HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 2 Linear Equations in One Variable Solutions

# Class 8 Maths Chapter 2 Linear Equations in One Variable Solutions

linear-equations

## Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Solutions

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Ex 2.1 Class 8 Maths Question 1.
Solve the equation: x – 2 = 7.
Solution:
Given: x – 2 = 7
⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides)
⇒ x = 9 (Required solution)

Ex 2.1 Class 8 Maths Question 2.
Solve the equation: y + 3 = 10.
Given: y + 3 = 10
⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side)
⇒ y = 7 (Required solution)

Ex 2.1 Class 8 Maths Question 3.
Solve the equation: 6 = z + 2
Solution:
We have 6 = z + 2
⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side)
⇒ 4 = z
Thus, z = 4 is the required solution.

Ex 2.1 Class 8 Maths Question 4.
Solve the equations: $\frac { 3 }{ 7 }$ + x = $\frac { 17 }{ 7 }$
Solution:

Ex 2.1 Class 8 Maths Question 5.
Solve the equation 6x = 12.
Solution:
We have 6x = 12
⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)
⇒ x = 2
Thus, x = 2 is the required solution.

Ex 2.1 Class 8 Maths Question 6.
Solve the equation $\frac { t }{ 5 }$ = 10.
Solution:
Given $\frac { t }{ 5 }$ = 10
⇒ $\frac { t }{ 5 }$ × 5 = 10 × 5 (multiplying both sides by 5)
⇒ t = 50
Thus, t = 50 is the required solution.

Ex 2.1 Class 8 Maths Question 7.
Solve the equation $\frac { 2x }{ 3 }$ = 18.
Solution:
We have $\frac { 2x }{ 3 }$ = 18
⇒ $\frac { 2x }{ 3 }$ × 3 = 18 × 3 (multiplying both sides by 3)
⇒ 2x = 54
⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)
⇒ x = 27
Thus, x = 27 is the required solution.

Ex 2.1 Class 8 Maths Question 8.
Solve the equation 1.6 = $\frac { y }{ 1.5 }$
Solution:
Given: 1.6 = $\frac { y }{ 1.5 }$
⇒ 1.6 × 1.5 = $\frac { y }{ 1.5 }$ × 1.5 (multiplying both sides by 1.5)
⇒ 2.40 = y
Thus, y = 2.40 is the required solution.

Ex 2.1 Class 8 Maths Question 9.
Solve the equation 7x – 9 = 16.
Solution:
We have 7x – 9 = 16
⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides)
⇒ 7x = 25
⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7)
⇒ x = $\frac { 25 }{ 7 }$
Thus, x = $\frac { 25 }{ 7 }$ is the required solution.

Ex 2.1 Class 8 Maths Question 10.
Solve the equation 14y – 8 = 13.
Solution:
We have 14y – 8 = 13
⇒ 14y – 8 + 8 = 13 + 8 (adding 8 to both sides)
⇒ 14y = 21
⇒ 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14)
⇒ y = $\frac { 21 }{ 14 }$
⇒ y = $\frac { 3 }{ 2 }$
Thus, y = $\frac { 3 }{ 2 }$ is the required solution.

Ex 2.1 Class 8 Maths Question 11.
Solve the equation 17 + 6p = 9.
Solution:
We have, 17 + 6p = 9
⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides)
⇒ 6p = -8
⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6)
⇒ p = $\frac { -8 }{ 6 }$
⇒ p = $\frac { -4 }{ 3 }$
Thus, p = $\frac { -4 }{ 3 }$ is the required solution.

Ex 2.1 Class 8 Maths Question 12.
Solve the equation $\frac { x }{ 3 }$ + 1 = $\frac { 7 }{ 15 }$
Solution:

# Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.2 (Exercises and Solutions)

## NCERT Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.2

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2

Ex 2.2 Class 8 Maths Question 1.
If you subtract $\frac { 1 }{ 2 }$ from a number and multiply the result by $\frac { 1 }{ 2 }$, you get $\frac { 1 }{ 8 }$. What is the number?
Solution:
Let the required number be x.

Ex 2.2 Class 8 Maths Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Let the breadth of the pool be x m.
Condition I: Length = (2x + 2) m.
Condition II: Perimeter = 154 m.
We know that Perimeter of rectangle = 2 × [length + breadth]
2 × [2x + 2 + x] = 154
⇒ 2 × [3x + 2] = 154
⇒ 6x + 4 = 154 (solving the bracket)
⇒ 6x = 154 – 4 [Transposing 4 from (+) to (-)]
⇒ 6x = 150
⇒ x = 150 ÷ 6 [Transposing 6 from (×) to (÷)]
⇒ x = 25
Thus, the required breadth = 25 m
and the length = 2 × 25 + 2 = 50 + 2 = 52 m.

