2. Linear Equations in One Variable Mathematics Exercise - 2.4 class 8 Maths in English - CBSE Study
NCERT Solutions for Class 8 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Linear Equations in One Variable with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 8 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 8 English Medium Mathematics All Chapters:
2. Linear Equations in One Variable
4. Exercise 2.4
Q1. Amna thinks of a number and subtract 5/2 from it. She multiplies result from 8 the result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let, the number be x
So, A.T.Q the equation will be: (x - 5/2 ) 8 = 3x
⇒ 8x - 40/2 = 3x
⇒ 8x – 3x = 20
⇒ 5x = 20
⇒ x = 20/5 = 4
So, the number is x = 4.
Q2. A positive number is 5 times another number. If 21 are added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let, the first number is x
And the second number is 5x
2(x + 21) = 5x + 21
⇒ 2x + 42 = 5x + 21
⇒ 5x – 2x = 42 – 21
⇒ 3x = 21
⇒ x = 21/7 =7
So, the both numbers will be x = 7 × 1 = 7 and 5x = 5 × 7 = 35 respectively.
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:
Let, the tens digit of number is = x
The once digit of number is = y
So, the sum of digits x + y = 9---------------- (1)
So, the number will be (10 × x) + y
= 10x + y ----------------- (2)
When we under change the digits then the numbers will be: 10y + x ------------ (3)
By equation (1) and (2)
(10x + y) – (10y + x) = 27
⇒ 10x + y – 10y – x = 27
⇒ 9x – 9y = 27
⇒ 9(x – y) = 27
⇒ x – y = 27/3 = 3
So, x – y = 3 -------------- (4)
Adding equation (3) and (4)
x + y = 9
+ x – y = 3 .
2x = 12
⇒ x = 12/2 = 6
Putting x = 6 in equation (1)
6 + y = 9
⇒ y = 9 – 6 = 3
At last the number = 10x + y = 10 × 6 + 3 = 63
Q4. 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, the first digit be x
So, the second digit will be 3x
So, the number will be 10 × x + 3x = 13x
Interchanging the digits 10 × 3x + x = 31x
A.T.Q the equation will be 13x + 31x = 88
⇒ 44x = 88
⇒ x = 44/88 = 2
So, the original number is 13x = 13 × 2 = 26
Q5. 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 the age of Shobo be x
So, the age of her mother is 6x
ATQ, the equation will be: 3(x + 5) = 6x
⇒ 3x + 15 = 6x
⇒ 6x – 3x = 15
⇒ 3x = 15
⇒ x = 15/3 = 5
so the age of shobo x = 5
and the age of shobo mother 6x = 6 × 5 = 30
Q6. 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 meter it will cost the village panchayat 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let, the length and the breadth of the plot is 11x and 4x respectively.
So, the perimeter of plot = 2(l+b)
= 2(11x+4x) = 22x + 8x= 30x
Perimeter of plot = the total cost of plot/per meter cost
So, the length of the plot 11x = 11×25 = 275m
The breadth of the plot 4x = 4×25 = 100m
Q7. Hasan buys two kinds of cloths materials for school uniforms, shirt materials that cost him R.s 50 per meter and trouser material that costs him R.s 90 per meter. For every 3 meters of the shirt material he buys 2 meters 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:
Q9. Grandfather is ten times older than his granddaughter.
He is also 54 years older than her. Find their present ages.
Solution:
Let, the age of granddaughter be x and the age of grandfather 10x
According to the question the equation will be:
⇒ 10x – x = 54
⇒ 9x = 54
so, the age of grandfather is 10x = 10 × 6 = 60
and the age of granddaughter x = 6
Q10. 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, the age of Aman’s son xv and the age of Aman be 3x
According to the question the equation will be
5(x-10) = 3x-10
⇒ 5x – 50 = 3x - 10
⇒ 5x – 3x = -10 + 50
⇒ 2x = 40
So, the age of Aman's son x = 20
and the age of Aman 3x = 3 × 20 = 60
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