Tag: RS Aggarwal Class 7 Maths Solutions

RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find each of the following products.
Question 1.
Solution:
(5x + 7) (3x + 4)
= 5x (3x + 4) + 7 (3x + 4)
= 5x x 3x + 5x x 4 + 7 x 3x + 7 x 4
= 15x² + 20x + 21x + 28
= 15x² + 41x + 28

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

Question 3.
Solution:
(x – 6) (4x + 9)
= x (4x + 9) – 6 (4x + 9)
= x x 4x + x x 9 – 6 x 4x – 6 x 9
= 4x² + 9x – 24x – 54
= 4x² – 15x – 54

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

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

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

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

Question 8.
Solution:
(0.8x – 0.5y) (1.5x – 3y)
= 0.8x (1.5x – 3y) – 0.5y (1.5x – 3y)
= 0.8x x 1.5x – 0.8x x 3y – 0.5y x 1.5x – 0.5y x (-3y)
= 1.20x² – 2.4xy – 0.75xy + 1.5y²
= 1.2x² – 3.15xy + 1.5y²

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(x² – a²) (x – a)
= x² (x – a) – a² (x – a)
= x² x x – x² x a – a² x x – a² (-a)
= x³ – x² a – xa² + a³

Question 13.
Solution:
(3p² + q²) (2p² – 3q²)
= 3p² (2p² – 3q²) + q² (2p² – 3q²)
= 3p² x 2p² – 3p² x 3q² + q² x 2p² – q² x 3q²
= 6q⁴ – 9p²q² + 2p²q² – 3q⁴
= 6p⁴ – 7p²q² – 3q⁴

Question 14.
Solution:
(2x² – 5y²) (x² + 3y²)
= 2x² (x² + 3y²) – 5y² (x² + 3y²)
= 2x² x x² + 2x² x 3y² – 5y² x x² – 5y² x 3y²
= 2x⁴ + 6x² y² – 5x² y² – 15y⁴
= 2y⁴ + x² y² – 15y⁴

Question 15.
Solution:
(x³ – y³) (x² + y²)
= x³ (x² + y²) – y³ (x² + y²)
= x³ x x² + x³ x y² – y³ x x² – y³ x y²
= x⁵ + x³ y² – x² y³ – y⁵

Question 16.
Solution:
(x⁴ + y⁴) (x² – y²)
= x⁴ (x² – y²) + y⁴ (x² – y²)
= x⁴ x x² – x⁴ x y² + y⁴ x x² – y⁴ x y²
= x⁶ – x⁴ y² + x² y⁴ – y⁶

Question 17.
Solution:

Question 18.
Solution:
(x² – y²) (x + 2y)
= x² (x + 2y) – y² (x + 2y)
= x² x x + x² x 2y – y² x x – y² x 2y
= x³ + 2x² y – xy² – 2y³

Question 19.
Solution:
(2x + 3y – 5) (x + y)
= 2x (x + y) + 3y (x + y) – 5 (x + y)
= 2x x x + 2x x y + 3y x x + 3y x y – 5 x x – 5 x y
= 2x² + 2xy + 3xy + 3y² – 5x – 5y
= 2x² + 5xy + 3y² – 5x – 5y

Question 20.
Solution:
(3x + 2y – 4) (x – y)
= 3x (x – y) + 2y (x – y) – 4 (x – y)
= 3x x x – 3x x y + 2y x x – 2y x y – 4 x x – 4 x (-y)
= 3x² – 3xy + 2xy – 2y² – 4x + 4y
= 3x² – xy – 2y² – 4x + 4y

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

Question 22.
Solution:
(3x² + 5x – 9) (3x – 9)
= 3x² (3x – 9) + 5x (3x – 9) – 9 (3x – 9)
= 3x² x 3x – 3×2 x 9 + 5x x 3x + 5x x (-9) – 9 x 3x – 9 x (-9)
= 9x³ – 27x² + 15x² – 45x – 27x + 81
= 9x³ – 12x² – 72x + 81

Question 23.
Solution:
(9x² – x + 15) (x² – 3)
= 9x² (x² – 3) – x (x² – 3) + 15 (x² – 3)
= 9x² x x² – 9x² x 3 – x x x² + x x 3 + 15 x x² – 15 x 3
= 9x⁴ – 27x² – x³ + 3x + 15x² – 45
= 9x⁴ – x³ – 12x² + 3x – 45

