Rd Sharma Solutions Class 7 Chapter 8 Linear Equations In One Variable
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 8, Linear Equations In One Variable. Here students can easily find Exercise wise solution for chapter 8, Linear Equations In One Variable. Students will find proper solutions for Exercise 8.1, 8.2, 8.3 and 8.4. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Linear Equations In One Variable Exercise 8.1 Solution
Question no – (1)
Solution :
According to the given question –
(i) x = 4 is the root of 3x – 5 = 7
∴ 3x – 5 = 7
= 3x = 7 + 5
= x = 12/3
∴ x = 4…(Verified)
(ii) x = 3 is the root of 5 + 3x = 14
∴ 5 + 3x = 14
= 3x = 14 – 5
= x = 9/3
∴ x = 3…(Verified)
(iii) x = 2 is the root of 3x – 2 = 8x – 12
∴ 3x – 2 = 8x – 12
= 3x – 8x = – 12 + 2
= x = 10/5
∴ x = 2…(Verified)
(iv) x = 4 is the root of 3x/2 = 6
∴ 3x/2 = 6
= x = 6 × 3/3
= x = 12/3
∴ x = 4…(Verified)
(v) y – 2 is the root of y – 3 = 2y – 5
∴ y – 3 = 2y – 5
= 2y – y = – 3 + 5
∴ y = 3…(Verified)
(vi) x = 8 is the root of 1/2x + 7 = 11
∴ 1/2x + 7 = 11
= 1/2x = 11- 7
= x = 4 × 2
∴ x = 8
Question no – (2)
Solution :
As per the question we can solve these problems as follows :
(i) x + 3 = 12
= x = 12 – 3
∴ x = 9
(ii) x – 7 = 10
= x = 10 + 7
∴ x = 17
(iii) 4x = 28
= x = 28/4
∴ x = 7
(iv) x/2 + 7= 11
= x/2 + 7 = 11
= x/2 = 11 – 7
∴ x = 8
(v) 2x + 4 = 3x
= 2x – 3x = – 4
∴ x = 4
(vi) x/4 = 12
= x = 12 × 4
∴ x = 48
(vii) 15/x = 3
= x = 15/3
∴ x = 5
(viii) x/18 = 20
= x = 20 × 18
∴ x = 360
Linear Equations In One Variable Exercise 8.2 Solution
Question no – (1)
Solution :
In the given question,
x – 3 = 5
∴ x – 3 = 5
= x = 5 + 3
So, x = 8
Question no – (2)
Solution :
In the given question,
x + 9 = 13
∴ x + 9 = 13
= x = 13 – 9
Thus, x = 4
Question no – (3)
Solution :
Given in the question,
x – 3/5 = 7/5
∴ x – 3/5 = 7/5
= x = 7/5 + 3/5
= x = 10/5
Therefore, x = 2
Question no – (4)
Solution :
Given, 3x = 0
∴ 3x = 0
So, x = 0
Question no – (5)
Solution :
Given, x/2 = 0
∴ x/2 = 0
= x = 0
Question no – (6)
Solution :
In the given question,
x – 1/3 = 2/3
∴ x – 1/3 = 2/3
= x = 2/3 + 1/3
= x = 2 + 1/3
Hence, = x = 1
Question no – (7)
Solution :
Given in the question,
x + 1/2 = – 7/2
∴ x + 1/2 = – 7/2
= x = 7/2 = 1/2
= x = 6/2
Thus, = x = 3
Question no – (8)
Solution :
In the question,
10 – y = 6
∴ 10 – y = 6
= 10 – 6 = y
Hence, y = 4
Question no – (9)
Solution :
Given in the question,
7 + 4y = – 5
∴ 7 + 4y = – 5
= 4y = – 5 – 7
Thus, y = – 3
Question no – (10)
Solution :
In the given question,
= 4/5 – x = 3/5
∴ 4/5 – x = 3/5
= 4/5 – 3/5 = x
= x = 4 – 3/5
Hence, x = 1/5
