NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

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

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

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2.

### Ex 2.2 Class 8 Maths Question 1.

If you subtract ½ from a number and multiply the result by ½, you get 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.
Then the length of the pool = (2x + 2) m
Perimeter of the pool = 154 m
We know that, perimeter of a 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 of the pool = 25 m
and the length of the pool = 2 × 25 + 2 = 50 + 2 = 52 m.

### Ex 2.2 Class 8 Maths Question 3.

The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 42/15 cm. What is the length of either of the remaining equal sides?

Solution:
Let the length of each of equal sides of the isosceles 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 is 95. If one exceeds the other by 15, find the numbers.

Solution:
Let one number be x.
Then the other number = x + 15
According to the question, we have
x + (x + 15) = 95
x + x + 15 = 95
2x + 15 = 95
2x = 95 – 15            [Transposing 15 from (+) to (-)]
2x = 80
x = 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.
According to the question, we have
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.
According to the question, we have
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 x, x + 8 and x + 16.
According to the question, we have
x + (x + 8) + (x + 16) = 888
x + x + 8 + x + 16 = 888
3x + 24 = 888
3x = 888 – 24            (Transposing 24 to RHS)
3x = 864
x = 864 ÷ 3                 (Transposing 3 to RHS)
x = 288
Thus, the required multiples of 8 are
288, 288 + 8 = 296 and 288 + 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.
According to the question, 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               (Transposing 9 to RHS)
x = 7

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 be 5x years and 7x years respectively.
Four years later, the age of Rahul will be (5x + 4) years.

Four years later, the age of Haroon will be (7x + 4) years.
According to the question, we have
(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           (Transposing 12 to RHS)

x = 4
Hence, the present age of Rahul = 5 × 4 = 20 years.
and the present 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 number of girls. What is the total class strength?

Solution:
Let the number of boys be 7x and the number of girls be 5x.
According to the question, we have
7x – 5x = 8
2x = 8
x = 8 ÷ 2        (Transposing 2 to RHS)
x = 4

Thus, the total number of boys = 7 × 4 = 28
and the total number of girls = 5 × 4 = 20
Hence, the total strength of the class = 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.
Then the age of his father = (x + 29) years,
and the age of his grandfather = (x + 29) + 26 = (x + 55) years.
According to the question, we have
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 his grandfather’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, Ravi’s age will be (x + 15) years.
According to the question, we have
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 is 5 years.

### Ex 2.2 Class 8 Maths Question 13.

A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get 7/12. What is the number?

Solution:
Let the rational number be x.
According to the question, we have

Hence, the required rational number is 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 get
2x × 100, 3x × 50 and 5x × 10, i.e., 200x, 150x and 50x
According to the question, we have
200x + 150x + 50x = 4,00,000
400x = 4,00,000
x = 4,00,000 ÷ 400           (Transposing 400 to RHS)
x = 1000
Hence, the required number of notes of ₹ 100 = 2 × 1000 = 2000,
the required number of notes of ₹ 50 = 3 × 1000 = 3000
and the required number of notes of ₹ 10 = 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.
Then the number of ₹ 2 coins = 3x
Total number of coins = 160
Therefore, the number of ₹ 1 coins = 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)
According to the question, 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, the number of ₹ 5 coins = 20,
the number of ₹ 2 coins = 3 × 20 = 60
and the 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 be x.
Number of participants who does not win = (63 – x)
The total amount got by winners = ₹ 100 × x = ₹ 100x
The total amount got by participants who does not win = ₹ (63 – x) × 25

= ₹ (1575 – 25x)
According to the question, we have
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 is 19.

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