Question 24.
Solution:
(x² + xy + y²) (x – y)
= x² (x – y) + xy (x – y) + y² (x – y)
= x² x x – x² x y + xy x x – xy x y + y² x x – y² x y
= x³ – x²y + x²y – xy² + xy² – y²
= x³ – y³

Question 25.
Solution:
(x² – xy + y²) (x + y)
= x² (x + y) – xy (x + y) + y² (x + y)
= x³ + x² y – x² y – xy² + xy² + y³
= x³ + y³

Question 26.
Solution:
(x² – 5x + 8) (x² + 2)
= x² (x² + 2) – 5x (x² + 2) + 8 (x² + 2)
= x² x x² + x² x 2 – 5x x x² – 5x x 2 + 8 x x² + 8 x 2
= x⁴ + 2x² – 5x³ – 10x + 8x² + 16
= x⁴ – 5x³ + 2x² + 8x² – 10x + 16
= x⁴ – 5x³ + 10x² – 10x + 16

Simplify:
Question 27.
Solution:
(3x + 4) (2x – 3) + (5x – 4) (x + 2)
= [3x (2x – 3) + 4 (2x – 3)] + [5x (x + 2) – 4 (x + 2)]
= [3x x 2x – 3x x 3 + 4 x 2x – 4 x 3] + [5x x x + 5x x 2 – 4 x x – 4 x 2]
= [6x² – 9x + 8x – 12] + [5x² + 10x – 4x – 8]
= (6x² – x – 12) + (5x² + 6x – 8)
= 6x² – x – 12 + 5x² + 6x – 8
= 6x² + 5x² – x + 6x – 12 – 8
= 11x² + 5x – 20

Question 28.
Solution:
(5x – 3) (x + 4) – (2x + 5) (3x – 4)
= [5x x x + 5x x 4 – 3 x x – 3 x 4] – [2x x 3x + 2x x (-4) + 5 x 3x + 5x (-4)]
= [5x² + 20x – 3x – 12] – [6x² – 8x + 15x – 20]
= (5x² + 17x – 12) – (6x² + 7x – 20)
= 5x² + 17x – 12 – 6x² – 7x + 20
= 5x² – 6x² + 17x – 7x – 12 + 20
= -x² + 10x + 8

Question 29.
Solution:
(9x – 7) (2x – 5) – (3x – 8) (5x – 3)
= [9x (2x – 5) -7 (2x – 5)] – [3x (5x – 3) -8 (5x – 3)]
= [9x x 2x – 9x x 5 – 7 x 2x – 7 x (-5)] – [3x x 5x – 3x x 3 – 8 x 5x – 8 x (-3)]
= [18x² – 45x – 14x + 35] – [15x² – 9x – 40x + 24]
= 18x² – 45x – 14x + 35 – 15x² + 9x + 40x – 24
= 18x² – 15x² – 45x – 14x + 9x + 40x + 35 – 24
= 3x² – 59x + 49x + 11
= 3x² – 10x + 11

Question 30.
Solution:
(2x + 5y) (3x + 4y) – (7x + 3y) (2x + y)
= [2x (3x + 4y) + 5y (3x + 4y)] – [7x (2x + y) + 3y (2x + y)]
= [2x x 3x + 2x x 4y + 5y x 3x + 5y x 4y] – [7x x 2x + 7x x y + 3y x 2x + 3y x y]
= [6x² + 8xy + 15xy + 20y²] – [14x² + 7xy + 6xy + 3y²]
= [6x² + 23xy + 20y²] – [14x² + 13xy + 3y^2]
= 6x² + 23xy + 20y² – 14x² – 13xy – 3y²
= 6x² – 14x² + 23xy – 13xy + 20y² – 3y²
= -8x² + 10xy + 11y²

Question 31.
Solution:
(3x² + 5x – 7) (x – 1) – (x² – 2x + 3) (x + 4)
= [3x² (x – 1) + 5x (x – 1) – 7 (x – 1)] – [x² (x + 4) – 2x (x + 4) + 3 (x + 4)]
= [3x² x x – 3x² x 1 + 5x x x – 5x x (-1) – 7 x x -7 x (-1)] -[x² x x + x² x 4 – 2x x x – 2x x 4 + 3 x x + 3 x 4]
= [3x³ – 3x² + 5x² – 5x – 7x + 7] – [x³ + 4x² – 2x² – 8x + 3x + 12]
= [3x³ + 2x² – 12x + 7] – [x³ + 2x² – 5x + 12]
= 3x³ + 2x² – 12x + 7 – x³ – 2x² + 5x – 12
= 3x³ – x³ + 2x² – 2x² – 12x + 5x – 12 + 7
= 2x³ – 7x – 5

Hope given RS Aggarwal Solutions Class 7 Chapter 6 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 7 Solutions Chapter 4 Rational Numbers Ex 4F

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:

(vii) Reciprocal of -1 = -1
(viii) Reciprocal of 0 does not exist.