Ex 2.2 Class 8 Maths Question 3.
The base of an isosceles triangle is $\frac { 4 }{ 3 }$ cm. The perimeter of the triangle is 4$\frac { 2 }{ 15 }$ cm. What is the length of either of the remaining equal sides?
Solution:
Let the length of each of equal sides of the triangle be x cm.
Perimeter of the triangle = sum of the three sides

Ex 2.2 Class 8 Maths Question 4.
Sum of two numbers be 95. If one exceeds the other by 15, find the numbers.
Solution:
Let one number be x
Other number = x + 15
As per the condition of the question, we get
x + (x + 15) = 95
⇒ x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15 [transposing 15 from (+) to (-)]
⇒ 2x = 80
⇒ x = $\frac { 80 }{ 2 }$ [transposing 2 from (×) to (÷)]
⇒ x = 40
Other number = 95 – 40 = 55
Thus, the required numbers are 40 and 55.

Ex 2.2 Class 8 Maths Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two numbers be 5x and 3x.
As per the conditions, we get
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18 ÷ 2 [Transposing 2 from (×) to (÷)]
⇒ x = 9.
Thus, the required numbers are 5 × 9 = 45 and 3 × 9 = 27

Ex 2.2 Class 8 Maths Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2.
As per the condition, we get
x + (x + 1) + (x + 2) = 51
⇒ x + x + 1 + x + 2 = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3 [transposing 3 to RHS]
⇒ 3x = 48
⇒ x = 48 ÷ 3 [transposing 3 to RHS]
⇒ x = 16
Thus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.

Ex 2.2 Class 8 Maths Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.
As per the conditions, we get
8x + (8x + 8) + (8x + 16) = 888
⇒ 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
⇒ 24x = 888 – 24 (transposing 24 to RHS)
⇒ 24x = 864
⇒ x = 864 ÷ 24 (transposing 24 to RHS)

⇒ x = 36
Thus, the required multiples are
36 × 8 = 288, 36 × 8 + 8 = 296 and 36 × 8 + 16 = 304,
i.e., 288, 296 and 304.

Ex 2.2 Class 8 Maths Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the three consecutive integers be x, x + 1 and x + 2.
As per the condition, we have
2x + 3(x + 1) + 4(x + 2) = 74
⇒ 2x + 3x + 3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 – 11 (transposing 11 to RHS)
⇒ 9x = 63
⇒ x = 63 ÷ 9
⇒ x = 7 (transposing 7 to RHS)
Thus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.

Ex 2.2 Class 8 Maths Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Let the present ages of Rahul and Haroon he 5x years and 7x years respectively.
4 years later, the age of Rahul will be (5x + 4) years.
4 years later, the age of Haroon will be (7x + 4) years.
As per the conditions, we get
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 (transposing 8 to RHS)
⇒ 12x = 48
⇒ x = 48 ÷ 12 = 4 (transposing 12 to RHS)
Hence, the required age of Rahul = 5 × 4 = 20 years.
and the required age of Haroon = 7 × 4 = 28 years.

Ex 2.2 Class 8 Maths Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the numbers of girls. What is the total class strength?
Solution:
Let the number of boys be 7x
and the number of girls be 5x
As per the conditions, we get
7x – 5x = 8
⇒ 2x = 8
⇒ x = 8 ÷ 2 = 4 (transposing 2 to RHS)
the required number of boys = 7 × 4 = 28
and the number of girls = 5 × 4 = 20
Hence, total class strength = 28 + 20 = 48

Ex 2.2 Class 8 Maths Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let the age of Baichung be x years.
The age of his father = x + 29 years,
and the age of his grandfather = x + 29 + 26 = (x + 55) years.
As per the conditions, we get
x + x + 29 + x + 55 = 135
⇒ 3x + 84 = 135
⇒ 3x = 135 – 84 (transposing 84 to RHS)
⇒ 3x = 51
⇒ x = 51 ÷ 3 (transposing 3 to RHS)
⇒ x = 17
Hence Baichung’s age = 17 years
Baichung’s father’s age = 17 + 29 = 46 years,
and grand father’s age = 46 + 26 = 72 years.

Ex 2.2 Class 8 Maths Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let the present age of Ravi be x years.
After 15 years, his age will be = (x + 15) years
As per the conditions, we get
⇒ x + 15 = 4x
⇒ 15 = 4x – x (transposing x to RHS)
⇒ 15 = 3x
⇒ 15 ÷ 3 = x (transposing 3 to LHS)
⇒ x = 5
Hence, the present age of Ravi = 5 years.

