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## Case Study Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable

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Here we are providing Case Study questions for Class 8 Maths Chapter 2 Linear Equations in One Variable.

## Case Study Questions

Related posts, learning outcomes.

- Linear Equation in One Variable
- Solving an Equation
- Solving Equation having variable on both sides
- Reducing Equation to Simpler Form

## Important Keywords

- Natural Numbers: Positive Counting number starting from 1.
- Whole Number: All natural number together with 0 .

## Fundamental Facts

- Equation: An equation is like a scale, while manipulating equations we have to keep the scale balanced by doing the same thing to both sides.

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- Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Last Updated on April 29, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 2 Linear Equations in One Variable. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

## Case Study Questions on Linear Equations in One Variable

Passage 1: It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered. Few situations are given below in the form questions. Find the correct option.

Difficulty Level: Medium

Q. 1. If the fixed charge in a city is ₹ x and charge per km is ₹5 and total fare is ₹60 then find the linear equation for the journey of 10 km. (a) x + 50 = 60 (b) x – 50 = 60 (c) x + 50 = 50 (d) None of these

Q. 2. In the above question, what is the value of fixed charge? (a) ₹20 (b) ₹5 (c) ₹10 (d) ₹15

Q. 3. If in a city a person has to pay ₹110 for a journey of 15 km and fixed charge is ₹20 then what is the charge per km is? (a) ₹12 (b) ₹6 (c) ₹8 (d) no fixed charge

Q. 4. If in a city fixed charge is double of the charge per km and a person paid ₹75 for a journey of 1 km, then the linear equation for the following situation is?

Q. 5. According to the given equation: 2x + 17 = 85, if ₹17 is the fixed charge and the total fare is ₹85 for a journey of 2km then what is the charge per kilometre?

## Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Rational numbers class 8 case study questions maths chapter 1, topics from which case study questions may be asked.

- Linear Equation in One Variable
- Solving an Equation
- Solving Equation having variable on both sides
- Reducing Equation to Simpler Form

Case study questions from the above given topic may be asked.

## Frequently Asked Questions (FAQs) on Linear Equations in One Variable Case Study

Q1: what is an algebraic expression.

A1: A combination of constants and variables connected by some or all of the four fundamental operators such as +, –, × and ÷ is called algebraic expression.

## Q2: What is an equation?

A2: An equation is like a scale, while manipulating equations we have to keep the scale balanced by doing the same thing to both sides.

## Q3: What is linear equation in one variable?

A3: An equation is called linear equation if it has only one degree i.e., the highest power of the variable appearing in equation is 1. e.g., ax + b = 0

## Q4: What is solution of an equation?

A4: A number which satisfies an equation is called the solution of the equation.

## Q5: How do you solve a linear equation in one variable?

A5: To solve a linear equation in one variable, isolate the variable on one side of the equation by performing inverse operations (such as addition, subtraction, multiplication, or division) on both sides until the variable is alone.

## Q6: What is the solution of a linear equation?

A6: The solution of a linear equation is the value of the variable that satisfies the equation. It is the value that, when substituted into the equation, makes the equation true.

## Q7: What does the term ‘one variable’ mean in linear equations?

A7: ‘One variable’ means that the equation contains only a single unknown quantity, typically represented by the letter ‘x’. Other terms or constants may be present, but there is only one variable.

## Q8: Can linear equations be represented graphically?

A8: Yes, linear equations can be represented graphically as straight lines on the coordinate plane. The slope-intercept form (y = mx + b) is particularly useful for graphing linear equations.

## Q9: Are there any online resources or tools available for practicing linear equations in one variable case study questions?

A9: We provide case study questions for CBSE Class 8 Maths on our website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.

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## CBSE Class 8th Maths Value Based Questions PDF Download

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CBSE Class 8th Maths Value Based Questions are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

## CBSE Class 8th Maths Value Based Questions

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

## CBSE Maths Value Based Questions Class 8th PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 8th download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

## CBSE Maths Value Based Questions for Class 8th Free PDF

Class 8th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions certainly help students to obtain good marks in the examinations.

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## NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations In One Variable

- NCERT Solutions
- Chapter 2 Linear Equations In One Variable

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - Free PDF

Suppose you are looking for the simplest and easiest way of understanding Class 8 Maths Chapter 2 concepts, then you must avail of Linear Equations in One Variable Class 8 CBSE Solutions created by the expert team of Vedantu. Keeping in mind the understanding level of class 8 students, the scholars have designed the NCERT Solutions for Class 8 Maths Chapter 2 after a lot of research. The solutions also adhere to the latest CBSE guidelines; hence students can expect to get good grades in maths after going through our Class 8 Maths NCERT Solution Chapter 2. You can also download NCERT Solutions for Class 8 Science on our website.

## Access NCERT solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

Exercise 2.1

1. Solve: $x-2=7$

\[x-2=7\]

Transposing $2$ to R.H.S

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - PDF Download

, we obtain

2. Solve:$y+3=10$

Transposing $3$ to R.H.S, we obtain

3. Solve: $6=z+2$

Transposing $2$ to L.H.S, we obtain

4. Solve: $\frac{3}{7}+x=\frac{17}{7}$

$\frac{3}{7}+x=\frac{17}{7}$

Transposing $\frac{3}{7}$ to R.H.S, we obtain

$x=\frac{17}{7}-\frac{3}{7}=\frac{14}{7}=2$

5. Solve:$6x=12$

Dividing both sides by $6$ , we obtain

$\frac{6x}{6}=\frac{12}{6}$

6. Solve: $\frac{t}{5}=10$

$\frac{t}{5}=10$

Multiplying both sides by $5$, we obtain

$\frac{t}{5}\times 5=10\times 5$

7. Solve: $\frac{2x}{3}=18$

$\frac{2x}{3}=18$

Multiplying both sides by$\frac{3}{2}$ , we obtain

$\frac{2x}{3}\times \frac{3}{2}=18\times \frac{3}{2}$

8. Solve: $1.6=\frac{y}{1.5}$

$1.6=\frac{y}{1.5}$

Multiplying both sides by$1.5$, we obtain

$1.6\times 1.5=\frac{y}{1.5}\times 1.5$

9. Solve:$7x-9=16$

Transposing$9$ to R.H.S, we obtain

Dividing both sides by$7$, we obtain

$\frac{7x}{7}=\frac{25}{7}$

$x=\frac{25}{7}$

10. Solve: $14y-8=13$

Transposing $8$ to R.H.S, we obtain

Dividing both sides by$14$, we obtain

$\frac{14y}{14}=\frac{21}{14}$

$y=\frac{3}{2}$

11. Solve: $17+6p=9$

Transposing $17$ to R.H.S, we obtain

Dividing both sides by$6$, we obtain

$\frac{6p}{6}=-\frac{8}{6}$

$p=-\frac{4}{3}$

12. Solve: $\frac{x}{3}+1=\frac{7}{15}$

$\frac{x}{3}+1=\frac{7}{15}$

Transposing $1$ to R.H.S, we obtain

$\frac{x}{3}=\frac{7}{15}-1$

$\frac{x}{3}=\frac{7-15}{15}$

$\frac{x}{3}=-\frac{8}{15}$

Multiplying both sides by $3$ , we obtain

$\frac{x}{3}\times 3=-\frac{8}{15}\times 3$

$x=-\frac{8}{5}$

Exercise 2.2

1. If you subtract$\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$ , $\frac{1}{8}$you get . What is the number?

Ans:

Let the number be x. According to the question,

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}=\frac{1}{8}$

On multiplying both sides by$2$, we obtain

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}\times 2=\frac{1}{8}\times 2$

$x-\frac{1}{2}=\frac{1}{4}$

On transposing $\frac{1}{2}$ to R.H.S, we obtain

$x=\frac{1}{4}+\frac{1}{2}$

$\,\,\,\,=\frac{1+2}{4}=\frac{3}{4}$

Therefore, the number is$\frac{3}{4}$ .

2. The perimeter of a rectangular swimming pool is $154$ m. Its length is $2$ m more than twice its breadth. What are the length and the breadth of the pool?

Let the breadth be x m. The length will be$\left( 2x+2 \right)$ m.

Perimeter of swimming pool $=2\left( l+b \right)=154$m

$2\left( 2x+2+x \right)=154$

$2\left( 3x+2 \right)=154$

Dividing both sides by$2$, we obtain

$\frac{2\left( 3x+2 \right)}{2}=\frac{154}{2}$

On Transposing $2$ to R.H.S, we obtain

On dividing both sides by$3$, we obtain

$\frac{3x}{3}=\frac{75}{3}$

$2x+2=2\times 25+2=52$

Hence, the breadth and length of the pool are $25$ m and$52$ m respectively.

3. The base of an isosceles triangle is $\frac{4}{3}$ cm. The perimeter of the triangle is $4\frac{2}{15}$ cm. What is the length of either of the remaining equal sides?

Let the length of equal sides be x cm.

Perimeter$=\text{x }cm+\,\text{x}\,cm+\,\text{Base}=4\frac{2}{15}\text{cm}$

$2x+\frac{4}{3}=\frac{62}{15}$

On Transposing $\frac{4}{3}$ to R.H.S, we obtain

$2x=\frac{62}{15}-\frac{4}{3}$

$2x=\frac{62-4\times 5}{15}=\frac{62-20}{15}$

$2x=\frac{42}{15}$

On dividing both sides by$2$, we obtain

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$

$x=\frac{7}{5}=1\frac{2}{5}$

Therefore, the length of equal sides is $1\frac{2}{5}$ cm.

4. Sum of two numbers is$95$. If one exceeds the other by $15$, find the numbers.

Let one number be x. Therefore, the other number will be$\text{x}+15$. According to the question,

$x+x+15=95$

On Transposing $15$ to R.H.S, we obtain

$\frac{2x}{2}=\frac{80}{2}$

$x+15=40+15=55$

Hence, the numbers are$40\text{ and }55$.

5. Two numbers are in the ratio $5:3$. If they differ by$18$, what are the numbers?

Let the common ratio between these numbers be x. Therefore, the numbers will be $5\text{x and }3\text{x}$respectively.

Difference between these numbers$=18$

$5\text{x}-3\text{x}=18$

Dividing both sides by$2$,

$\frac{2x}{2}=\frac{18}{2}$

First number $=5x=5\times 9=45$

Second number $=3x=3\times 9=27$

6. Three consecutive integers add up to$51$. What are these integers?

Let three consecutive integers be$x,x+1\text{ and }x+2$.

Sum of these numbers$=x+x+1+x+2=51$

On Transposing $3$ to R.H.S, we obtain

$\frac{3x}{3}=\frac{48}{3}$

Hence, the consecutive integers are$16,17,\,\text{and }18$ .

7. The sum of three consecutive multiples of $8\,\,\,\text{is }888$. Find the multiples.

Let the three consecutive multiples of $8\,\text{be }8x,\,8\left( x+1 \right),\,8\left( x+2 \right)$.

Sum of these numbers$=8x+8\left( x+1 \right)+8\left( x+2 \right)=888$

$8\left( x+x+1+x+2 \right)=888$

$8\left( 3x+3 \right)=888$

On dividing both sides by$8$, we obtain

$\frac{8\left( 3x+3 \right)}{8}=\frac{888}{8}$

On dividing both sides by $3$ , we obtain

$\frac{3x}{3}=\frac{108}{3}$

First multiple $=8x=8\times 36=288$

Second multiple $=8\left( x+1 \right)=8\times \left( 36+1 \right)=8\times 37=296$

Third multiple $=8\left( x+2 \right)=8\times \left( 36+2 \right)=8\times 38=304$

Hence, the required numbers are$288,296,\text{and }304$ .

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by $2,3\,\text{and }4$respectively, they add up to $74$. Find these numbers.

Let three consecutive integers be $\text{x,}\,\text{x}+1\text{,}\,\text{x}+2$. According to the question,

$2x+3\left( x+1 \right)+4\left( x+2 \right)=74$

$2x+3x+3+4x+8=74$

On Transposing $11$ to R.H.S, we obtain

On dividing both sides by$9$, we obtain

$\frac{9x}{9}=\frac{63}{9}$

$x+1=7+1=8$

$x+2=7+2=9$

Hence, the numbers are $7,8\,\text{and }9$.

9. The ages of Rahul and Haroon are in the ratio$5:7$. Four years later the sum of their ages will be $56$ years. What are their present ages?

Ans: Let common ratio between Rahul’s age and Haroon’s age be x. Therefore, the age of Rahul and Haroon will be $5x$ years and $7x$ years respectively. After $4$ years, the age of Rahul and Haroon will be $\left( 5x+4 \right)$ years and $\left( 7x+4 \right)$ years respectively.

According to the given question, after$4$ years, the sum of the ages of Rahul and Haroon is $56$ years.

$\therefore \left( 5x+4+7x+4 \right)=56$

On Transposing $8$ to R.H.S, we obtain

On dividing both sides by $12$, we obtain

$\frac{12x}{12}=\frac{48}{12}$

Rahul’s age $=5x$ years $=\left( 5\times 4 \right)$ years $=20$ years

Haroon’s age $=7x$ years $=\left( 7\times 4 \right)$years $=28$ years

10. The number of boys and girls in a class are in the ratio$7:5$. The number of boys is $8$ more than the number of girls. What is the total class strength?

Ans: Let the common ratio between the number of boys and numbers of girls be x.

