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

NCERT Solutions Class 8 Mathematics 
Chapter – 2 (Linear Equations in One Variable) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 2 Linear Equations in One Variable Exercise 2.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2: Linear Equations in One Variable

Exercise – 2.2 

1. If you subtract ½ from a number and multiply the result by \mathbf{\frac{1}{2}}, you get \mathbf{\frac{1}{8}}. What is the number?

Solution –
Let the required number be x.
According to the question,
(x – \frac{1}{2}) × \frac{1}{2} = \frac{1}{8}

\frac{x}{2} – \frac{1}{4} = \frac{1}{8}

\frac{x}{2} = \frac{1}{8} + \frac{1}{4}

\frac{x}{2} = \frac{1+2}{8}

\frac{x}{2} = \frac{3}{8}

x = \frac{3}{8} × 2

x = \frac{3}{4}

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 breadth of the pool?

Solution –
Let the breadth of the rectangle swimming pool be = x
The perimeter of the rectangular swimming pool = 154 m.
According to the question,
Length of the rectangle = 2x + 2
Perimeter = 2(length + breadth)
⇒ 2(2x + 2 + x) = 154 m
⇒ 2(3x + 2) = 154

⇒ 3x +2 = \frac{154}{2}

⇒ 3x = 77 – 2
⇒ 3x = 75

⇒ x = \frac{75}{3}
⇒ x = 25 m
Therefore, Breadth = x = 25 cm
Length = 2x + 2
= (2 × 25) + 2
= 50 + 2
= 52 m

3. The base of an isosceles triangle is \mathbf{\frac{4}{3}} cm. The perimeter of the triangle is \mathbf{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.
Base of isosceles triangle = \frac{4}{3} cm
Perimeter of triangle = 4\frac{2}{15} cm = \frac{62}{15}
According to the question,
Perimeter of the triangle = sum of the three sides

\frac{4}{3} + x + x = \frac{62}{15} cm

⇒ 2x = \frac{62}{15} – \frac{4}{3}

⇒ 2x = \frac{62-20}{15}

⇒ 2x = \frac{42}{15}

⇒ x = \frac{42}{15} × \frac{1}{2}

⇒ x = \frac{42}{30} cm

⇒ x = \frac{7}{5} cm = 1\frac{2}{5} cm
The length of either of the remaining equal sides is 1\frac{2}{5}  cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution –
Let one number be x
Other number = x + 15
According to the question
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15
⇒ 2x = 80
⇒ x = \frac{80}{2}
⇒ x = 40
First number = x = 40
And, other number = x + 15
= 40 + 15 = 55

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,
5x – 3x = 18
⇒ 2x = 18
⇒ x = \frac{18}{2}
⇒ x = 9
Thus,
The numbers are 5x = 5 × 9 = 45
And 3x = 3 × 9 = 27.

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,
x + (x+1) + (x+2) = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3
⇒ 3x = 48
⇒ x = \frac{48}{3}
⇒ x = 16
Thus, the required integers are
x = 16,
x + 1 = 17 and
x + 2 = 18

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.
According to the question,
8x + (8x + 8) + (8x + 16) = 888
⇒ 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
⇒ 24x = 888 – 24
⇒ 24x = 864
⇒ x = \frac{864}{24}
⇒ x = 36
Thus, the required multiples are
8x = 8 × 36 = 288,
8x + 8 = 8 × 36 + 8 = 288 + 8 = 296 and
8x + 16 = 8 × 36 + 16 = 288 + 16 = 304

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,
2x + 3(x+1) + 4(x+2) = 74
⇒ 2x + 3x +3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 – 11
⇒ 9x = 63
⇒ x = \frac{63}{9}
⇒ x = 7
Thus, the numbers are:
x = 7,
x + 1 = 8 and
x + 2 = 9

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 ages of Rahul and Haroon be 5x and 7x.
Four years later,
The ages of Rahul and Haroon will be (5x + 4) and (7x + 4), respectively.
According to the question,
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8
⇒ 12x = 48
⇒ x = \frac{48}{12}
⇒ x = 4
Therefore, Present age of Rahul = 5x = 5 × 4 = 20
And, present age of Haroon = 7x = 7 × 4 = 28

10. The number of boys and girls in a class is in the ratio of 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,
7x = 5x + 8
⇒ 7x – 5x = 8
⇒ 2x = 8
⇒ x = \frac{8}{2}
⇒ x = 4
Therefore, number of boys = 7 × 4 = 28
and, number of girls = 5×4 = 20
Total number of students = 20 + 28 = 48

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.
According to the question,
x + x + 29 + x + 55 = 135
⇒ 3x + 84 = 135
⇒ 3x = 135 – 84
⇒ 3x = 51
⇒ x = \frac{51}{3}
⇒ 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.

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.
After 15 years, his age will be = (x + 15) years
According to the question,
x + 15 = 4x
⇒ 4x – x = 15
⇒ 3x = 15
⇒ x = \frac{15}{3}
⇒ x = 5
Therefore, the present age of Ravi = 5 years.

13. A rational number is such that when you multiply it by \mathbf{\frac{5}{2}} and add \mathbf{\frac{2}{3}} to the product, you get \mathbf{\frac{-7}{12}}. What is the number?

Solution –
Let the rational be x.
According to the question,

x × \frac{5}{2} + \frac{2}{3} = -\frac{7}{12}

⇒ \frac{5x}{2} + \frac{2}{3}  = -\frac{7}{12}

⇒ \frac{5x}{2} = -\frac{7}{12} – \frac{2}{3}  

⇒ \frac{5x}{2} = \frac{-7-8}{12}

⇒ \frac{5x}{2} = \frac{-15}{12}

⇒ \frac{5x}{2} = \frac{-5}{4}

⇒ x = \frac{-5}{4} × \frac{2}{5}  

⇒ x = \frac{-10}{20}

⇒ x = -\frac{1}{2}

Therefore, the rational number is -\frac{1}{2}.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10, respectively. The ratio of the number 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 numbers of notes of ₹100, ₹50 and ₹10 be 2x, 3x and 5x, respectively.
Value of ₹100 = 2x × 100 = 200x
Value of ₹50 = 3x × 50 = 150x
Value of ₹10 = 5x × 10 = 50x
According to the question,
200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000

⇒ x = \frac{4,00,000}{400}

⇒ x = 1000
Numbers of ₹100 notes = 2x = 2000
Numbers of ₹50 notes = 3x = 3000
Numbers of ₹10 notes = 5x = 5000

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
Now,
Value of ₹5 coins = x × 5 = 5x
Value of ₹2 coins = 3x × 2 = 6x
Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)
According to the question,
5x + 6x + (160 – 4x) = 300
⇒ 11x + 160 – 4x = 300
⇒ 7x = 140

⇒ x = \frac{140}{7}
⇒ x = 20
Number of ₹5 coins = x = 20
Number of ₹2 coins = 3x = 60
Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80

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)
According to the question,
100x + 1575 – 25x = 3000
⇒ 75x + 1575 = 3000
⇒ 75x = 3000 – 1575
⇒ 75x = 1425

⇒ x = \frac{1425}{75}
⇒ x = 19
Therefore, the numbers of winners are 19.

 

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