +5)² – (2x – 5)²

(iii) (7m – 8n)² + (7m + 8n)² (iv) (4m + 5n)² + (5m + 4n)²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

(vi) (ab + bc)² – 2ab²c (vii) (m² – n²m)² + 2m³n²

Answer:

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

= a⁴ – 2a²b² + b⁴

(ii) (2x +5)² – (2x – 5)² = (2x)² + 2(2x) (5) + (5)² – [(2x)² – 2(2x) (5) + (5)²]

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

= 4x² + 20x + 25 – [4x² – 20x + 25]

= 4x² + 20x + 25 – 4x² + 20x – 25 = 40x

(iii) (7m – 8n)² + (7m + 8n)²

= (7m)² – 2(7m) (8n) + (8n)² + (7m)² + 2(7m) (8n) + (8n)²

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

= 49m² – 112mn + 64n² + 49m² + 112mn + 64n²

= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²

= (4m)² + 2(4m) (5n) + (5n)² + (5m)² + 2(5m) (4n) + (4n)²

[ (a + b)² = a² + 2ab + b²]

= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²

= 41m² + 80mn + 41n²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

= (2.5p)² – 2(2.5p) (1.5q) + (1.5q)² – [(1.5p)² – 2(1.5p)(2.5q) + (2.5q)²]

[(a – b)² = a² – 2ab + b² ]

= 6.25p² – 7.5pq + 2.25q² – [2.25p² – 7.5pq + 6.25q²]

= 6.25p² – 7.5pq + 2.25q² – 2.25p² + 7.5pq – 6.25q²]

= 4p² – 4q²

6. (ab + bc)² – 2ab²c

= (ab)² + 2(ab)(bc) + (bc)² – 2ab²c [(a + b)² = a² + 2ab + b² ]

= a²b² + 2ab²c + b²c² – 2ab²c

= a²b² + b²c²

7. (m² – n²m)² + 2m³n²

= (m²)² – 2(m²) (n²m) + (n²m)² + 2m³n² [(a – b)² = a² – 2ab + b² ]

= m⁴ – 2m³n² + n⁴m² + 2m³n²

= m⁴ + n⁴m²

Q5: Show that

(i) (3x + 7)² – 84x = (3x – 7)²

(ii) (9p – 5q)² + 180pq = (9p + 5q)²

(iii)

(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Answer:

(i) L.H.S = (3x + 7)² – 84x

= (3x)² + 2(3x)(7) + (7)² – 84x

= 9x² + 42x + 49 – 84x

= 9x² – 42x + 49

R.H.S = (3x – 7)² = (3x)² – 2(3x)(7) +(7)²

= 9x² – 42x + 49

L.H.S = R.H.S

(ii) L.H.S = (9p – 5q)² + 180pq

= (9p)² – 2(9p)(5q) + (5q)² – 180pq

= 81p² – 90pq + 25q² + 180pq

= 81p² + 90pq + 25q²

R.H.S = (9p + 5q)²

= (9p)² + 2(9p)(5q) + (5q)²

= 81p² + 90pq + 25q²

L.H.S = R.H.S

(iii) L.H.S =

(iv) L.H.S = (4pq + 3q)² – (4pq – 3q)²

= (4pq)² + 2(4pq)(3q) + (3q)² – [(4pq)² – 2(4pq) (3q) + (3q)²]

= 16p²q² + 24pq² + 9q² – [16p²q² – 24pq² + 9q²]

= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²

= 48pq² = R.H.S

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

= (a² – b²) + (b² – c²) + (c² – a²) = 0 = R.H.S.

Q6: Using identities, evaluate.