Number of boys $=7x$

Number of girls $=5x$

According to the given question,

Number of boys $=$ Number of girls $+8$

$\therefore 7x=5x+8$

On Transposing $5x$ to L.H.S, we obtain

On dividing both sides by\[2\] , we obtain

\[\frac{2x}{2}=\frac{8}{2}\]

Number of boys \[=7x=7\times 4=28\]

Number of girls \[=5x=5\times 4=20\]

Hence, total class strength \[=28+20=48\] students

11. Baichung’s father is $26$ years younger than Baichung’s grandfather and $29$ years older than Baichung. The sum of the ages of all the three is $135$ years. What is the age of each one of them?

Let Baichung’s father’s age be x years. Therefore, Baichung’s age and Baichung’s grandfather’s age will be \[\left( x-29 \right)\] years and \[\left( x+26 \right)\] years respectively. According to the given question, the sum of the ages of these 3 people is \[135\] years.

\[\therefore x+x-29+x+26=135\]

\[3x-3=135\]

On Transposing \[3\] to R.H.S, we obtain

\[3x=135+3\]

On dividing both sides by \[3\], we obtain

\[\frac{3x}{3}=\frac{138}{3}\]

Baichung’s father’s age \[=x\] years \[=46\] years

Baichung’s age \[=\left( x-29 \right)\] years \[=\left( 46-29 \right)\] years = \[17\] years

Baichung’s grandfather’s age \[=\left( x+26 \right)\] years \[=\left( 46+26 \right)\] years \[=72\] years

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age \[=4\times \] His present age

\[x+15=4x\]

On Transposing x to R.H.S, we obtain

\[15=4x-x\]

On dividing both sides by\[3\] , we obtain

\[\frac{15}{3}=\frac{3x}{3}\]

Hence, Ravi’s present age \[=5\]years

13. A rational number is such that when you multiply it by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, you get$-\frac{7}{12}$ . What is the number?

Let the number be x.

According to the given question,

\[\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}\]

On Transposing\[\frac{2}{3}\] to R.H.S, we obtain

\[\frac{5}{2}x=-\frac{7}{12}-\frac{2}{3}\]

\[\frac{5}{2}x=\frac{-7-\left( 2\times 4 \right)}{12}\]

\[\frac{5}{2}x=-\frac{15}{12}\]

On multiplying both sides by\[\frac{2}{5}\] , we obtain

\[x=-\frac{15}{12}\times \frac{2}{5}=-\frac{1}{2}\]

Hence, the rational number is \[-\frac{1}{2}\].

14. Lakshmi is a cashier in a bank. She has currency notes of denominations$\text{Rs }100,\,\text{Rs }50\,\,\,\text{and Rs}\,10$, respectively. The ratio of the number of these notes is$2:3:5$. The total cash with Lakshmi is$\text{Rs }4,00,000$. How many notes of each denomination does she have?

Let the common ratio between the numbers of notes of different denominations be x. Therefore, numbers of \[\text{Rs }100\] notes, \[\text{Rs }50\] notes, and \[\text{Rs }10\]notes will be \[2x,3x,\,\text{and }5x\] respectively.

Amount of \[\text{Rs }100\]notes \[=\text{Rs}\left( 100\times 2\text{x} \right)=\text{Rs}\,200\text{x}\]

Amount of \[\text{Rs }50\]notes \[=\text{Rs}\left( 50\times 3\text{x} \right)=\text{Rs}\,150\text{x}\]

Amount of \[\text{Rs }10\]note\[=\text{Rs}\left( 10\times 5\text{x} \right)=\text{Rs}\,50\text{x}\]

It is given that the total amount is \[\text{Rs}\,400000\].

\[\therefore 200\text{x+}150\text{x+}50\text{x=400000}\]

\[\Rightarrow \text{400x=400000}\]

On dividing both sides by\[400\], we obtain

Number of Rs 100 notes \[=2x=2\times 1000=2000\]

Number of Rs 50 notes \[=3x=3\times 1000=3000\]

Number of Rs 10 notes \[=5x=5\times 1000=5000\]

15. I have a total of $\text{Rs}\,300$in coins of denomination$\text{Re }1,\text{Re }2\,\text{and Re }5$. The number of $\text{Rs }2$coins is$3$ times the number of $\text{Rs }5$ coins. The total number of coins is$160$. How many coins of each denomination are with me?

Let the number of \[\text{Rs }5\] coins be x.

Number of \[\text{Rs }2\] coins \[=3\times \]Number of \[\text{Rs }5\] coins\[=3\text{x}\]

Number of \[\text{Re }1\] coins \[=160-\] (Number of coins of \[\text{Rs }5\] and of\[\text{Rs 2}\])

Amount \[\text{Re }1\]of coins \[=\,\text{Rs }\left[ 1\times \left( 160-4x \right) \right]=\,\text{Rs}\,\left( 160-4x \right)\]

Amount of \[\text{Rs }2\] coins \[=\,\text{Rs}\,\,\left( 2\times 3x \right)=\,\text{Rs}\,\,\text{6x}\]

Amount of \[\text{Rs 5}\]coins \[=\,\text{Rs}\,\,\left( 5\times x \right)=\,\text{Rs}\,\,5\text{x}\]

It is given that the total amount is \[\text{Rs 300}\].

\[\therefore 160-4x+6x+5x=300\]

\[160+7x=300\]

On Transposing \[160\] to R.H.S, we obtain

\[7x=300-160\]

On dividing both sides by \[7\], we obtain

\[\frac{7x}{7}=\frac{140}{7}\]

Number of \[\text{Re }1\] coins \[=160-4x=160-4\times 20=160-80=80\]

Number of \[\text{Rs }2\] coins\[=3x=3\times 20=60\]

Number of \[\text{Rs 5}\] coins \[=x=20\]

16. The organizers of an essay competition decide that a winner in the competition gets a prize of $\text{Rs 100}$ and a participant who does not win gets a prize of$\text{Rs 25}$. The total prize money distributed is $\text{Rs 3000}$. Find the number of winners, if the total number of participants is $63$ .

Let the number of winners be x. Therefore, the number of participants who did not win will be \[63-x\].

Amount given to the winners \[=\text{Rs }\left( 100\times x \right)=~~\text{Rs 100}x\]

Amount given to the participants who did not win \[=\text{Rs }\left[ 25\left( 63-x \right) \right]\]

\[=\text{Rs }\left( 1575-25x \right)\]

\[100x+1575-25x=3000\]

On Transposing \[1575\] to R.H.S, we obtain

\[75x=3000-1575\]

\[75x=1425\]

On dividing both sides by \[75\] , we obtain

\[x=19\]

Hence, number of winners \[=19\]

Exercise 2.3

1. Solve and check result: $3x=2x+18$

\[3x=2x+18\]

On Transposing \[2x\] to L.H.S, we obtain

\[3x-2x=18\]

L.H.S \[=3x=3\times 18=54\]

R.H.S \[=2x+18=2\times 18+18=36+18=54\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

2. Solve and check result: $5t-3=3t-5$

\[5t-3=3t-5\]

On Transposing \[3t\] to L.H.S and \[-3\] to R.H.S, we obtain

\[5t-3=-5-\left( -3 \right)\]

On dividing both sides by\[2\], we obtain

L.H.S \[=5t-3=5\times \left( -1 \right)-3=-8\]

R.H.S \[=3t-5=3\times \left( -1 \right)-5=-3-5=-8\]

Hence, the result obtained above is correct.

3. Solve and check result: $5x+9-5+3x$

\[5x+9=5+3x\]

On Transposing \[3x\] to L.H.S and \[9\] to R.H.S, we obtain

\[5x-3x=5-9\]

L.H.S \[=5x+9=5\times \left( -2 \right)+9=-10+9=-1\]

R.H.S \[=5+3x=5+3\times \left( -2 \right)=5-6=-1\]

4. Solve and check result: $4z+3=6+2z$

\[4z+3=6+2z\]

On Transposing \[2z\] to L.H.S and \[3\] to R.H.S, we obtain

\[4z-2z=6-3\]

Dividing both sides by\[2\] , we obtain

L.H.S \[=4z+3=4\times \left( \frac{3}{2} \right)+3=6+3=9\]

R.H.S \[=6+2z=6+2\times \left( \frac{3}{2} \right)=6+3=9\]

5. Solve and check result: $2x-1=14-x$

\[2x-1=14-x\]

Transposing x to L.H.S and $1$ to R.H.S, we obtain

\[2x+x=14+1\]

Dividing both sides by \[3\], we obtain

L.H.S \[=2x-1=2\times \left( 5 \right)-1=10-1=9\]

R.H.S \[=14-x=14-5=9\]

6. Solve and check result: $8x+4=3\left( x-1 \right)+7$

\[8x+4=3\left( x-1 \right)+7\]

\[8x+4=3x-3+7\]

Transposing \[3x\] to L.H.S and $4$ to R.H.S, we obtain

\[8x-3x=-3+7-4\]

\[5x=-7+7\]

L.H.S \[=8x+4=8\times \left( 0 \right)+4=4\]

R.H.S \[=3\left( x-1 \right)+7=3\left( 0-1 \right)+7=-3+7=4\]

7. Solve and check result: $x=\frac{4}{5}\left( x+10 \right)$

\[x=\frac{4}{5}\left( x+10 \right)\]

Multiplying both sides by\[5\], we obtain

\[5x=4\left( x+10 \right)\]

\[5x=4x+40\]

Transposing \[4x\] to L.H.S, we obtain

L.H.S \[=x=40\]

R.H.S \[=\frac{4}{5}\left( x+10 \right)=\frac{4}{5}\left( 40+10 \right)=\frac{4}{5}\times 50=40\]

8. Solve and check result: $\frac{2x}{3}+1=\frac{7x}{15}+3$

\[\frac{2x}{3}+1=\frac{7x}{15}+3\]

Transposing \[\frac{7x}{15}\] to L.H.S and $1$ to R.H.S, we obtain

\[\frac{2x}{3}-\frac{7x}{15}=3-1\]

\[\frac{5\times 2x-7x}{15}=2\]

\[\frac{3x}{15}=2\]

\[\frac{x}{5}=2\]

Multiplying both sides by\[5\] , we obtain

L.H.S \[=\frac{2x}{3}+1=\frac{2\times 10}{3}+1=\frac{2\times 10+1\times 3}{3}=\frac{23}{3}\]

R.H.S\[=\frac{7x}{15}+3=\frac{7\times 10}{15}+3=\frac{7\times 2}{3}+3=\frac{14}{3}+3=\frac{14+3\times 3}{3}=\frac{23}{3}\]

9. Solve and check result: $2y+\frac{5}{3}=\frac{26}{3}-y$

\[2y+\frac{5}{3}=\frac{26}{3}-y\]

Transposing y to L.H.S and \[\frac{5}{3}\] to R.H.S, we obtain

\[2y+y=\frac{26}{3}-\frac{5}{3}\]

\[3y=\frac{21}{3}=7\]

Dividing both sides by$3$, we obtain

\[y=\frac{7}{3}\]

L.H.S \[=2y+\frac{5}{3}=2\times \frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}\]

R.H.S = \[\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}=\frac{19}{3}\]

L.H.S. = R.H.S. Hence, the result obtained above is correct.

10. Solve and check result: $3m=5m-\frac{8}{5}$

\[3m=5m-\frac{8}{5}\]

Transposing \[5m\] to L.H.S, we obtain

\[3m-5m=-\frac{8}{5}\]

\[-2m=-\frac{8}{5}\]

Dividing both sides by\[-2\] , we obtain

\[m=\frac{4}{5}\]

L.H.S \[=3m=3\times \frac{4}{5}=\frac{12}{5}\]

R.H.S \[5m-\frac{8}{5}=5\times \frac{4}{5}-\frac{8}{5}=\frac{12}{5}\]

Exercise 2.4

1. Amina thinks of a number and subtracts $\frac{5}{2}$ 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?

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

\[8x-20=3x\]

Transposing \[3x\]to L.H.S and \[-20\] to R.H.S, we obtain

\[8x-3x=20\]

Dividing both sides by \[5\], we obtain

Hence, the number 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?

Let the numbers be \[x\] and\[5x\] . According to the question,

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

\[21+5x=2x+42\]

Transposing \[2x\] to L.H.S and 21 to R.H.S, we obtain

\[5x-2x=42-21\]

Dividing both sides by \[3\], we obtain

\[x=7\]

\[5x=5\times 7=35\]

Hence, the numbers are $7\,\text{and }35$ respectively.

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?

Let the digits at tens place and ones place be \[x\,\,\text{and }9-x\] respectively.

Therefore, original number \[=10x+\left( 9-x \right)=9x+9\]

On interchanging the digits, the digits at ones place and tens place will be \[x\,\,\text{and }9-x\] respectively.

Therefore, new number after interchanging the digits \[=10\left( 9-x \right)+x\]

\[=90-10x+x\]

New number = Original number\[+27\]

\[90-9x=9x+9+27\]

\[90-9x=9x+36\]

Transposing \[9x\]to R.H.S and \[36\] to L.H.S, we obtain

\[90-36=18x\]

Dividing both sides by \[18\], we obtain

\[3=x\text{ and }9-x=6\]

Hence, the digits at tens place and ones place of the number are \[3\text{ and }6\] respectively. Therefore, the two-digit number is\[9x+9=9\times 3+9=36\]

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$88$. What is the original number?

Let the digits at tens place and ones place be \[x\text{ and }3x\] respectively.

Therefore, original number \[=10x+3x=13x\]

On interchanging the digits, the digits at ones place and tens place will be \[x\text{ and }3x\]respectively.

Number after interchanging \[=10x+3x=13x\]

Original number $+$ New number\[=88\]

\[13x+31x=88\]

Dividing both sides by \[44\], we obtain

Therefore, original number \[=13x=13\times 2=26\]

By considering the tens place and ones place as \[3x\text{ and }x\]respectively, the two-digit number obtained is \[62\].