Question no – (11)
Solution :
In the given question,
2y – 1/2 = -1/3
∴ 2y – 1/2 = -1/3
= 2y = 1/2 – 1/3
= 2y = 3 – 2/6
∴ y = 1/12
Question no – (12)
Solution :
Given in the question,
14 = 7x/10 – 8
∴ 14 = 7x/10 – 8
= 7x/10 = 14 + 8
= x = 22 × 10/7
Thus, x = 220/7
Question no – (13)
Solution :
In the question,
3 (x + 2) = 15
∴ 3 (x + 2) = 15
= x + 2 = 5
= x = 5 – 2
Therefore, x = 3
Question no – (14)
Solution :
In the given question,
x/4 = 7/8
∴ x/4 = 7/8
= x = 7 × 4/8
Hence, x = 7/2
Question no – (15)
Solution :
Given, 1/3 – 2x = 0
∴ 1/3 – 2x = 0
= 2x = 1/3
Thus, x = 1/6
Question no – (16)
Solution :
Here we have,
3 (x + 6) = 24
∴ 3 (x + 6) = 24
= x + 6 = 8
= x = 8 – 6
Thus, x = 2
Question no – (17)
Solution :
Here we have,
3 (x + 2) – 2 (x – 1) = 7
∴ 3 (x + 2) – 2 (x – 1) = 7
= 3x + 6 – 2x + 2 = 7
= x = 7 – 8
Thus, x = -1
Question no – (18)
Solution :
In the question we have,
8 (2x – 5) – 6 (3x – 7) = 1
∴ 8 (2x – 5) – 6 (3x – 7) = 1
= 16x – 40 – 18x + 42 = 1
= -2x + 2 = 1
= 2x = 1 – 2
Hence, x = 1/2
Question no – (19)
Solution :
Given in the question,
6 (1 – 4x) + 7 (2 + 5x) = 53
∴ 6 (1 – 4x) + 7 (2 + 5x) = 53
= 6 – 24x + 14 + 35x = 53
= 11x = 53 – 20
= x = 33/11
Therefore, x = 3
Question no – (20)
Solution :
In the question we have,
5 (2 – 3x) – 17 (2x – 5) = 16
∴ (2 – 3x) – 17 (2x – 5) = 16
= 10 – 15x – 34x + 85 = 16
= – 49x = 16 – 95
Hence, x = 79/49
Question no – (21)
Solution :
Given in the question,
x – 3/5 – 2 = – 1
∴ x – 3/5 – 2 = – 1
x – 3/5 = – 1 + 2
= x – 3 = 5
= x = 5 + 3
Thus, x = 8
Question no – (22)
Solution :
In the question we have,
5 (x – 2) + 3 (x + 1) = 25
∴ 5 (x – 2) + 3 (x + 1) = 25
= 5x – 10 + 3 x + 3 = 25
= 8x = 25 + 7
= x = 32/8
Therefore, x = 4
Linear Equations In One Variable Exercise 8.3 Solution
Question no – (1)
Solution :
In the given question,
6x + 5 = 2x + 17
∴ 6x + 5 = 2x + 17
= 6x – 2x = 17 – 5
= 4x = 12
= x = 12/4
= x = 3
Hence, x = 3…(Verified)
Question no – (2)
Solution :
In the question we have,
2 (5x – 3) – 3 (2x – 1) = 9
∴ 2(5x – 3) – 3 (2x – 1) = 9
= 10x – 6 – 6x + 3 = 9
= 4x = 9 + 3
= x = 12/4
Thus, x = 3…(Verified)
Question no – (3)
Solution :
Given in the question,
x/2 = x/3 + 1
∴ x/2 = x/3 + 1
= x/2 – x/3 = + 1
= 3x – 2x/6 = + 1
Therefore, x = 6…(Verified)
Question no – (4)
Solution :
In the question we have,
x/2 + 3/2 = 2x/5 – 1
∴ x/2 + 3/2 = 2x/5 – 1
= x/2 – 2x/5 = -1 – 3/2
= 5x – 4x/10 = -2-3/2
= x = 5 × 10/2
Thus, x = -25…(Verified)
Question no – (5)
Solution :
Given in the question,
3/4 (x – 1) = x – 3
∴ 3/4 (x – 1) = x – 3
= 3x – 3 = 4x – 12
= 3x – 4x = – 12 + 3
= – x = – 9
