NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4 – Linear Equation in one variable, has been designed by the NCERT to test the knowledge of the student on the topic – Some More Applications of Linear Equation in one variable

### NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4

**NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4**

**Q.1 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:**

Let, the number be x

As given in the question,

[x – (5/2)] × 8 = 3 × x

8x – 20 = 3x

8x – 3x = 20

5x = 20

x = 20/5

x = 4

Hence, the number is 4

**Q.2 ****A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?**

**Solution:**

Let, the numbers be x and 5x

As given in the question,

2 × (x+21) = (5x+21)

2x + 42 = 5x + 21

5x – 2x = 42 – 21

3x = 21

x = 7

Hence, the numbers are 7 and 5 x 7 = 35

**Q.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 27. What is the two-digit number?**

**Solution:**

Let, the units place digit be x

Therefore, the tens place digit = 9 – x

For a 2 digit number,

Number = 10 x tens place digit + units place digit

Original number = 10 × (9-x) + x

As given in the question,

New number = Original number + 27

10x + (9-x) = 10(9-x) + x + 27

10x + 9 – x = 90 – 10x + x + 27

9x + 9 = -9x + 117

18x = 108

x = 6

Hence, the two-digit number = 10 × (9-6) + 6 = 36

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

**Solution:**

Let, Units place of the 2-digit number be x

Therefore, Tens place = 3x

Then, two-digit Number = 10 × 3x + x

Number after interchanged = 10 × x + 3x

As given in the question,

10 × 3x + x + 10 × x + 3x = 88

30x + x + 10x + 3x = 88

44x = 88

x = 2

If, Unit digit = x and Tens place = 3x

Then, two-digit number =10 × 3x + x = 62

If, Unit digit = 3x and Tens place = x

Then, two-digit number = 10 × x + 3x = 10 x 2 + 3 x 2 = 20 + 6 = 26

Hence, the two-digit Number may be 62 or 26.

**Q.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?**

**Solution:**

Let, Shobos’s present age be x

Therefore, Shobo’s mother’s present age = 6x

As given in the question,

x+5 = (1/3) × 6x

x+5 = 2x

2x-x = 5

x = 5

Hence, Shobo’s present age is 5 years and Shobo’s mother’s present age is 6 x 5 = 30 years

**Q.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 ₹ 75000 to fence the plot. What are the dimensions of the plot?**

**Solution:**

Let, length and breadth of the narrow rectangular plot be 11x and 4x respectively

Perimeter of a rectangle = 2(l+b) = 2(11x+4x)

Cost of fencing the plot at the rate of ₹ 100 per metre = ₹ 75000

As given in the question,

Perimeter x 100 = 75000

2 × (11x+4x) × 100 = 75000

2 × 15x × 100 = 75000

3000x = 75000

x = (75000/3000)

x = 25

Hence, length of the plot = 11x = 11 × 25 = 275m

and breadth of plot = 4x = 4 × 25 = 100m

**Q.7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?**

**Solution:**

Let, the ratio of Shirt material and trouser material be 3x : 2x

Cost of shirt material = 3x × 50 = 150x

Cost of trouser material = 2x × 90 = 180x

Selling price of shirt material at 12% gain =(100+ Profit Percentage)/100 × Cost Price

= [(100+12)/100] × 150x

= 168x

Selling price of trouser material at 10% gain = (100 + Profit Percentage)/100 × Cost Price

= [(100+10)/100] × 180x

= 198x

As given in the question,

168x + 198x = 36600

366x = 36600

x = 100 meters

Hence, trouser material = 2x = 2 x 100 = 200 meters

**Q.8 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:**

Let, total number of deer in the herd be x

Half of the deer are grazing in the field (x/2)

Three-fourths of the remaining are playing nearby = (3/4) × [x – (x/2)] = (3/4) × (x/2) = (3x/8)

Rest 9 are drinking water from the pond

x – [(3x)/8 +(x/2)] = 9

x – (7x/8) = 9

(x/8) = 9

x = 9 × 8 = 72

Hence, the total number of deer in herd is 72

**Q.9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.**

**Solution:**

Let, age of granddaughter be x years

Age of grandfather = 10x years

As given in question,

10x = x + 54

10x – x = 54

9x = 54

x = 6

Hence, present age of granddaughter = 6 years and present age of grandfather = 10 x 6 = 60 years

**Q.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.**

**Solution:**

Let, present age of Aman’s son be x years

Aman’s present age = 3x years

As given in the question,

Ten years ago Aman’s age = 3x – 10

Ten years ago Aman’s son age = x – 10

Ten years ago he was five times his son’s age

3x – 10 = 5(x-10)

3x – 10 = 5x – 50

5x – 3x = 50 – 10

2x = 40

x = 20

Hence, present age of Aman’s son is 20 years and present age of Aman = 3 x 20 = 60 years

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.5 – Linear Equation in one variable can be accessed by clicking here**

**Download NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4**