NCERT Exemplar Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 8 PDF Download

Exercise Page: 110

In questions 1 to 15 out of the four options only one is correct, write the correct answer.
Q.1. The solution of which of the following equations is neither a fraction nor an integer. 
(a) 3x + 2 = 5x + 2 
(b) 4x – 18 = 2
(c) 4x + 7 = x + 2 
(d) 5x – 8 = x + 4

Ans: (c)
Explanation: Transposing 7 to RHS and it becomes -7 and x to LHS it becomes -x
4x – x = 2 – 7
3x = – 5
X = -5/3
So, -5/3 is neither a fraction nor an integer.

Q.2. The solution of the equation ax + b = 0 is
(a) x = a/b 
(b) x = -b
(c) x = -b/a 
(d) x = b/a

Ans: (c)
Explanation: 
Given, ax + b = 0
Transposing b to RHS and it becomes -b
Then,
ax = -b
x = -b/a

Q.3. If 8x – 3 = 25 +17x, then x is
(a) a fraction 
(b) an integer
(c) a rational number 
(d) cannot be solved

Ans: (c)
Explanation: 
Given, 8x – 3 = 25 + 17x
Transposing -3 to RHS and it becomes 3 and 17x to LHS it becomes -17x.
8x – 17x = 25 + 3
-9x = 28
X = -28/9
Therefore x is a rational number.

Q.4. The shifting of a number from one side of an equation to other is called
(a) Transposition 
(b) Distributivity
(c) Commutativity 
(d) Associativity

Ans: (a)
Explanation: The shifting of a number from one side of an equation to other is called Transposition

Q.5. If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is
(a) 19/13 
(b) -13/19 
(c) 0 
(d) 13/19

Ans: (b)
Explanation: 
Given, (5x/3) – 4 = (2x/5)
(5x/3) – (2x/5) = 4
LCM of 3 and 5 is 15
(25x – 6x)/15 = 4
19x = 4 × 15
19x = 60
X = 60/19
Then, Substitute the value of x in 2x -7
= (2 × (60/19)) – 7
= (120/19) – 7
= (120 – 133)/19
= – 13/19

Q.6. The value of x for which the expressions 3x – 4 and 2x + 1 become equal is
(a) -3 
(b) 0 
(c) 5 
(d) 13/19

Ans: (c)
Explanation:
Given, 3x – 4 = 2x + 1
Transposing -4 to RHS and it becomes 4 and 2x to LHS it becomes -2x.
3x – 2x = 1 + 4
X = 5

Q.7. If a and b are positive integers, then the solution of the equation ax = b has to be always
(a) positive 
(b) negative 
(c) one 
(d) zero

Ans: (a)
Explanation:
Let a = 3, b = 4
Then, ax = b
3x = 4
X = 4/3

Q.8. Linear equation in one variable has
(a) only one variable with any power.
(b) only one term with a variable.
(c) only one variable with power 1.
(d) only constant term.

Ans: (c)

Q.9. Which of the following is a linear expression:
(a) x + 1 
(b) y + y² 
(c) 4 
(d) 1 + z²

Ans: (a)
Explanation: The linear expressions is one which having highest power as 1Standard form of linear equation

Q.10. A linear equation in one variable has
(a) Only one solution
(b) Two solutions
(c) More than two solutions
(d) No solution

Ans: (a)

Q.11. Value of S in (1/3) + S = 2/5
(a) 4/5 
(b) 1/15 
(c) 10 
(d) 0

Ans: (b)
Explanation: 
Given,
1/3 + S = 2/5
S = 2/5 – 1/3
S = (6 – 5)/15
S = 1/15

Q.12. (-4/3)y = – ¾, then y =
(a) -(¾)² 
(b) -(4/3)² 
(c) (¾ )² 
(d) (4/3)²

Ans: (c)
Explanation: 
Given,
(-4/3)y = -¾
Y = – ¾ × -¾
Y = 9/16
Y = (3 × 3)/(4 × 4)
Y = 3²/4²
Y = (¾)²

Q.13. The digit in the tens place of a two digit number is 3 more than the digit in the units place. Let the digit at units place be b. Then the number is
(a) 11b + 30 
(b) 10b + 30 
(c) 11b + 3 
(d) 10b + 3