Hence, x = 9…(Verified)
Question no – (6)
Solution :
In the question we have,
3 (x – 3) = 5 (2x + 1)
∴ 3 (x – 3) = 5 (2x + 1)
= 3x – 9 = 10x + 5
= 3x – 10x = + 5 + 9
= – 7x = 14
Therefore, x = – 2…(Verified)
Question no – (7)
Solution :
Given in the question,
3x – 2 (2x – 5) = 2 (x + 3) – 8
∴ 3x – 2 (2x – 5) = 2 (x + 3) – 8
= 3x – 4x + 10 = 2x + 6 – 8
= -x – 2x = – 2 – 10
= -3x = – 12
= x = 12/3
Hence, x = 4…(Verified)
Question no – (8)
Solution :
In the question,
x – x/4 – 1/2 = 3 + x/4
∴ x – x/4 – 1/2 = 3 + x/4
= 4x – x – 2/4 = 12 + x/4
= 3x – 2 = 12 + x
= 3x – x = 12 + 2
= x = 14/2
Hence, x = 7…(Verified)
Question no – (9)
Solution :
In the question we have,
6x – 2/9 + 3x + 5/18 = 1/3
∴ 6x – 2/9 + 3x + 5/18 = 1/3
= 12x – 4 + 3x + 5/18 = 1/3
= 15x + 1 = 18/3
= 15x + 1 = 6
= 15x = 6 – 1
= x = 5/15
Therefore, x = 1/3…(Verified)
Question no – (10)
Solution :
In the question,
m – m – 1/2 = 1 – m – 2/3
∴ m – m – 1/2 = 1 – m – 2/3
= 2m – m + 1/2 = 3 – m + 2/3
= m + 1/2 = 5 – m/3
= 3m + 3 = 10 – 2m
= 3m + 2m = 10 – 3
= 3m = + 7
Thus, m = + 7/5
Question no – (11)
Solution :
In the given question,
(5x – 1)/3 – (2x – 2)/3 = 1
∴ (5x – 1)/3 – (2x – 2)/3 = 1
= 5x – 1 – 2x + 2/3 = 1
= 3x + 1 = 3
= 3x = 3 – 1
Thus, x = 2/3
Question no – (12)
Solution :
In the question we have,
0.6x + 4/5 = 0.28x + 1.16
∴ 0.6x + 4/5 = 0.28x + 1.16
= 0.6x – 0.28x = 1.16 – 4/5
= 0.32x = 5.8 – 4/5
= x = 10.8/5 × 0.32
= x = 108 × 100/5 × 32 × 10
= x = 18/16
Hence, x = 9/8
Question no – (13)
Solution :
Given in the question,
0.5x + x/3 = 0.25x + 7
∴ 0.5x + x/3 = 0.25x + 7
= 0.5x + x/3 – 0.25x = 7
= 1.5x + x – 0.75x/3 = 7
= 1.75x = 21
= x = 21/1.75
= x = 21 × 100/175
Therefore, x = 12
Linear Equations In One Variable Exercise 8.4 Solution
Question no – (1)
Solution :
Let, the number be x.
∴ 3x – 5 = 16
= 3x = 16 + 5
= x – 21/3
∴ x = 7
Therefore, the number is 7.
Question no – (2)
Solution :
Let the number = x
∴ 7x = x + 78
= 7x – x = 78
= 6x = 78
= x = 13
Thus, the required number is 13.
Question no – (3)
Solution :
Let, Three numbers be x, x + 1, x + 2
∴ x + x + 1 = x + 2 + 15
= 2x – x = 17 – 1
= x = 16
∴ Three numbers are,
x = 16,
16 + 1 = 17
16 + 2 = 18
Therefore, the three consecutive natural numbers are 16, 17, 18.
Question no – (4)
Solution :
Let, two numbers be x and y,
x – y = 7
x – y = 7 = x = y + 7
and = 6y + x = 77
= 6y + y + 7 = 77
= 7y = 77 – 7
= 7y = 70
∴ y = 10
Required number be 10 and 10 + 7 = 17
Question no – (5)
Solution :
Let, the number x
∴ x/3 + 5 = 2x
= x + 15/3 = 2x
= 6x = x + 15
= 6x – x = 15
= 5x = 15
= x = 15/5
= x = 3
Therefore, the number will be 3.
