Category: CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Add:

Question 1.
Solution:
8ab + ( – 5ab) + (3ab) + ( – ab)
= 8ab – 5ab + 3ab – ab = 11 ab – 6ab
= 5ab Ans.

Question 2.
Solution:
7x + ( – 3x) + 5x + ( – x) + ( – 2x)
= 7x – 3x + 5x – x – 2x
= 12x – 6x = 6x Ans.

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Subtarct:

Question 9.
Solution:
-5a²b – 3a²b = – 8a²b Ans.

Question 10.
Solution:
6pq – ( – 8pq) = 6pq + 8pq

Question 11.
Solution:
– 8abc – ( – 2abc)
= – 8abc + 2abc
= – 6abc Ans.

Question 12.
Solution:
– 11p – ( – 16p)
= – 11p + 16p
= 5p Ans.

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:
Let length of rectangle = 5x² – 3y²
and breadth = x² + 2xy
Perimeter = 2(Length + Breadth)
= 2(5x² – 3y² + x² + 2xy)
= 2(6x² – 3y² + 2xy)
= 12x² – 6y² + 4xy Ans.

Question 20.
Solution:
Perimeter of a triangle = 6p² – 4p + 9
Sum of two sides of it = 3p² – 5p + 3 + p² – 2p + 1 = 4p² – 7p – 4
Third side = (6p² – 4p + 9) – (4p² – 7p + 4)
= 6p² – 4p + 9 – 4p² + 7p – 4
= 2p² + 3p + 5 Ans

 

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Question 1.
Solution:
(i)(x + 6)(x + 6)
= (x + 6)²

You can also Download NCERT Solutions for Class 8 English to help you to revise complete Syllabus and score more marks in your examinations.

Question 2.
Solution:
(i) (x – 4)(x – 4)
= (x – 4)²

Question 3.
Solution:
(i)(8a + 3b)²
= (8a)² + 2 x 8a x 3b + (3b)²

Question 4.
Solution:
(i)(x + 3)(x – 3)
= (x)² – (3)²

Question 5.
Solution:
(i) (54)²
= (50 + 4)²

Question 6.
Solution:
(i) (69)²
= (70 – 1)²

Question 7.
Solution:
(i)(82)² – (18)²
= (82 – 18)(82 – 18)

Question 8.
Solution:
9x² + 24x + 16
= (3x)² + 2 x 3x x 4 + (4)²
= (3x + 4)²
= (3 x 12 + 4)²
= (36 + 4)²
= (40)²
= 1600 Ans.

Question 9.
Solution:
64x² + 81y² + 144xy = (8x)² + (9y)² + 2 x 8x x 9y
= (8x + 9y)²
= \({ \left( 8\times 11+9\times \frac { 4 }{ 3 } \right) }^{ 2 }\)
= (88 + 12)²
= (100)²
= 10000 Ans.

Question 10.
Solution:
(36x² + 25y² – 60xy)
= (6x)² + (5y)² – 2 x 6x x 5y
= (6x – 5y)²
= \({ \left( 6\times \frac { 2 }{ 3 } -5\times \frac { 1 }{ 5 } \right) }^{ 2 } \)
= (4 – 1)²
= (3)² = 9

Question 11.
Solution:
\(\left( x+\frac { 1 }{ 4 } \right) =4 \)
Squaring on both sides

Question 12.
Solution:
\(\left( x-\frac { 1 }{ x } \right) =5 \)
(i) Squaring on both sides

Question 13.
Solution:
(i) (x + 1) (x – 1) (x² + 1)
= {(x)² – (1)²} (x² + 1)
= (x² – 1) (x² + 1)
= (x²)² – (1)² = x⁴ – 1 Ans.
(ii) (x – 3) (x + 3) (x² + 9)
= {(x)² – (3)² } (x² + 9)
= (x² – 9) (x² + 9)
= (x²)² – (9)² = x⁴ – 81 Ans.
(iii) (3x – 2y) (3x + 2y) (9x² + 4y²)
= {(3x)² – (2y)²} (9x² + 4y²)
= (9x² – 4y²) (9x² + 4y²)
= (9x²)² – (4y²)²
= 81x⁴ – 16y⁴ Ans.
(iv) (2p + 3) (2p – 3) (4p² + 9)
= {(2p)² – (3)²} (4p² + 9)
= (4p² – 9) (4p² + 9)
= (4p²)² – (9)² = 16p⁴ – 81 Ans.