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

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?

Let Shobo’s age be x years. Therefore, his mother’s age will be $6x$ years.

After \[5\text{ years, Shobo }\!\!'\!\!\text{ s age}=\frac{\text{Shobo }\!\!'\!\!\text{ s mother }\!\!'\!\!\text{ s present age}}{3}\]

\[x+5=\frac{6x}{3}\]

Transposing x to R.H.S, we obtain

\[6x=6\times 5=30\]

Therefore, the present ages of Shobo and Shobo’s mother will be \[5\text{ years and }30\,\text{years}\] respectively.

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 $\text{Rs }100$ per metre it will cost the village panchayat $\text{Rs }75,000$ to fence the plot. What are the dimensions of the plot?

Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be $11x$ m and $4x$ m respectively.

Perimeter of the plot \[=2\] (Length + Breadth) \[=\left[ 2\left( 11x+4x \right) \right]\text{m}=30x\text{ m}\]

It is given that the cost of fencing the plot at the rate of \[\text{Rs }100\] per metre is\[\text{Rs 75,000}\].

\[\therefore 100\times \text{Perimeter}=75000\]

\[100\times 30x=75000\]

\[3000x=75000\]

Dividing both sides by \[3000\], we obtain

Length \[=11\times \text{m}=\left( 11\times 25 \right)\text{m}=275\,\text{m}\]

Breadth \[=4\times \text{m}=\left( 4\times 25 \right)\text{m}=100\,\text{m}\]

Hence, the dimensions of the plot are 2\[275\text{ m and }100\text{ m}\] respectively.

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him $\text{Rs 50}$ per metre and trouser material that costs him $\text{Rs 90}$ per metre. For every $2$ meters of the trouser material he buys $3$metres of the shirt material. He sells the materials at $12%\text{ and }10%$ profit respectively. His total sale is $\text{Rs 36660}$. How much trouser material did he buy?

Let $2x$ m of trouser material and $3x$ m of shirt material be bought by him.

Per metre selling price of trouser material \[\text{=}~\text{Rs }\left( 90+\frac{90\times 12}{100} \right)\text{=Rs 100}\text{.80}\]

Per metre selling price of shirt material \[\text{=}~\text{Rs }\left( 50+\frac{50\times 12}{100} \right)\text{=Rs 55}\]

Given that, total amount of selling \[\text{=}~\text{Rs }36660\]

\[100.80\times \left( 2x \right)+55\times \left( 3x \right)=36660\]

\[201.60x+165x=36660\]

\[366.60x=36660\]

Dividing both sides by \[366.60\], we obtain

Trouser material \[=2\times m=\left( 2\times 100 \right)m=200m\]

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.

Let the number of deer be x.

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

Number of deer playing nearby\[=\frac{3}{4}\times \text{Number of remaining deer}\]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{4}\times \left( x-\frac{x}{2} \right)=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}\]

Number of deer drinking water from the pond $=9$

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

\[x-\left( \frac{4x+3x}{8} \right)=9\]

\[x-\frac{7x}{8}=9\]

\[\frac{x}{8}=9\]

Multiplying both sides by\[8\], we obtain

Hence, the total number of deer in the herd is \[72\].

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

Let the granddaughter’s age be x years. Therefore, grandfather’s age will be \[10x\] years.

According to the question,

Grandfather’s age$=$ Granddaughter’s age \[+\,54\] years

\[10x=x+54\]

Transposing x to L.H.S, we obtain

\[10x-x=54\]

\[9x=54\]

Granddaughter’s age \[=x\] years \[=6\] years

Grandfather’s age \[=10x\] years \[=\left( 10\times 6 \right)\] years \[=60\] 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.

Let Aman’s son’s age be x years. Therefore, Aman’s age will be $3x$ years. Ten years ago, their age was \[\left( x-10 \right)\] years and \[\left( 3x-10 \right)\] years respectively.

\[10\]years ago, Aman’s age \[=5\times \]Aman’s son’s age \[10\] years ago

\[3x-10=5\left( x-10 \right)\]

\[3x-10=5x-50\]

Transposing \[3x\] to R.H.S and \[50\] to L.H.S, we obtain

\[50-10=5x-3x\]

Dividing both sides by\[2\], we obtain

Aman’s son’s age \[=x\] years \[=20\] years

Aman’s age \[=3x\] years \[=\left( 3\times 20 \right)\] years \[=60\] years

Exercise 2.5

1. Solve the linear equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

\[\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\]

L.C.M. of the denominators, \[2,3,4,\text{and 5,}\]is.

Multiplying both sides by \[60\] , we obtain

\[60\left( \frac{x}{2}-\frac{1}{5} \right)=60\left( \frac{x}{3}+\frac{1}{4} \right)\]

\[\Rightarrow 30x-12=20x+15\] (Opening the brackets)

\[\Rightarrow 30x-20x=15+12\]

\[\Rightarrow 10x=27\]

\[\Rightarrow x=\frac{27}{10}\]

2. Solve the linear equation$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

\[\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21\]

L.C.M. of the denominators, \[2,4,\text{ and }6\text{ is }12\].

Multiplying both sides by\[12\], we obtain

\[6n-9n+10n=252\]

\[\Rightarrow 7n=252\]

\[\Rightarrow n=\frac{252}{7}\]

\[\Rightarrow n=36\]

3. Solve the linear equation $x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

\[x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}\]

L.C.M. of the denominators, \[2,3,\text{ and }6,\text{is }6\].

Multiplying both sides by \[6\], we obtain

\[6x+42-16x=17-15x\]

\[\Rightarrow 6x-16x+15x=17-42\]

\[\Rightarrow 5x=-25\]

\[\Rightarrow x=\frac{-25}{5}\]

\[\Rightarrow x=-5\]

4. Solve the linear equation $\frac{x-5}{3}=\frac{x-3}{5}$

\[\frac{x-5}{3}=\frac{x-3}{5}\]

L.C.M. of the denominators, \[3\text{ and }5,\text{ is }15\].

Multiplying both sides by\[15\], we obtain

\[5\left( x-5 \right)=3\left( x-3 \right)\]

\[\Rightarrow 5x-25=3x-9\] (Opening the brackets)

\[\Rightarrow 5x-3x=25-9\]

\[\Rightarrow 2x=16\]

\[\Rightarrow x=\frac{16}{2}\]

\[\Rightarrow x=8\]

5. Solve the linear equation $\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

\[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\]

L.C.M. of the denominators, \[3\text{ and }4,\text{is}\,12\].

Multiplying both sides by \[12\], we obtain

\[3\left( 3t-2 \right)-4\left( 2t+3 \right)=8-12t\]

\[\Rightarrow 9t-6-8t-12=8-12t\] (Opening the brackets)

\[\Rightarrow 9t-8t+12t=8+6+12\]

\[\Rightarrow 13t=26\]

\[\Rightarrow t=\frac{26}{13}\]

\[\Rightarrow t=2\]

6. Solve the linear equation$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

\[m-\frac{m-1}{2}=1-\frac{m-2}{3}\]

L.C.M. of the denominators, \[2\text{ and }3,\text{ is}\,\text{ }6\].

\[6m-3\left( m-1 \right)=6-2\left( m-2 \right)\]

\[\Rightarrow 6m-3m+3=6-2m+4\] (Opening the brackets)

\[\Rightarrow 6m-3m+2m=6+4-3\]

\[\Rightarrow 5m=7\]

\[\Rightarrow m=\frac{7}{5}\]

7. Simplify and solve the linear equation $3\left( t-3 \right)=5\left( 2t+1 \right)$

\[3\left( t-3 \right)=5\left( 2t+1 \right)\]

\[\Rightarrow 3t-9=10t+5\] (Opening the brackets)

\[\Rightarrow -9-5=10t-3t\]

\[\Rightarrow -14=7t\]

\[\Rightarrow t=\frac{-14}{7}\]

\[\Rightarrow t=-2\]

8. Simplify and solve the linear equation$15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0$

\[15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0\]

\[\Rightarrow 15y-60-2y+18+5y+30=0\] (Opening the brackets)

\[\Rightarrow 18y-12=0\]

\[\Rightarrow 18y=12\]

\[\Rightarrow y=\frac{12}{8}=\frac{2}{3}\]

9.Simplify and solve the linear equation $3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17$

\[3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17\]

\[\Rightarrow 15z-21-18z+22=32z-52-17\] (Opening the brackets)

\[\Rightarrow -3z+1=32z-69\]

\[\Rightarrow -3z-32z=-69-1\]

\[\Rightarrow -35z=-70\]

\[\Rightarrow z=\frac{70}{~35}=2\]

10. Simplify and solve the linear equation $0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)$

\[0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)\]

$\frac{1}{4}\left( 4f-3 \right)=\frac{1}{20}\left( 10f-9 \right)$

Multiplying both sides by\[20\], we obtain

\[5\left( 4f-3 \right)=10f-9\]

\[\Rightarrow 20f-15=10f-9\] (Opening the brackets)

\[\Rightarrow 20f-10f=-9+15\]

\[\Rightarrow 10f=6\]

\[\Rightarrow f=\frac{3}{5}=0.6\]

Exercise 2.6

1. Solve: $\frac{8x-3}{3x}=2$

\[\frac{8x-3}{3x}=2\]

On multiplying both sides by\[3x\] , we obtain

\[8x-3=6x\]

\[\Rightarrow 8x-6x=3\]

\[\Rightarrow 2x=3\]

\[\Rightarrow x=\frac{3}{2}\]

2. Solve: $\frac{9x}{7-6x}=15$

\[\frac{9x}{7-6x}=15\]

On multiplying both sides by\[7-6x\], we obtain

\[9x=15\left( 7-6x \right)\]

\[\Rightarrow 9x=105-90x\]

\[\Rightarrow 9x+90x=105\]

\[\Rightarrow 99x=105\]

\[\Rightarrow x=\frac{105}{99}=\frac{35}{33}\]

3. Solve:

$\frac{z}{z+15}=\frac{4}{9}$

\[\frac{z}{z+15}=\frac{4}{9}\]

On multiplying both sides by \[9\left( z+15 \right)\], we obtain

\[9z=4\left( z+15 \right)\]

\[\Rightarrow 9z=4z+60\]

\[\Rightarrow 9z-4z=60\]

\[\Rightarrow 5z=60\]

\[\Rightarrow z=12\]

4. Solve: $\frac{3y+4}{2-6y}=\frac{-2}{5}$

\[\frac{3y+4}{2-6y}=\frac{-2}{5}\]

On multiplying both sides by\[5\left( 2-6y \right)\], we obtain

\[5\left( 3y+4 \right)=-2\left( 2-6y \right)\]

\[\Rightarrow 15y+20=-4+12y\]

\[\Rightarrow 15y-12y=-4-20\]

\[\Rightarrow 3y=-24\]

\[\Rightarrow y=-8\]

5. Solve: $\frac{7y+4}{y+2}=-\frac{4}{3}$

\[\frac{7y+4}{y+2}=-\frac{4}{3}\]

On multiplying both sides by\[3\left( y+2 \right)\], we obtain

\[3\left( 7y+4 \right)=-4\left( y+2 \right)\]

\[\Rightarrow 21y+12=-4y-8\]

\[\Rightarrow 21y+4y=-8-12\]

\[\Rightarrow 25y=-20\]

\[\Rightarrow y=-\frac{4}{5}\]

6. The ages of Hari and Harry are in the ratio $5:7$. Four years from now the ratio of their ages will be$3:4$. Find their present ages.

Let the common ratio between their ages be x. Therefore, Hari’s age and Harry’s age will be\[5x\]years and \[7x\] years respectively and four years later, their ages will be \[\left( 5x+4 \right)\] years \[\left( 7x+4 \right)\] years respectively.

According to the situation given in the question,

\[\frac{5x+4}{7x+4}=\frac{3}{4}\]

\[\Rightarrow 4\left( 5x+4 \right)=3\left( 7x+4 \right)\]

\[\Rightarrow 20x+16=21x+12\]

\[\Rightarrow 16-12=21x-20x\]

\[\Rightarrow 4=x\]

Hari’s age \[=5x\] years \[=\left( 5\times 4 \right)\] years \[=20\] years

Harry’s age \[=7x\] years \[=\left( 7\times 4 \right)\] years \[=28\] years

Therefore, Hari’s age and Harry’s age are \[20\] years and \[28\] years respectively.

7. The denominator of a rational number is greater than its numerator by $8$. If the numerator is increased by $17$ and the denominator is decreased by$1$, the number obtained is. Find the rational number.

Let the numerator of the rational number be\[x\]. Therefore, its denominator will be\[x+8\]

The rational number will be\[\frac{x}{x+8}\]. According to the question,

\[\frac{x+17}{x+8-1}=\frac{3}{2}\]$$

\[\Rightarrow \frac{x+17}{x+7}=\frac{3}{2}\]

\[\Rightarrow 2\left( x+17 \right)=3\left( x+7 \right)\]

\[\Rightarrow 2x+34=3x+21\]

\[\Rightarrow 34-21=3x-2x\]

\[\Rightarrow 13=x\]

Numerator of the rational number \[=x=13\]

Denominator of the rational number \[=x+8=13+8=21\]

Rational number\[=\frac{13}{21}\]

Obtaining a PDF file for all of the solutions in Linear Equations in One Variable Class 8 is now straightforward. Simply browse to Vedantu's official website and download the NCERT Answers Class 8 Mathematics Chapter 2 PDF on your devices at any time. After you've downloaded them, you won't need an internet connection to rapidly review some formulae or important Chapter 2 Class 8 Mathematics subjects. The solutions are also printable, so you may keep a tangible copy with you at the test centre.

