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Joy of Mathematics Class 6 Solutions Chapter 11 Linear Equations in One Variable
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Joy of Mathematics Class 6 Book, Chapter 11 Linear Equations in One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations in One Variable. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Students will get chapter 11 solutions. Here students will find exercise wise solutions for chapter 11 Exercise 11.1 and 11.2
Linear Equations in One Variable Exercise 11.1 Solution :
Question no – (1)
Solution :
(a) x + 3
= 19
(b) 2x – 2
= 15
(c) 1/5 x
= 8
(d) 2x/3 + x
= 12
Question no – (2)
Solution :
(a) x – 7 = 10
=> x – 7 + 7 = 10 + 7
=> x = 17
(b) 8x + 2 = 10x
=> 8x + 2 – 2 = 10x – 2
=> 8x = 10x – 2
=> 8x – 10x = – 2
=> – 2x = – 2
=> x = 1
(c) 10 – x = 16x
=> 10 + x – 10 = 16x – 10
=> x = 16x – 10
=> x – 16x = – 10
=> – 15x = – 10
=> x = 10/15
=> 2/3
(d) a – 1 = 1
=> a – 1 + 1 = 1 + 1
=> a = 2
Question no – (3)
Solution :
(a) y + 3 = 7
=> y = 7 – 3
= 4
Now, 4 + 3 = 7 (Verified)
(b) y – 7 = 10
=> y = 10 + 7
= 17
(c) x + 8 = 11
=> x = 11 – 8
= 3
Now, 3 + 8 = 11 (Verified)
(d) x + 4 = 8
=> x = 8 – 4
= 4
Now, 4 + 4 = 8 (Verified)
(e) x + 7 = 11
=> x = 11 – 7
= 4
Now, 4 + 7 = 11 (Verified)
(f) 2y + 4 = 3
=> 2y = 3 – 4
= – 1
=> y = – 1/2
Now, 2 × (- 1/2) + 4
= – 1 + 4
= 3 (Verified)
(g) 3x + 4 = 8
=> 3x = 8 – 4
= 4
=> x = 4/3
(h) x + 7 = 24
=> x = 24 – 7
= 17
Now, 17 + 7 = 24 (Verified)
(i) x + 7 = 15
=> x = 15 – 7
= 8
Now, 8 + 7 = 15 (Verified)
(j) 2x/5 = 14
=> 2x = 14 × 5
=> x = 14 × 5/2
= 35
Now, 2 × 35/5
= 14 (Verified)
(k) x/3 = 2
=> x = 6
Now, 6/3 = 2 (Verified)
(l) x + 0.5 = 21.5
=> x = 21.5 – 0.5
= 21
Now, 21 + 0.5 = 21.5 (Verified)
(m) x – 2/5 = 4/10
=> x = 4/10 + 2/5
= 4 + 4/10
= 8/10
= 4/5
Now, 4/5 – 2/5
= 4 – 2/5
= 2/5
= 4/10 (Verified)
(n) 2x – 0.8 = 23.2
=> 2x = 23.2 + 0.8
= 24
=> x = 24/2
= 12
Now, 2 × 12 – 0.8
= 24 – 0.8
= 23.2
(o) x – 3 2/9 = 5 7/9
=> x – 29/9 = 52/9
=> x = 52/9 + 29/9
= 52 + 29/9
= 81/9
= 9
Question no – (4)
Solution :
(a) 2x + 2 = 11 (2)
for x = 2
L.H.S. 2 × 2 + 2
= 4 + 2
= 6
∴ L.H.S. ≠ R.H.S.
(b) 2x – 3 = 7 (2)
For x = 2
L.H.S. 2x – 3
= 2 × 2 – 3
= 4 – 3
= 1 ≠ 7
∴ L.H.S. ≠ R.H.S.
(c) 2x/3 = 2 (3)
For x = 3
L.H.S. 2x/3
= 2 × 3/3
= 2
= R.H.S.
(d) x/4 = 9 (20)
For x = 20
L.H.S. 20/4 = 5 ≠ 9
∴ L.H.S. ≠ R.H.S.