Question 14.
Solution:
x + y = 12
Squaring both sides,
(x + y)² = (12)²
=> x² + y² + 2xy = 144
=> x² + y² + 2 x 14 = 144
=> x² + y² + 28 = 144
=> x² + y² = 144 – 28 = 116
x² + y² = 116 Ans.

Question 15.
Solution:
x – y = 7
Squaring both sides,
(x – y)² = (7)²
=> x² + y² – 2xy = 49
=> x² + y² – 2 x 9 = 49
=> x² + y² – 18 = 49
=> x² + y² = 49 + 18 = 67
x² + y² = 67 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Find each of the following products?

Question 1.
Solution:
(5x + 7) X (3x + 4) = 5x(3x + 4) + 7(3x + 4)
= 15x² + 20x + 21x + 28 = 15x² + 41x + 28 Ans.

Question 2.
Solution:
(4x + 9) X (x – 6) = 4x(x – 6) + 9(x – 6)
= 4x² – 24x + 9x – 54
= 4x² – 15x – 54 Ans.

Question 3.
Solution:
(2x + 5) X (4x – 3) = 2x(4x – 3) + 5(4x – 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15 Ans.

Question 4.
Solution:
(3y – 8) X (5y – 1) = 3y(5y – 1) – 8(5y – 1)
= 15y² – 3y – 40y + 8
= 15y² – 43y + 8 Ans.

Question 5.
Solution:
(7x + 2y) X (x + 4y)
= 7x(x + 4y) + 2y(x + 4y)
= 7x² + 28xy + 2xy + 8y²
= 7x² + 30xy + 8y² Ans.

Question 6.
Solution:
(9x + 5y) X (4x + 3y) = 9x(4x + 3y) + 5y(4x + 3y)
= 36x² + 27xy + 20xy + 15y²
= 36x² + 47xy + 15y² Ans.

Question 7.
Solution:
(3m – 4n) X (2m – 3n)
= 3m(2m – 3n) – 4n(2m – 3n)
= 6m² – 9mn – 8mn + 12n²
= 6m² – 17mn + 12n² Ans.

Question 8.
Solution:
(x² – a²) X (x – a)
= x²(x – a) – a²(x – a)
= x³ – x²a – xa² + a³ Ans.

Question 9.
Solution:
(x² – y²) X (x + 2y)
= x²(x + 2y) – y²(x + 2y)
= x³ + 2x²y – xy² – 2y³ Ans.

Question 10.
Solution:
(3p² + q²) X (2p² – 3q²)
= 3p²(2p² – 3q²) + q²(2p² – 3q²)
= 6p⁴ – 9p²q² + 2p²q² – 3q⁴
= 6p⁴ – 7p²q² – 3q⁴ Ans.

Question 11.
Solution:
(2x² – 5y²) X (x² + 3y²)
= 2x²(x² + 3y²) – 5y²(x² + 3y²)
= 2x⁴ + 6x²y² – 5x²y² – 15y⁴)
= 2x⁴ + x2y² – 15y⁴ Ans.

Question 12.
Solution:
(x³ – y³) X (x² + y²)
= x³(x² + y²) – y³(x² + y²)
= x⁵ + x³y² – x²y³ – y⁵ Ans.

Question 13.
Solution:
(x⁴ + y⁴) X (x² – y²)
= x⁴(x² – y²) + y⁴(x² – y²)
= x⁶ – x⁴y² + x²y⁴ – y⁶ Ans.

Question 14.
Solution:
\(\left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } \right) \times \left( { x+ }\frac { 1 }{ { x } } \right)\)

Find each of the following products:

Question 15.
Solution:
(x² – 3x + 7) X (2x + 3)
= (x² – 3x + 7) (2x) + (x² – 3x + 7) X 3
= 2x³ – 6x + 14x + 3x² – 9x + 21
= 2x³ – 3x² + 5x + 21 Ans.

Question 16.
Solution:
(3x² + 5x – 9) X (3x – 5)
= 3x²(3x – 5x) + 5x(3x – 5) – 9(3x – 5)
= 9x³ – 15x² + 15x² – 25x – 27x + 45
= 9x³ – 52x + 45 Ans.

Question 17.
Solution:
(x² – xy + y²) X (x + y)
= (x² – xy + y²) X x + (x² – xy + y²) X y
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³ Ans.

