NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable

Exercise 2.1  Exercise 2.2  Exercise 2.3  Exercise 2.4  Exercise 2.5  Exercise 2.6

EXERCISE 2.2

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 ?

sol :

Let the number be x . According to the question , 

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

On multiplying both sides by 2 , we obtain

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

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

On transposing ​\( \dfrac{1}{2} \)​ to R.H.S , we obtain 

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

\( =\dfrac{1+2}{4}=\dfrac{3}{4} \)

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 ?

sol :

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

On transposing 2 to R.H.S , we obtain

3x = 77 – 2

3x = 75 

On dividing both sides by 3 , we obtain 

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

x = 25

\( \begin{align*}&=2x+2\\&=2\times{25}+2\\&=52\end{align*} \)

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

 

Question 3 :

The base of an isosceles triangle is ​\( \dfrac{4}{3} \)​ cm . The perimeter of the triangle is ​\( 4\dfrac{2}{15} \)​ cm . What is the length of either of the remaining equal sides ?

Sol :

Let the length of equal sides be x cm .

Perimeter = x cm +x cm + Base

\( \begin{align*}&=4\dfrac{2}{15}\\&=2x+\dfrac{4}{3}=\dfrac{62}{15}\end{align*} \)

On transposing ​\( \dfrac{4}{3} \)​ to R.H.S , we obtain

\( \begin{align*}&2x=\dfrac{62}{15}-\dfrac{4}{3}\\&2x=\dfrac{62-4\times5}{15}\\&2x=\dfrac{62-20}{15}\\&2x=\dfrac{42}{15}\end{align*} \)

On dividing both sides by 2 , we obtain

\( \dfrac{2x}{2}=\dfrac{42}{15}\times\dfrac{1}{2} \)

\( x=\dfrac{7}{5} \)

\( x=1\dfrac{2}{5} \)

 

Question 4 :

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

Sol :

Let one number be x . There fore , the other number will be x+15.

According to the question,

x + x +15 = 95

2x + 15 = 95

On transposing 15 to R.H.S , we obtain

2x = 95 – 15

2x = 80

On dividing both sides by 2 , we obtain 

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

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 ?

Sol :

Let the common ration between these numbers be x . Therefore , the numbers will be 5x and 3 x respectively.

Difference between these numbers = 18

5x – 3x = 18

2x = 18

Dividing both sides by 2 ,

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

x = 9

First number be 5x = ​\( 5\times{9}=45 \)

Second number be 3x = ​\( 3\times{9}=27 \)

 

Question 6 :

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

Sol : 

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

Sum of these numbers = x+x+1+x+2 = 51

3x + 3 = 51

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

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

x = 16

x+1 = 17

x+2 = 18

Hence , the consecutive integers re 16 , 17 and 18.

 

Question 7 :

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

Sol :

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

On dividing both sides by 8 , we obtain

\( \dfrac{8(3x+3)}{8}=\dfrac{888}{8} \)

3x+3 = 111

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

3x = 111-3

3x = 108

On dividing both sides by 3 , we obtain

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

x = 36

First multiple ​\( \begin{align*}&=8x\\&=8\times{36}\\&=288\end{align*} \)

Second multiple ​\( \begin{align*}&=8(x+1)\\&=8\times(36+1)\\&=8\times{37}\\&=296\end{align*} \)

Third multiple ​\( \begin{align*}&=8(x+2)\\&=8\times(36+2)\\&=8\times{38}\\&=304\end{align*} \)

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.

Sol :

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

On transposing 11 to R.H.S , we obtain

9x = 74 – 11

9x = 63

On dividing both sides by 9 , we obtain

\( \dfrac{9x}{9}=\dfrac{63}{9} \)

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 sum of their ages will be 56 years. what are their present ages ?

Sol :

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.

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

On transposing 8 to R.H.S , we obtain

12x = 56 – 8

12x = 48

On dividing both sides by 12 , we obtain

\( \dfrac{12x}{21}=\dfrac{48}{12} \)

x = 4

Rahul’s age =​\( \begin{align*}&=5x~years\\&=(5\times{4})~years\\&=20~years\end{align*} \)

Haroon”s age =​\( \begin{align*}&=7x~years\\&=(7\times{4})~years\\&=28~years\end{align*} \)

 

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 strenght ?

