NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.4

The second chapter of the class 8 maths syllabus is on Linear Equations in One Variable. It is a significant chapter in maths that is covered in class 8. NCERT Class 8 Chapter 2 Linear Equations in One Variable is divided into 5 major parts that are required to be covered to understand the chapter properly. The following is a list of the 5 important topics that are covered under the chapter on Linear Equations in One Variable. We recommend that students go through these individual topics carefully to get a proper understanding of the chapter.

Key Learnings from NCERT Solutions for Class 8 Math Chapter 2 Linear Equations in One Variable (EX 2.4) Exercise 2.4

• Dividing and multiplying algebraic expressions where integers are coefficients

• Some common errors faced by students

• Factorisation

• Introduction to Identities

• The process of solving linear equations in one variable.

Importance of Linear Equations in One Variable

In a case where there are two or more integers with one of them being unknown, linear equations in one variable can be used to calculate the value of this integer. It is possible to easily find the value of an unknown integer using the expression in the equation form. It is encouraged that students learn as much as they can from this chapter to be able to solve problems based on linear equations easily in their exams.

Access NCERT solutions for Math Chapter 2 – Linear Equations in One

Exercise 2.4

1. Amina thinks of a number and subtracts \[\frac{\text{5}}{\text{2}}\] from it. She multiplies the result by \[\text{8}\]. The result now obtained is \[\text{3}\] times the same number she thought of. What is the number?

Ans: Let us assume the number be \[\text{x}\]

According to the question,

\[\text{8}\left( \text{x-}\frac{\text{5}}{\text{2}} \right)\text{=3x}\]

\[\Rightarrow \text{8}\left( \frac{\text{2x-5}}{\text{2}} \right)\text{=3x}\]

\[\Rightarrow \text{4}\left( \text{2x-5} \right)\text{=3x}\]

\[\Rightarrow \text{8x-20=3x}\]

Shifting \[3x\] to left hand side and \[20\] to right hand side,

\[\Rightarrow \text{8x-3x=20}\]

\[\Rightarrow \text{5x=20}\]

\[\Rightarrow \text{x=4}\]

Therefore, the required number is \[\text{4}\].

2. A positive number is \[\text{5}\] times another number. If \[\text{21}\] is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Ans: Let us assume the one number be \[\text{x}\]

Other number would be \[\text{5x}\]

According to the question,

\[\text{21+5x=2}\left( \text{x+21} \right)\]

\[\Rightarrow \text{21+5x=2x+42}\]

Shifting \[2x\] to left hand side and \[21\] to right hand side

\[\Rightarrow \text{5x-2x=42-21}\]

\[\Rightarrow \text{3x=21}\]

\[\Rightarrow \text{x=7}\]

\[\Rightarrow \]other number would be \[\text{5x=35}\]

Therefore, the required numbers are \[7\] and \[35\].

3. Sum of the digits of a two-digit number is \[\text{9}\]. When we interchange the digits it is found that the resulting new number is greater than the original number by \[\text{27}\]. What is the two-digit number?

Ans: Let the digit at tens place be \[x\] and digit at ones place be \[\text{9-x}\]

Now, original number would be \[\text{10x+}\left( \text{9-x} \right)\text{=9x+9}\]

The new number after interchanging the digit would be \[\text{10}\left( \text{9-x} \right)\text{+x=90-9x}\]

According to the question,

\[\text{90-9x=9x+9+27}\]

\[\Rightarrow \text{90-9x=9x+36}\]

Shifting \[\text{9x}\] to right hand side and \[\text{36}\] to left hand side,

\[\Rightarrow \text{90-36=9x+9x}\]

\[\Rightarrow 5\text{4=18x}\]

\[\Rightarrow 3\text{=x}\]

\[\Rightarrow \text{9-x=9-3}\]

\[\Rightarrow \text{9-3=6}\]

Therefore, the number at tens place is \[\text{3}\] and the number at ones place is \[\text{6}\]

Now, the two-digit number \[\left( \text{9x+9} \right)\] is \[\text{36}\].

4. One of the two digits of a two-digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get \[\text{88}\]. What is the original number?

Ans: Let the digit at tens place be \[x\] and digit at ones place be \[\text{3x}\]

Now, original number would be \[\text{10x+3x=13x}\]

The new number after interchanging the digit would be \[\text{10}\left( \text{3x} \right)\text{+x=31x}\]

According to the question, sum of original number and the number formed after interchanging the digits is \[\text{88}\] 

\[\Rightarrow \text{13x+31x=88}\]

\[\Rightarrow \text{44x=88}\]

Dividing the equation by \[\text{44}\], we get

\[\Rightarrow \text{x=2}\]

Therefore, the original number is \[\text{13x=26}\] and the number obtained after interchanging the digits would be \[\text{62}\].