Ex 2.2 Class 8 Maths Question 13.
A rational number is such that when you multiply it by $\frac { 5 }{ 2 }$ and add $\frac { 2 }{ 3 }$ to the product, you get $\frac { -7 }{ 12 }$. What is the number?
Solution:
Let the required rational number be x.
As per the condition, we get

Hence, the required rational number is $\frac { -1 }{ 2 }$

Ex 2.2 Class 8 Maths Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of ₹ 100, ₹ 50 and ₹ 10 notes be 2x, 3x and 5x respectively.
Converting all the denominations into rupees, we have
2x × 100, 3x × 50 and 5x × 10 i.e. 200x, 150x and 50x
As per the conditions, we have
200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000
⇒ x = 4,00,000 ÷ 400 (transposing 400 to RHS)
⇒ x = 1,000
Hence, the required number of notes of
₹ 100 notes = 2 × 1000 = 2000
₹ 50 notes = 3 × 1000 = 3000
and ₹ 10 notes = 5 × 1000 = 5000

Ex 2.2 Class 8 Maths Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x.
Number of ₹ 2 coins = 3x
Total number of coins = 160
Number of ₹ 1 coin = 160 – (x + 3x) = 160 – 4x
Converting the number of coins into rupees, we have
x coins of ₹ 5 amount to ₹ 5x
3x coins of ₹ 2 amount to ₹ 3x × 2 = ₹ 6x
and (160 – 4x) coins of ₹ 1 amount to ₹ 1 × (160 – 4x) = ₹ (160 – 4x)
As per the conditions, we have
5x + 6x + 160 – 4x = 300
⇒ 7x + 160 = 300
⇒ 7x = 300 – 160 (transposing 160 to RHS)
⇒ 7x = 140
⇒ x = 140 ÷ 7 (transposing 7 to RHS)
⇒ x = 20
Thus, number of ₹ 5 coins = 20
Number of ₹ 2 coins = 3 × 20 = 60
and Number of ₹ 1 coins = 160 – 4 × 20 = 160 – 80 = 80

Ex 2.2 Class 8 Maths Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners = x
Number of participants who does not win the prize = (63 – x)
Amount got by winners = ₹ 100 × x = ₹ 100x
Amount got by loosers = ₹ (63 – x) × 25 = ₹ (1575 – 25x)
As per the conditions, we get
100x + 1575 – 25x = 3000
⇒ 75x + 1575 = 3000
⇒ 75x = 3000 – 1575 (transposing 1575 to RHS)
⇒ 75x = 1425
⇒ x = 1425 ÷ 75 (Transposing 75 to RHS)

⇒ x = 19
Thus, the number of winners = 19

# Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.3 (Exercises and Solutions)

## NCERT Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.3

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.
Ex 2.3 Class 8 Maths Question 1.
3x = 2x + 18
Solution:
We have 3x = 2x + 18
⇒ 3x – 2x = 18 (Transposing 2x to LHS)
⇒ x = 18
Hence, x = 18 is the required solution.
Check: 3x = 2x + 18
Putting x = 18, we have
LHS = 3 × 18 = 54
RHS = 2 × 18 + 18 = 36 + 18 = 54
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 2.
5t – 3 = 3t – 5
Solution:
We have 5t – 3 = 3t – 5
⇒ 5t – 3t – 3 = -5 (Transposing 3t to LHS)
⇒ 2t = -5 + 3 (Transposing -3 to RHS)
⇒ 2t = -2
⇒ t = -2 ÷ 2
⇒ t = -1
Hence t = -1 is the required solution.
Check: 5t – 3 = 3t – 5
Putting t = -1, we have
LHS = 5t – 3 = 5 × (-1)-3 = -5 – 3 = -8
RHS = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 3.
5x + 9 = 5 + 3x
Solution:
We have 5x + 9 = 5 + 3x
⇒ 5x – 3x + 9 = 5 (Transposing 3x to LHS) => 2x + 9 = 5
⇒ 2x = 5 – 9 (Transposing 9 to RHS)
⇒ 2x = -4
⇒ x = -4 ÷ 2 = -2
Hence x = -2 is the required solution.
Check: 5x + 9 = 5 + 3x
Putting x = -2, we have
LHS = 5 × (-2) + 9 = -10 + 9 = -1
RHS = 5 + 3 × (-2) = 5 – 6 = -1
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 4.
4z + 3 = 6 + 2z
Solution:
We have 4z + 3 = 6 + 2z
⇒ 4z – 2z + 3 = 6 (Transposing 2z to LHS)
⇒ 2z + 3 = 6
⇒ 2z = 6 – 3 (Transposing 3 to RHS)
⇒ 2z = 3
⇒ z = $\frac { 3 }{ 2 }$
Hence z = $\frac { 3 }{ 2 }$ is the required solution.
Check: 4z + 3 = 6 + 2z
Putting z = $\frac { 3 }{ 2 }$, we have
LHS = 4z + 3 = 4 × $\frac { 3 }{ 2 }$ + 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 × $\frac { 3 }{ 2 }$ = 6 + 3 = 9
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 5.
2x – 1 = 14 – x
Solution:
We have 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)
⇒ 3x = 15
⇒ x = 15 ÷ 3 = 5
Hence x = 5 is the required solution.
Check: 2x – 1 = 14 – x
Putting x = 5
LHS we have 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
RHS = 14 – x = 14 – 5 = 9
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
We have 8x + 4 = 3(x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7 (Solving the bracket)
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4 [Transposing 3x to LHS and 4 to RHS]
⇒ 5x = 0
⇒ x = 0 ÷ 5 [Transposing 5 to RHS]
or x = 0
Thus x = 0 is the required solution.
Check: 8x + 4 = 3(x – 1) + 7
Putting x = 0, we have
8 × 0 + 4 = 3(0 – 1) + 7
⇒ 0 + 4 = -3 + 7
⇒ 4 = 4
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 7.
x = $\frac { 4 }{ 5 }$ (x + 10)
Solution:
We have x = $\frac { 4 }{ 5 }$ (x + 10)
⇒ 5 × x = 4(x + 10) (Transposing 5 to LHS)
⇒ 5x = 4x + 40 (Solving the bracket)
⇒ 5x – 4x = 40 (Transposing 4x to LHS)
⇒ x = 40
Thus x = 40 is the required solution.
Check: x = $\frac { 4 }{ 5 }$ (x + 10)
Putting x = 40, we have
40 = $\frac { 4 }{ 5 }$ (40 + 10)
⇒ 40 = $\frac { 4 }{ 5 }$ × 50
⇒ 40 = 4 × 10
⇒ 40 = 40
LHS = RHS
Hence verified.

