Category: CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5B

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

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

Question 1.
Solution:
We can write in standard form
(i) 538 = 5.38 x 10²
(ii) 6428000 = 6.428000 x 10⁶ = 6.428 x 10⁶
(iii) 82934000000 = 8.2934000000 x 10¹⁰ = 8.2934 x 10¹⁰
(iv) 940000000000 = 9.40000000000 x 10¹¹ = 9.4 x 10¹¹
(v) 23000000 = 2.3000000 x 10⁷ = 2.3 x 10⁷

Question 2.
Solution:
(i) Diameter of Earth = 12756000 m = 1.2756000 x 10⁷m = 1.2756 x 10⁷ m
(ii) Distance between Earth and Moon = 384000000 m = 3.84000000 x 10⁸ = 3.84 x 10⁸
(iii) Population of India in March 2001 = 1027000000 = 1.027000000 x 10⁹ = 1.027 x 10⁹
(iv) Number of stars in a galaxy = 100000000000 = 1.00000000000 x 10¹¹ = 1 x 10¹¹
(v) The present age of universe = 12000000000 years = 1.2000000000 x 10¹⁰ years = 1.2 x 10¹⁰ years

Question 3.
Solution:
The expanded form will be
(i) 684502 = 6 x 10⁵ + 8 x 10⁴ + 4 x 10³ + 5 x 10² + 2
(ii) 4007185 = 4 x 10⁶ + 0 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 1 x 10² + 8 x 10¹ + 5 x 10⁰
(iii) 5807294 = 5 x 10⁶ + 8 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 2 x 10² + 9 x 10¹ + 4 x 10⁰
(iv) 50074 = 5 x 10⁴ + 0 x 10³ + 0 x 10² + 7 x 10¹ + 4 x 10⁰

Question 4.
Solution:
(i) 6 x 10⁴ + 3 x 10³ + 0 x 10² + 7 x 10¹ + 8 x 10⁰
= 60000 + 3000 + 0 + 70 + 8
= 63078
(ii) 9 x 10⁶ + 7 x 10⁵ + 0 x 10⁴ + 3 x 10³ + 4 x 10² + 6 x 10¹ + 2 x 10⁰
= 9000000 + 700000 + 0 + 3000 + 400 + 60 + 2
= 9703462
(iii) 8 x 10⁵ + 6 x 10⁴ + 4 x 10³ + 2 x 10² + 9 x 10¹ + 6 x 10⁰
= 800000 + 60000 + 4000 + 200 + 90 + 6
= 864296

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B 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 7 Linear Equations in One Variable Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Question 1.
Solution:
Let the required number = x
Then 2x – 7 = 45
2x = 45 + 7 = 52
x = 26
Required number = 26

Question 2.
Solution:
Let the required number = x Then
3x + 5 = 44
⇒ 3x = 44 – 5 = 39
x = 13
Required number = 13

Question 3.
Solution:
Let the required fraction = x
then 2x + 4 = \(\frac { 26 }{ 5 }\)

Question 4.
Solution:
Let the required number = x
and half of .the number = \(\frac { x }{ 2 }\)

Question 5.
Solution:
Let the required number = x
Two third of the number = \(\frac { 2 }{ 3 }\) x

Question 6.
Solution:
Let the required number = x
Then, 4x = x + 45
⇒ 4x – x = 45
⇒ 3x = 45
⇒ x = 15
Required number = 15

Question 7.
Solution:
Let the required number = x
Then x – 21 = 71 – x
⇒ x + x = 71 + 21
⇒ 2x = 92
⇒ x = 46

Question 8.
Solution:
Let the original number = x
Then \(\frac { 2 }{ 3 }\) of the number = \(\frac { 2 }{ 3 }\) x

Question 9.
Solution:
Let the second number = x
then first number = \(\frac { 2 }{ 5 }\) x
their sum = 70

Question 10.
Solution:
Let the required number = x

Question 11.
Solution:
Let the required number = x
Fifth part of the number = \(\frac { x }{ 5 }\)
Fourth part of the number = \(\frac { x }{ 4 }\)

Question 12.
Solution:
Let first natural number = x then
next number = x + 1
x + x + 1 = 63
⇒ 2x = 63 – 1 = 62
x = 31
first number = 31
and second number = 31 + 1 = 32
Numbers are 31, 32

Question 13.
Solution:
Let first odd number = 2x + 1
second odd number = 2x + 3
2x + 1 + 2x + 3 = 76
⇒ 4x + 4 = 76
⇒ 4x = 76 – 4 = 72
x = 18
First number = 2x + 1 = 2 x 18 + 1 = 36 + 1 = 37
Second number = 2x + 3 = 2 x 18 + 3 = 36 + 3 = 39
Numbers are 37, 39

Question 14.
Solution:
Let first positive even number = 2x
Second number = 2x + 2
Third number = 2x + 4
2x + 2x +2 + 2x + 4 = 90
⇒ 6x + 6 = 90
⇒ 6x = 90 – 6 = 84
x = 14
First even number = 2x = 2 x 14 = 28
Second number = 2x + 2 = 2 x 14 + 2 = 28 + 2 = 30
Third number = 30 + 2 = 32
Required numbers are 28, 30, 32