Question on – (5)
Solution :
(a) 11(2x + 11) – 5 (x + 4) = 9
=> 22x + 11 – 5x – 20 = 9
=> 17x – 9 = 9
=> 17x = 9 + 9
= 18
=> x = 18/17
(b) 2(x – 5) + 3(x + 6) = 12
=> 2x – 10 + 3x + 18 = 12
=> 5x + 8 = 12
=> 5x = 12 – 8
= 4
=> x = 4/5
(c) 2x + 4 = x + 9
=> 2x – x = 9 – 4
=> x = 5
(d) x/5 – 6 = 1/5x
=> x/5 – x/15 = 6
=> 3x – x/15 = 6
=> 2x = 6 × 15
=> x = 6 × 15/2
= 45
(e) 2x + 5 = 9x – 8
=> 2x – 9x = – 8 – 5
=> – 7x = – 13
=> x = 13/7
(f) 2x – 6 (x – 7) = 40
=> 2x – 6x + 42 = 40
=> – 4x = 40 – 42
= – 2
=> x = 2/4
= 1/2
Linear Equations in One Variable Exercise 11.2 Solution :
Question no – (1)
Solution :
Let, number ‘x’
∴ 3x – 7 = 56
=> 3x = 56 + 7
=> x = 63/3
= 21
∴ Number is 21
Question no – (2)
Solution :
Let, one number ‘x’
other, (x + 9)
∴ x + x + 9 = 75
=> 2x = 75 – 9
= 66
=> x = 66/2
= 33
∴ One number is 33
Other is = (33 + 9)
= 42
Question no – (3)
Solution :
Let, three consecutive number be x, x + 1, x + 2
∴ x + x + 1 + x + 2 = 66
=> 3x + 3 = 66
=> 3x = 66 – 3
= 63
=> x = 63/3
= 21
∴ Number will be 21, 22, 23
Question – (4)
Solution :
Let, three consecutive odd number be,
x, x + 2, x + 4
∴ x + x + 2 + x + 4 = 69
=> 3x = 69 – 6
= 63
=> x = 63/3
= 21
∴ Number is be 21, (21 + 2) = 23
= (23 + 4)
= 27
Question no – (5)
Solution :
Let, father’s age x
Son’s age (x – 30)
∴ x + x – 30 = 60
=> 2x = 60 + 30
= 90
=> x = 90/2
= 45
Father’s age = 45 years
Son’s age = (45 – 30)
= 15 years.
Question no – (6)
Solution :
Let, William’s age x yr.
William’s father’s age = 3x yr.
∴ x + 3x = 84
=> 4x = 84
=> x = 84/4
= 21 yr.
∴ William’s age = 21 yr.
∴ Father’s age = (21 × 3)
= 63 yr.
Question no – (7)
Solution :
Let, cost of chair x Rs
Cost of table (x + 150) Rs
∴ (x + 150) + 2 (x) = 1500
=> x + 150 + 2x = 1500
=> 3x = 1500 – 150
=> x = 1350/3
= 450 Rs
Chair’s cost 450 Rs
Table cost (450 + 150)
= 600 Rs.
Question no – (8)
Solution :
Let, breadth of rectangle x m
Length of rectangle 4x m
∴ Perimeter = 2 (x + 4x)
= 2 × 5x
= 10x
∴ 10x = 60
=> x = 60/10
= 6m
∴ Breadth 6 m
Length = (6 × 4)x
= 24m
Question no – (9)
Solution :
Let, Number be ‘x’
∴ x/6 + x/10 = 64
=> 5x + 3x/30 = 64
=> 8x = 30 × 64
=> x = 30 × 64/8
= 240
∴ Number is 240
Question no – (10)
Solution :
Let, length 5x, breadth 3x
∴ 2(5x + 3x) = 144
=> 2 × 8x = 144
=> x = 144/16
= 9
∴ Breadth = (3 × 9) = 27 m
∴ Length = (5 × 9) = 45 m.
Area will be,
= 45 × 27 [∵ area of rectangle = length × breadth]
= 1215 m2
Therefore, the area of the rectangle will be 1215 m2
Next Chapter Solution :
👉 Chapter 12 👈