Sol :

Let the common ratio between the number of boys and number 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

\( \dfrac{2x}{2}=\dfrac{8}{2} \)

x = 4

Number of boys = ​\( \begin{align*}&=7x\\&=7\times4\\&=28\end{align*} \)

Number of girls =​\( \begin{align*}&=5x\\&=5\times4\\&=20\end{align*} \)

Hence , total class strength = ​\( \begin{align*}&=28+20\\&=48~students\end{align*} \)

 

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 the each one of them ?

Sol :

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

According to the given question, 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

\( \dfrac{3x}{3}=\dfrac{138}{3} \)

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 ?

Sol :

Let Ravi’s present age be x years.

Fifteen years later , Ravi’s age = 4 times his present age

x + 15 = 4x

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

15 = 4x – x

15 = 3x

On dividing both sides by 3 , we obtain

\( \dfrac{15}{3}=\dfrac{3x}{3} \)

5 = x

Hence , Ravi’s present age = 5 years

 

Question 13 :

A rational number is such that when you multiply is by ​\( \dfrac{5}{2} \)​ and  ​\( \dfrac{2}{3} \)​  to the product, you get ​\( -\dfrac{7}{12} \)​ . What is the number ?

Sol :

Let the number be x .

 According to the given question,

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

On transposing ​\( \dfrac{2}{3} \)​ to R.H.S , we obtain,

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

\( \dfrac{5}{2}x=\dfrac{-7-(2\times4)}{12} \)

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

On multiplying both sides by ​\( \dfrac{2}{5} \)​ , we obtain

\( x=-\dfrac{15}{12}\times\dfrac{2}{5}=-\dfrac{1}{2} \)

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

 

Question 14 :

Lakshmi is a cashier in a bank. She has currency notes of denominations ₹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 ?

Sol :

Let have common ratio between the numbers of notes of different denominations be x. Therefore , numbers of ₹ 100 notes , ₹ 50 notes and ₹ 10 notes will be 2x , 3x and 5x respectively.

Amount of ₹ 100 notes ​\( \begin{align*}&=100\times2x\\&=200x\end{align*} \)
 
 
Amount of ₹ 50 notes ​\( \begin{align*}&=50\times3x\\&=150x\end{align*} \)
 
 
Amount of ₹ 10 notes ​\( \begin{align*}&=10\times5x\\&=50x\end{align*} \)
 
It is given that total amount is ₹ 400000.
200x + 150x + 50x = 400000
400x = 400000
On dividing both sides by 400 , we obtain
x = 1000
Number of ₹ 100 notes ​\( \begin{align*}&=2x\\&=2\times1000\\&=2000\end{align*} \)
Number of ₹ 50 notes ​\( \begin{align*}&=3x\\&=3\times1000\\&=3000\end{align*} \)
Number of ₹ 10 notes ​\( \begin{align*}&=5x\\&=5\times1000\\&=5000\end{align*} \)

 

Question 15 :

I have a total of ₹ 300 in coins of denominations ₹ 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 ?

Sol :

Let the number of ₹ 5 coins be x 

Number of ₹ 2 coins ​\( \begin{align*}&=3\times~number~of~₹~5~coins\\&=3x\end{align*} \)

Number of ₹ 1 coins = 160 – (Number of coins of ₹ 5 and ₹ 2 )

= 160 – (3x + x) = 160 – 4x

Amount of ₹ 1 coins ​\( \begin{align*}&=₹[1\times(160-4x)]\\&=₹(160-4x)\end{align*} \)

Amount of ₹ 2 coins ​\( \begin{align*}&=₹~(2\times3x)\\&=₹~6x\end{align*} \)

Amount of ₹ 5 coins ​\( \begin{align*}&=₹~(5\times~x)\\&=₹~5x\end{align*} \)

It is given that the total amount is ₹ 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

\( \dfrac{7x}{7}=\dfrac{140}{7} \)

x = 20

Number of ₹ 2 coins ​\( \begin{align*}&=3x\\&=3\times{20}\\&=60\end{align*} \)

Number of ₹ 5 coins = x = 20

 

Question 16 :

The organizers 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 ₹ 3000. Find the number of winners , if the total number of participants is 63 .

Sol :

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 ​\( \begin{align*}&=100\times x\\&=₹~100x\end{align*} \)

Amount given to the participants who did not win ​\( \begin{align*}&=₹[25(63-x)]\\&=₹(1575-25x)\end{align*} \)

According to the given question, 

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

\( \dfrac{75x}{75}=\dfrac{1425}{75} \)

x = 19

Hence, number of winners = 19

 

3 thoughts on “NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable

Leave a Reply

Your email address will not be published. Required fields are marked *