Question 15.
Solution:
Sum of two numbers = 184
Let first number = x
Then second number = 184 – x

First part = 72
Second part = 184 – 72 = 112
Hence parts are 72, 112

Question 16.
Solution:
Total number of notes = 90
Let number of notes of Rs. 5 = x
Then number of notes of Rs.10 = 90 – x
Then x x 5 + (90 – x) x 10 = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900 = -400
x = 8
Number of 5 rupees notes = 80
and ten rupees notes = 90 – 80 = 10

Question 17.
Solution:
Amount of coins = Rs. 34
Let 50 paisa coins = x
then 25 paisa coins = 2x

Number of 50 paisa coins = 34
and number of 25 paisa coins = 2x = 2 x 34 = 68

Question 18.
Solution:
Let present age of Raju’s cousin = x years
then age of Raju = (x – 19) years
After 5 years,
Raju’s age = x – 19 + 5 = (x – 14) years
and his cousin age = x + 5
(x – 14) : (x + 5) = 2 : 3
⇒ \(\frac { x – 14 }{ x + 5 }\) = \(\frac { 2 }{ 3 }\)
⇒ 3(x – 14) = 2 (x + 5) (By cross multiplication)
⇒ 3x – 42 = 2x + 10
⇒ 3x – 2x = 10 + 42
⇒ x = 52
Raju’s age = x – 19 = 52 – 19 = 33 years
and his cousin age = 52 years.

Question 19.
Solution:
Let present age of son = x years
Age of father = (x + 30) years
12 years after,
Father’s age = x + 30 + 12 = (x + 42) years
and son’s age = (x + 12) years
(x + 42) = 3(x + 12)
⇒ x + 42 = 3x + 36
⇒ 3x + 36 = x + 42
⇒ 3x – x = 42 – 36
⇒ 2x = 6
⇒ x = 3
Son’s age = 3 years
Father’s age = 3 + 30 = 33 years

Question 20.
Solution:
Ratio in present ages of Sonal and Manoj = 7 : 5
Let Sonal’s age = 7x
then Manoj’s age = 5x
10 years hence,
Sonal’s age will be = 7x + 10
and Manoj’s age = 5x + 10

Sonal’s present age = 7x = 7 x 5 = 35 years
and Manoj’s age = 5x = 5 x 5 = 25 years

Question 21.
Solution:
Five years ago,
Let Son’s age = x years
and father’s age = 7x years
Present age of son = (x + 5) years
and age of father = (7x + 5) years
5 years hence,
father’s age = 7x + 5 + 5 = 7x + 10
and Son’s age = x + 5 + 5 = x + 10
(7x + 10) = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30 – 10
⇒ 4x = 20
⇒ x = 5
Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years
and son’s age = x + 5 = 5 + 5 = 10 years

Question 22.
Solution:
Let age of Manoj 4 years ago = x
then his present age = x + 4
After 12 years his age will be = x + 4 + 12 = x + 16
x + 16 = 3(x)
x + 16 = 3x
⇒ 16 = 3x – x
⇒ 2x = 16
x = 8
His present age = 8 + 4 = 12 years

Question 23.
Solution:
Let total marks = x
Pass marks = 40% of x = \(\frac { 40x }{ 100 }\) = \(\frac { 2 }{ 5 }\) x
No. of marks got by Rupa = 185
No. of marks by which she failed = 15
Pass marks = 185 + 15 = 200
\(\frac { 2 }{ 5 }\) x = 200
⇒ x = \(\frac { 200 x 5 }{ 2 }\) x
⇒ x = 500
Hence total marks = 500

Question 24.
Solution:
Sum of digits = 8
Let units digit = x
Then tens digit = 8 – x
and number will be x + 10 (8 – x) ….(i)
By adding 18, the digits are reversed then
units digit = 8 – x
and tens digit = x
Number = (8 – x) = 10x
According to the condition,
(8 – x) + 10x = 18 + x + 10 (8 – x)
⇒ 8 – x + 10x = 18 + x + 80 – 10x
⇒ 10x – x – x + 10x = 18 + 80 – 8
⇒ 18x = 90
⇒ x = 5
Number is
x + 10(8 – x) = 5 + 10(8 – 5) = 5 + 10 x 3 = 35

Question 25.
Solution:
Cost of 3 tables and 2 chairs = 1850
Cost of table = Rs. 75 + cost of a chair
Let cost of chair = Rs. x,
then Cost of table = Rs. 75 + x
According to the condition,
3 (75 + x) + 2x = 1850
⇒ 225 + 3x + 2x = 1850
⇒ 225 + 5x = 1850
⇒ 5x = 1850 – 225 = 1625
x = 325
Cost of chair = Rs. 325
and cost of table = Rs. 325 + 75 = Rs. 400

Question 26.
Solution:
S.P of article = Rs. 495
gain = 10%
Let cost price = Rs. x