Ex 2.3 Class 8 Maths Question 8.
$\frac { 2x }{ 3 }$ + 1 = $\frac { 7x }{ 15 }$ + 3
Solution:
We have $\frac { 2x }{ 3 }$ + 1 = $\frac { 7x }{ 15 }$ + 3
15($\frac { 2x }{ 3 }$ + 1) = 15($\frac { 7x }{ 15 }$ + 3)
LCM of 3 and 15 is 15
$\frac { 2x }{ 3 }$ × 15 + 1 × 15 = $\frac { 7x }{ 15 }$ × 15 + 3 × 15 [Multiplying both sides by 15]
⇒ 2x × 5 + 15 = 7x + 45
⇒ 10x + 15 = 7x + 45
⇒ 10x – 7x = 45 – 15 (Transposing 7x to LHS and 15 to RHS)
⇒ 3x = 30
⇒ x = 30 ÷ 3 = 10 (Transposing 3 to RHS)
Thus the required solution is x = 10

Ex 2.3 Class 8 Maths Question 9.
2y + $\frac { 5 }{ 3 }$ = $\frac { 26 }{ 3 }$ – y
Solution:

Ex 2.3 Class 8 Maths Question 10.
3m = 5m – $\frac { 8 }{ 5 }$
Solution:
We have

# Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.4 (Exercises and Solutions)

## NCERT Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.4 Solutions

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4

Ex 2.4 Class 8 Maths Question 1.
Amina thinks of a number and subtracts $\frac { 5 }{ 2 }$ from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the required number be x.
Condition I: x – $\frac { 5 }{ 2 }$
Condition II: 8 × (x – $\frac { 5 }{ 2 }$)
Condition III: 8 × (x – $\frac { 5 }{ 2 }$) = 3x
⇒ 8x – $\frac { 5 }{ 2 }$ × 8 = 3x (Solving the bracket)
⇒ 8x – 20 = 3x
⇒ 8x – 3x = 20 (Transposing 3x to LHS and 20 to RHS)
⇒ 5x = 20
⇒ x = 20 ÷ 5 = 4 (Transposing 5 to RHS)
Thus, x = 4 is the required number.

Ex 2.4 Class 8 Maths Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other number. What are the numbers?
Solution:
Let the positive number be x.
Other number = 5x
Condition I: x + 21 and 5x + 21
Condition II: 5x + 21 = 2 (x + 21)
⇒ 5x + 21 = 2x + 42 (Solving the bracket)
⇒ 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)
⇒ 3x = 21
⇒ x = 21 ÷ 3 = 7 (Transposing 3 to RHS)
Thus, the required numbers are 7 and 7 × 5 = 35.