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Product of two number = 10
One number = -8
Second number = 10 ÷ (-8)

Question 8.
Solution:
Product of two rational numbers = – 9
One number = -12
Second number = (-9) ÷ (-12)

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Cloth required for 24 pairs of trousers =54 m
Cloth required for one pair = (54 ÷ 24) m

Question 12.
Solution:
Total length of rape = 30 m

Question 13.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Cost of 1 metre of cloth

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find each of the following products:
Question 1.
Solution:
4a(3a + 7b) = 4a x 3a + 4a x 7b = 12a² + 28ab

Question 2.
Solution:
5a(6a – 3b) = 5a x 6a – 5a x 3b = 30a² – 15ab

Question 3.
Solution:
8a² (2a + 5b) = 8a² x 2a + 8a² x 5b = 16a³ + 40a²b

Question 4.
Solution:
9x² (5x + 7) = 9x² x 5x + 9x² x 7 = 45x³ + 63x²

Question 5.
Solution:
ab(a² – b²) = ab x a² – ab x b² = a³b – ab³

Question 6.
Solution:
2x² (3x – 4x²) = 2x² x 3x – 2x² x 4x² = 6x³ – 8x⁴

Question 7.
Solution:

Question 8.
Solution:
-1 7x² (3x – 4) = -17x² x 3x – 17x² x (-4) = -51x³ + 68x²

Question 9.
Solution:

Question 10.
Solution:
-4x²y (3x² – 5y)
= -4x² y x 3x² – 4x² y x (-5y)
= -12x² y + 20x² y²

Question 11.
Solution:

Question 12.
Solution:
9t² (t + 7t³) = 9t² x t + 9t² x 7t³ = 9t³ + 63t⁵

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
24x² (1 – 2x)
= 24x² x 1 – 24x² x 2x
= 24x² – 48x³
If x = 2, then
24x² – 48x³
= 24(2)² – 48(2)³
= 24 x 4 – 48 x 8
= 96 – 384
= -288

Question 17.
Solution:

Question 18.
Solution:
s (s² – st) = s x s² – s x st = s³ – s²t
If s = 2, t = 3, then
s³ – s²t = (2)³ – (2)² x 3 = 8 – 4 x 3 = 8 – 12 = -4

Question 19.
Solution:
-3y (xy + y²) = -3y x xy + (-3y) x y² = -3xy² – 3y³
if x = 4, y = 5, then
-3xy² – 3y³
= -3(4)(5)² – 3(5)³ = -3 x 4 x 25 – 3 x 125 = -300 – 375 = -675

Simplify each of the following:
Question 20.
Solution:
a(b – c) + b(c – a) + c(a – b) = ab – ac + bc – ab + ac – bc = 0

Question 21.
Solution:
a(b – c) – b(c – a) – c(a – b) = ab – ac – bc + ab – ac + bc = 2ab – 2ac

Question 22.
Solution:
3x² + 2(x + 2) – 3x (2x + 1)
= 3x² + 2x + 4 – 6x² – 3x = 3x² – 6x² + 2x – 3x + 4 = -3x² – x + 4

Question 23.
Solution:
x (x + 4) + 3x (2x² – 1) + 4x² + 4
= x² + 4x + 6x³ – 3x + 4x² + 4
= 6x³ + x² + 4x² + 4x – 3x + 4
= 6x³ + 5x² + x + 4

Question 24.
Solution:
2x² + 3x (1 – 2x³) + x (x + 1)
= 2x² + 3x – 6x⁴ + x² + x
= – 6x⁴ + 2x² + x² + 3x + x
= – 6x⁴ + 3x² + 4x

Question 25.
Solution:
a²b (a – b²) + ab(4ab – 2a²) – a³b (1 – 2b)
= a³b – a²b³ – 2a³b² + 4a²b³ – 2a³b² – a³b + 2a³b²
= a³b – a³b + 2a³b² – a²b³ + 4a²6³
= 3a²b³