## Chapter 2 - Linear Equations in One Variable

2.1 Introduction

In the introduction part of NCERT Maths Class 8 Chapter 2, you will be reminded of algebraic equations and expressions of earlier days. Those are of the format shown below:

Expresions - 5x + y, x + y, y + y 2 , etc.

Equations - 5x + y = 10, x + y = -2, y + y 2 = 9 etc.

From these examples, you can see that equations have an equality sign (=), which is not present in expressions. Algebraic expression involves variables, constants, and some mathematical operations like addition or multiplication. An equation is an expression that equates two expressions.

Students would learn the history of algebraic equations and their definitions. They would also get an idea of what is coming up in the chapter in this section.

## 2.2 Solving Equations With Linear Expression on One SIde and Numbers on the Other Side

This section of Maths NCERT Solutions Class 8 Chapter 2 starts by recalling the method for solving equations like the one shown below:

5x - 3 = 8

We solve this by adding 3 on both sides so we get

5x - 3 + 3 = 8 + 3

5x = 11, hence x = 2.2

The above example is a linear expression where the highest power of a variable is only 1. A linear equation that involves one variable can be marked on a straight line. A linear equation can be in one or more than one variable.

You are then presented with a few examples which are based on the method of solving linear equations as described above.

## 2.3 Some Applications

In this topic, students will go through some applications of linear equations which are in the form of puzzles. There are many wordy samples solved which involve some real-life situations like calculating age, counting money, etc.

## 2.4 Solving Equations that Have Variables on Both Sides

Till now, the chapter has dealt with equations where the value on the right-hand side of the equality sign has been only numbers. In this section, we will look into problems where there are variables on both sides of the equation. An example of such an equation is:

2x - 5 = x +3

Solution -> 2x = x + 3 + 5

Subtracting x from both sides we get 2x - x = x + 8 - x

So, x = 8

You are then given many solved examples of such equations and an exercise on it with 10 questions.

## 2.6 Reducing Equations to Simpler Forms

A complex linear equation that has fractions in it can be reduced into a simpler form by the process described below:

Take the LCM(Least common multiple) of the denominator.

Now multiply both RHS and LHS of the equation with this LCM.

By applying the above multiplication, the equation gets reduced to a form without the denominator in it.

Then you can apply the methods learned in the sections above to solve such problems.

We will explain this with an example:

x/2 - 1/5 = x/3 + 1/4 + 1

To reduce the above equation into a simpler form, let us take the LCM of the denominators, i.e. 2, 5, 3, and 4, which is 60. Now multiply each term on both sides of the equation with 60.

60* x/2 - 1/5 * 60 = x/3 * 60 + 1/4 * 60 + 1 * 60

30x - 12 = 20x + 15 + 60

30x - 12 = 20x + 75

Now we will move all the expressions with variables to the LHS and all the constant to the RHS, so we get:

30x - 20x = 75 + 12

X = 87/10 = 8.7

## NCERT Solutions Class 8 Maths Chapter 2 Exercises

Important topics under ncert solutions for class 8 maths chapter 2 linear equations in one variable.

Chapter 2 of the class 8 maths syllabus includes Linear Equations in One Variable, which is a very important chapter in mathematics that is covered in class 8. The chapter on Linear Equations in One Variable is divided into 5 major sections. The following is a list of the important topics covered under Linear Equations in One Variable. We recommend that students go through every topic carefully to make sure they don’t miss out on learning the concept of Linear Equations in One Variable, properly.

Introduction

Solving equations where linear expressions are on one side and numbers are on the other side of the equation

Applications of linear equations in one variable

Solving equations with variables on both the sides

Reducing the given equations to a simpler form

## Benefits of NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

Students would benefit greatly from Vedantu's NCERT Answers for Class 8 Maths Chapter 2 Linear Equations in One Variable if they want to achieve more than 90 in their Class 8 Mathematics examinations. Our solutions are best-in-class and can assist students in understanding the answers to the problems in this chapter of the NCERT textbook.

## Quick Revision

The following is a list of key points that have been covered in the solutions and are required to be remembered while studying the chapter on Linear Equations in One Variable.

The multiplication and division of algebraic expressions

Factorisation

Some Common Errors

Methods to solve linear equations in one variable in contextual problems that involve multiplication and division operations

## Important Formula Related to Linear Equations in One Variable

The standard form of a linear equation in one variable is ax + b = 0, where, a ≠ 0 and x is the variable.

## Key Learnings from NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

Division and multiplication of algebraic expressions with integers as coefficients

Common errors faced by students

Introduction to identities

Process of solving linear equations in one variable

## Importance of Class 8 Maths Chapter 2 Linear Equations in One Variable Linear Equations

When there are two or more integers and one of them is unknown, then linear equations in one variable can be used for calculating the value of this integer. It is possible to easily find the value of an unknown integer using the expression in the equation form. We encourage students to learn as much as they can from this chapter to be able to solve problems linear equations easily in exams.

## Key Features of NCERT Solutions for Class 8 Maths Chapter 2

The Class 8 Maths Chapter 2 Solution provided by Vedantu is the most reliable online resource for the revision and preparation of CBSE exams. The key benefits of these solutions are:

The solutions are simple to understand and broken into steps so that students can grasp the concept perfectly.

You can also download the solutions in PDF format or print them for a group study, making exam time revisions quick and convenient.

The solutions are based entirely on the CBSE curriculum; hence you are well prepared for your exams once you go through our solutions.

## Conclusion

When there are two or more integers and one of them is unknown, linear equations in one variable can be utilized to determine its value. With the statement in equation form, it is simple to find the value of an unknown number. We recommend students to absorb as much as they can from this chapter in order to be able to answer linear equation problems efficiently in tests.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations In One Variable

1. Give an example of an equation that is not linear but can be reduced to a linear form.

Sometimes we come across equations that are not linear as per the definition of linear equations but can be reduced to a linear form and then the method of solving linear equations can be applied to them to solve them. The example below illustrates one such equation and how to solve it:

(X + 1)/(2x + 6) = 3/8

To reduce this nonlinear equation into a linear equation we multiply both sides by the denominator of LHS which is 2x + 6.

(X + 1)/(2x + 6) * (2x + 6) = 3/8 * (2x + 6)

X + 1 = 6x + 18/8 - This is a linear equation now. We can solve it by multiplying both sides with LCM of denominators which is 8

8 * (x + 1) = 8 * (6x + 18/8)

8x + 8 = 6x + 18

Now moving all variable to LHS and all constant to RHS we get:

8x - 6x = 18 - 8

2. Mention some of the important features of a linear equation.

A linear equation is characterized by the following key properties:

The highest power of the variable involved in a linear equation is 1.

The linear equation can have one or two variables in it.

A linear equation has an equality sign. The expression on the left side of the equality is called LHS (left-hand side), and the expression on the right side of the equality sign is termed as RHS (right-hand side).

The two expressions, LHS and RHS, are equal for only certain values of the variables. These values of the variables are the solutions of the linear equation.

The points of a linear equation with just one variable can be marked on the number line.

3. What are the sub-topics covered in Chapter 2 of Class 8 Maths?

The concepts covered in Chapter 2 of Class 8 Maths are :

2.2 Solving equations which have a Linear expression on one side and numbers on the other side

2.3 Some applications

2.4 Solving equations having the variables on both sides

2.5 Some more applications

2.6 Reducing equations to a simpler form

2.7 Equations reducible to linear form.

4. How many exercises are there in Chapter 2 of Class 8 Maths?

Chapter 2, “Linear Equations In One Variable”, consists of 6 exercises -

Exercise 2.1 contains 12 questions.

Exercise 2.2 contains 16 questions.

Exercise 2.3 contains 10 questions.

Exercise 2.4 contains 10 questions.

Exercise 2.5 contains 10 questions.

Exercise 2.6 contains 10 questions.

Exercise 2.7 contains 7 questions.

Students can practise all the questions in the NCERT solutions designed by the experts at Vedantu to get well versed in this chapter. Download these solutions on the Vedantu website or the app for free of cost to prepare for the exams. These questions will help the students to secure a perfect score in the exams.

5. Do I need to practice all the questions given in NCERT Solutions?

Yes, you need to practice all the questions given in the NCERT Solutions. NCERT Solutions prepared by the experts at Vedantu will help the students clear all their doubts in the chapter and practice more to get a perfect score in the exam. You can download the solutions of this chapter by clicking on NCERT Solutions for Class 8 Maths Chapter 2 to ace your exams. These solutions are available on Vedantu (vedantu.com) free of cost. You can download it through the Vedantu app as well.

6. Are NCERT Solutions Chapter 2 Class 8 Maths important for your exams?

Undoubtedly, NCERT Answers Chapter 2 Class 8 Mathematics are significant for your examinations because they include the majority of the problems. Vedantu's NCERT Answers Chapter 2 Class 8 Mathematics assist students in gaining a thorough understanding of all subjects. These solutions assist students in revising and practising all of the issues handled by professionals before to tests.

7. Where to download NCERT Solutions for Chapter 2 Class 8 Maths?

Students of Class 8 can download the NCERT Solutions for Chapter 2 using the link NCERT Solutions for Class 8 Maths Chapter 2 from Vedantu’s app or website (vedantu.com). Since the solutions are based on the CBSE syllabus, students will be able to practice all the problems. The experts at Vedantu prepare these solutions, keeping in mind the students so that they can excel in their exams.

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## Chapter 2 Class 8 Linear Equations in One Variable

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Get NCERT Solutions of all Exercise Questions and Examples of Chapter 2 Class 8 Linear Equations in One Variable free at Teachoo. Answers to each and every question with detailed explanation for your understanding. Best answers guaranteed!

We studied Equations and Expressions in Algebra ( Chapter 9 Class 8 ). Equation has an = sign, expression does not have it

In this chapter, we will learn

- What is an Equation
- What is a Linear Equation
- What is a Linear Equation in one variable
- Solving Linear Equations - by taking variables on one side, numbers on the side
- Making Equations from Statements , and then Solving.
- Perimeter and Area Questions
- Fraction questions - where we need to find Numerator or Denominator
- Cost, Sale, Profit Questions
- Questions where we have Consecutive Numbers
- Questions involving Consecutive Multiples
- Coins and Currency Notes Questions
- Questions where numbers are added or subtracted
- Questions where a Two Digit number is given, and we reverse its digits
- Questions involving Age

Click on an exercise or topic link below to start doing the chapter.

Note: Important Questions of this chapter have been marked. When you click on a link, at the bottom of the page there is a list with arrows. In that list, you can find all the questions of that exercise and find Important Questions marked as blue.

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## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 2 are in accordance with the latest CBSE guidelines and marking schemes.

## Class 8 Maths Chapter 2 Exercise 2.1 Solutions

## Class 8 Maths Chapter 2 Exercise 2.2 Solutions

## Class 8 Maths Chapter 2 Exercise 2.3 Solutions

## Class 8 Maths Chapter 2 Exercise 2.4 Solutions

## Class 8 Maths Chapter 2 Exercise 2.5 Solutions

## Class 8 Maths Chapter 2 Exercise 2.6 Solutions

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## Unit 2: Linear equations in one variable

Solving equations with variable on one side.

- Worked example: two-step equations (Opens a modal)
- Equation with variables on both sides: fractions (Opens a modal)
- Two-step equations Get 5 of 7 questions to level up!
- Equations with variables on both sides: decimals & fractions Get 3 of 4 questions to level up!

## Solving equations with variables on both sides

- Equations with parentheses (Opens a modal)
- Equations with variables on both sides Get 3 of 4 questions to level up!
- Equations with parentheses: decimals & fractions Get 3 of 4 questions to level up!

## Equations reducible to linear form

- No videos or articles available in this lesson
- Solving equations reducible to linear form Get 3 of 4 questions to level up!

## Linear equations word problems

- Two-step equation word problem: garden (Opens a modal)
- Sums of consecutive integers (Opens a modal)
- Two-step equation word problem: oranges (Opens a modal)
- Word problems linear equations (basic) Get 3 of 4 questions to level up!
- Sums of consecutive integers Get 3 of 4 questions to level up!
- Linear equations word problems (advanced) Get 3 of 4 questions to level up!

- CBSE-Linear Equations in One Variable
- Important Questions

## Linear Equations in One Variable-Important Questions

- STUDY MATERIAL FOR CBSE CLASS 8 MATH
- Chapter 1 - Algebraic Expressions and Identities
- Chapter 2 - Comparing Quantities
- Chapter 3 - Cubes and Cube Roots
- Chapter 4 - Data handling
- Chapter 5 - Direct and Inverse Proportions
- Chapter 6 - Exponents and Powers
- Chapter 7 - Factorization
- Chapter 8 - Introduction to Graphs
- Chapter 9 - Mensuration
- Chapter 10 - Playing with Numbers
- Chapter 11 - Practical Geometry
- Chapter 12 - Squares and Square Roots
- Chapter 13 - Visualizing Solid Shapes
- Chapter 14 - Linear Equations in One Variable
- Chapter 15 - Rational Numbers
- Chapter 16 - Understanding Quadrilaterals

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## Important Questions for CBSE Class 8 Maths Chapter 2 – Linear Equations in One Variable

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## CBSE Class 8 Maths Important Questions for Linear Equations in One Variable

Maths is used worldwide as a necessary element in many different fields. There are several uses of Maths, and having a solid foundation in the subject will benefit a variety of careers. Since Maths is entirely based on numbers and is not a language-based subject, you can either get the answer right or wrong, which increases your chance of receiving full marks for this subject.

