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 2.4

Question 1:

Amina thinks a number and subtracts ​\( \dfrac{5}{2} \)​ 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 ?

Sol :

Let the number be x .

According to the given question,

\( 8\Bigg(x-\dfrac{5}{2}\Bigg)=3x \)

5x = 20

Dividing both sides by 5 , we obtain

x = 4

Hence, the number is 4

 

Question 2:

A positive number is 5 times another number. If 21 is added to both the numbers , then one of the numbers becomes twice the other number. What are the numbers ?

Sol :

Let the numbers be x and 5x . According to the question,

21 + 5x = 2(x + 21)

21 + 5x = 2x +42

On transposing 2x to L.H.S and 21 to R.H.S , we obtain

5x – 2x = 42 – 21

3x = 21

Dividing both sides by 3, we obtain

x = 7

5x = ​\( 5 \times{7}=35 \)​ 

Hence , the numbers are 7 and 35 respectively

 

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 2. What is the two digit number ?

Sol :

Let the digits at tens place and once place be x and 9 – x respectively.

Therefore , original number = 10x + (9 – x) = 9x + 9

On interchanging the digits, the digits at ones place and tens place will be x and 9 – x respectively.

Therefore , new number after interchanging the digits = 10(9 – x) + x

= 90 – 10x + x

= 90 – 9x

According to the given question,

New number = Original number + 27

90 – 9x = 9x +9 +27

90 – 9x = 9x + 36

On transposing 9x to R.H.S and 36 to L.H.S , we obtain

90 – 36 = 18x

54 = 18x

Dividing both sides by 18 , we obtain

3 = x and 9 – x = 6

Hence, the digits at tens place and ones place of the number are 3 and  6 respectively.

Therefore, the two digit number is

= 9x + 9

= ​\( 9\times{3}+9= 36 \)​ 

 

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 the resulting number to the original number, you get 88. What is the original number ?

Sol :

Let the digits at tens place and once place be x and 3x respectively.

Therefore , original number = 10x + 3x = 13x

On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.

Therefore , new number after interchanging the digits

\( \begin{align*}&=10\times3x+x\\&=30x+x\\&=31x\end{align*} \)

According to the given question,

New number + Original number = 88

13x + 31x = 88

44x = 88

Dividing both sides by 44 , we obtain

x = 2

Therefore,the original number ​\( \begin{align*}&=13x\\&=13\times2\\&=26\end{align*} \)

By considering the tens place and ones place as 3x and x respectively, the two digit number obtained is 62

Therefore , the two digit number may be 26 or 62

 

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 ?

Sol :

Let Shobo’s age be x years. Therefore, his mother’s age will be 6x years.

According to the given question,

After 5 years Shobo’s age ​\( =\dfrac{Shobo’s~mother’s~present~age}{3} \)

\( x+5=\dfrac{6x}{3} \)

x + 5 = 2x

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

5 = 2x – x

5 = x

6x = ​\( 6\times5=30 \)

Therefore, the present ages of Shobo and Shobo’s mother will be 5 years and 30 years respectively.

 

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 ₹ 100 per metre it will cost the village panchayat ₹ 75,000 to fence the plot. What are the dimensions of the plot ?

Sol :

Let the common ratio between the length and breadth of the rectangular plot be x . Hence , the length and breadth of the rectangular plot will be 11x m and 14x m respectively.

Perimeter of the plot 

\( \begin{align*}&=2(l+b)\\&=2(11x +4x)m\\&=30x~m\end{align*} \)

It is given that the cost of fencing the plot at the rate of ₹ 100 per meter is ₹ 75,000

\( \begin{align*}100\times30x&=75000\\3000x&=75000\end{align*} \)

Dividing both sides by 3000, we obtain

x = 25

Length ​\( \begin{align*}&=11x~m\\&=(11\times25)m\\&=275m\end{align*} \)

 

Breadth ​\( \begin{align*}&=4x~m\\&=(4\times25)m\\&=100~m\end{align*} \)

 

Hence, the dimensions of the plot are 275 m and 100 m respectively.

 

Question 7:

Hasan buys two kinds of cloth materials for school uniforms, shirt material that cost him ₹ 50 per meter and trouser material that costs him ₹ 90 per meter. For every 2 meters of the trouser material he buys 3 meters of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36660. How much trouser material did he buy ?

Sol :

Let 2x m of trouser material and 3x m of shirt material be bought by him.

Per meter selling price of trouser material 

\( =₹~\Bigg(90+\dfrac{90\times12}{100}\Bigg)\\₹~100.80 \)

Per meter selling price of shirt material

\( =₹~\Bigg(50+\dfrac{50\times10}{100}\Bigg)\\₹~55 \)

Given that, the total amount of selling = ₹ 36660

\( \begin{align*}100.80\times(2x)+55\times(3x)&=36660\\201.60x+165x&=36660\end{align*} \)

 

366.60x = 36660

Dividing both sides by 366.60, we obtain

x = 100

Trouser material =2x m

\( =(2\times100)m\\=200~m \)

 

Question 8:

Half of a heard 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 heard.

Sol :

Let the number of deer be x.

Number of deer grazing in the field = ​\( \dfrac{x}{2} \)

\[ \begin{align*}{Number~of~deer~plaing~nearby}~&=\dfrac{3}{4}\times{Number~of~remaining~deer}\\&=\dfrac{3}{4}\times\bigg(x-\dfrac{x}{2}\bigg)\\&=\dfrac{3}{4}\times\dfrac{x}{2}\\&=\dfrac{3x}{8}\end{align*} \]

​Number of deer drinking water from the pond = 9

\( x-\bigg(\dfrac{x}{2}+\dfrac{3x}{8}\bigg)=9 \)

\( x-\bigg(\dfrac{4x+3x}{8}\bigg)=9 \)

\( x-\dfrac{7x}{8}=9 \)

\( \dfrac{x}{8}=9 \)

Multiplying both sides by 8, we obtain

x = 72

Hence, the total number of deer in the herd is 72

 

Question 9:

A grand father is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Sol :

Let the granddaughter’s age be x years. Therefore , granddaughter’ s age will be 10x years.

According to the question,

Grandfather’s age = Granddaughter’s age +54 years

10x – x = 54

9x = 54 

x = 6 

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years

=  ​\( 10\times6~years=60~years \)

 

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.

Sol :

Let Aman’s son’s age be x years. Therefore, Aman’s age will be 3x years. Ten years ago, their age was (x – 10) years and (3x – 10) years respectively.

10 years ago, Aman’s age = 5 times Aman’s son’s age 10 yeasr ago

3x 10 = 5(x – 10)

3x – 10 = 5x – 50

On transposing 3x to R.H.S and 50 to L.H.S, we obtain

50 – 10 = 5x – 3x

40 = 2x

Dividing both sides by 2 , we obtain

20 = x

Aman’s son’s age = x years = 20 years

Aman’s age = 3x years ​\( =3\times20=60~years \)

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