NCERT Solution for Class 8 Maths Chapter – 2 Exercise 2.4

Q1. Amina thinks of a number and subtracts 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?

Solution:-

( x – 5/2 ) x 8 = 3x

8x – 40/2 = 3x

8x – 3x = 40/2

5x = 20

X = 4

Q2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new number. What are the numbers?

Solution:-

5x + 21 = 2( x + 21)

5x + 21 = 2x + 42

5x – 2x = 42 – 21

3x = 21

X = 7

One number = x = 7

Other number = 5x = 5 x 7 = 35

Q3. 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 27, What is the two – digit number?

Solution:-

10x + ( 9 – x ) +27 = 10( 9 – x ) + x

10x + 9 – x + 27 = 90 – 10x + x

9x + 36 = 90 – 9x

9x + 9x = 90-36

18x = 54

X = 3

The number is 36.

Q4. One of the two digits of a two – digit number is three times the other digit. If you interchange the digits numbers and add the resulting number to the original number, you get 88. What is the original number ?

Solution:-

( 30x + x) + ( 10x + 3x ) = 88

31x + 13x = 88

44x = 88

X = 2

The number is 26.

Q5. Shobo mother present age is six times shobo present age. Shobo age five years from now will be one third of his mothers present age . What are their present age?

Solution:-

(x+ 5) = (1/3) x 6x

X + 5 = 2x

2x – x = 5

X = 5

Present age of shobo mother is = 6x = 6×5 = 30 years.

Q6. There is a narrow rectangular plots, reserved for a school, in mahuli village. The length and breadth of the plot are in the ratio 11:4 At the rate Rs 100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution:-

(30x x 100)= 75000

3000 x = 75000

X = 75000/3000

X = 25

11x = 11 x 25 = 275 m

4x = 4 x 25 = 100 m.

Q7. Hasan buys two kind of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per meter . For every 3 meters of the shirt material he buys 2 meters of the trouser material . He sells of the material at 12% and 10% profit respectively. His total sale is Rs 36000 . How much trouser material did he buy?

Solution:-

( 2x x 99) + ( 3 x 56) = 36600

= 198x + 168x = 36600

366x = 36600/366

366x = 100

2x = 2 x 100 = 200m.

Q8. Half of a herd of 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 herd.

Solution:-

X/2 + 3x/8 + 9 = x

( 4x + 2x)/8 + 9 = x

7x/8 + 9 = x

X – 7x/8 = 9

(8x – 7x)/8 = 9

X= 9 x 8

X = 72

Q9. A grandfather is ten times older than his granddaughter He is also 54 years older than her. Find their present ages.

Solution:-

10 – x = 54

9x = 54

X = 6

10x = 10 x 6 = 60

X = 6

Q10. Aman age is three times his son age Ten years ago he was five times his son age Find their present ages.

NCERT Solution for Class 8 Maths Chapter – 2 Exercise 2.4

Q6. There is a narrow rectangular plots, reserved for a school, in mahuli village. The length and breadth of the plot are in the ratio 11:4 At the rate Rs 100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution:-

(30x x 100)= 75000

3000 x = 75000

X = 75000/3000

X = 25

11x = 11 x 25 = 275 m

4x = 4 x 25 = 100 m.

Solution:-

( 2x x 99) + ( 3 x 56) = 36600

= 198x + 168x = 36600

366x = 36600/366

366x = 100

2x = 2 x 100 = 200m.

Q8. Half of a herd of 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 herd.

Solution:-

X/2 + 3x/8 + 9 = x

( 4x + 2x)/8 + 9 = x

7x/8 + 9 = x

X – 7x/8 = 9

(8x – 7x)/8 = 9

X= 9 x 8

X = 72

Q9. A grandfather is ten times older than his granddaughter He is also 54 years older than her. Find their present ages.

Solution:-

10 – x = 54

9x = 54

X = 6

10x = 10 x 6 = 60

X = 6

Q10. Aman age is three times his son age Ten years ago he was five times his son age Find their present ages.

Solution:-

5(x – 10) = 3x – 10

5x – 50 = 3x – 10

5x – 3x = -10 + 50

2x = 40

X = 20

Aman son age = x = 20 years

Aman age = 3x = 3 x 20 = 60 years