Quick Links

Chapter 2 of Class 8 Maths is called ‘Linear Equations in One Variable’. These are the basic equations used to represent and resolve an unknown quantity. It is represented with the help of graphs by using straight lines. A linear equation is a simple way of representing maths equations. The unknown quantities can be represented using the variable ‘X’. A linear equation can be solved in various simple methods. All the variables are placed on one side, and the rest is isolated on the other to obtain the value of the unknown quantity. In a linear equation, the degree of the equation is exactly equal to 1. Similarly, there is only one variable, and we can only obtain one solution. It is drawn in a graph, and it happens to appear either vertically or horizontally.

Extramarks is one of the best learning platforms for students to achieve excellent scores. It helps students by providing them with important questions from the CBSE syllabus. The Extramarks team refers to NCERT books, CBSE sample papers, CBSE revision notes, CBSE past year’s questions, etc., and provides you with the solutions to important questions. Students can get a good hold on this chapter, by solving our class 8 linear equations extra questions .

To ace their exams, students can register on the Extramarks website to access Class 8 Maths Chapter 2 important questions, CBSE extra questions, Maths formulas, and much more.

## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

## Linear Equations Class 8 Extra Questions w ith Solution

Mentioned below are a few of the questions and their answers from our Chapter 2 Class 8 Maths important questions.

Question 1: The perimeter of a rectangular swimming pool is 154m. Its length is

2m, more than twice its breadth. What is the length and the breadth of the pool?

Answer 1: Let the breadth of the swimming pool be x m.

The length of the swimming pool will be = (2x + 2) m.

Perimeter of swimming pool:- 2 (l + b) =154

2 (2x + 2 + x)=154

2 (3x + 2)=154

∴Dividing both sides by 2, we obtain

(3x + 2)=77

On transporting two on the R.H.S., we get

3x = 77 – 2

3x = 75

x= 75/3

x= 25 m

Hence, the breadth of the swimming pool is x= 25m

The length of the swimming pool will be= (2x + 2) m.

=(2 х 25 + 2) m

=(50 + 2) m

=52 m

Thus, the length of the swimming pool is 52m, and the breadth of the swimming pool is 25m.

Question 2: What is the share of A when Rs 25 are divided between A and B so that A gets Rs 8 more than B is 16.5?

Answer 2: Let the share of B be x.

Let the share of A be (x + 8).

From this, we get,

x + x + 8 = 25

2x = 25 – 8

2x= 17

x = 17/2

x = 8.5

Therefore, A’s share will be 8.5.

Question 3: Find three consecutive odd numbers whose sum is 147.

Answer 3: Let the first, second, and third consecutive odd numbers be (2x +1),(2x + 3) and (2x + 5), respectively.

Hence the sum of the consecutive odd numbers is

(2x + 1) + (2x + 3) + (2x + 5)= 147.

On further simplifying, we get

2x + 2x + 2x + 1 + 3 + 5=147

6x + 9= 147.

On rearranging, we obtain

6x= 147 – 9

X= 138/6=23,

So the three consecutive odd numbers are (2x + 1)= 47

(2x + 3)= 49

(2x + 5)= 51.

Question 4: Ram’s father is 26 years younger than Ram’s grandfather and 29 years older than Ram. The sum of the ages of all three is 135 years. What is the age of each one of them?

Answer 4: Let Ram’s present age be x years

Rams father’s present age is = (x + 29) years

Rams grandfather’s present age =(x + 29 + 26) years

The sum of all three ages adds up to 135 years

Hence,

x + (x + 29) + (x + 29 + 26)= 135

x + x + x + 29 + 29 + 26 =135

3x + 84= 135

3x = 135-84

3x = 51

x= 51/3

x= 17

Hence, Ram’s present age is x=17 years

Ram’s father’s present age =(x + 29)

=(17 + 29)

=46 years

Ram’s grandfather’s age =(x + 29 + 26)

=(17 + 29 + 26)= 72 years

Question 5: If 8x – 3 +17x, then x ________.

- is a fraction
- is an integer
- is a rational number
- cannot be solved

Answer 5: (C) A rational number

Given:- 8x-3=25+17x

Moving -3 to R.H.S. and becomes 3 and 17x to L.H.S.

8x – 17x = 25 +3

Thus, x is a rational number.

Question 6: 3x+2/3=2x+1

Answer 6: 3x+ 2/3 = 2x +1

By transposing the above equation, we get

3x+2=3(2x+1)

By moving all the variables on the L.H.S., we get,

Question 7: The angles of a triangle are in the ratio 2 : 3: 4. Find the angles of the triangle.

Answer 7: Let the angles of the triangle be 2x°, 3x° and 4x°.

From the given question, we get,

2x + 3x + 4x = 180

∵ The sum of all the angles of a triangle is 180°)

⇒ x = 20……….. (Transposing 9 to R.H.S.)

Hence, The angles of the given triangle are

2× 20 = 40°,

3 × 20 = 60°,

4 × 20 = 80°.

Question 8: The sum of the two numbers is 95. If one exceeds the other by 15, find the numbers.

Answer 8: Let the smaller number be x.

Then, the larger number =x +15.

According to the question,

the sum of the two numbers is 95

x + (x + 15) =95

2x + 15 =95 ………..(transposing 15 to the R.H.S.)

2x= 80

x=80/2

x= 40

Hence, the smaller number is 40

The larger number is (x + 15)= 40 +15=55

Hence, the required numbers are 40 and 55

Question 9: If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is

Answer 9: (B) -13/19

Given :- (5x/3) – 4 = (2x/5)

(5x/3) – (2x/5) = 4

L.C.M. of 3 and 5 is 15

(25x – 6x)/15 = 4

19x = 4 × 15

Substituting x=60/19 in the given equation,

= (2 × (60/19)) – 7

= (120/19) – 7

= (120 – 133)/19

= – 13/19

Question 10: 9x + 5 = 4(x – 2)+ 8

Answer 10: 9x + 5= 4(x – 2) + 8,

By transposing the above equation, we get,

9x + 5= 4x – 8 + 8

9x – 4x =5

Again by transposing

5x=5

X=5/5

X=1

Question 11: The sum of three consecutive multiples of 8 is 888. Find the multiple.

Answer 11: Let the three consecutive multiples be x, x + 8, x + 16

According to the given question,

The sum of three consecutive multiples of 8 is 888

x + x + 8 + x + 16 = 888

3x + 24 = 888

3x= 888 – 24

3x =864

x =864/3

x =288

Therefore the three consecutive multiples are:

x =288

x + 8=296

x + 16=304, respectively.

Question 12: A rational number is such that when you multiply it by 5/2 and add 2/3 to get -7/12. What is the number?

Answer 12: Let the rational number be x

According to the question,

X x (5/2) + 2/3 =-7/12

5x/2 +2/3 =-7/12

5x/2= -7/12 – 2/3

Taking L.C.M. on the R.H.S.

5x/2 = (-7-8)/12

5x/2 = -15/12

5x/2 = -5/4

x= (-5/4) x (2/5)

x=-10/20

x= -1/2

Therefore, the rational number is -1/2

Question 13: Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.

Answer 13: Let the number be x.

According to the question, we get

(1/5)x + 5 = (1/4) x – 5

On rearranging the given equation,

(1/5) x – (1/4) x = -5-5

(1/5) x – (1/4)x =-10

By taking L.C.M., we will get,

(4x-5x)/20=-10

Again by transposing

-x= -200

x= 200

Question 14: The sum of two numbers is 11, and their difference is 5. Find the numbers.

Answer 14: Let one of the numbers from the two numbers be x.

Let the other number = 11 – x.

As per the given conditions, we have

x – (11 – x) = 5

⇒ x – 11 + x =5

⇒ 2x – 11 = 5

⇒ 2x = 5 + 11……………… (Transposing 11 to R.H.S.)

⇒ 2x = 16

⇒ x = 8

Hence, the required numbers for the given question are 8 and 11 – 8 = 3, respectively.

Question 15: The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Answer 15: Let the number of boys in a class be 7x

Let the number of girls in a class be 5x

7x = 5x + 8

7x -5x = 8

2x = 8

x=8/2

x = 4

Therefore, Number of boys = 7 x 4 =28

Number of girls = 5 x 4 =20

Total number of students = 20+ 28= 48

Question 16:- Linear equation in one variable has ________.

- only one variable with any power
- only one term with a variable
- only one variable with power 1
- only constant term

Answer 16:- (C) Only one variable with power 1

Question 17:- Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x. Then Arpita’s present age is

- 3(x – 3)
- 3x – 9

Answer 17:- (D) 3(X + 3)

Given:- Shilpa’s age three years ago was x

Then, Shilpa’s present age is= x + 3

Arpita’s present age is thrice of Shilpa’s age =3(x + 3)

Question 18: Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.

Answer 18 : 4.4x – 3.8 = 5

Putting x = 2, we have

4.4 × 2 – 3.8 = 5

⇒ 8.8 – 3.8 = 5

⇒ 5 = 5

L.H.S. = R.H.S.

Hence, it is verified.

## Benefits Of Solving Important Questions Class 8 Maths Chapter 2

The subject of Maths requires a lot of practice. The Maths of Classes 8, 9, and 10 are the milestones of fundamental knowledge. We suggest students access our platform Extramarks to access linear equations class 8 extra questions . By systematically solving questions and going through the required solutions, students will gain confidence to solve complex questions from linear equations with one variable.

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Q.1 Five years ago, the age of Neeraj and Neera was in the ratio 4:5. The ratio of their present ages is 5:6. Find their present ages.

Marks: 4 Ans

Let the present age of Neeraj and Neera be 5x and 6x years respectively. F ive years before, Neeraj ‘ s age=5x-5 years Neera ‘ s age=6x-5 years According to the question, 5x-5 6x-5 = 4 5 5 5 x 5 = 4 6 x 5 25 x 25 = 24 x 20 25 x 24 x = 25 20 x = 5 Therefore , the present age of Neeraj=5x=5 × 5=25 years and the present age of Neera=6x=6 × 5=30 years.

Q.2 The sum of two numbers is 184. One-third of one number exceeds one-seventh of the other number by 8. Find the two numbers.

Marks: 3 Ans

Let one number be x , then the other be 184 x . According to the question , 1 3 x = 1 7 184 x + 8 x 3 = 184 x + 56 7 7 x = 3 240 x 7 x = 720 3 x 10 x = 720 x = 720 10 x = 72 So , One number = 72 Secondnumber=184 72 = 112 .

Solve the equation: 3 4 y + 7 2 5 y 4 = 5 4

Marks: 2 Ans

3 y + 28 4 × 5 2 y 20 = 5 4 20 3 y + 28 = 20 2 y 20 3 y + 28 = 2 y 20 y = 48

Q.4 The sum of three consecutive multiples of 7 is 777, find the numbers.

Let the first multiple of 7 be x.

Other two will be (x + 7) and (x + 14).

According to the question,

x + x + 7 + x + 14 = 777 3 x = 756 x = 252 Hence, the three consecutive multiples of 7 are 252, 259 and 266.

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Important questions for class 8 maths, chapter 1. rational numbers.

## Chapter 3. Understanding Quadrilaterals

Chapter 4. practical geometry, chapter 5. data handling, chapter 6. squares and square roots, chapter 7. cubes and cube roots, chapter 8. comparing quantities, chapter 9. algebraic expressions and identities, chapter 10. visualising solid shapes, chapter 11. mensuration, chapter 12. exponents and powers, chapter 13. direct and inverse proportions, chapter 14. factorisation, chapter 15. introduction to graphs, chapter 16. playing with numbers, faqs (frequently asked questions), 1. how many solutions does a linear equation in one variable have.

A linear equation with one variable has only one solution.

## 2. What is the formula of a linear equation in one variable?

The standard equation of a linear equation in one variable is given by ax+b=0.

There is only a single variable which is x.

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- RD Sharma Solutions
- Chapter 9 Linear Equation In One Variable

## RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equation in One Variable

Mathematics is one of the scoring subjects where students secure maximum marks in the exam. When it comes to preparing for the annual exam, it is the toughest time when most students struggle to solve problems. So, here at BYJU’S, our expert faculty have formulated RD Sharma Class 8 Maths Solutions , which help students prepare for their exams effortlessly. All the solutions are well designed, keeping in mind the latest CBSE syllabus and exam pattern. Also, students can learn easy tricks and shortcut methods by practising these solutions on a regular basis. The PDFs of this chapter are available here, and students can download them for free from the links provided below.

Chapter 9 – Linear Equation in One Variable contains four exercises, and the RD Sharma Solutions available on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts covered in this chapter.