Therefore, the two-digit number might be \[26\]or \[\text{62}\].

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 this mother’s present age. What are their present ages?

Ans: Let us assume the present age of Shobo be \[x\] years and Shobo’s mother age be \[6x\] years

According to the question,

\[\text{x+5=}\frac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ 6x}\]

\[\Rightarrow \text{x+5=2x}\]

Shifting \[\text{x}\] to the right side, 

\[\Rightarrow \text{5=2x-x}\]

\[\Rightarrow \text{5=x}\]

Therefore, the present age of Shobo is \[\text{5}\] years and Shobos’s mother age is \[\text{30}\] 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 \[\text{11:4}\]. At the rate Rs \[\text{100}\] per metre it will cost the village panchayat Rs \[\text{75,000}\] to fence the plot. What are the dimensions of the plot?

Ans: Let the ratio between length and breadth be \[\text{x}\]

Length of the plot is \[\text{11x}\] and the breadth of the plot is \[\text{4x}\]

Perimeter of the plot \[\text{=2}\left( \text{l+b} \right)\]

\[\Rightarrow \text{2}\left( \text{11x+4x} \right)\]

\[\Rightarrow \text{2}\left( \text{15x} \right)\]

\[\Rightarrow \text{30x}\]

The cost of fencing the plot is Rs. \[\text{100}\] per metre

Total cost of fencing\[\text{=75,000}\]

\[\Rightarrow \text{100 }\!\!\times\!\!\text{ 30x=75,000}\]

\[\Rightarrow \text{3000x=75,000}\]

Dividing the equation by \[\text{3000}\]

\[\Rightarrow \text{x=25}\]

Length \[\text{=11 }\!\!\times\!\!\text{ 25=275}\] m 

Breadth \[\text{=4 }\!\!\times\!\!\text{ 25=100}\] m

Therefore, the length of the plot is \[\text{275}\] metre and the breadth of the plot is \[\text{100}\] metre.

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs \[\text{50}\] per metre and trouser material that costs him Rs \[\text{90}\] per metre. For every \[\text{2}\] meters of the trouser material he buys \[\text{3}\] metres of the shirt material. He sells the materials at \[\text{12 }\!\!%\!\!\text{ }\] and \[\text{10 }\!\!%\!\!\text{ }\] profit respectively. His total sale is Rs \[\text{36,660}\]. How much trouser material did he buy?

Ans: Let us assume for-\[\text{2x}\] m of trouser material he bought \[\text{3x}\] m of shirt material

Selling price of trouser material per metre \[\text{=}\] Rs. \[\left( \text{90+}\frac{\text{90 }\!\!\times\!\!\text{ 12}}{\text{100}} \right)\]

\[\Rightarrow \left( \text{90+}\frac{\text{90 }\!\!\times\!\!\text{ 12}}{\text{100}} \right)\text{=100}\text{.80}\]

Selling price of trouser material per meter \[\text{=}\] Rs. \[\left( \text{50+}\frac{\text{50 }\!\!\times\!\!\text{ 10}}{\text{100}} \right)\]

\[\Rightarrow \left( \text{50+}\frac{\text{50 }\!\!\times\!\!\text{ 10}}{\text{100}} \right)\text{=55}\]

According to the question, total amount of selling \[\text{=36660}\]

\[\Rightarrow \text{100}\text{.80 }\!\!\times\!\!\text{ }\left( \text{2x} \right)\text{+55 }\!\!\times\!\!\text{ }\left( \text{3x} \right)\text{=36660}\]

\[\Rightarrow \text{201}\text{.60x+165x=36660}\]

\[\Rightarrow \text{366}\text{.60x=36660}\]

\[\Rightarrow \text{x=100}\]

Trouser material \[\text{=2x}\]

\[\Rightarrow \text{2 }\!\!\times\!\!\text{ 100=200}\] m

Therefore, he bought \[\text{200}\] a meter of trouser material.

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest \[\text{9}\] are drinking water from the pond. Find the number of deer in the herd.