Ans: (a)
Explanation:
From the question it is given that,
Let the digit at units place be b.
The digit in the tens place of a two digit number is 3 more than the digit in the units place
= 3 + b
So, the number = 10 (3 + b) + b
= 30 + 10b + b
= 30 + 11b

Q.14. Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x. Then Arpita’s present age is
(a) 3(x – 3) 
(b) 3x + 3 
(c)3x – 9 
(d) 3(x + 3)

Ans: (d)
Explanation:
Given,
Shilpa’s age three years ago was x
Then, Shilpa’s present age is = x + 3
Arpita’s present age is thrice of Shilpa = 3 (x + 3)

Q.15. The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.
(a) 112 
(b) 126 
(c) 119 
(d) 116

Ans: (a)
Explanation:
Let us assume the three consecutive multiples of 7 be 7x, (7x + 7), (7x + 14) where x is a natural number.
As per the condition in the question,
7x + (7x + 7) + (7x + 14) = 357
21x + 21 = 357
21(x + 1) = 357
(21(x + 1))/21 = 357/21
X + 1 = 17
X = 17 – 1
X = 16
Therefore, the smallest multiple of 7 is,
7 × 16 = 112.
In questions 16 to 32, fill in the blanks to make each statement true.

Q.16. In a linear equation, the _________ power of the variable appearing in the equation is one.

Solution: In a linear equation, the highest power of the variable appearing in the equation is one.

Q.17. The solution of the equation 3x – 4 = 1 – 2 x is .

Solution:
The solution of the equation 3x – 4 = 1 – 2 x
3x + 2x = 1 +4
5x =  5
 x = 1

Q.18. The solution of the equation 2y = 5y – 18/5 is .

Solution:
The solution of the equation 2y = 5y – 18/5
2y = 5y – (18/5)
(18/5) = 5y – 2y
(18/5) = 3y
y = (18/5) × (1/3)
y = (6/5)

Q.19. Any value of the variable which makes both sides of an equation equal is known as a _________ of the equation.

Solution: Any value of the variable which makes both sides of an equation equal is known as a solution of the equation.

Q.20. 9x – y = –21 has

Solution:
9x – y = –21 has the solution (–2)
In the question it is given that, x = -2
Then, let us assume the missing number be y
(9 × (-2)) – y = -21
-18 – y = -21
– y = -21 + 18
– y = – 3
y = 3

Q.21. Three consecutive numbers whose sum is 12 are _________, _________ and _________.

Solution: Three consecutive numbers whose sum is 12 are 3, 4 and 5.
3 + 4 + 5 = 12

Q.22. The share of A when Rs 25 are divided between A and B so that A gets Rs. 8 more than B is _________.

Solution: The share of A when Rs 25 are divided between A and B so that A gets Rs. 8 more than B is Rs 16.50.
Let us assume B share be x
As per the condition in the question A share be x + 8
Then,
x + (x + 8) = 25
x + x + 8 = 25
2x + 8 = 25
2x = 25 – 8
2x = 17
x = 17/2
x = 8.5
So, A gets x + 8
= 8.5 + 8
= Rs 16.5

Q.23. A term of an equation can be transposed to the other side by changing its _________.

Solution: A term of an equation can be transposed to the other side by changing its sign.
For example:- 2x + 3 = 0
Transposing 3 to RHS and it becomes -3
2x = -3
X = -3/2

Q.24. On subtracting 8 from x, the result is 2. The value of x is _________.

Solution: On subtracting 8 from x, the result is 2. The value of x is 10.
From the question,
On subtracting 8 from x, the result is 2,
= x – 8 = 2
Transposing -8 to RHS and it becomes 8
X = 2 + 8
X = 10

Q.25. (x/5) + 30 = 18 has the solution as .

Solution:
(x/5) + 30 = 18 has the solution as -60.
Given, (x/5) + 30 = 18
Transposing 30 to RHS and it becomes -30.
(x/5) = 18 – 30
(x/5) = -12
X = -12 × 5
X = -60

Q.26. When a number is divided by 8, the result is –3. The number is _________.

Solution: When a number is divided by 8, the result is –3. The number is -24.
Let the number be x,
Then,
x/8 = -3
x = -3 × 8
x = -24

Q.27. 9 is subtracted from the product of p and 4, the result is 11. The value of p is _________.