- Linear equation and its definitions.
- A solution of a linear equation.
- Solving equations having variable terms on one side and number(s) on the other side.
- Transposition method for solving linear equations in one variable.
- Cross-multiplication method for solving equations.
- Applications of linear equations to practical problems.
- RD Sharma Solutions for Class 8 Maths Chapter 1 Rational Numbers
- RD Sharma Solutions for Class 8 Maths Chapter 2 Powers
- RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots
- RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots
- RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers
- RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
- RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization
- RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions
- RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable
- RD Sharma Solutions for Class 8 Maths Chapter 10 Direct and Inverse Variations
- RD Sharma Solutions for Class 8 Maths Chapter 11 Time and Work
- RD Sharma Solutions for Class 8 Maths Chapter 12 Percentage
- RD Sharma Solutions for Class 8 Maths Chapter 13 Profit, Loss, Discount and Value Added Tax (VAT)
- RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest
- RD Sharma Solutions for Class 8 Maths Chapter 15 Understanding Shapes – I (Polygons)
- RD Sharma Solutions for Class 8 Maths Chapter 16 Understanding Shapes – II (Quadrilaterals)
- RD Sharma Solutions for Class 8 Maths Chapter 17 Understanding Shapes – II (Special Types of Quadrilaterals)
- RD Sharma Solutions for Class 8 Maths Chapter 18 Practical Geometry (Constructions)
- RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
- RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
- RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
- RD Sharma Solutions for Class 8 Maths Chapter 22 Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
- RD Sharma Solutions for Class 8 Maths Chapter 23 Data Handling – I (Classification and Tabulation of Data)
- RD Sharma Solutions for Class 8 Maths Chapter 24 Data Handling – II (Graphical Representation of Data as Histograms)
- RD Sharma Solutions for Class 8 Maths Chapter 25 Data Handling – III (Pictorial Representation of Data as Pie Charts)
- RD Sharma Solutions for Class 8 Maths Chapter 26 Data Handling – IV (Probability)
- RD Sharma Solutions for Class 8 Maths Chapter 27 Introduction to Graphs
- Exercise 9.1 Chapter 9 Linear Equations in One Variable
- Exercise 9.2 Chapter 9 Linear Equations in One Variable
- Exercise 9.3 Chapter 9 Linear Equations in One Variable
- Exercise 9.4 Chapter 9 Linear Equations in One Variable

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## Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable

Exercise 9.1 page no: 9.5.

Solve each of the following equations and also verify your solution:

1. 9 ¼ = y – 1 1/3

9 ¼ = y – 1 1/3

37/4 = y – 4/3

Upon solving, we get,

y = 37/4 + 4/3

By taking LCM for 4 and 3, we get 12

y = (37×3)/12 + (4×4)/12

= 111/12 + 16/12

= (111 + 16)/12

∴ y = 127/12

Verification

RHS = y – 1 1/3

= 127/12 – 4/3

= (127 – 16)/12

= 9 ¼

2. 5x/3 + 2/5 = 1

5x/3 + 2/5 = 1

5x/3 = 1 – 2/5 (by taking LCM)

By using cross-multiplication, we get,

5x = (3×3)/5

x = 9/(5×5)

∴ x = 9/25

LHS = 5x/3 + 2/5

= 5/3 × 9/25 + 2/5

= 3/5 + 2/5

= (3 + 2)/5

3. x/2 + x/3 + x/4 = 13

x/2 + x/3 + x/4 = 13

let us take LCM for 2, 3 and 4, which is 12

(x×6)/12 + (x×4)/12 + (x×3)/12 = 13

6x/12 + 4x/12 + 3x/12 = 13

(6x+4x+3x)/12 = 13

13x/12 = 13

13x = 12×13

∴ x = 12

LHS = x/2 + x/3 + x/4

= 12/2 + 12/3 + 12/4

= 6 + 4 + 3

4. x/2 + x/8 = 1/8

x/2 + x/8 = 1/8

let us take LCM for 2 and 8, which is 8

(x×4)/8 + (x×1)/8 = 1/8

4x/8 + x/8 = 1/8

∴ x = 1/5

LHS = x/2 + x/8

= (1/5)/2 + (1/5)/8

= 1/10 + 1/40

= (4 + 1)/40

5. 2x/3 – 3x/8 = 7/12

2x/3 – 3x/8 = 7/12

By taking LCM for 3 and 8, we get 24

(2x×8)/24 – (3x×3)/24 = 7/12

16x/24 – 9x/24 = 7/12

(16x-9x)/24 = 7/12

7x/24 = 7/12

7x×12 = 7×24

x = (7×24)/(7×12)

∴ x = 2

LHS = 2x/3 – 3x/8

= 2(2)/3 – 3(2)/8

= 4/3 – 6/8

= 4/3 – 3/4

= (16 – 9)/ 12

6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Upon expansion, we get,

x 2 + 5x + 6 + x 2 – 5x +6 – 2x 2 – 2x =0

-2x + 12 = 0

By dividing the equation using -2, we get,

x – 6 = 0

∴ x = 6

LHS = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1)

= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6) (6 + 1)

= (8) (9) + (3) (4) – 12(7)

= 72 + 12 – 84

= 84 – 84

7. x/2 – 4/5 + x/5 + 3x/10 = 1/5

x/2 – 4/5 + x/5 + 3x/10 = 1/5

x/2 + x/5 + 3x/10 = 1/5 + 4/5

by taking LCM for 2, 5 and 10, which is 10

(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5

5x/10 + 2x/10 + 3x/10 = 1

(5x+2x+3x)/10 = 1

∴ x = 1

LHS = x/2 – 4/5 + x/5 + 3x/10

= ½ – 4/5 + 1/5 + 3(1)/10

= (5 – 8 + 2 + 3)/10

= (10 – 8)/10

8. 7/x + 35 = 1/10

7/x + 35 = 1/10

7/x = 1/10 – 35

= ((1×1) – (35×10))/10

= (1 – 350)/10

7/x = -349/10

x = -70/349

∴ x = -70/349

LHS = 7/x + 35

= 7/(-70/349) + 35

= (-7 × 349)/70 + 35

= -349/10 + 35

= (-349 + 350)/ 10

9. (2x-1)/3 – (6x-2)/5 = 1/3

(2x-1)/3 – (6x-2)/5 = 1/3

By taking LCM for 3 and 5, which is 15

((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3

(10x – 5)/15 – (18x – 6)/15 = 1/3

(10x – 5 – 18x + 6)/15 = 1/3

(-8x + 1)/15 = 1/3

(-8x + 1)3 = 15

-24x + 3 = 15

-24x = 15 – 3

∴ x = -1/2

LHS = (2x – 1)/3 – (6x – 2)/5

= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5

= (- 1 – 1)/3 – (-3 – 2)/5

= – 2/3 – (-5/5)

= (-2 + 3)/3

10. 13(y – 4) – 3(y – 9) – 5(y + 4) = 0

13(y – 4) – 3(y – 9) – 5(y + 4) = 0

13y – 52 – 3y + 27 – 5y – 20 = 0

13y – 3y – 5y = 52 – 27 + 20

∴ y = 9

LHS = 13(y – 4) – 3 (y – 9) – 5 (y + 4)

= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)

= 13 (5) – 3 (0) – 5 (13)

= 65 – 0 – 65

11. 2/3(x – 5) – 1/4(x – 2) = 9/2

2/3(x – 5) – 1/4(x – 2) = 9/2

2x/3 – 10/3 – x/4 + 2/4 = 9/2

2x/3 – 10/3 – x/4 + 1/2 = 9/2

2x/3 – x/4 = 9/2 + 10/3 – 1/2

By taking LCM for (3 and 4 is 12) (2 and 3 is 6)

(2x×4)/12 – (x×3)/12 = (9×3)/6 + (10×2)/6 – (1×3)/6

8x/12 – 3x/12 = 27/6 + 20/6 – 3/6

(8x-3x)/12 = (27+20-3)6

5x/12 = 44/6

5x×6 = 44×12

LHS = 2/3 (x – 5) – ¼ (x – 2)

= 2/3 [(88/5) – 5] – ¼ [(88/5) – 2]

= 2/3 [(88 – 25)/5] – ¼ [(88 – 10)/5]

= 2/3 × 63/5 – ¼ × 78/5

= 42/5 – 39/10

= (84 – 39)/10

## EXERCISE 9.2 PAGE NO: 9.11

Solve each of the following equations and also check your results in each case:

1. (2x+5)/3 = 3x – 10

(2x+5)/3 = 3x – 10

Let us simplify,

(2x+5)/3 – 3x = – 10

By taking LCM

(2x + 5 – 9x)/3 = -10

(-7x + 5)/3 = -10

-7x + 5 = -30

-7x = -30 – 5

Let us verify the given equation now,

By substituting the value of ‘x’, we get,

(2×5 + 5)/3 = 3(5) – 10

(10+5)/3 = 15-10

Hence, the given equation is verified

2. (a-8)/3 = (a-3)/2

(a-8)/3 = (a-3)/2

(a-8)2 = (a-3)3

2a – 16 = 3a – 9

2a – 3a = -9 + 16

By substituting the value of ‘a’ we get,

(-7 – 8)/3 = (-7 – 3)/2

-15/3 = -10/2

3. (7y + 2)/5 = (6y – 5)/11

(7y + 2)/5 = (6y – 5)/11

(7y + 2)11 = (6y – 5)5

77y + 22 = 30y – 25

77y – 30y = -25 – 22

By substituting the value of ‘y’, we get,

(7(-1) + 2)/5 = (6(-1) – 5)/11

(-7 + 2)/5 = (-6 – 5)/11

-5/5 = -11/11

4. x – 2x + 2 – 16/3x + 5 = 3 – 7/2x

x – 2x + 2 – 16/3x + 5 = 3 – 7/2x

Let us rearrange the equation

x – 2x – 16x/3 + 7x/2 = 3 – 2 – 5

By taking LCM for 2 and 3, which is 6

(6x – 12x – 32x + 21x)/6 = -4

-17x/6 = -4

By cross-multiplying

-17x = -4×6

x = -24/-17

24/17 – 2(24/17) + 2 – (16/3)(24/17) + 5 = 3 – (7/2)(24/17)

24/17 – 48/17 + 2 – 384/51 + 5 = 3 – 168/34

By taking 51 and 17 as the LCM we get,

(72 – 144 + 102 – 384 + 255)/51 = (102 – 168)/34

-99/51 = -66/34

-33/17 = -33/17

5. 1/2x + 7x – 6 = 7x + 1/4

1/2x + 7x – 6 = 7x + 1/4

1/2x + 7x – 7x = 1/4 + 6 (by taking LCM)

1/2x = (1+ 24)/4

1/2x = 25/4

4x = 25 × 2

(1/2) (25/2) + 7(25/2) – 6 = 7(25/2) + 1/4

25/4 + 175/2 – 6 = 175/2 + 1/4

By taking LCM for 4 and 2 is 4

(25 + 350 – 24)/4 = (350+1)/4

351/4 = 351/4

6. 3/4x + 4x = 7/8 + 6x – 6

3/4x + 4x = 7/8 + 6x – 6

3/4x + 4x – 6x = 7/8 – 6

By taking 4 and 8 as LCM

(3x + 16x – 24x)/4 = (7 – 48)/8

-5x/4 = -41/8

-5x(8) = -41(4)

