The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4 will help you practice solving word problems based on linear equations. In this exercise, you will read a small story or situation and learn how to convert it into equations. Once the equation is made you will be able to solve it like you have learnt in the previous exercises.
Exercise 2.4 helps in real life application. It also helps you learn how to read, think, and convert a word problem into an equation. The exercise aligns with the NCERT syllabus for Class 8 students and the solutions are provided in a step by step manner. Practicing these questions can make you feel more confident regarding application based questions in your examinations.
The NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.4 is available for download. Click the link below to download the PDF for free:
This exercise is about using linear equations to solve word problems. The key concepts covered in this exercise include:
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?
Sol.
Let Amina think of a number x.
She subtracts 5/2 from it: (x − 5/2).
She multiplies the result by 8: 8(x − 5/2).
The result obtained is 3 times the number she thought of: 3x.
So, the equation is 8(x − 5/2) = 3x.
Distribute 8 on the LHS:
=> 8x − (8 × 5/2) = 3x
=> 8x − 20 = 3x
Subtract 3x from both sides:
=> 8x − 3x = 20
=> 5x = 20
Divide by 5:
=> x = 20/5
=> x=4.
Hence, the number thought of by Amina is 4.
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?
Sol.
Let the smaller positive number be x.
The other number is 5 times x, so it is 5x.
If 21 is added to both numbers:
New smaller number = x+21.
New larger number = 5x+21.
One of the new numbers becomes twice the other new number.
Since 5x+21 is clearly the larger new number, it must be twice the smaller new number.
So, 5x+21 = 2(x+21).
Distribute 2 on the RHS:
=> 5x+21 = 2x+42
Subtract 2x from both sides:
=> 5x−2x = 42−21
=> 3x = 21
Divide by 3:
=> x = 21/3
=> x=7.
Hence, the required numbers are:
Smaller number = x = 7.
Larger number = 5x = 5 × 7 = 35.
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?
Sol.
Let the unit's digit be x.
The sum of the digits is 9, so the ten's digit = (9−x).
The original two-digit number = 10 × (ten's digit) + (unit's digit)
= 10(9−x) + x
= 90 − 10x + x
= 90 − 9x.
When the digits are interchanged:
The new unit's digit is (9−x).
The new ten's digit is x.
The new number = 10 × (new ten's digit) + (new unit's digit)
= 10x + (9−x)
= 9x + 9.
It is found that the resulting new number is greater than the original number by 27.
So, New Number = Original Number + 27.
=> 9x + 9 = (90 − 9x) + 27
=> 9x + 9 = 117 − 9x
Add 9x to both sides:
=> 9x + 9x + 9 = 117
=> 18x + 9 = 117
Subtract 9 from both sides:
=> 18x = 117 − 9
=> 18x = 108
Divide by 18:
=> x = 108/18
=> x=6.
So, the unit's digit = 6.
The ten's digit = 9−x = 9−6 = 3.
Hence, the original two-digit number = 36.
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?
Sol.
Let one's digit be x.
Then, the other digit is 3x. (There are two possibilities: either unit's digit is x and ten's is 3x, or unit's is 3x and ten's is x).
Case 1: Unit's digit = x, Ten's digit = 3x.
Original Number = 10(3x) + x = 30x + x = 31x.
Number obtained by reversing the digits: Unit's digit = 3x, Ten's digit = x.
Reversed Number = 10x + 3x = 13x.
Sum of original number and reversed number is 88.
31x + 13x = 88
44x = 88
x = 88/44
x = 2.
In this case: Unit's digit = 2, Ten's digit = 3(2) = 6.
Original Number = 62.
Case 2: Unit's digit = 3x, Ten's digit = x.
Original Number = 10x + 3x = 13x.
Number obtained by reversing the digits: Unit's digit = x, Ten's digit = 3x.
Reversed Number = 10(3x) + x = 30x + x = 31x.
Sum of original number and reversed number is 88.
13x + 31x = 88
44x = 88
x = 2.
In this case: Unit's digit = 3(2) = 6, Ten's digit = 2.
Original Number = 26.
The problem doesn't specify which digit is three times the other, so both 62 and 26 are valid original numbers. However, usually, questions like this imply choosing a common variable for one digit. The solution only gives 62. Both should be accepted.
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 the present age of Shobo be x years.
Shobo's mother's present age = 6x years.
Shobo's age five years from now = (x+5) years.
According to the problem, Shobo's age five years from now will be one third of his mother's present age.
So, (x+5) = (1/3) × (6x).
=> x+5 = 2x.
