ncert solutions for class 8 maths chapter 2 exercise 2.2

ncert solutions for class 8 maths chapter 2 exercise 2.2 involve complete answers for each question in the exercise 2.1. The solutions provide students a strategic methods to prepare for their exam. ncert solutions for class 8 maths chapter 2 exercise 2.2 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.ncert solutions for class 8 maths chapter 2 exercise 2.2 prepared by our subject matter experts in very delicate, easy and creative way.

Question 1:

If you subtract \(\dfrac{1}{2}\) from a number and multiply the result by \(\dfrac{1}{2}\), you get \(\dfrac{1}{8}\). What is the number?

Answer:

Let the number be \(x\).

According to the question,

\((x - \dfrac{1}{2}) × \dfrac{1}{2} = \dfrac{1}{8}\)

On multiplying both sides by 2, we obtain

\((x - \dfrac{1}{2}) × \dfrac{1}{2} × 2 = \dfrac{1}{8} × 2\)

\((x - \dfrac{1}{2})= \dfrac{1}{4}\)

\(x = \dfrac{1}{4} + \dfrac{1}{2}\)

\(=\dfrac{3}{4}\)

Therefore, the number is \(=\dfrac{3}{4}\).

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?

Answer:

Let the breadth be \(x\) m. The length will be \((2x + 2)\) m.

Perimeter of swimming pool \(= 2(l + b) = 154\) m

\(2(2x + 2 + x) = 154\)

\(2(3x + 2) = 154\)

Dividing both sides by \(2\), we obtain

\(\dfrac{2(3x+2)}{2} = \dfrac{154}{2}\)

\(3x + 2 = 77\)

\(3x = 77 − 2\)

\(3x = 75\)

\(x=\dfrac{75}{3}\)

\(x = 25\)

\(2x + 2 = 2 × 25 + 2 = 52\)

Hence, the breadth and length of the pool are \(25\) m and \(52\) m respectively.

Question 3:

The base of an isosceles triangle is ⁴/₃ cm. The perimeter of the triangle is 4 ²/₁₅ cm.What is the length of either of the remaining equal sides?

Answer:


Let the length of equal sides be \(x\) cm.

Perimeter \(= x  + x  + \text{Base} = 4\dfrac{2}{15}\) cm

\(2x + \dfrac{4}{3} = \dfrac{62}{15}\)


\(2x = \dfrac{62}{15 }- \dfrac{4}{3}\)

\(2x = \dfrac{62 - ( 4 × 5 ) }{15} \)

\(2x = \dfrac{42}{15 }\)

\(x= \dfrac{14}{5\times 2}\)



Therefore, the length of equal sides is \(1\dfrac{2}{5}\) cm.

Question 4:

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

Answer:

Let one number be \(x\). Therefore, the other number will be \(x + 15\).

According to the question,

\(x + x + 15 = 95\)

\(2x + 15 = 95\)

\(2x = 95 − 15\)

\(2x = 80\)

\(x = \dfrac{80}{2}\)

\(x = 40\)

\(x + 15 = 40 + 15 = 55\)

Hence, the numbers are \(40\) and \(55\).

Question 5:

Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Answer:

Let the common ratio between these numbers be \(x\). Therefore, the numbers will be \(5x\) and \(3x\)
respectively.

Difference between these numbers = \(18\)

\(5x − 3x = 18\)

\(2x = 18\)

\(x = \dfrac{18}{2}\)

\(x = 9\)

First number \(= 5x = 5 × 9 = 45\)

Second number \(= 3x = 3 × 9 = 27\)

Question 6:

Three consecutive integers add up to 51. What are these integers?

Answer:

Let three consecutive integers be \(x, x + 1\), and \(x + 2\).

Sum of these numbers \(= x+ x + 1 + x + 2 = 51\)

\(3x + 3 = 51\)

\(3x = 51 − 3\)

\(3x = 48\)

\(x = \dfrac{48}{3}\)

\(x = 16\)

\(x + 1 = 17\)

\(x + 2 = 18\)

Hence, the consecutive integers are \(16, 17\), and \(18\).

Question 7:

The sum of three consecutive multiples of 8 is 888. Find the multiples.

Answer:

Let the three consecutive multiples of \(8\) be \(8x, 8(x + 1), 8(x + 2)\).

Sum of these numbers \(= 8x + 8(x + 1) + 8(x + 2) = 888\)

\(8(x + x + 1 + x + 2) = 888\)

\(8(3x + 3) = 888\)

\(3x + 3 = 111\)

\(3x = 111 − 3\)

\(3x = 108\)

\(x = \dfrac{108}{3}\)

\(x = 36\)

First multiple \(= 8x = 8 × 36 = 288\)

Second multiple \(= 8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296\)

Third multiple \(= 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304\)

Hence, the required numbers are \(288, 296\), and \(304\).

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.

Answer:

Let three consecutive integers be x, x + 1, 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

On dividing both sides by 9, we obtain

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Hence, the numbers are 7, 8, and 9.