-40x = -164

x = -164/-40

(3/4)(41/10) + 4(41/10) = 7/8 + 6(41/10) – 6

123/40 + 164/10 = 7/8 + 246/10 – 6

(123 + 656)/40 = (70 + 1968 – 480)/80

779/40 = 1558/80

779/40 = 779/40

7. 7x/2 – 5x/2 = 20x/3 + 10

7x/2 – 5x/2 = 20x/3 + 10

7x/2 – 5x/2 – 20x/3 = 10

By taking LCM for 2 and 3 is 6

(21x – 15x – 40x)/6 = 10

-34x/6 = 10

(7-/2)(-30/17) – (5/2)(-30/17) = (20/3)(-30/17) + 10

-210/34 +150/34 = -600/51 + 10

-30/17 = (-600+510)/51

-30/17 = -30/17

8. (6x+1)/2 + 1 = (7x-3)/3

(6x+1)/2 + 1 = (7x-3)/3

(6x + 1 + 2)/2 = (7x – 3)/3

(6x + 3)3 = (7x – 3)2

18x + 9 = 14x – 6

18x – 14x = -6 – 9

(6(-15/4) + 1)/2 + 1 = (7(-15/4) – 3)/3

(3(-15/2) + 1)/2 + 1 = (-105/4 -3)/3

(-45/2 + 1)/2 + 1 = (-117/4)/3

(-43/4) + 1 = -117/12

(-43+4)/4 = -39/4

-39/4 = -39/4

9. (3a-2)/3 + (2a+3)/2 = a + 7/6

(3a-2)/3 + (2a+3)/2 = a + 7/6

(3a-2)/3 + (2a+3)/2 – a = 7/6

((3a-2)2 + (2a+3)3 – 6a)/6 = 7/6

(6a – 4 + 6a + 9 – 6a)/6 = 7/6

(6a + 5)/6 = 7/6

By substituting the value of ‘a’, we get,

(3(1/3)-2)/3 + (2(1/3) + 3)/2 = 1/3 + 7/6

(1-2)/3 + (2/3 + 3)/2 = (2+7)/6

-1/3 + (11/3)/2 = 9/6

-1/3 + 11/6 = 3/2

(-2+11)/6 = 3/2

10. x – (x-1)/2 = 1 – (x-2)/3

x – (x-1)/2 = 1 – (x-2)/3

x – (x-1)/2 + (x-2)/3 = 1

(6x – (x-1)3 + (x-2)2)/6 = 1

(6x – 3x + 3 + 2x – 4)/6 = 1

(5x – 1)/6 = 1

5x – 1 = 6

7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3

7/5 – (2/5)/2 = 1 – (-3/5)/3

7/5 – 2/10 = 1 + 3/15

(14 – 2)/10 = (15+3)/15

12/10 = 18/15

11. 3x/4 – (x-1)/2 = (x-2)/3

3x/4 – (x-1)/2 = (x-2)/3

3x/4 – (x-1)/2 – (x-2)/3 = 0

By taking LCM for 4, 2 and 3, which is 12

(9x – (x-1)6 – (x-2)4)/12 = 0

(9x – 6x + 6 – 4x + 8)/12 = 0

(-x + 14)/12 = 0

-x + 14 = 0

3(14)/4 – (14-1)/2 = (14-2)/3

42/4 – 13/2 = 12/3

(42 – 26)/4 = 4

12. 5x/3 – (x-1)/4 = (x-3)/5

5x/3 – (x-1)/4 = (x-3)/5

5x/3 – (x-1)/4 – (x-3)/5 = 0

By taking LCM for 3, 4 and 5, which is 60

((5x×20) – (x-1)15 – (x-3)12)/60 = 0

(100x – 15x + 15 -12x + 36)/60 = 0

(73x + 51)/60 = 0

73x + 51 = 0

(20x – (x-1)3)/12 = (-51/73 – 3)/5

(20x – 3x + 3)/12 = (-270/73)/5

(17x + 3)/12 = -270/365

(17(-51/73) + 3)/12 = -54/73

(-867/73 + 3)/12 = -54/73

((-867 + 219)/73)/12 = -54/73

(-648)/876 = -54/73

-54/73 = -54/73

13. (3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14

(3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14

(3x+1)/16 + (2x-3)/7 – (x+3)/8 – (3x-1)/14 = 0

By taking LCM for 16, 7, 8 and 14, which is 112

((3x+1)7 + (2x-3)16 – (x+3)14 – (3x-1)8)/112 = 0

(21x + 7 + 32x – 48 – 14x – 42 – 24x + 8)/112 = 0

(21x + 32x – 14x – 24x + 7 – 48 – 42 + 8)/112 = 0

(15x – 75)/112 = 0

15x – 75 = 0

(3(5)+1)/16 + (2(5)-3)/7 = (5+3)/8 + (3(5)-1)/14

(15+1)/16 + (10-3)/7 = 8/8 + (15-1)/14

16/16 + 7/7 = 8/8 + 14/14

1 + 1 = 1 + 1

14. (1-2x)/7 – (2-3x)/8 = 3/2 + x/4

(1-2x)/7 – (2-3x)/8 = 3/2 + x/4

(1-2x)/7 – (2-3x)/8 – x/4 = 3/2

By taking LCM for 7, 8 and 4, which is 56

((1-2x)8 – (2-3x)7 – 14x)/56 = 3/2

(8 – 16x – 14 + 21x – 14x)/56 = 3/2

(-9x – 6)/56 = 3/2

2(-9x-6) = 3(56)

-18x – 12 = 168

-18x = 168+12

x = 180/-18

(1-2(-10))/7 – (2-3(-10))/8 = 3/2 + (-10)/4

(1+20)/7 – (2+30)/8 = 3/2 – 5/2

21/7 – 32/8 = 3/2 – 5/2

3 – 4 = -2/2

15. (9x+7)/2 – (x – (x-2)/7) = 36

(9x+7)/2 – (x – (x-2)/7) = 36

Let us simplify the given equation into a simple form

(9x+7)/2 – (7x-x+2)/7 = 36

(9x+7)/2 – (6x+2)/7 = 36

By taking LCM for 2 and 7 is 14

(7(9x+7) – 2(6x+2))/14 = 36

(63x+49 – 12x – 4)/14 = 36

(51x + 45)/14 = 36

51x + 45 = 36(14)

51x + 45 = 504

51x = 504-45

(9(9)+7)/2 – (6(9)+2)/7 = 36

(81+7)/2 – (54+2)/7 = 36

88/2 – 56/7 = 36

44 – 8 = 36

16. 0.18(5x – 4) = 0.5x + 0.8

0.18(5x – 4) = 0.5x + 0.8

0.18(5x – 4) – 0.5x = 0.8

0.90x – 0.72 – 0.5x = 0.8

0.90x – 0.5x = 0.8 + 0.72

0.40x = 1.52

x = 1.52/0.40

0.18(5(3.8)-4) = 0.5(3.8) + 0.8

0.18(19-4) = 1.9 + 0.8

17. 2/3x – 3/2x = 1/12

2/3x – 3/2x = 1/12

By taking LCM for 3x and 2x, which is 6x

((2×2) – (3×3))/6x = 1/12

(4-9)/6x = 1/12

-5/6x = 1/12

2/3(-10) – 3/2(-10) = 1/12

-2/30 + 3/20 = 1/12

((-2×2) + (3×3))/60 = 1/12

(-4+9)/60 = 1/12

5/60 = 1/12

1/12 = 1/12

18. 4x/9 + 1/3 + 13x/108 = (8x+19)/18

4x/9 + 1/3 + 13x/108 = (8x+19)/18

4x/9 + 13x/108 – (8x+19)/18 = -1/3

By taking LCM for 9, 108 and 18, which is 108

((4x×12) + 13x×1 – (8x+19)6)/108 = -1/3

(48x + 13x – 48x – 114)/108 = -1/3

(13x – 114)/108 = -1/3

(13x – 114)3 = -108

39x – 342 = -108

39x = -108 + 342

4(6)/9 + 1/3 + 13(6)/108 = (8(6)+19)/18

24/9 + 1/3 + 78/108 = 67/18

8/3 + 1/3 + 13/18 = 67/18

((8×6) + (1×6) + (13×1))/18 = 67/18

(48 + 6 + 13)/18 = 67/18

67/18 = 67/18

19. (45-2x)/15 – (4x+10)/5 = (15-14x)/9

(45-2x)/15 – (4x+10)/5 = (15-14x)/9

By rearranging

(45-2x)/15 – (4x+10)/5 – (15-14x)/9 = 0

By taking LCM for 15, 5 and 9, which is 45

((45-2x)3 – (4x+10)9 – (15-14x)5)/45 = 0

(135 – 6x – 36x – 90 – 75 + 70x)/45 = 0

(28x – 30)/45 = 0

28x – 30 = 0

(45-2(15/14))/15 – (4(15/14) + 10)/5 = (15 – 14(15/14))/9

(45- 15/7)/15 – (30/7 + 10)/5 = (15-15)/9

300/105 – 100/35 = 0

(300-300)/105 = 0

20. 5(7x+5)/3 – 23/3 = 13 – (4x-2)/3

5(7x+5)/3 – 23/3 = 13 – (4x-2)/3

(35x + 25)/3 + (4x – 2)/3 = 13 + 23/3

(35x + 25 + 4x – 2)/3 = (39+23)/3

(39x + 23)/3 = 62/3

(39x + 23)3 = 62(3)

39x + 23 = 62

39x = 62 – 23

(35x + 25)/3 – 23/3 = 13 – (4x-2)/3

(35+25)/3 – 23/3 = 13 – (4-2)/3

60/3 – 23/3 = 13 – 2/3

(60-23)/3 = (39-2)/3

37/3 = 37/3

21. (7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3

(7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3

Upon expansion

(7x-1)/4 – (4x-1+x)/6 = 10/3

(7x-1)/4 – (5x-1)/6 = 10/3

By taking LCM for 4 and 6, we get 24

((7x-1)6 – (5x-1)4)/24 = 10/3

(42x – 6 – 20x + 4)/24 = 10/3

(22x – 2)/24 = 10/3

22x – 2 = 10(8)

22x – 2 = 80

(7(41/11)-1)/4 – (5(41/11)-1)/6 = 10/3

(287/11 – 1)/4 – (205/11 – 1)/6 = 10/3

(287-11)/44 – (205-11)/66 = 10/3

276/44 – 194/66 = 10/3

69/11 – 97/33 = 10/3

((69×3) – (97×1))/33 = 10/3

(207 – 97)/33 = 10/3

110/33 = 10/3

10/3 = 10/3

22. 0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1

0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1

Let us simplify

(0.5/0.35)(x – 0.4) – (0.6/0.42)(x – 2.71) = x + 6.1

(x – 0.4)/0.7 – (x – 2.71)/0.7 = x + 6.1

(x – 0.4 – x + 2.71)/0.7 = x + 6.1

-0.4 + 2.71 = 0.7(x + 6.1)

0.7x = 2.71 – 0.4 – 4.27

x = -1.96/0.7

0.5(-2.8 – 0.4)/0.35 – 0.6(-2.8 – 2.71)/0.42 = -2.8 + 6.1

-1.6/0.35 + 3.306/0.42 = 3.3

-4.571 + 7.871 = 3.3

23. 6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2

6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2

6.5x + (19.5x – 32.5)/2 – 6.5x – (13x – 26)/2 = 13

(19.5x – 32.5)/2 – (13x – 26)/2 = 13

(19.5x – 32.5 – 13x + 26)/2 = 13

(6.5x – 6.5)/2 = 13

6.5x – 6.5 = 13×2

6.5x – 6.5 = 26

6.5x = 26+6.5

6.5x = 32.5

x = 32.5/6.5

6.5(5) + (19.5(5) – 32.5)/2 = 6.5(5) + 13 + (13(5) – 26)/2

32.5 + (97.5 – 32.5)/2 = 32.5 + 13 + (65 – 26)/2

32.5 + 65/2 = 45.5 + 39/2

(65 + 65)/2 = (91+39)/2

130/2 = 130/2

24. (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 = x 2 + 7x – 3x – 21

9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 – x 2 – 7x + 3x + 21 = 0

9x 2 – 8x 2 – x 2 + 6x – 24x – 4x + 22x – 7x + 3x – 16 + 21 + 11 = 0

-4x + 16 = 0

(3(4) – 8) (3(4) + 2) – (4(4) – 11) (2(4) + 1) = (4 – 3) (4 + 7)

(12-8) (12+2) – (16-11) (8+1) = 1(11)

4 (14) – 5(9) = 11

56 – 45 = 11

25. [(2x+3) + (x+5)] 2 + [(2x+3) – (x+5)] 2 = 10x 2 + 92

Let us simplify the given equation

By using the formula (a+b) 2

9x 2 + 48x + 64 + x 2 – 4x + 4 = 10x 2 + 92

9x 2 – 10x 2 + x 2 + 48x – 4x = 92 – 64 – 4

(106/11) 2 + (-16/11) 2 = (360 + 11132)/121

11236/121 + 256/121 = 11492/121

11492/121 = 11492/121

## EXERCISE 9.3 PAGE NO: 9.17

Solve the following equations and verify your answer:

1. (2x-3) / (3x+2) = -2/3

(2x-3) / (3x+2) = -2/3

Let us perform cross-multiplication we get,

3(2x – 3) = -2(3x + 2)

6x – 9 = -6x – 4

When rearranged,

6x + 6x = 9 – 4

Now let us verify the given equation,

(2(5/12) – 3) / (3(5/12) + 2) = -2/3

((5/6)-3) / ((5/4) + 2) = -2/3

((5-18)/6) / ((5+8)/4) = -2/3

(-13/6) / (13/4) = -2/3

(-13/6) × (4/13) = -2/3

-4/6 = -2/3

-2/3 = -2/3

2. (2-y) / (y+7) = 3/5

(2-y) / (y+7) = 3/5

Let us perform cross-multiplication, we get,

5(2-y) = 3(y+7)

10 – 5y = 3y + 21

10 – 21 = 3y + 5y

8y = – 11

(2 – (-11/8)) / ((-11/8) + 7) = 3/5

((16+11)/8) / ((-11+56)/8) = 3/5

(27/8) / (45/8) = 3/5

(27/8) × (8/45) = 3/5

27/45 = 3/5

3. (5x – 7) / (3x) = 2

(5x – 7) / (3x) = 2

5x – 7 = 2(3x)

5x – 7 = 6x

5x – 6x = 7

(5(-7) – 7) / (3(-7)) = 2

(-35 – 7) / -21 = 2

-42/-21 = 2

4. (3x+5) / (2x + 7) = 4

(3x+5) / (2x + 7) = 4

3x + 5 = 4(2x+7)

3x + 5 = 8x + 28

3x – 8x = 28 – 5

(3(-23/5) + 5) / (2(-23/5) + 7) = 4

(-69/5 + 5) / (-46/5 + 7) = 4

(-69+25)/5 / (-46+35)/5 = 4

-44/5 / -11/5 = 4

-44/5 × 5/-11 = 4

5. (2y + 5) / (y + 4) = 1

(2y + 5) / (y + 4) = 1

2y + 5 = y + 4

2y – y = 4 – 5

(2(-1) + 5) / (-1 + 4) = 1

(-2+5) / 3 = 1

6. (2x + 1) / (3x – 2) = 5/9

(2x + 1) / (3x – 2) = 5/9

9(2x + 1) = 5(3x – 2)

18x + 9 = 15x – 10

18x – 15x = -10 – 9

(2(-19/3) + 1) / (3(-19/3) – 2) = 5/9

(-38/3 + 1) / (-57/3 – 2) = 5/9

(-38 + 3)/3 / (-57 – 6)/3 = 5/9

-35/3 / -63/3 = 5/9

-35/3 × 3/-63 = 5/9

-35/-63 = 5/9

7. (1 – 9y) / (19 – 3y) = 5/8

(1 – 9y) / (19 – 3y) = 5/8

8(1- 9y) = 5(19-3y)

8 – 72y = 95 – 15y

8 – 95 = 72y – 15y

(1 – 9(-29/19)) / (19 – 3(-29/19)) = 5/8

(19+261)/19 / (361+87)/19 = 5/8

280/19 × 19/448 = 5/8

280/ 448 = 5/8

8. 2x / (3x + 1) = 1

2x / (3x + 1) = 1

2x = 1(3x + 1)

2x = 3x + 1

2x – 3x = 1

2(-1) / (3(-1) + 1) = 1

-2 /(-3+1) = 1

9. y – (7 – 8y)/9y – (3 + 4y) = 2/3

y – (7 – 8y)/9y – (3 + 4y) = 2/3

(y – 7 + 8y) / (9y – 3 – 4y) = 2/3

(-7 + 9y) / (5y – 3) = 2/3

3(-7 + 9y) = 2(5y – 3)