Question 18.
Solution:
(x² + xy + y²) X (x – y)
(x² + xy + y²) X x + (x² + xy + y²) X ( – y)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³ Ans.

Question 19.
Solution:
(x³ – 2x² + 5) X (4x – 1)
= (x³ – 2x² + 5) X 4x + (x³ – 2x² + 5) X ( – 1)
= 4x⁴ – 8x³ + 20x – x³ + 2x² – 5
= 4x⁴ – 9x³ + 2x² + 20x – 5 Ans.

Question 20.
Solution:
(9x² – x + 15) X (x² – 3)
(9x² – X + 15) X x² + (9x² – x + 15) X ( – 3)
= 9x⁴ – x³ + 15x² – 27x² + 3x – 45
= 9x⁴ – x³ – 12x² + 3x – 45 Ans.

Question 21.
Solution:
(x² – 5x + 8) X (x² + 2)
= (x² – 5x + 8) X x² + (x² – 5x + 8) X 2
= x⁴ – 5x³ + 8x² + 2x² – 10x + 16
= x⁴ – 5x³ + 10x² – 10x + 16 Ans.

Question 22.
Solution:
(x³ – 5x² + 3x + 1) X (x² – 3)
= (x³ – 5x² + 3x + 1) X x² + (x³ – 5x² + 3x + 1) X ( – 3)
= x⁵ – 5x⁴ + 3x³ + x² – 3x³ + 15x² – 9x – 3
= x⁵ – 5x⁴ + 16x² – 9x – 3 Ans.

Question 23.
Solution:
(3x + 2y – 4) X (x – y + 2)

Question 24.
Solution:
(x² – 5x + 8) X (x² + 2x – 3)

Question 25.
Solution:
(2x² + 3x – 7) X (3x² – 5x + 4)

Question 26.
Solution:
(9x² – x + 15) X (x² – x – 1)

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D

Evaluate:

Question 1.
Solution:
\(\sqrt [ 3 ]{ 64 } \)
= \(\sqrt [ 3 ]{ 4X4X4 } \)
= \(\sqrt [ 3 ]{ { 4 }^{ 3 } } \)
= 4

Question 2.
Solution:
\(\sqrt [ 3 ]{ 343 } \)
= \(\sqrt [ 3 ]{ 7X7X7 } \)
= \(\sqrt [ 3 ]{ { 7 }^{ 3 } } \)
= 7

Question 3.
Solution:
\(\sqrt [ 3 ]{ 729 } \)

Question 4.
Solution:
\(\sqrt [ 3 ]{ 1728 } \)

Question 5.
Solution:
\(\sqrt [ 3 ]{ 9261 } \)

Question 6.
Solution:
\(\sqrt [ 3 ]{ 4096 } \)

Question 7.
Solution:
\(\sqrt [ 3 ]{ 8000 } \)

Question 8.
Solution:
\(\sqrt [ 3 ]{ 3375 } \)

Question 9.
Solution:
\(\sqrt [ 3 ]{ -216 } \)

Question 10.
Solution:
\(\sqrt [ 3 ]{ -512 } \)

Question 11.
Solution:
\(\sqrt [ 3 ]{ -1331 } \)

Question 12.
Solution:
\(\sqrt [ 3 ]{ \frac { 27 }{ 64 } } \)

Question 13.
Solution:
\(\sqrt [ 3 ]{ \frac { 125 }{ 216 } } \)

Question 14.
Solution:
\(\sqrt [ 3 ]{ \frac { -27 }{ 125 } } \)

Question 15.
Solution:
\(\sqrt [ 3 ]{ \frac { -64 }{ 343 } } \)

Question 16.
Solution:
\(\sqrt [ 3 ]{ 64\times 729 }\)

Question 17.
Solution:
\(\sqrt [ 3 ]{ \frac { 729 }{ 1000 } } \)

Question 18.
Solution:
\(\sqrt [ 3 ]{ \frac { -512 }{ 343 } } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² – 36
= (x)² – (6)² { ∵ a² – b² = (a + b) (a – b)}
= (x + 6) (x – 6) Ans.