(i) 71² (ii) 99² (iii) 102² (iv) 998²

(v) (5.2)² (vi) 297 x 303 (vii) 78 x 82

(viii) 8.9² (ix) 1.05 x 9.5

Answer:

(i) 71² = (70 + 1)²

= (70)² + 2(70) (1) + (1)² [(a + b)² = a² + 2ab + b² ]

= 4900 + 140 + 1 = 5041

(ii) 99² = (100 – 1)²

= (100)² – 2(100) (1) + (1)² [(a – b)² = a² – 2ab + b² ]

= 10000 – 200 + 1 = 9801

(iii) 102² = (100 + 2)²

= (100)² + 2(100)(2) + (2)² [(a + b)² = a² + 2ab + b² ]

= 10000 + 400 + 4 = 10404

(iv) 998² = (1000 – 2)²

= (1000)² – 2(1000)(2) + (2)² [(a – b)² = a² – 2ab + b² ]

= 1000000 – 4000 + 4 = 996004

(v) (5.2)² = (5.0 + 0.2)²

= (5.0)² + 2(5.0) (0.2) + (0.2)² [(a + b)² = a² + 2ab + b² ]

= 25 + 2 + 0.04 = 27.04

(vi) 297 x 303 = (300 – 3) x (300 + 3)

= (300)² – (3)² [(a + b) (a – b) = a² – b²]

= 90000 – 9 = 89991

(vii) 78 x 82 = (80 – 2) (80 + 2)

= (80)² – (2)² [(a + b) (a – b) = a² – b²]

= 6400 – 4 = 6396

(viii) 8.9² = (9.0 – 0.1)²

= (9.0)² – 2(9.0) (0.1) + (0.1)² [(a – b)² = a² – 2ab + b² ]

= 81 – 1.8 + 0.01 = 79.21

(ix) 1.05 x 9.5 = 1.05 x 0.95 x 10

= (1 + 0.05) (1- 0.05) x 10

= [(1)² – (0.05)²] x 10

= [1 – 0.0025] x 10 [(a + b) (a – b) = a² – b²]

= 0.9975 x 10 = 9.975

Q7:

Using a² – b² = (a + b) (a – b), find

(i) 51² – 49² (ii) (1.02)² – (0.98)² (iii) 153² – 147²

(iv) 12.1² – 7.9²

Answer:

(i) 51² – 49² = (51 + 49) (51 – 49)

= (100) (2) = 200

(ii) (1.02)² – (0.98)² = (1.02 + 0.98) (1.02 – 0.98)

= (2) (0.04) = 0.08

(iii) 153² – 147² = (153 + 147) (153 – 147)

= (300) (6) = 1800

(iv) 12.1² – 7.9² = (12.1 + 7.9) (12.1 – 7.9)

= (20.0) (4.2) = 84

Q8: Using (x + a) (x + b) = x² + (a + b) x + ab, find

(i) 103 x 104 (ii) 5.1 x 5.2 (iii) 103 x 98 (iv) 9.7 x 9.8

Answer:

(i) 103 x 104 = (100 + 3) (100 + 4)

= (100)² + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 x 5.2 = (5 + 0.1) (5 + 0.2)

= (5)² + (0.1 + 0.2) (5) + (0.1) (0.2)

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 x 98 = (100 + 3) (100 – 2)

= (100)² + [3 + (- 2)] (100) + (3) (- 2)

= 10000 + 100 – 6

= 10094

(iv) 9.7 x 9.8 = (10 – 0.3) (10 – 0.2)

= (10)² + [(- 0.3) + (- 0.2)] (10) + (- 0.3) (- 0.2)

= 100 + (- 0.5)10 + 0.06 = 100.06 – 5 = 95.06.

NCERT Class 8 Maths All Chapters Solution 

Chapter 1: Rational Numbers

Chapter 2: Linear Equations in One Variable

Chapter 3: Understanding Quadrilaterals

Chapter 4: Practical Geometry

Chapter 5: Data Handling

Chapter 6: Squares and Square root

Chapter 7: Cubes and Cube Roots

Chapter 8: Comparing Quantities

Chapter 9: Arithmetic Expressions

Chapter 10: Visualising Solid Shapes

Chapter 11: Mensuration

Chapter 12: Exponents and Powers

Chapter 13: Direct and Inverse Proportions

Chapter 14: Factorisation

Chapter 15: Introduction to Graphs

Chapter 16: Playing With Numbers