Solution:
9 is subtracted from the product of p and 4, the result is 11. The value of p is 5.
From the question, it is given that,
9 is subtracted from the product of p and 4, the result is 11 = 4p – 9 = 11
4p – 9 = 11
Transposing -9 to RHS and it becomes 9.
4p = 11 + 9
4p = 20
P = 20/4
P = 5

Q.28. If (2/5)x – 2 = 5 – (3/5)x, then x = ?

Solution:
If (2/5)x – 2 = 5 – (3/5)x,
Given,
(2/5)x – 2 = 5 – (3/5)x
Transposing -2 to RHS and it becomes 2 and (3/5)x to LHS it becomes –(3/5)x.
(2/5)x + (3/5)x = 5 + 2
(2x + 3x)/5 = 7
5x = 7 × 5
X = 35/5
X = 7

Q.29. After 18 years, Swarnim will be 4 times as old as he is now. His present age is _________.

Solution:
After 18 years, Swarnim will be 4 times as old as he is now. His present age is 6 years.
Let us assume swarnim’s parent age be x year.
Then, after 18 year, Swarnim’s age = (x + 18) year
According to the question,
X + 18 = 4x
X – 4x = -18
– 3x = -18
– 3x/3 = (-18/3)
X = 6
Therefore, swarnim’s present age is 6 year.

Q.30. Convert the statement adding 15 to 4 times x is 39 into an equation _________

Solution: Convert the statement Adding 15 to 4 times x is 39 into an equation 4x + 15 = 39.

Q.31. The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 and the denominator is decreased by 1, then expression for new denominator is _________.

Solution: The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 and the denominator is decreased by 1, then expression for new denominator is x + 9.
Let us assume numerator be x,
So, denominator = x + 10
Rational number = x/(x + 10)
As per the condition given in the question, the numerator is increased by 1 and the denominator is decreased by 1.
New rational number = Numerator + 1/ (denominator – 1)
= (x + 1)/(x + 10 – 1)
= (x + 1)/(x + 9)
∴ the new denominator is x + 9.

Q.32. The sum of two consecutive multiples of 10 is 210. The smaller multiple is _________.

Solution: The sum of two consecutive multiples of 10 is 210. The smaller multiple is 100.
Let us assume the two consecutive multiples of 10 be 10x and 10x + 10.
So,
Sum of two consecutive multiples of 10 = 10x + 10x + 10 = 210
20x + 10 = 210
20x = 210 – 10
20x = 200
x = 200/20
x = 10
∴ the two consecutive multiples of 10 are 10x = 10 × 10 = 100
10x + 10 = (10 × 10) + 10
= 110
Hence, the smaller multiple of 10 is 100.

In questions 33 to 48, state whether the statements are true (T) or false (F).
Q.33. 3 years ago, the age of a boy was y years. His age 2 years ago was (y – 2) years

Solution: False.

Given, 3 yr ago, age of boy = y yr
Then, present age of boy = (y + 3)yr
2 yr ago, age of boy = y + 3-2 = (y + 1)yr

Q.34. Shikha’s present age is p years. Reemu’s present age is 4 times the present age of Shikha. After 5 years Reemu’s age will be 15p years.

Solution: False.
Given,
Shikha’s present age is p years
Reemu’s present age is 4 times the present age of Shikha = 4p
After 5 years Reemu’s age will be = (4p + 5)

Q.35. In a 2 digit number, the units place digit is x. If the sum of digits be 9, then the number is (10x – 9).

Solution: False.
From the question it is given that,
The unit’s place digit is = x
Sum of two digits = 9
Then,
Ten’s digit = 9 – x
So,
The number = 10 (9 – x) + x
= 90 – 10x + x
= 90 – 9x

Q.36. Sum of the ages of Anju and her mother is 65 years. If Anju’s present age is y years then her mother’s age before 5 years is (60 – y) years.

Solution: True.
From the question it is given that,
Anju’s present age = y years
Present age of Anju’s mother = (65 – y) years
Then,
Before 5 years, Anju’s mother age = 65 – y – 5
= (60 – y) years

Q.37. The number of boys and girls in a class are in the ratio 5:4. If the number of boys is 9 more than the number of girls, then number of boys is 9.