Question 26.
Solution:
4st (s – t) – 6s² (t – t²) – 3t² (2s² – s) + 2st(s – t)
= 4s²t – 4st² – 6s²t + 6s²t² – 6s²t² + 3st² + 2s²t – 2st²
= 4s²t – 6s²t + 2s²t – 4st² + 3st² – 2st² + 6s²t² – 6s²t²
= 6s²t – 6s²t – 6st² + 3st² + 6s²t² – 6s²t²
= – 3st²

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C 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 7 Solutions Chapter 6 Algebraic Expressions Ex 6B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find the products:
Question 1.
Solution:
3 x 8 a²⁺⁴ = 24a⁶

Question 2.
Solution:
(-6x³) x (5x²) = -6 x 5x²⁺³ = -30x⁵

Question 3.
Solution:
(-4ab) x (-3a²bc)
= (-4) x (-3) a. a².b. b. c
= 12.a²⁺¹. b¹⁺¹.c
= 12a³b²c

Question 4.
Solution:
= (2a²b³) x (-3a³b)
= 2 x (-3) a². a³. b³. b. b
= -6a²⁺³.b³⁺¹
= -6a⁵.b⁴

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:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Find the products given below and in each case verify the result for a = 1, b = 2 and c = 3 :
Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:
Sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)

Now subtract (2a – 3b + 4c) from 4a + 8c
= 4a + 8c – (2a – 3b + 4c)
= 4a + 8c – 2a + 3b – 4c
= 4a – 2a + 3b + 8c – 4c
= 2a + 3b + 4c

Question 4.
Solution:

Question 5.
Solution:
Sum of (8a – 6a² + 9) and (-10a – 8 + 8a²)
= 8a – 6a² + 9 + (-10a) – 8 + 8a²
= 8a – 10a – 6a² + 8a² + 9 – 8
= -2a + 2a² + 1
Now -3 – (-2a + 2a² + 1)
= (-3 + 2a – 2a² – 1)
= -4 + 2a – 2a²

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 6 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 7 Solutions Chapter 4 Rational Numbers Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Draw a number line and locate a point O on it. Let it represent 0 Now \(\frac { 1 }{ 3 }\) has been presented on the number line given below.

(ii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 2 }{ 7 }\) has been represented on the number line given below:

(iii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 7 }{ 3 }\) has been represented on the number line given below:

(iv) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 7 }{ 3 }\) has been represented on it as given below:

(v) Draw a number line and locate a point O on it. Let it represented 0. The number \(\frac { 37 }{ 8 }\) has been represented on it as given below:
\(\frac { 37 }{ 8 }\) = 4\(\frac { 5 }{ 8 }\)

(vi) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { -1 }{ 3 }\) has been represented on it as given below:

(vii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { -3 }{ 4 }\) has been represented on it is as given below:

(viii) Draw a number line and locate a point on it. Let it represent 0. The number \(\frac { -12 }{ 7 }\) has been represented on it as given below:

(ix) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 36 }{ -5 }\) has been represented on it as given below:

(x) Draw a number line and locate is point O on it. Let is represent 0. The number \(\frac { -43 }{ 9 }\) has been represented on it as given below:

Question 2.
Solution:
(i) \(\frac { 5 }{ 6 }\) or 0, \(\frac { 5 }{ 6 }\) is greater as any positive number is always greater than 0.

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
(i) True: All negative numbers lie on the left of 0.
(ii) False: All negative numbers lie on the left of 0.
(iii) True: All positive numbers lie on the right of 0 and all negative numbers on the left of 0.
(iv) False: \(\frac { -18 }{ -13 }\) = \(\frac { 18 }{ 13 }\) which is positive and positive number lie on the left of 0.
(v) True: \(\frac { -5 }{ -8 }\) = \(\frac { 5 }{ 8 }\) which is positive and all positive number lie on the right of negative numbers.
(i), (iii) and (iv) are true.

Question 8.
Solution:
5 rational numbers between -3 and -2.