Question 27.
Solution:
Perimeter of field = 150 m
Length + Breadth = \(\frac { 150 }{ 2 }\) = 75 m
[Perimeter = 2(l + b)]
Let length = x Then breadth = 75 – x
Then x = 2(75 – x)
⇒ x = 150 – 2x
⇒ x + 2x = 150
⇒ 3x = 150
⇒ x = \(\frac { 150 }{ 3 }\) = 50
Length = 50 m
and breadth = 75 – 50 = 25 m

Question 28.
Solution:
Perimeter of an isosceles triangle = 55 m
Let the third side of an isosceles triangle = x
Then each equal side = (2x – 5) m
According to the condition,
x + 2 (2x – 5) = 55
⇒ x + 4x – 10 = 55
⇒ 5x = 55 + 10
⇒ 5x = 65
⇒ x = 13
and 2x – 5 = 2 x 13 – 5 = 21 m
Sides will be 13m, 21m, 21m

Question 29.
Solution:
Sum of two complementary angles = 90°
Let first angle = x
then second = 90° – x
x – (90 – x) = 8
⇒ x – 90 + x = 8
⇒ 2x = 8 + 90
⇒ 2x = 98
⇒ x = 49
first angle = 49°
and second angle = 90° – 49° = 41°
Hence angles are 41°, 49°

Question 30.
Solution:
Sum of two supplementary angles = 180°
Let first angle = x
Then second angle = 180° – x
x – (180° – x) = 44°
⇒ x – 180° + x = 44°
⇒ 2x = 44° + 180° = 224°
⇒ 2x = 224°
⇒ x = 112°
First angle = 112°
and second angle = 180° – 112° = 68°
Hence angles are 68°, 112°

Question 31.
Solution:
In an isosceles triangle
Let each equal base angles = x
Then vertex angle = 2x
According to the condition,
x + x + 2x = 180° (sum of angles of a triangle)
⇒ 4x = 180°
⇒ x = 45°
Then vertex angle = 2x = 2 x 45° = 90°
Angles of the triangle are 45°, 45° and 90°

Question 32.
Solution:
Let length of total journey = x km
According to the condition,

⇒ 39x + 80 = 40x
⇒ 40x – 39x = 80
⇒ x = 80
Total journey = 80km

Question 33.
Solution:
No. of days = 20 Let no. of days he worked = x
Then he will receive amount = x x Rs. 120 = Rs. 120x
No. of days he did not work = 20 – x
Fine paid = (20 – x) x Rs. 10 = Rs. 10(20 -x)
120x – 10 (20 – x) = 1880
⇒ 120x – 200 + 10x = 1880
⇒ 130x = 1880 + 200 = 2080
x = 16
No. of days he remained absent = 20 – x = 20 – 16 = 4 days

Question 34.
Solution:
Let value of property = x

Question 35.
Solution:
Solution = 400 mL
Quantity of alcohol = 15% of 400 mL
= \(\frac { 400 x 15 }{ 100 }\) = 60 mL
Let pure alcohol added = x mL
Total solution = 400 + x
and total alcohol = (x + 60)
Now (400 + x) x 32% = x + 60
⇒ (400 + x) x \(\frac { 32 }{ 100 }\) = x + 60
⇒ 32 (400 + x) = 100 (x + 60)
⇒ 12800 + 32x = 100x + 6000
⇒ 12800 – 6000 = 100x – 32x
⇒ 6800 = 68x
⇒ x = 6800
Pure alcohol added = 100 mL

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable 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 7 Solutions Chapter 5 Exponents Ex 5A

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

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5B
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5C
• RS Aggarwal Solutions Class 7 Chapter 5 Exponents CCE Test Paper

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:
Product of two numbers = 4

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 5 Exponents Ex 5A 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 7 Linear Equations in One Variable Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = \(\frac { 5 }{ 3 }\)
Check:
L.H.S. = 3x – 5
= 3 x \(\frac { 5 }{ 3 }\) – 5
= 5 – 5
= 0
= R.H.S.
Hence x = \(\frac { 5 }{ 3 }\)

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = \(\frac { 7 }{ 14 }\) = \(\frac { 1 }{ 2 }\)
Check : L.H.S. = 2(x – 2) + 3 (4x -1)

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
\(\frac { 2m + 5 }{ 3 }\) = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:

R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:

Question 32.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A 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 CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper.

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:
Let the required number be x.

Question 4.
Solution:
Let the other number be x.

Question 5.
Solution:

Question 6.
Solution:
Let the required number be x

Question 7.
Solution:
Length of rope = 45 m

Question 8.
Solution:

Mark (✓) against the correct answer in each of the following :
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:
(i) False.
This is because \(\frac { -15 }{ -11 }\) = \(\frac { 15 }{ 11 }\), which lies to the right of 0.
(ii) True
(iii) True
(iv) False
L.C.M. of 5 and 3 is 15.

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper  are helpful to complete your math homework.

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

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

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

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(b)

Question 3.
Solution:
(a)

Question 4.
Solution:
(c)

Question 5.
Solution:
(b)

Question 6.
Solution:
(a)

Question 7.
Solution:
(a)

Question 8.
Solution:
(c)

Question 9.
Solution:
(b)

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:

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

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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.

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 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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