2. Linear Equations in One Variable Mathematics class 8 in English Medium ncert book solutions Exercise 2.4
2. Linear Equations in One Variable Exercise 2.4 – Complete NCERT Book Solutions for Class 8 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 2. Linear Equations in One Variable Exercise 2.4 to help you master concepts and score higher.
2. Linear Equations in One Variable Mathematics class 8 in English Medium ncert book solutions Exercise 2.4
NCERT Solutions for Class 8 Mathematics play an important role in helping students understand the concepts of the chapter 2. Linear Equations in One Variable clearly. This chapter includes the topic Exercise 2.4, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 8 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 2. Linear Equations in One Variable is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 2.4. By studying these updated NCERT Solutions for Class 8 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
2. Linear Equations in One Variable
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
See other sub-topics of this chapter:
1. Exercise 2.1 2. Exercise 2.2 3. Exercise 2.3 4. Exercise 2.4 5. Exercise 2.5 6. Exercise 2.6
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