Question 2.
Solution:
4a² – 9
= (2a)² – (3)²
= (2a + 3) (2a – 3)
{ ∵ a² – b² = (a + b) (a – b)}

You can also Download NCERT Solutions for Class 8 English to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
81 – 49x²
= (9)² – (7x)²
= (9 + 7x) (9 – 7x) { ∵ a² – b² = (a + b) (a – b)}

Question 4.
Solution:
= (2x)² – (3y)²
= (2x + 3y)(2x-3y)
{∵ a² – b² = (a + b) (a – b)}

Question 5.
Solution:
Using a² – b²
= (a + b) (a – b)
= 16a² – 225b²
= (4a)² – (15b)²
= (4a + 15b) (4b – 5b)

Question 6.
Solution:
Using a² – b²
= (a + b) (a – b)
= (3ab)² – (5)²
= (3ab + 5) (3ab – 5)

Question 7.
Solution:
Using a² – b² = (a + b) (a – b)
16a² – 144 = (4a)² = (12)²
= (4a + 12) (4a – 12)
= 4 (a + 3) x 4 (a – 3)
= 16 (a + 3) (a – 3)

Question 8.
Solution:
63a² – 112b²
= 7 (9a² – 16b²)
= 7 [(3a)² – (4b)²
= 7 (3a + 4b) (3a – 4b)

Question 9.
Solution:
20a² – 45b²
= 5 {4a² – 9b²}
= 5{(2a)² – (3b)²)
= 5(2a + 3b) (2a – 3b) Ans.

Question 10.
Solution:
12x² – 27
= 3(4x² – 9)
= 3{(2x)² – (3)²}
= 3(2x + 3) (2x – 3) Ans.

Question 11.
Solution:
x³ – 64x
= x(x² – 64)
= x{(x)² – (8)²}
= x(x + 8) (x – 8) Ans.

Question 12.
Solution:
16x⁵ – 144x³
= 16x³ [x² – 9]
= 16x³ [(x)² – (3)²]
= 16x³ (x + 3) (x – 3)

Question 13.
Solution:
3x⁵ – 48x³
= 3x³ {x² – 16}
= 3x³{(x)² – (4)²}
= 3x³ (x + 4) (x – 4) Ans.

Question 14.
Solution:
16p³ – 4p
= 4p [4p² – 1]
= 4p ((2p)² – (1)²]
= 4p(2p + 1)(2p – 1)

Question 15.
Solution:
63a²b² – 7
= 7(9a²b² – 1)
= 7{(3ab)² – (1)²)
= 7(3ab + 1) (3ab – 1) Ans.

Question 16.
Solution:
1 – (b – c)²
= (1)² – (b – c)²
= (1 + b + c) (1 – b + c) Ans.
{ ∵ a² – b² = (a + b) (a – b)}

Question 17.
Solution:
(2a + 3b)² – 16c²
= (2a + 3b)² – (4c)²
=(2a + 3b + 4c)(2a + 3b – 4c)Ans.
{ ∵ a² – b² = (a + b)(a – b)}

Question 18.
Solution:
(l + m)² – (l – m)²
= (l + m + l – m)(l + m – l + m)
{ ∵ a² – b² = (a + b)(a – b)}
= 2l x 2m = 4lm

Question 19.
Solution:
(2x + 5y)² – (1)²
=(2x + 5y + 1)(2x + 5y – 1)
{ ∵ a² – b² = (a + b)(a – b)}

Question 20.
Solution:
36c² – (5a + b)²
= (6c)² – (5a + b)²
{ ∵ a² – b² = (a + b)(a – b)}
= (6c + 5a + b)(6c – 5a – b)

Question 21.
Solution:
(3x – 4y)² – 25z²
= (3x – 4y)² – (5z)²
= (3x – 4y + 5z) (3x – 4y – 5z) Ans.

Question 22.
Solution:
x² – y² – 2y – 1
= x² – (y² + 2y + 1)
= (x)² – (y + 1)²
= (x + y + 1)(x – y – 1)Ans.

Question 23.
Solution:
25 – a² – b² – 2ab
= 25 – (a² + b² + 2ab)
= (5)² – (a + b)²
= (5 + a + b)(5 – a – b)Ans.

Question 24.
Solution:
25a² – 4b² + 28bc – 49c²
= 25a² – [4b² – 28bc + 49c²]
{ ∵ a² – 2ab + b² = (a – b)²}
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
{ ∵ (a² – b² = (a + b)(a – b)}
= (5a + 2b – 7c) (5a – 2b + 7c)

Question 25.
Solution:
9a² – b² + 4b – 4
= 9a² – (b² – 4b + 4)
= (3a)² – [(b)² – 2 x b x 2 + (2)²]
= (3a)² – (b – 2)²
{ ∵ a² – 2ab + b² = (a – b)²}
= (3a + b – 2)(3a – b + 2)
{ ∵ a² – b² = (a + b)(a – b)}

Question 26.
Solution:
(10)² – (x – 5)²
= (10)² – (x – 5)²
= (10 + x – 5)(10 – x + 5)
= (5 + x) (15 – x) Ans.