Ex 2.4 Class 8 Maths Question 3.
Sum of the digits of a two digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution:
Let unit place digit be x.
Ten’s place digit = 9 – x
Original number = x + 10(9 – x)
Condition I: 10x + (9 – x) (Interchanging the digits)
Condition II: New number = original number + 27
⇒ 10x + (9 – x) = x + 10(9 – x) + 27
⇒ 10x + 9 – x = x + 90 – 10x + 27 (solving the brackets)
⇒ 9x + 9 = -9x + 117 (Transposing 9x to LHS and 9 to RHS)
⇒ 9x + 9x = 117 – 9
⇒ 18x = 108
⇒ x = 108 ÷ 18 (Transposing 18 to RHS)
⇒ x = 6
Unit place digit = 6
Ten’s place digit = 9 – 6 = 3
Thus, the required number = 6 + 3 × 10 = 6 + 30 = 36

Ex 2.4 Class 8 Maths Question 4.
One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution:
Let unit place digit be x.
Ten’s place digit = 3x
Original number = x + 3x × 10 = x + 30x = 31x
Condition I: 10x + 3x = 13x (interchanging the digits)
Condition II: New number + original number = 88
13x + 31x = 88
⇒ 44x = 88
⇒ x = 88 ÷ 44 (Transposing 44 to RHS)
⇒ x = 2
Thus, the original number = 31x = 31 × 2 = 62
Hence the required number = 62

Ex 2.4 Class 8 Maths Question 5.
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution:
Let Shobo’s present age be x years.
Shobo’s mother’s age = 6x years.
After 5 years Shobo’s age will be (x + 5) years.
As per the condition, we have
x + 5 = $\frac { 1 }{ 3 }$ × 6x
⇒ x + 5 = 2x
⇒ 5 = 2x – x (Transposing x to RHS)
⇒ 5 = x
Hence Shobo’s present age = 5 years
and Shobo’s mother’s present age 6x = 6 × 5 = 30 years.

Ex 2.4 Class 8 Maths Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11 : 4. At the rate of ₹ 100 per metre, it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the length and breadth of the plot be 11x m and 4x m respectively.
Fencing all around = perimeter of the rectangular plot
Perimeter of the plot = $\frac { 75000 }{ 100 }$ = 750 m
2(l + b) = 750
⇒ 2(11x + 4x) = 750
⇒ 2(15x) = 750
⇒ 30x = 750
⇒ x = 750 ÷ 30 = 25
length = 11 × 25 m = 275 m
and breadth = 4 × 25 m = 100 m

Ex 2.4 Class 8 Maths Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trousers material that costs him ₹ 90 per metre. For every 3 metres of the shirt material, he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?
Solution:
Ratio of shirt material bought to the trouser material bought = 3 : 2
Let the shirt material bought = 3x m
and trouser material bought = 2x m
Cost of shirt material = 50 × 3x = ₹ 150x
Cost of trouser material = 90 × 2x = ₹ 180x

As per the conditions, we have
168x + 198x = 36,600
⇒ 366x = 36,600
⇒ x = 36600 ÷ 366 = 100
Length of trouser material bought = 2 × 100 = 200 m.

Ex 2.4 Class 8 Maths Question 8.
Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution:
Let the number of deer be x.
As per the condition, we have
$\frac { x }{ 2 }$ deer are grazing in the field.

⇒ 4x + 3x + 72 = 8x
⇒ 7x + 72 = 8x
⇒ 72 = 8x – 7x (Transposing 7x to RHS)
⇒ x = 12
Hence, the required number of deer = 72.

Ex 2.4 Class 8 Maths Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let the present age of granddaughter = x years.
the present age of grandfather = 10x years.
As per the conditions, we have
10x – x = 54
⇒ 9x = 54
⇒ x = 54 ÷ 9 = 6 [Transposing 9 to RHS]
Hence, the present age of the granddaughter = 6 years
and the present age of grandfather = 6 × 10 = 60 years.

Ex 2.4 Class 8 Maths Question 10.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution:
Let the present age of the son be x years.
Present age of Aman = 3x years
10 years ago, the son’s age was = (x – 10) years
10 years ago, the father’s age was = (3x – 10) years
As per the conditions, we have
5(x – 10) = 3x – 10
⇒ 5x – 50 = 3x – 10
⇒ 5x – 3x = 50 – 10(Transposing 3x to LHS and 50 to RHS)
⇒ 2x = 40
⇒ x = 40 ÷ 2 = 20
Hence, the son’s age = 20 years.
and the age of Aman = 20 × 3 = 60 years.

# Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.5 (Exercises and Solutions)

## NCERT Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.5 Solutions

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5

Solve the following linear equations.
Ex 2.5 Class 8 Maths Question 1.

Solution:

⇒ 30x – 12 = 20x + 15
⇒ 30x – 20x = 15 + 12 (Transposing 20x to LHS and 12 to RHS)
⇒ 10x = 27
⇒ x = $\frac { 27 }{ 10 }$

Ex 2.5 Class 8 Maths Question 2.

Solution:
LCM of 2, 4 and 6 = 12

(Multiplying both sides by 12)
⇒ 6n – 9n + 10n = 252
⇒ 7n = 252
⇒ n = 252 ÷ 7
⇒ n = 36

Ex 2.5 Class 8 Maths Question 3.

