Collins Maths Solutions Class 7 Chapter 12 Linear Equations and Inequations
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 12, Linear Equations and Inequations. Here students can easily find chapter 7 solutions with exercise wise explanation. Students will find proper solutions for Exercise 12.1, 12.2 and 12.3 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Linear Equations and Inequations Exercise 12.1 Solution :
Question no – (1)
Solution :
2x + 6 = 12
= 2x = 12 – 6
= x = 6/2
∴ x = 3
Question no – (2)
Solution :
3x – 7 = 11
= 3x = 11 + 7 = 18
= x = 18/3
∴ x = 6
Question no – (3)
Solution :
2y – 5 = 9
= 2y = 9 + 5 = 14
= y = 14/2
∴ y = 7
Question no – (4)
Solution :
13 – x = 1
= – x = 1 – 13
= -x = – 12
∴ x = 12
Question no – (5)
Solution :
a/5 = – 15
= a = 5 × (-15)
∴ a = -75
Question no – (6)
Solution :
3x/5 = 18
= x = 5 × 18/3
∴ x = 30
Question no – (7)
Solution :
7c + 15 = 40
= c = 40 – 15/7
= c = 25/7
Question no – (8)
Solution :
4v + 2 = – 74
= 4v = – 4 – 2
= v = -6/4
∴ v = -3/2
Linear Equations and Inequations Exercise 12.2 Solution :
Question no – (1)
Solution :
Let, the number is = x
= 4x + 5 = 10
= 4x = 10 – 5
= x = 10 – 5/4
= x = 5/4
= x = 1 1/4
Therefore, the number is 1 1/4.
Question no – (2)
Solution :
Let, three consecutive number be,
= x, x + 2, x + 4
∴ 2 (x + x + 2 + x + 4) = 36
= 2 (3x + 6) = 36
= 6x + 12 = 36
= 6x = 36 – 12
= x = 24/6
= x = 4
∴ The required numbers are,
= x = 4
= x + 2 = 4 + 2 = 6
= x + 4 = 4 + 4 = 8
Hence, the numbers are 4, 6, 8.
Question no – (3)
Solution :
Let, the sisters age = x years.
My age = 3x years.
∴ 1 1/2 (3x + 9) = (x + 9)
= 3/2 (3x + 9) = (x + 9)
= 1.5 × 3x + 13.5 = x + 9
= 4.5x = x + 9 – 13.5
= 4.5x – x = 9
Therefore, my present age will be 9 years
Question no – (4)
Solution :
Sum of sons age = (8 + 3) = 11 years
Let, after x year
Twice of sum = 11 + 2 × 5
= 21 years
∴ (21 × 2)= 42 years
Now, (42 – 37)
= 5 years
Therefore, in 5 years.
Question no – (5)
Solution :
Let, one number is = x
and other number is = (x + 8)
∴ x + x + 8 = 46
= 2x = 46 – 8 = 38
= x = 38/2
= x = 19
∴ Numbers are, 19 and (19 + 8) = 27 years.
Question no – (6)
Solution :
Let, the numbers are x and (x + 6)
∴ x + x + 6 = 78
= 2x = 78 – 6 = 72
= x = 72/2
= x = 36
∴ Multiples are,
= x = 36
= x + 6 = 36 + 6 = 42
Therefore, the multiples are 36 and 42
Question no – (7)
Solution :
Let, pen cost = a Rs
book cost = (a + 50) Rs
∴ 5(a + 50) + 3a = 730
= 5a + 250 + 3a = 730
= 8a = 730 – 250
= a = 480/8 = 60
∴ Pen cost = 60 Rs
∴ Book cost = ( 60 + 50) = 110 Rs.
Question no – (8)
Solution :
Let, the number be = ‘x’
∴ 4x + 5 = 3x
= 4x – 3x k= 5
= x = 5
Therefore, the required number will be 5.
Question no – (10)
Solution :
Bill = (150 – 30)
= 120 Rs.
∴ Travel = 120/6
= 20 km.
Therefore, a person can be travel 20 km.
Linear Equations and Inequations Exercise 12.3 Solution :
Question no – (1)
Solution :
x – 3 ≥ 2; x ∈ N
= x – 3 + 3 ≥ 2 + 3
= x = 5
∴ x = 5, 6, 7, ……….
Question no – (2)
Solution :
15 < 2x + 3 < 9; y ∈ Z
∴ 15 < 2x – 3
= 15 – 3 < 2x + 3 – 3
= 12 < 2x
= x > 6
And, 2x + 3 < 9
= 2x + 3 – 3 <9 – 3
= 2x < 6
= x < 3
∴ 6 < x < 3
Question no – (3)
Solution :
1st, 17 ≤ 5x + 3
= 17 – 3 ≤ 5x + 3 – 3
= 14 ≤ 5x
= 14/5 ≤ x
2nd, 5x + 3 ≤ 18
= 5x + 3 – 3 ≤ 18 – 3
= 5x ≤ 15
= x ≤ 3
∴ x = 3
Question no – (4)
Solution :
3x + 9 < 12; x ∈ Z
= 3x + 9 – 9 < 12 – 9
= 3x < 3
= x < 1
∴ x = {0}
Question no – (5)
Solution :
4y – 6 ≤ 10; y ∈ N
= 4y – 6 + 6 ≤ 10 + 6
= 4y ≤ 16
= y ≤ 16/4
= y = 4
∴ y = {1, 2, 3, 4}
Question no – (6)
Solution :
2p + 8 ≥ 14
= 2p + 8 – 8 ≥ 14 – 8
= 2p ≥ 6
= p ≥ 3
∴ p = {3, 4, 5, …..}
And, 2p + 8 ≤ 18
= 2p + 8 – 8 ≤ 18 – 8
= 2p ≤ 10
= p ≤ 10/2
= p ≤ 5
∴ {3, 4, 5}
Next Chapter Solution :
👉 Chapter 13 👈