Question 27.
Solution:
{(405)² – (395)²}
= (405)² – (395)²
= (405 + 395) (405 – 395)
{ ∵ a² – b²(a + b) (a – b)}
= 800 x 10 = 8000

Question 28.
Solution:
(7.8)² – (2.2)²
= (7.8 + 2.2) (7.8 – 2.2)
= 10.0 x 5.6
= 56 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Find the square root of each of the following numbers by using the method of prime factorization:

Question 1.
Solution:
225 = 3 x 3 x 5 x 5

Question 2.
Solution:
441 = 3 x 3 x 7 x 7

Question 3.
Solution:
729 =

Question 4.
Solution:
1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

Question 5.
Solution:
2025 =

Question 6.
Solution:
4096 =

Question 7.
Solution:
7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7

Question 8.
Solution:
8100 =

Question 9.
Solution:
9216 =

Question 10.
Solution:
11025 =

Question 11.
Solution:
15876 =

Question 12.
Solution:
17424 =

Question 13.
Solution:
Factorizing 252, we get :
252 = \(\overline { 2X2 } X\overline { 3X3 } X7\)

To make it is a perfect square it must by multiplied by 7.
Square root of 252 x 7 = 1764
= 2 x 3 x 7 = 42 Ans.

Question 14.
Solution:
Factorizing 2925, we get :
2925 = \(\overline { 3X3 } X\overline { 5X5 } X13\)

To make it a perfect square it must be divided by 13.
2925 ÷ 13 = 225
Square root of 225 = 3 x 5 = 15 Ans.

Question 15.
Solution:
Let no. of rows = x
Then the number of plants in each row = x

Hence no. of rows = 35
and no. of plants in each row = 35

Question 16.
Solution:
Let no. of students in the class = x
Then contribution of each student = x

No. of students of the class = 34 Ans.

Question 17.
Solution:
L.C.M. of 6, 9, 15 and 20
= 2 x 3 x 5 x 2 x 3 = 180
180 = \(\overline { 2X2 } X\overline { 3X3 } X5\)

To make it a perfect square, it must be multiplied by 5
Product = 180 x 5 = 900
Hence, the least square number = 900 Ans.

Question 18.
Solution:
L.C.M. of 8, 12, 15, 20 = 2 x 2 x 2 x 3 x 5= 120
120 = \(\overline { 2X2 }\)X2X3X5
To make it a perfect square it must be multiplied by 2 x 3 x 5 = 30
Then least perfect square = 120 x 30 = 3600

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 2 Exponents Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2A
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2B
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C

Question 1.
Solution:
(i) 57.36 = 5.736 x 10¹
(ii) 3500000 = 3.5 x 10⁶
(iii) 273000 = 2.73 x 10⁵
(iv) 168000000 = 1.68 x 10⁸
(v) 4630000000000 = 4.63 x 10¹²
(vi) 345 x 10⁵ = 3.45 x 10² x 10⁵ = 3.45 x 10⁷

Question 2.
Solution:
(i) 3.74 x 10⁵
= \(\\ \frac { 374 }{ 100 } \) x 10⁵

Question 3.
Solution:
(i) Height of Mount Everest = 8848 m
= 8.848 x 1000
= 8.848 x 10³
(ii) Speed of light = 300000000 m/sec.
= 3.00000000 x 100000000
= (3 x 10⁸) m/sec.
(iii) Distance between the earth and the sun = 149600000000 m
= (1.49600000000 x 100000000000)m = (1.496 x 10¹¹) m

Question 4.
Solution:
Mass of earth = (15.97 x 10²⁴) kg
and mass of moon = (7.35 x 10²²) kg
= Total mass of earth and moon

Question 5.
Solution:
(i) 0.0006
= \(\\ \frac { 6 }{ 10000 } \) = \(\frac { 6 }{ { 10 }^{ 4 } }\) = 6.10^-4

Question 6.
Solution:

Question 7.
Solution:
(i) 2.06 x 10^-5

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1H

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Objective Questions :
Tick the correct answer in each of the following :

Question 1.
Solution:
Answer = (c)

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Answer = (b)

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
Product of two numbers = \(\\ \frac { -28 }{ 81 } \)
One number = \(\\ \frac { 14 }{ 27 } \)

Question 13.

