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

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

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:
8a(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:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 1

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

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 2

Question 10.
Solution:
-4x2y (3x2 – 5y)
= -4xy x 3x2 – 4xy x (-5y)
= -12xy + 20xy2

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 3

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

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 4

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 5

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 6

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

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

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

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:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 1

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 2

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 3

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 4

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 5

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 6

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 7
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 8

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 9

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 10

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 11

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 12
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 13

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 14

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 15

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 16
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 17

Question 19.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 18
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 19

Question 20.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 20

Question 21.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 21
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 22

Find the products given below and in each case verify the result for a = 1, b = 2 and c = 3 :
Question 22.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 23

Question 23.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 24
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 25
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 26

Question 24.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 27
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 28

Question 25.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 29
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 30

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

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

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 1
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 2
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 3
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 4
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 5
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 6

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 7

Question 3.
Solution:
Sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 8
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:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 9

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:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 10
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 11

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 5 Exponents CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 1

Question 2.
Solution:
Let the required number be x
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 2

Question 3.
Solution:
Let the required number be x.
(-20)-1 ÷ x = (-10)-1
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 3

Question 4.
Solution:
(i) 2000000 = 2.000000 x 106 [Since the decimal point is moved 6 to the left]
= 2 x 106
(ii) 6.4 x 105 = 6.4 x 100000 = 640000

Question 5.
Solution:
We have :
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 4

Question 6.
Solution:
We have :
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 5

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 6

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 7

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 8
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 9

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 10

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 11
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 12

Question 12.
Solution:
(c) 3.263 x 105
A given number is said to be in standard form if it can be expressed as
k x 10n, where k is a real number such that 1 < k < 10 and n is a positive integer.
For example : 3.263 x 105

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 13
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 14

Question 14.
Solution:
(i) True
645 = 6.45 x 102
[Since the decimal point is moved 2 places to the left]
(ii) False
27000 = 2.7 x 104
[Since the decimal point is moved 4 places to the left]
(iii) False
(3° + 4° + 5°) = 1
(iv) False
Reciprocal of 56 = Reciprocal of
RS Aggarwal Class 7 Solutions Chapter 5 Exponents CCE Test Paper 15

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper 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 5 Exponents Ex 5C

RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C.

Other Exercises

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 1

Question 2.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 2

Question 3.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 3

Question 4.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 4

Question 5.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 5

Question 6.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 6

Question 7.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 7

Question 8.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 8

Question 9.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 9

Question 10.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 10

Question 11.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 11

Question 12.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 12

Question 13.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 13

Question 14.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 14

Question 15.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 15

Question 16.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 16

Question 17.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 17

Question 18.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 18

Question 19.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 19

Question 20.
Solution:
(c)
Required number = 10-1 ÷ (-8)-1
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 20

Question 21.
Solution:
(c)
The number which is in standard form is 2.156 x 106

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C 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.