Solution:

⇒ -10x + 42 = 17 – 15x
⇒ -10x + 15x = 17 – 42 [Transposing 15x to LHS and 42 to RHS]
⇒ 5x = -25
⇒ x = -25 ÷ 5 [Transposing 5 to RHS]
⇒ x = -5

Ex 2.5 Class 8 Maths Question 4.

Solution:

⇒ (x – 5) × 5 = (x – 3) × 3
⇒ 5x – 25 = 3x – 9 (Solving the brackets)
⇒ 5x – 3x = 25 – 9 (Transposing 3x to LHS and 25 to RHS)
⇒ 2x = 16
⇒ x = 16 ÷ 2 = 8 (Transposing 2 to RHS)
⇒ x = 8

Ex 2.5 Class 8 Maths Question 5.

Solution:

⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t (Solving the brackets)
⇒ t – 18 = 8 – 12t
⇒ t + 12t = 8 + 18 (Transposing 12t to LHS and 18 to RHS)
⇒ 13t = 26
⇒ t = 2 (Transposing 13 to RHS)
Hence t = 2 is the required solution.

Ex 2.5 Class 8 Maths Question 6.

Solution.

⇒ 6m – (m – 1) × 3 = 6 – (m – 2) × 2
⇒ 6m – 3m + 3 = 6 – 2m + 4 (Solving the brackets)
⇒ 3m + 3 = 10 – 2m
⇒ 3m + 2m = 10 – 3 (Transposing 2m to LHS and 3 to RHS)
⇒ 5m = 7
⇒ m = $\frac { 7 }{ 5 }$ (Transposing 5 to RHS)

Simplify and solve the following linear equations.
Ex 2.5 Class 8 Maths Question 7.
3(t – 3) = 5(21 + 1)
Solution:
We have
3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5 (Solving the brackets)
⇒ 3t – 10t = 9 + 5 (Transposing 10t to LHS and 9 to RHS)
⇒ -7t = 14
⇒ t = -2(Transposing -7 to RHS)
Hence, t = -2 is the required solution.

Ex 2.5 Class 8 Maths Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
We have 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
⇒ 15y – 60 – 2y + 18 + 5y + 30 = 0 (Solving the brackets)
⇒ 8y – 12 = 0
⇒ 8y = 12 (Transposing 12 to RHS)
⇒ y = $\frac { 2 }{ 3 }$
Hence, y = $\frac { 2 }{ 3 }$ is the required solution.

Ex 2.5 Class 8 Maths Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
We have
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17 (Solving the bracket)
⇒ -3z + 1 = 32z – 69
⇒ -3z – 32z = – 69 – 1 (Transposing 322 to LHS and 1 to RHS)
⇒ -35z = -70
⇒ z = 2
Hence, z = 2 is the required solution.

Ex 2.5 Class 8 Maths Question 10.
0.25(4f – 3) = 0.05(10f – 9)
Solution:
We have
0.25(4f – 3) = 0.05(10f – 9)
⇒ 0.25 × 4f – 3 × 0.25 = 0.05 × 10f – 9 × 0.05 (Solving the brackets)
⇒ 1.00f – 0.75 = 0.5f – 0.45
⇒ f – 0.5f = -0.45 + 0.75 (Transposing 0.5 to LHS and 0.75 to RHS)
⇒ 0.5f = 0.30
⇒ f = 0.6
Hence, f = 0.6 is the required solution.

# Class 8 Maths Chapter 2: Linear Equations in One Variable Ex 2.6 (Exercises and Solutions)

## NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Solutions

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6

Solve the following equations.
Ex 2.6 Class 8 Maths Question 1.
$\frac { 8x-3 }{ 3x } =2$
Solution:
We have $\frac { 8x-3 }{ 3x } =2$
⇒ $\frac { 8x-3 }{ 3x }$ = $\frac { 2 }{ 1 }$
⇒ 8x – 3 = 2 × 3x (Cross-multiplication)
⇒ 8x – 3 = 6x
⇒ 8x – 6x = 3 (Transposing 6x to LHS and 3 to RHS)
⇒ 2x = 3
⇒ x = $\frac { 3 }{ 2 }$

Ex 2.6 Class 8 Maths Question 2.
$\frac { 9x }{ 7-6x }$ = 15
Solution:
we have $\frac { 9x }{ 7-6x }$ = 15
⇒ $\frac { 9x }{ 7-6x }$ = $\frac { 15 }{ 1 }$
⇒ 9x = 15(7 – 6x) (Cross-multiplication)
⇒ 9x = 105 – 90x (Solving the bracket)
⇒ 9x + 90x = 105 (Transposing 90x to LHS)
⇒ 99x = 105
⇒ x = $\frac { 105 }{ 99 }$
⇒ x = $\frac { 35 }{ 33 }$