Question 9.
Solution:

Question 10.
Solution:
L.C.M. of 5 and 2 = 10

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:
We know that a decimal divided by 10, the decimal point is shifted to the left by one place. Therefore
(i) 131.6 ÷ 10 = 13.16
(ii) 32.56 ÷ 10 = 3.256
(iii) 4.38 ÷ 10 = 0.438
(iv) 0.34 ÷ 10 = 0.034
(v) 0.08 ÷ 10 = 0.008
(vi) 0.062 ÷ 10 = 0.0062

Question 2.
Solution:
We know that decimal divided by 100, the decimal point is shifted to the left by two place. Therefore
(i) 137.2 ÷ 100 = 1.372
(ii) 23.4 ÷ 100= 0.234
(iii) 4.1 ÷ 100 = 0.047
(iv) 0.3 ÷ 100 = 0.003
(v) 0.58 ÷ 100 = 0.0058
(vi) 0.02 ÷ 100 = 0.0002

Question 3.
Solution:
We know that a decimal divided by 1000, the decimal point is shifted to the left by three places. Therefore:
(i) 1286.5 ÷ 1000= 1.2865
(ii) 354.16 ÷ 1000 = 0.35416
(iii) 38.9 ÷ 1000 = 0.0389
(iv) 4.6 ÷ 1000 = 0.0046
(v) 0.8 ÷ 1000 = 0.0008
(vi) 2 ÷ 1000 = 0.002

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Cost of 24 chairs = Rs. 9255.60
Cost of one chair = Rs. \(\frac { 9255.60 }{ 24 }\) = Rs. 385.65

Question 9.
Solution:
Length of cloth for one shirt = 1.8 m
Total length of piece of cloth = 45 m
Number of shirts will be = 45 ÷ 1.8

Question 10.
Solution:
A car covers in 2.4 litre = 22.8 km
It will cover in 1 litre = \(\frac { 22.8 }{ 2.4 }\) km = 9.5 km

Question 11.
Solution:
Oil in one tin = 16.5 l
Total oil = 478.5 l

Question 12.
Solution:
Weight of 37 bags of sugar=3644.5 kg
Weight of one bag of sugar = 3644.5 ÷ 37

Question 13.
Solution:
Capacity of 69 buckets = 586.5 litres
Capacity of 1 bucket = 586.5 ÷ 69

Question 14.
Solution:
Number of pieces in 1.15 m = 1

Question 15.
Solution:
Total weight of cement = 1792.8 kg
Cement in one bag = 49.8 kg
Number of bags = 1792.8 ÷ 49.8

Question 16.
Solution:
Total thickness = 1.89 m = 189 cm
Thickness of one piece = 0.3 5 cm
Number of pieces = 189 ÷ 0.35

Question 17.
Solution:
Product of two decimals = 261.36
One decimal = 17.6
Second decimal = 261.36 ÷ 17.6

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Rational numbers: The numbers of the form \(\frac { p }{ q }\) where p and q are integers and q ≠ 0, are called rational numbers.

(iv) Yes, there is one rational number (0) which is neither positive nor negative.

Question 2.
Solution:

(viii) \(\frac { 0 }{ 1 }\) are all rational number but \(\frac { 1 }{ 0 }\) and \(\frac { 0 }{ 0 }\) are not rational number as their denominator is zero.

Question 3.
Solution:
(i) Numerator = 8, denominator =19
(ii) Numerator = 5, denominator = – 8
(iii) Numerator =-13, denominator =15
(iv) Numerator = – 8, denominator = -11
(v) Numerator = 9, denominator = 1

Question 4.
Solution:

Question 5.
Solution:
According to the definition, a rational number is positive if both of numerator and denominator have same signs. Therefore
(iii), (iv) and (vi) 8 are positive rational numbers.

Question 6.
Solution:
According to the definition, a rational number is negative if numerator and denominator have opposite sign. Therefore.
(iii), (iv), (v), (vi) are all negative rational numbers.

Question 7.
Solution:
Equivalent rational numbers of each are given below:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:
(i) False, as there is no end of smallest and largest rational number,
(ii) True.
(iii) False, as zero is a rational number but the division of zero is meaningless.
(iv) True.
(v) False, every rational is not a fraction
In a fraction, numerator and denominators is a whole number but the denominator can’t be zero

 

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
Reenu get \(\frac { 2 }{ 7 }\) of an apple while Sonal gets \(\frac { 4 }{ 5 }\) part of it

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
For finding the required fraction, we have to subtract 7\(\frac { 3 }{ 5 }\) from 18

Question 12.
Solution:
For finding the required fraction we should subtract 7\(\frac { 4 }{ 15 }\) from 8\(\frac { 2 }{ 5 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C are helpful to complete your math homework.

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