Solution:
Answer = (c)
Let x be the required number, then

Question 14.
Solution:
Answer = (d)
Let x is to be subtracted then

Question 15.
Solution:
Answer = (c)
sum = -3,one number = \(\\ \frac { -10 }{ 3 } \)

Question 16.
Solution:
Answer = (c)
We know that a number is called in standard form if the numerator and denominator has no common divisor except 1.
\(\\ \frac { -9 }{ 6 } \) is in standard form.

Question 17.
Solution:

Question 18.
Solution:
Answer = (b)

Question 19.
Solution:
Answer = (d)
Let x is required rational

Question 20.
Solution:
Additive inverse of \(\\ \frac { -5 }{ 9 } \) is = – \(\left( \frac { -5 }{ 9 } \right) \)

Question 21.
Solution:
Reciprocal of \(\\ \frac { -3 }{ 4 } \) is \(\\ \frac { -4 }{ 3 } \)

Question 22.
Solution:
A rational number between = \(\\ \frac { -2 }{ 3 } \)

Question 23.

Solution:
Answer: (b)
The reciprocal of a negative rational
the number is also a negative rational number.

Hope given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
Product of two numbers = – 9
one number = – 12
Let second number = x

Question 5.
Solution:
Product of two rational numbers = \(\\ \frac { -16}{ 9 } \)
One number = \(\\ \frac { -4 }{ 3 } \)

Question 6.
Solution:
Let x be multiplied

Question 7.
Solution:
Let x be multiplied

Question 8.
Solution:
Let required number = x

Question 9.
Solution:
sum of \(\\ \frac { 13 }{ 5 } \) and \(\\ \frac { -12 }{ 7 } \)
= \(\\ \frac { 13 }{ 5 } \) + \(\\ \frac { -12 }{ 7 } \)

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(i) No, not always closed under division.
(ii) No, not always commutative.
(iii) No, not always associative.
(iv) No. It is not possible to divide any number by zero.

Hope given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:

Question 2.
Solution:

\(\sqrt { 33.64 } \) = 5.8

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

\(\sqrt { 10.0489 } \) = 3.17

Question 7.
Solution:

\(\sqrt { 1.0816 } \) = 1.04

Question 8.
Solution:

\(\sqrt { 0.2916 } \) = 0.54

Question 9.
Solution:
\(\sqrt { 3 } \) = 1.73

Question 10.
Solution:
\(\sqrt { 2.8 } \) = 1.6733 = 1.67

Question 11.
Solution:
\(\sqrt { 0.9 } \) = 0.948
= 0.95

Question 12.
Solution:
Length of rectangle (l) = 13.6 m
and width (b) = 3.4 m

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Finding the square root of 2509 by division we find that 9 is left as remainder

9 must be subtracted to get the perfect square 100.
Least number to be subtracted = 9

Question 14.
Solution:
Finding the square root of 7581 by division method, we find that 12 is left as remainder.
12 must be subtracted from 7581 to get a perfect square i.e., 7581 – 12 = 7569

(i) The least number to be subtracted = 12
(ii) Perfect square = 7569
(iii) and square root = 87 Ans.

Question 15.
Solution:
Finding the square root of 6203 by division method, we find that 38 is to be added to get a perfect square.
(i) Least number to be added = 38
(ii) Perfect square = 6241
(iii) Square root = 79 Ans.

Question 16.
Solution:
Finding the square root of 8400 by long division method, we find that 64 is to be added to 8400,
We, get 8400 + 64 = 8464

Least number to be added = 64
Perfect square = 8464
Square root = 92 Ans.

Question 17.
Solution:
Least four-digit number = 1000

Finding the square root of 1000 by the division method, we find that 24 must be added to get a perfect square of 4 digits.
Perfect square = 1000 + 24 = 1024 Ans.
square root of 1024 = 32

Question 18.
Solution:
Greatest number of five digits = 99999
Finding the square root of 99999
We get remainder = 143

Required perfect square = 99999 – 143 = 99856
and square root = 316 Ans

Question 19.
Solution:
Area of a square field = 60025 m²
Let its side = a

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E are helpful to complete your math homework.

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