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?

Answer:

Let common ratio between Rahul’s age and Haroon’s age be x.

Therefore, age of Rahul and Haroon will be 5x years and 7x years respectively. After 4 years,

The age of Rahul and Haroon will be (5x + 4) years and (7x + 4) years respectively.

According to the given question,

After 4 years, the sum of the ages of Rahul and Haroon is 56 years.

∴ (5x + 4 + 7x + 4) = 56

12x + 8 = 56

12x = 56 − 8

12x = 48

x = 4

Rahul’s age = 5x years = (5 × 4) years = 20 years

Haroon’s age = 7x years = (7 × 4) years = 28 years

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?

Answer:

Let the common ratio between the number of boys and numbers of girls be x.

Number of boys = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

∴ 7x = 5x + 8

On transposing 5x to L.H.S, we obtain

7x − 5x = 8

2x = 8

On dividing both sides by 2, we obtain

x = 4

Number of boys = 7x = 7 × 4 = 28

Number of girls = 5x = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48 students

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?

Answer:

Let Baichung’s father’s age be x years.

Therefore, Baichung’s age and Baichung’s grandfather’s age will be (x − 29) years and (x + 26) years respectively.

ATQ,

the sum of the ages of these 3 people is 135 years.

∴ x + x − 29 + x + 26 = 135

3x − 3 = 135

On transposing 3 to R.H.S, we obtain

3x = 135 + 3

3x = 138

On dividing both sides by 3, we obtain

x = 46

Baichung’s father’s age = x years = 46 years

Baichung’s age = (x − 29) years = (46 − 29) years = 17 years

Baichung’s grandfather’s age = (x + 26) years = (46 + 26) years = 72 years

Question 12:

Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer:

Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age = 4 × His present age

x + 15 = 4x

15 = 4x − x

15 = 3x

On dividing both sides by 3, we obtain

5 = x

Hence, Ravi’s present age = 5 years

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?

Answer:

Let the number be \(x\).

According to the given question,

\(\dfrac{5x}{2} +\dfrac{2}{3} = \dfrac{-7}{12}\)

\(\dfrac{5x}{2} = \dfrac{-7}{12} - \dfrac{2}{3}\)

\(\dfrac{5x}{2} =-\dfrac{7-(2×4)}{12} = \dfrac{-15}{12}\)

\(x=\dfrac{-15}{12} \times \dfrac{2}{5}\)

\(=\dfrac{-1}{2}\)

Hence, the rational number is \(\dfrac{-1}{2}\).

Question 14:

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Answer:

Let the common ratio between the numbers of notes of different denominations be x.

Therefore, numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be 2x, 3x, and 5x respectively.

Amount of Rs 100 notes = Rs (100 × 2x) = Rs 200x

Amount of Rs 50 notes = Rs (50 × 3x)= Rs 150x

Amount of Rs 10 notes = Rs (10 × 5x) = Rs 50x

It is given that total amount is Rs 400000.

∴ 200x + 150x + 50x = 400000

⇒ 400x = 400000

On dividing both sides by 400, we obtain

x = 1000

Number of Rs 100 notes = 2x = 2 × 1000 = 2000

Number of Rs 50 notes = 3x = 3 × 1000 = 3000

Number of Rs 10 notes = 5x = 5 × 1000 = 5000

Question 15:

I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160.  How many coins of each denomination are with me?

Answer:

Let the number of Rs 5 coins be x.

Number of Rs 2 coins = 3 × Number of Rs 5 coins = 3x

Number of Re 1 coins = 160 − (Number of coins of Rs 5 and of Rs 2)

= 160 − (3x + x) = 160 − 4x

Amount of Re 1 coins = Rs [1 × (160 − 4x)] = Rs (160 − 4x)

Amount of Rs 2 coins = Rs (2 × 3x)= Rs 6x

Amount of Rs 5 coins = Rs (5 × x) = Rs 5x

It is given that the total amount is Rs 300.

∴ 160 − 4x + 6x + 5x = 300

160 + 7x = 300

On transposing 160 to R.H.S, we obtain

7x = 300 − 160

7x = 140

On dividing both sides by 7, we obtain

x = 20

Number of Re 1 coins = 160 − 4x = 160 − 4 × 20 = 160 − 80 = 80

Number of Rs 2 coins = 3x = 3 × 20 = 60

Number of Rs 5 coins = x = 20

Question 16:

The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.

Answer:

Let the number of winners be x.

Therefore, the number of participants who did not win will be 63 − x.

Amount given to the winners = Rs (100 × x) = Rs 100x

Amount given to the participants who did not win = Rs [25(63 − x)]

= Rs (1575 − 25x)

ATQ,

100x + 1575 − 25x = 3000

On transposing 1575 to R.H.S, we obtain

75x = 3000 − 1575

75x = 1425

On dividing both sides by 75, we obtain

x = 19

Hence, number of winners = 19