Solution: False.
Let us assume number of boys be 5y and the number of girls be 4y.
From the question,
5x – 4x = 9
X = 9
∴ Number of boys = 5x = 5 × 9 = 45 boys
Number of girls = 4x = 4 × 9 = 36 girls.

Q.38. A and B are together 90 years old. Five years ago A was thrice as old as B was. Hence, the ages of A and B five years back would be (x – 5) years and (85 – x) years respectively.

Solution: True.
Let us assume age of A be y years.
So, age of B = (90 – y) years.
Then,
Before 5 years A’s age = (x – 5) years
and B’s age = (90 – x – 5) = (85 – x) years.

Q.39. Two different equations can never have the same answer.

Solution: False.
Two different equations can have the same answer
For example: – (i) 4x + 2 = 3
4x = 3 – 2
X = ¼
(ii) 4x – 6 = – 5
4x = – 5 + 6
X = ¼

Q.40. In the equation 3x – 3 = 9, transposing –3 to RHS, we get 3x = 9.

Solution: False.
Given, 3x – 3 = 9
Transposing –3 to RHS it becomes 3
3x = 9 + 3
3x = 12

Q.41. In the equation 2x = 4 – x, transposing –x to LHS, we get x = 4.

Solution: False.
Given, 2x = 4 – x
Transposing –x to LHS it becomes x
2x + x = 4
3x = 4
X = 3/4

Q.42. If (15/8) – 7x = 9, then -7x = 9 + (15/8)

Solution: False.
Given, (15/8) – 7x = 9
Transposing 15/8 to RHS it becomes – (15/8)
– 7x = 9 – (15/8)

Q.43. If (x/3) + 1 = (7/15), then x/3 = 6/15

Solution: False.
Given, (x/3) + 1 = (7/15)
Transposing 1 to RHS it becomes – 1
(x/3) = (7/15) – 1
(x/3) = (7 – 15)/15
(x/3) = -8/15

Q.44. If 6x = 18, then 18x = 54

Solution: True.
Given, 6x = 18
Multiplying both LHS and RHS by 3, we get
6x × 3 = 18 × 3
18x = 54

Q.45. If x/11 = 15, then x = 11/15

Solution: False.
Given, x/11 = 15
Multiplying both LHS and RHS by 11, we get
(x/11) × 11 = 15 × 11
X = 165

Q.46. If x is an even number, then the next even number is 2(x + 1).

Solution: False.
If x is an even number, then the next even number is (x + 2)

Q.47. If the sum of two consecutive numbers is 93 and one of them is x, then the other number is 93 – x.

Solution: True
Given, one of the consecutive number = x
Two consecutive number are = x and 93 – x
Then, sum of two consecutive numbers = x + (93 – x) = 93
X + 93 – x = 93
Transposing 93 to RHS it becomes – 93
x – x = 93 -93
0 = 0

Q.48. Two numbers differ by 40, when each number is increased by 8, the bigger becomes thrice the lesser number. If one number is x, then the other number is (40 – x).

Solution: False.
From the question it is given that,
One number = x
Other number = 40 –x
Let us assume (40 – x) > x
So, 40 – x + 8 = 3 (x + 8)
48 – x = 3x + 24
– x – 3x = 24 -48
– 4x = -24
X = -24 × (-1/4)
X = 6
∴ One number is x = 6
Other number is = 40 – x
= 40 – 6
= 34
Difference between numbers = 34 – 6 = 28

Solve the following:
Q.49. ((3x – 8)/2x) = 1

Solution: We have,
((3x – 8)/2x) = 1
By cross multiplication, we get
(3x – 8) = 2x
Transposing -8 to RHS it becomes 8 and 2x to LHS it becomes – 2x
3x – 2x = 8
x = 8

Q.50. (5x/(2x – 1)) = 2

Solution: We have,
(5x/(2x – 1)) = 2
By cross multiplication, we get
5x = 2 × (2x – 1)
5x = 4x – 2
Transposing 4x to LHS it becomes – 4x
5x – 4x = -2
x = -2

Q.51. ((2x – 3)/(4x + 5)) = (1/3)