Ex 2.6 Class 8 Maths Question 3.
$\frac { z }{ z+15 } =\frac { 4 }{ 9 }$
Solution:
We have $\frac { z }{ z+15 } =\frac { 4 }{ 9 }$
⇒ 9z = 4 (z + 15) (Cross-multiplication)
⇒ 9z = 4z + 60 (Solving the bracket)
⇒ 9z – 42 = 60
⇒ 5z = 60
⇒ z = 12

Ex 2.6 Class 8 Maths Question 4.
$\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }$
Solution:
we have $\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }$
⇒ 5(3y + 4) = -2(2 – 6y) (Cross-multiplication)
⇒ 15y + 20 = -4 + 12y (Solving the bracket)
⇒ 15y – 12y = -4 – 20 (Transposing 12y to LHS and 20 to RHS)
⇒ 3y = -24 (Transposing 3 to RHS) -24
⇒ y = -8

Ex 2.6 Class 8 Maths Question 5.
$\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }$
Solution:
we have $\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }$
⇒ 3(7y + 4) = -4 (y + 2) (Corss-multiplication)
⇒ 21y + 12 = -4y – 8 [Solving the bracket]
⇒ 21y + 4y = -12 – 8 [Transposing 4y to LHS and 12 to RHS]
⇒ 25y = -20 [Transposing 25 to RHS]
⇒ y = $\frac { -4 }{ 5 }$

Ex 2.6 Class 8 Maths Question 6.
The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages.
Solution:
Let the present ages of Hari and Harry be 5x years and 7x years respectively.
After 4 years Hari’s age will be (5x + 4) years and Harry’s age will be (7x + 4) years.
As per the conditions, we have
$\frac { 5x+4 }{ 7x+4 } =\frac { 3 }{ 4 }$
⇒ 4(5x + 4) = 3(7x + 4) (Cross-multiplication)
⇒ 20x + 16 = 21x + 12 (Solving the bracket)
⇒ 20x – 21x = 12 – 16 (Transposing 21x to LHS and 16 to RHS)
⇒ -x = -4
⇒ x = 4
Hence the present ages of Hari and Harry are 5 × 4 = 20years and 7 × 4 = 28years respectively.

Ex 2.6 Class 8 Maths Question 7.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is $\frac { 3 }{ 2 }$. Find the rational number.
Solution:
Let the numerator of the rational number be x.
Denominator = (x + 8)
As per the conditions, we have

⇒ 2(x + 17) = 3(x + 7) (Cross-multiplication)
⇒ 2x + 34 = 3x + 21 (Solving the bracket)
⇒ 2x – 3x = 21 – 34 (Transposing 3x to LHS and 34 to RHS)
⇒ -x = -13
⇒ x = 13
Thus, numerator = 13
and denominator = 13 + 8 = 21
Hence the rational number is $\frac { 13 }{ 21 }$.

# Extra Questions

## Linear Equations in One Variable Extra Questions

Class 8 Maths Chapter 2 Linear Equations in One Variable

Question 2.
Check whether the linear equation 3x + 5 = 11 is true for x = 2.
Solution:
Given that 3x + 5 = 11
For x = 2, we get
LHS = 3 × 2 + 5 = 6 + 5 = 11
LHS = RHS = 11
Hence, the given equation is true for x = 2

Question 3.
Form a linear equation from the given statement: ‘When 5 is added to twice a number, it gives 11.’
Solution:
As per the given statement we have
2x + 5 = 11 which is the required linear equation.

Question 4.
If x = a, then which of the following is not always true for an integer k. (NCERT Exemplar)
(a) kx = ak
(b) $\frac { x }{ k }$ = $\frac { a }{ k }$
(c) x – k = a – k
(d) x + k = a + k
Solution:

Question 5.
Solve the following linear equations:
(a) 4x + 5 = 9
(b) x + $\frac { 3 }{ 2 }$ = 2x
Solution:
(a) We have 4x + 5 = 9
⇒ 4x = 9 – 5 (Transposing 5 to RHS)
⇒ 4x = 4
⇒ x = 1 (Transposing 4 to RHS)
(b) We have x + $\frac { 3 }{ 2 }$ = 2x
⇒ $\frac { 3 }{ 2 }$ = 2x – x
⇒ x = $\frac { 3 }{ 2 }$

Question 6.
Solve the given equation 3$\frac { 1 }{ x }$ × 5$\frac { 1 }{ 4 }$ = 17$\frac { 1 }{ 2 }$
Solution:
We have 3$\frac { 1 }{ x }$ × 5$\frac { 1 }{ 4 }$ = 17$\frac { 1 }{ 2 }$

Question 7.
Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.
Solution:
We have 4.4x – 3.8 = 5
Putting x = 2, we have
4.4 × 2 – 3.8 = 5
⇒ 8.8 – 3.8 = 5
⇒ 5 = 5
L.H.S. = R.H.S.
Hence verified.

Question 8.