Subtract x from both sides:
=> 5 = 2x−x
=> x=5.
Therefore, their present ages are:
Present age of Shobo = 5 years.
Present age of his mother = 6x = 6 × 5 = 30 years.
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?
Sol.
Let the length of the rectangular plot be 11x metres.
Let the breadth of the rectangular plot be 4x metres.
Perimeter of the plot = 2 × (Length + Breadth)
Perimeter = 2 × (11x + 4x) metres
Perimeter = 2 × (15x) metres
Perimeter = 30x metres.
The cost of fencing the plot is at the rate of ₹100 per metre.
Total cost of fencing = Perimeter × Rate per metre
Total cost = 30x × 100 = ₹3000x.
It is given that the total cost is ₹75000.
So, 3000x = 75000.
Divide by 3000:
=> x = 75000/3000
=> x=25.
Therefore, the dimensions of the plot are:
Length = 11x = 11 × 25 m = 275 metres.
Breadth = 4x = 4 × 25 m = 100 metres.
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 2 metres of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹36,660. How much trouser material did he buy?
Sol.
Let the quantity of trouser material bought be 2x metres.
Let the quantity of shirt material bought be 3x metres.
For trouser material:
Cost price (CP) of 2x metres of trouser material = 2x metres × ₹90/metre = ₹180x.
Profit on trouser material = 10%.
Selling price (SP) of trouser material = CP × (100 + Profit%)/100
= 180x × (100 + 10)/100
= 180x × (110/100)
= 180x × 1.1 = ₹198x.
For shirt material:
Cost price (CP) of 3x metres of shirt material = 3x metres × ₹50/metre = ₹150x.
Profit on shirt material = 12%.
Selling price (SP) of shirt material = CP × (100 + Profit%)/100
= 150x × (100 + 12)/100
= 150x × (112/100)
= 150x × 1.12 = ₹168x.
Total selling price = SP of trouser material + SP of shirt material
Total selling price = ₹198x + ₹168x = ₹366x.
It is given that his total sale (total selling price) is ₹36,660.
So, 366x = 36660.
Divide by 366:
=> x = 36660/366
=> x=100.16 (approximately).
The question asks for how much trouser material he bought.
Trouser material bought = 2x metres = 2 × 100.16 = 200.32 metres.
(Rounding to the nearest whole metre, it is approximately 200 metres).
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.
Sol.
Let the total number of deer in the herd be x.
Number of deer grazing in the field = x/2.
Remaining deer = x − x/2 = x/2.
Number of deer playing nearby = (3/4) of the remaining = (3/4) × (x/2) = 3x/8.
The rest 9 deer are drinking water.
So, the sum of deer in all activities must equal the total number of deer:
(Deer grazing) + (Deer playing) + (Deer drinking) = Total deer
x/2 + 3x/8 + 9 = x.
To clear denominators, multiply all terms by the LCM of 2 and 8, which is 8:
=> 8 × (x/2) + 8 × (3x/8) + 8 × 9 = 8 × x
=> 4x + 3x + 72 = 8x
=> 7x + 72 = 8x
Subtract 7x from both sides:
=> 72 = 8x − 7x
=> x=72.
Hence, the total number of deer in the herd is 72.
9. A grandfather is ten times older than his grand daughter. He is also 54 years older than her. Find their present ages.
Sol.
Let the present age of the granddaughter be x years.
The grandfather is ten times older than his granddaughter, so grandfather's age = 10x years.
The grandfather is also 54 years older than his granddaughter. This means the difference in their ages is 54 years.
So, (Grandfather's age) − (Granddaughter's age) = 54.
=> 10x − x = 54
=> 9x = 54
Divide by 9:
=> x = 54/9
=> x=6.
Therefore, their present ages are:
Granddaughter's age = 6 years.
Grandfather's age = 10x = 10 × 6 = 60 years.
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 the present age of the son be x years.
Aman's present age = 3x years.
Ten years ago:
Son's age = (x−10) years.
Aman's age = (3x−10) years.
Ten years ago, Aman was five times his son's age.
So, (3x−10) = 5(x−10).
Distribute 5 on the RHS:
=> 3x−10 = 5x−50
Subtract 3x from both sides:
=> −10 = 5x−3x−50
=> −10 = 2x−50
Add 50 to both sides:
=> −10+50 = 2x
=> 40 = 2x
Divide by 2:
=> x = 40/2
=> x=20.
Therefore, their present ages are:
Son's age = 20 years.
Aman's age = 3x = 3 × 20 = 60 years.
(Session 2025 - 26)