Solution: We have,
((2x – 3)/(4x + 5)) = (1/3)
By cross multiplication, we get
3 × (2x – 3) = 1 × (4x + 5)
6x – 9 = 4x + 5
Transposing -9 to RHS it becomes 9 and 4x to LHS it becomes – 4x.
6x – 4x = 5 + 9
2x = 14
x = 14/2
x = 7

Q.52. (8/x) = (5/(x – 1))

Solution: We have,
(8/x) = (5/(x – 1))
By cross multiplication, we get
8 × (x – 1) = 5 × x
8x – 8 = 5x
Transposing -8 to RHS it becomes 8 and 5x to LHS it becomes – 5x.
8x – 5x = 8
3x = 8
X = 8/3

Q.53. [(5(1 – x) + 3(1 + x))/ (1 – 2x)] = 8

Solution: We have,
[(5(1 – x) + 3(1 + x))/ (1 – 2x)] = 8
By cross multiplication, we get
(5(1 – x) + 3(1 + x)) = 8 × (1 – 2x)
5 – 5x + 3 + 3x = 8 – 16 x
8 – 2x = 8 – 16x
Transposing 8 to RHS it becomes – 8 and -16x to LHS it becomes 16x.
16x – 2x = 8 – 8
14x = 0
x = 0/14
x = 0

Q.54. ((0.2x + 5)/ (3.5x – 3)) = (2/5)

Solution: We have,
((0.2x + 5)/ (3.5x – 3)) = (2/5)
By cross multiplication, we get
5 × (0.2x + 5) = 2 × (3.5x – 3)
x + 25 = 7x – 6
Transposing x to RHS it becomes – x and -6 to LHS it becomes 6.
25 + 6 = 7x – x
31 = 6x
x = 31/6

Q.55. [(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5

Solution: We have,
[(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5
(y – 4 + 3y)/ (2y – 3 – 4y) = 1/5
(-4y – 4)/ (2y – 3) = 1/5
By cross multiplication, we get
5 × (-4y – 4) = 1 × (2y – 3)
20y – 20 = 2y – 3
Transposing – 20 to RHS it becomes 20 and 6y to LHS it becomes -6y.
20y – 2y = 20 – 3
22 y = 17
y = 17/22

Q.56. (x/5) = (x – 1)/6

Solution: We have,
(x/5) = (x – 1)/6
By cross multiplication, we get
6 × x = 5 × (x – 1)
6x = 5x – 5
Transposing 5x to RHS it becomes -5x
6x – 5x = -5
x = -5

Q.57. 0.4(3x –1) = 0.5x + 1

Solution: We have,
0.4(3x –1) = 0.5x + 1
1.2x – 0.4 = 0.5x + 1
Transposing – 0.4 to RHS it becomes 0.4 and 0.5x to LHS it becomes -0.5x.
1.2x – 0.5x = 1 + 0.4
0.7x = 1.4
x = 1.4/0.7
x = 14/7
x = 2

Q.58. 8x – 7 – 3x = 6x – 2x – 3

Solution: We have,
8x – 7 – 3x = 6x – 2x – 3
5x – 7 = 4x – 3
Transposing – 7 to RHS it becomes 7 and 4x to LHS it becomes -4x.
5x – 4x = 7 – 3
x = 4

Q.59. 10x – 5 – 7x = 5x + 15 – 8

Solution: We have,
10x – 5 – 7x = 5x + 15 – 8
3x – 5 = 5x + 7
Transposing – 5 to RHS it becomes 5 and 5x to LHS it becomes -5x.
3x – 5x = 7 + 5
– 2x = 12
x = -12/2
x = – 6

Q.60. 4t – 3 – (3t +1) = 5t – 4

Solution: We have,
4t – 3 – (3t +1) = 5t – 4
4t – 3 – 3t – 1 = 5t – 4
t – 4 = 5t – 4
Transposing t to RHS it becomes -t and -4 to LHS it becomes 4.
4 – 4 = 5t – t
0 = 4t
t = 0/4
t = 0

Q.61. 5(x – 1) – 2(x + 8) = 0

Solution: We have,
5(x – 1) – 2(x + 8) = 0
5x – 5 – 2x – 16 = 0
3x – 21 = 0
Transposing -21 to RHS it becomes 21.
3x = 21
x = 21/3
x = 7