Question no – (6)
Solution :
Let, the number be = x
∴ 3x + 5 = 50
= 3x = 50 – 5
= x = 45/3
= x = 15
Thus, the number will be 15.
Question no – (7)
Solution :
Let, Shikha age = x
and Ravish age = x + 3
∴ x + x + 3 = 37
= 2x = 37 – 3
= x = 34/2
∴ x = 17
So, Shikha age will be 17 year
And, Ravish’ s age will be = 17 + 3 = 20 years.
Question no – (8)
Solution :
Let, Nilu’s age = x
Jain’s age = x + 27
After 8 years Nilu’s age = x + 8
and Jain’s age,
= x + 27 + 8
= x + 35
∴ x + 35 = 2 (x + 8)
= x + 35 = 2x + 16
= x – 2x = 16 – 35
= – x = – 19
∴ x = 19
Nilu’s age will be 19 years,
And, Nilu’s Jain’s age will be,
= 19 + 27
= 46
Question no – (9)
Solution :
According to the question,
Son’s age = x
Man’s age = 4x
16 years after,
Man’s age = 4x + 16
Son’s age = x + 16
∴ 2 (x + 16) = 4x + 16
= 2x + 32 = 4x + 16
= 2x – 4x = + 16 – 32
= – 2x = – 16
∴ x = 8
Therefore, Son’s age will be = 8 years
And, Man’s age = 4 × 8 = 32 years.
Question no – (10)
Solution :
According to the question,
Girl age = x
Sister age = x – 4
Brother age,
= (x – 4) – 4
= x – 8
∴ x – 4 + x – 8 = 16
= 22x – 12 = 16
= 2x = 16 + 12
= x = 28/2 = 14
Girl age = 14 years
∴ Sister age,
= 14 – 4
= 10 years
∴ Brother age,
= 14 – 8
= 6 years
Question no – (11)
Solution :
Let, Anita find = x
Shella find = 2x
∴ 2x + x – 5 = 16
= 3x = 16 + 5
= x = 21/3 = 7
So, Anita find 7 shells
∴ Sandy find,
= 7 – 5
= 2 shells
∴ Shella find,
= 2 × 7
= 14 shells
Therefore, Anita find 7 shells, Sandy find 2 shells, and Shella find 14 shells.
Question no – (12)
Solution :
Let, Pandy has x marbles,
Andy has = 2x marbles
Sandy has = 2x + x/2 marbles
∴ 2x = 2x + x/2
= 2x – 3x/2 = + 75
= 4x – 3x/2 = 75
= x = 150
∴ Pandy has = 150 marbles
∴ Andy has,
= 150 × 2
= 300 marbles
∴ Sandy has,
= 450/2
= 225 marbles.
Hence, Pandy has 150 marbles, Andy has 300 marbles and Sandy has 225 marbles.
Question no – (13)
Solution :
Let, 50 paise coins x
and 25 paise coin = 4x
We know, 50 paise = 0.5 Rupee
₹30 = 30 × 100 = 3000 paise
∴ 50x + 4x × 25 = 3000
= 50x + 100x = 3000
= x = 20
∴ 50 paise coin = 20
∴ 25 paise coin,
= 4 × 20
= 80
Question no – (14)
Solution :
Let, Breadth = x
Length = 2x
∴ 2 (x + 2x) = 228
= 6x = 228
= x = 228/6
= x = 38
∴ Breadth = 38
∴ Length = 2 × 38 = 76
Therefore, the dimension of the field will be length 76 meter and breadth 38 metre.
Question no – (15)
Solution :
Let, 25 coins = x
₹ 17,50 = 1750 × 100
= 1750 paise
∴ x × 25 = 1750
= x = 70
Therefore, the number of coins in the purse is 70.
Question no – (16)
Solution :
Let, the number of students = x
50 kg = 50 × 1000 gm
= 400 × x = 50000
= 4x = 500
= x = 125
Therefore, there are 125 students in the hostel mess.
Next Chapter Solution :
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