Ans: Let us assume the number of deer in the herd be \[\text{x}\]

Number of deer grazing the field \[\text{=}\frac{\text{x}}{\text{2}}\] 

Number of deer playing nearby \[\text{=}\frac{\text{3x}}{\text{8}}\]

Number of deer drinking water \[\text{=9}\]

\[\Rightarrow \text{x-}\left( \frac{\text{x}}{\text{2}}\text{+}\frac{\text{3x}}{\text{8}} \right)\text{=9}\]

\[\Rightarrow \frac{\text{8x-4x-3x}}{\text{8}}\text{=9}\]

Multiplying the equation by \[\text{8}\]

\[\Rightarrow \text{x=72}\]

Therefore, there are a total of \[\text{72}\] deer in the herd.

9. A grandfather is ten times older than his granddaughter. He is also \[\text{54}\] years older than her. Find their present ages.

Ans: Let us assume the age of granddaughter be \[\text{x}\] years

Age of grandfather would be \[\text{10x}\] years

According to the question, grandfather is \[\text{54}\] years older than granddaughter

\[\Rightarrow \text{10x=x+54}\]

Shifting \[\text{x}\] to left hand side

\[\Rightarrow \text{10x-x=54}\]

\[\Rightarrow \text{9x=54}\]

Dividing the equation by \[\text{9}\]

\[\Rightarrow \text{x=6}\]

\[\Rightarrow \text{10x=60}\]

Therefore, the age of the granddaughter is \[\text{6}\] years and the age of grandfather is \[\text{10}\] 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.

Ans: Let us assume the age of Aman’s son be \[\text{x}\] years

Age of Aman would be \[\text{3x}\] years

Age of Aman \[\text{10}\] years ago would be \[\text{3x-10}\] years and Aman’s son age would be \[\text{x-10}\] years

According to the question, ten years ago, Aman was five times his son’s age

\[\Rightarrow \text{3x-10=5}\left( \text{x-10} \right)\]

\[\Rightarrow \text{3x-10=5x-50}\]

Shifting \[\text{-10}\] to left hand side and \[\text{3x}\] to right hand side

\[\Rightarrow \text{-10+50=5x-3x}\]

\[\Rightarrow \text{40=2x}\]

Dividing the equation by \[\text{2}\]

\[\Rightarrow \text{x=20}\]

\[\Rightarrow 3\text{x=60}\]

Therefore, the age of Aman’s son is \[\text{20}\] years and the age of Aman is \[\text{60}\] years.

Exercise 2.4 NCERT Solutions Class 8 Maths Chapter 2

Vedantu offers Math Class 8 Chapter 2 Exercise 2.4 Linear equations in two variable NCERT solutions with a detailed explanation and a summary giving formulae and concepts used in the chapter. The expert teachers have provided the step by step solution in the easiest way possible. The revision notes of subject specialists are easily understandable. It is easy to grab the concept of Class 8 NCERT Maths Chapter 2 Exercise 2.4 when you have reliable solutions. The team of expert teachers have prepared solutions to help students score good marks. The highly qualified and experienced faculty develops NCERT solutions following the Latest NCERT guidelines. 

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NCERT Solutions Class 8 Math Chapter 2 All Exercises

Chapter 2 - Linear Equations in One Variable Exercises in PDF Format

Understanding Concepts of Linear Equations

Linear equations in two variables is the chapter that focuses on critical thinking of the students. The questions in the NCERT textbooks have to be solved with concentration and focus. NCERT Solution of Class 8 Math Exercise 2.4 are solved step by step to ensure students easily understand the word problems. Students develop accuracy in the questions after practicing the number of questions in the sample paper. If your focus is on scoring good marks, NCERT solutions are the key.

Solutions for Linear Equations in One Variable NCERT Questions

To solve Exercise 2.4, students should know basic linear equations and algebra concepts. Students will grasp the basic concepts of linear equations to solve the questions. Students can excel in the exams by studying extra questions provided. Even if the questions are a bit difficult, the step by step approach is easy to get through. Variables with coefficients in the form of ratio are included in NCERT solutions free PDF.

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With the NCERT Solutions for Class 8 Math Chapter 2 Exercise 2.4, multiple topics, and questions for important sub-topics are provided. The solutions are updated according to the changes in the latest NCERT textbook so that students don’t miss even a topic. Every question is explained in a brief and easy way to provide students help in exam preparation. Students can view Vedantu notes or free PDF downloads are helpful for exam preparation. To practice the questions of the respective chapter, go through the sample papers provided. Vedantu offers NCERT Solutions for all subjects Class 12, so that students don’t suffer because of lack of study material. Students can easily download NCERT solutions to clear their queries of any other question of the textbook. Apart from Class 8 NCERT Math Chapter 2 Exercise 2.4, our teachers provide sample papers, revision notes and important questions to prepare students for the other exams as well. 

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