Solution:

⇒ 3x × 3 – (2x + 5) × 4 = 5 × 6
⇒ 9x – 8x – 20 = 30 (Solving the bracket)
⇒ x – 20 = 30
⇒ x = 30 + 20 (Transposing 20 to RHS)
⇒ x = 50
Hence x = 50 is the required solution.

Question 9.
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Solution:
Let the angles of a given triangle be 2x°, 3x° and 4x°.
2x + 3x + 4x = 180 (∵ Sum of the angles of a triangle is 180°)
⇒ 9x = 180
⇒ x = 20 (Transposing 9 to RHS)
Angles of the given triangles are
2 × 20 = 40°
3 × 20 = 60°
4 × 20 = 80°

Question 10.
The sum of two numbers is 11 and their difference is 5. Find the numbers.
Solution:
Let one of the two numbers be x.
Other number = 11 – x.
As per the conditions, we have
x – (11 – x) = 5
⇒ x – 11 + x = 5 (Solving the bracket)
⇒ 2x – 11 = 5
⇒ 2x = 5 + 11 (Transposing 11 to RHS)
⇒ 2x = 16
⇒ x = 8
Hence the required numbers are 8 and 11 – 8 = 3.

Question 11.
If the sum of two consecutive numbers is 11, find the numbers.
Solution:
Let the two consecutive numbers be x and x + 1.
As per the conditions, we have
x + x + 1 = 11
⇒ 2x + 1 = 11
⇒ 2x = 11 – 1 (Transposing 1 to RHS)
⇒ 2x = 10
x = 5
Hence, the required numbers are 5 and 5 + 1 = 6.

### Linear Equations in One Variable Class 8 Extra Questions Short Answer Type

Question 12.
The breadth of a rectangular garden is $\frac { 2 }{ 3 }$ of its length. If its perimeter is 40 m, find its dimensions.
Solution:
Let the length of the garden be x m
its breadth = $\frac { 2 }{ 3 }$ × m.
Perimeter = 2 [length + breadth]

Question 13.
The difference between two positive numbers is 40 and the ratio of these integers is 1 : 3. Find the integers.
Solution:
Let one integer be x.
Other integer = x – 40
As per the conditions, we have
$\frac { x-40 }{ x }$ = $\frac { 1 }{ 3 }$
⇒ 3(x – 40) = x
⇒ 3x – 120 = x
⇒ 3x – x = 120
⇒ 2x = 120
⇒ x = 2
Hence the integers are 60 and 60 – 40 = 20.

Question 14.
Solve for x:

Solution:

Question 15.
The sum of a two-digit number and the number obtained by reversing its digits is 121. Find the number if it’s unit place digit is 5.
Solution:
Unit place digit is given as 5
Let x be the tens place digit
Number formed = 5 + 10x
Number obtained by reversing the digits = 5 × 10 + x = 50 + x
As per the conditions, we have
5 + 10x + 50 + x = 121
⇒ 11x + 55 = 121
⇒ 11x = 121 – 55 (Transposing 55 to RHS)
⇒ 11x = 66
⇒ x = 6
Thus, the tens place digit = 6
Hence the required number = 5 + 6 × 10 = 5 + 60 = 65

### Linear Equations in One Variable Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 16.
If the length of the rectangle is increased by 40% and its breadth is decreased by 40%, what will be the percentage change in its perimeter?
Solution:
Let the length of the rectangle be x m and its breadth be y m
Perimeter = 2(x + y)
Now the length of the rectangle becomes after a 40% increase

Question 17.
A fruit seller buys some oranges at the rate of ₹ 5 per orange. He also buys an equal number of bananas at the rate of ₹ 2 per banana. He makes a profit of 20% on oranges and a profit of 15% on bananas. In the end, he sold all the fruits. If he earned a profit of ₹ 390, find the number of oranges.
Solution:
Let the number of oranges bought by him be x and also the number of bananas be x.
Cost of x oranges at the rate of ₹ 5 per orange = ₹ 5x
Cost of x bananas at the rate of ₹ 2 per banana = ₹ 2x

Question 18.
A steamer goes downstream from one point to another in 7 hours. It covers the same distance upstream in 8 hours. If the speed of stream be 2 km/h, find the speed of the steamer in still water and the distance between the ports. (NCERT Exemplar)
Solution:
Let speed of steamer in still water = x km/h
Speed of stream = 2 km/h
Speed downstream = (x + 2) km/h
Speed upstream = (x – 2) km/h
Distance covered in 7 hours while downstream = 7(x + 2)
Distance covered in 8 hours while upstream = 8(x – 2)
According to the condition,
7(x + 2) = 8(x – 2)
⇒ 7x + 14 = 8x – 16
⇒ x = 30 km/h
Total Distance = 7(x + 2) km = 7(30 + 2) km = 7 × 32 km = 224 km.

## NCERT Solutions for Class 8 Maths

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