-21 + 27y = 10y – 6

27y – 10y = 21 – 6

15/17 – (7-8(15/17))/ 9(15/17) – (3 + 4(15/17)) = 2/3

15/17 – (7 – 120/17) / 135/17 – (3 + 60/17) = 2/3

15/17 – ((119-120)/17) / 135/17 – ((51+60)/17) = 2/3

15/17 – (-1/17) / 135/17 – (111/17) = 2/23

((15 + 1)/17) / ((135-111)/17) = 2/3

16/17 / 24/17 = 2/3

16/24 = 2/3

10. 6/ 2x – (3 – 4x) = 2/3

6/ 2x – (3 – 4x) = 2/3

6/(2x – 3 + 4x) = 2/3

6/(6x – 3) = 2/3

3(6) = 2(6x – 3)

18 = 12x – 6

12x = 18 + 6

6/ (6(2) – 3) = 2/3

6/(12-3) = 2/3

11. 2/3x – 3/2x = 1/12

4-9/6x = 1/12

By cross-multiplying, we get,

12(-5) = 1 (6x)

2/-30 – 3/-20 = 1/12

-4+6/60 = 1/12

12. (3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)

(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)

(3x + 5)/ (4x + 2) – (3x + 4)/(4x + 7) = 0

By taking LCM as (4x + 2) (4x + 7)

((3x + 5) (4x + 7) – (3x + 4) (4x + 2)) / (4x + 2) (4x + 7) = 0

(3x + 5) (4x + 7) – (3x + 4) (4x + 2) = 0

12x 2 + 21x + 20x + 35 – 12x 2 – 6x – 16x – 8 = 0

19x + 35 – 8 = 0

(3(-27/19) +5) / (4(-27/19) + 2) = (3(-27/19) + 4) / (4(-27/19) + 7)

(-81/19 + 5) / (-108/19 + 2) = (-81/19 + 4) / (-108/19 + 7)

((-81+95)/19) / ((-108+38)/19) = ((-81+76)/19) / ((-108+133)/19)

14/19 / -70/19 = -5/19 / 25/19

-14/70 = -5/25

-1/5 = -1/5

13. (7x – 2) / (5x – 1) = (7x +3)/(5x + 4)

(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)

(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0

By taking LCM as (5x – 1) (5x + 4)

((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0

(7x-2) (5x+4) – (7x+3)(5x-1) = 0

Upon simplification,

35x 2 + 28x – 10x – 8 – 35x 2 + 7x – 15x + 3 = 0

10x – 5 = 0

(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)

(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)

((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)

(3/2) / (3/2) = (13/2) / (13/2)

14. ((x+1)/(x+2)) 2 = (x+2) / (x + 4)

((x+1)/(x+2)) 2 = (x+2) / (x + 4)

(x+1) 2 / (x+2) 2 – (x+2) / (x + 4) = 0

By taking LCM as (x+2) 2 (x+4)

((x+1) 2 (x+4) – (x+2) (x+2) 2 ) / (x+2) 2 (x+4) = 0

(x+1) 2 (x+4) – (x+2) (x+2) 2 = 0

Let us expand the equation

(x 2 + 2x + 1) (x + 4) – (x + 2) (x 2 + 4x + 4) = 0

x 3 + 2x 2 + x + 4x 2 + 8x + 4 – (x 3 + 4x 2 + 4x + 2x 2 + 8x + 8) = 0

x 3 + 2x 2 + x + 4x 2 + 8x + 4 – x 3 – 4x 2 – 4x – 2x 2 – 8x – 8 = 0

-3x – 4 = 0

(x+1) 2 / (x+2) 2 = (x+2) / (x + 4)

(-4/3 + 1) 2 / (-4/3 + 2) 2 = (-4/3 + 2) / (-4/3 + 4)

((-4+3)/3) 2 / ((-4+6)/3) 2 = ((-4+6)/3) / ((-4+12)/3)

(-1/3) 2 / (2/3) 2 = (2/3) / (8/3)

1/9 / 4/9 = 2/3 / 8/3

15. ((x+1)/(x-4)) 2 = (x+8)/(x-2)

((x+1)/(x-4)) 2 = (x+8)/(x-2)

(x+1) 2 / (x-4) 2 – (x+8) / (x-2) = 0

By taking LCM as (x-4) 2 (x-2)

((x+1) 2 (x-2) – (x+8) (x-4) 2 ) / (x-4) 2 (x-2) = 0

(x+1) 2 (x-2) – (x+8) (x-4) 2 = 0

(x 2 + 2x + 1) (x-2) – ((x+8) (x 2 – 8x + 16)) = 0

x 3 + 2x 2 + x – 2x 2 – 4x – 2 – (x 3 – 8x 2 + 16x + 8x 2 – 64x + 128) = 0

x 3 + 2x 2 + x – 2x 2 – 4x – 2 – x 3 + 8x 2 – 16x – 8x 2 + 64x – 128 = 0

45x – 130 = 0

(x+1) 2 / (x-4) 2 = (x+8) / (x-2)

(26/9 + 1) 2 / (26/9 – 4) 2 = (26/9 + 8) / (26/9 – 2)

((26+9)/9) 2 / ((26-36)/9) 2 = ((26+72)/9) / ((26-18)/9)

(35/9) 2 / (-10/9) 2 = (98/9) / (8/9)

(35/-10) 2 = (98/8)

(7/2) 2 = 49/4

49/4 = 49/4

16. (9x-7)/(3x+5) = (3x-4)/(x+6)

(9x-7)/(3x+5) = (3x-4)/(x+6)

(9x-7)/(3x+5) – (3x-4)/(x+6) = 0

By taking LCM as (3x+5) (x+6)

((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0

(9x-7) (x+6) – (3x-4) (3x+5) = 0

9x 2 + 54x – 7x – 42 – (9x 2 + 15x – 12x – 20) = 0

44x – 22 = 0

(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)

(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)

((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)

-5/2 / 13/2 = -5/2 / 13/2

-5/13 = -5/13

17. (x+2)/(x+5) = x/(x+6)

(x+2)/(x+5) = x/(x+6)

(x+2)/(x+5) – x/(x+6) = 0

By taking LCM as (x+5) (x+6)

((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0

(x+2) (x+6) – x(x+5) = 0

Upon expansion,

x 2 + 8x + 12 – x 2 – 5x = 0

3x + 12 = 0

(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)

-2/1 = -4 / (2)

18. 2x – (7-5x) / 9x – (3+4x) = 7/6

2x – (7-5x) / 9x – (3+4x) = 7/6

(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6

(7x – 7) / (5x – 3) = 7/6

6(7x – 7) = 7(5x – 3)

42x – 42 = 35x – 21

42x – 35x = -21 + 42

(7(3) -7) / (5(3) – 3) = 7/6

(21-7) / (15-3) = 7/6

14/12 = 7/6

19. (15(2-x) – 5(x+6)) / (1-3x) = 10

15(2-x) – 5(x+6) / (1-3x) = 10

(30-15x) – (5x + 30) / (1-3x) = 10

(30-15x) – (5x + 30) = 10(1- 3x)

30- 15x – 5x – 30 = 10 – 30x

30- 15x – 5x – 30 + 30x = 10

(15(2-x) – 5(x+6)) / (1-3x) = 10

(15(2-1) – 5(1+6)) / (1- 3) = 10

(15 – 5(7))/-2 = 10

(15-35)/-2 = 10

-20/-2 = 10

20. (x+3)/(x-3) + (x+2)/(x-2) = 2

(x+3)/(x-3) + (x+2)/(x-2) = 2

By taking LCM as (x-3) (x-2)

((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2

(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))

x 2 + 3x – 2x – 6 + x 2 – 3x + 2x – 6 = 2(x 2 – 3x – 2x + 6)

2x 2 – 12 = 2x 2 – 10x + 12

2x 2 – 2x 2 + 10x = 12 + 12

(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2

((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2

(27/5)/(-3/5) + (22/5)/(2/5) = 2

-27/3 + 22/2 = 2

((-27×2) + (22×3))/6 = 2

(-54 + 66)/6 = 2

21. ((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2

((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2

(x+2) (2x-3) – 2x 2 + 6) = 2(x-5)

2x 2 – 3x + 4x – 6 – 2x 2 + 6 = 2x – 10

x = 2x – 10

x – 2x = -10

((10+2) (2(10) – 3) – 2(10) 2 + 6)/ (10-5) = 2

(12(17) – 200 + 6)/5 = 2

(204 – 194)/5 = 2

22. (x 2 – (x+1) (x+2))/(5x+1) = 6

(x 2 – (x+1) (x+2))/(5x+1) = 6

(x 2 – (x+1) (x+2)) = 6(5x+1)

x 2 – x 2 – 2x – x – 2 = 30x + 6

-3x – 2 = 30x + 6

30x + 3x = -2 – 6

((-8/33) 2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6

(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6

(64/1089 – (25/33) (58/33)) / (-7/33) = 6

(64/1089 – 1450/1089) / (-7/33) = 6

((64-1450)/1089 / (-7/33)) = 6

-1386/1089 × 33/-7 = 6

1386 × 33 / 1089 × -7 = 6

23. ((2x+3) – (5x-7))/(6x+11) = -8/3

((2x+3) – (5x-7))/(6x+11) = -8/3

3((2x+3) – (5x-7)) = -8(6x+11)

3(2x + 3 – 5x + 7) = -48x – 88

3(-3x + 10) = -48x – 88

-9x + 30 = -48x – 88

-9x + 48x = -88 – 30

x = -118/39

((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3

((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3

(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3

(-219+863)/39 / (-279)/39 = -8/3

644/-279 = -8/3

-8/3 = -8/3

24. Find the positive value of x for which the given equation is satisfied:

(i) (x 2 – 9)/(5+x 2 ) = -5/9

(x 2 – 9)/(5+x 2 ) = -5/9

9(x 2 – 9) = -5(5+x 2 )

9x 2 – 81 = -25 – 5x 2

9x 2 + 5x 2 = -25 + 81

x 2 = 56/14

x = √4

(ii) (y 2 + 4)/(3y 2 + 7) = 1/2

(y 2 + 4)/(3y 2 + 7) = 1/2

2(y 2 + 4) = 1(3y 2 + 7)

2y 2 + 8 = 3y 2 + 7

3y 2 – 2y 2 = 7 – 8

y = √-1

## EXERCISE 9.4 PAGE NO: 9.29

1. Four-fifth of a number is more than three-fourths of the number by 4. Find the number.

Let us consider the number as ‘x’

So, Three-fourth of the number is 3x/4

Fourth-fifth of the number is 4x/5

4x/5 – 3x/4 = 4

By taking LCM of 5 and 4, we get 20

(16x – 15x)/20 = 4

16x – 15x = 4(20)

∴ The number is 80.

2. The difference between the squares of two consecutive numbers is 31. Find the numbers.

Let the two consecutive numbers be x and (x – 1)

x 2 – (x-1) 2 = 31

By using the formula (a-b) 2 = a 2 + b 2 – 2ab

x 2 – (x 2 – 2x + 1) = 31

x 2 – x 2 + 2x – 1 = 31

2x – 1 = 31

Two consecutive numbers are, x and (x-1) : 16 and (16-1) =15

∴ The two consecutive numbers are 16 and 15.

3. Find a number whose double is 45 greater than its half.

2x – x/2 = 45

(4x-x)/2 = 45

∴ The number is 30.

4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4, more than twice the number.

Then, five times the number will be 5x

And, two times, the number will be 2x

5x – 5 = 2x + 4

5x – 2x = 5 + 4

∴ The number is 3.

5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

x/5 + 5 = x/4 – 5

x/5 – x/4 = -5 – 5

By taking LCM for 5 and 4, which is 20

(4x-5x)/20 = -10

4x – 5x = -10(20)

∴ The number is 200.

6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.

We know that one of the digits be ‘x’

The other digit is 9-x

So, the two digit number is 10(9-x) + x

The number obtained after interchanging the digits is 10x + (9-x)

10(9-x) + x – 27 = 10x + (9-x)

90 – 10x + x – 27 = 10x + 9 – x

-10x + x – 10x + x = 9 – 90 + 27

The two-digit number is 10(9-x) + x

Substituting the value of x, we get,

10(9-x) + x

10(9 – 3) + 3

∴ The number is 63.

7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Let one of the numbers be ‘x’

The other number is 184 – x

So, One-third of one part may exceed one-seventh of another part by 8.

x/3 – (184-x)/7 = 8

LCM for 3 and 7 is 21

(7x – 552 + 3x)/21 = 8

(7x – 552 + 3x) = 8(21)

10x – 552 = 168

10x = 168 + 552

∴ One of the numbers is 72, and the other number is 184 – x => 184 – 72 = 112.

8. The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?

Let us consider the denominator as x and numerator as (x-6)

By using the formula,

Fraction = numerator/denominator = (x-6)/x

(x – 6 + 3)/x = 2/3

(x – 3)/x = 2/3

3(x-3) = 2x

3x – 9 = 2x

3x – 2x = 9

∴ The denominator is x = 9, numerator is (x-6) = (9-6) = 3

And the fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3

9. A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.

Let the number of 10Rs notes be x

Number of 20Rs notes be 50 – x

Amount due to 10Rs notes = 10 × x = 10x

Amount due to 20Rs notes = 20 × (50 – x) = 1000 – 20x

So the total amount is Rs 800

10x + 1000 – 20x = 800

-10x = 800 – 1000

-10x = -200

x = -200/-10

∴ The number of 10Rs notes is 20

Number of 20Rs notes are 50 – 20 = 30

10. Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty- five paise coins as she has fifty-paise coins. How many coins of each kind does she have?

Let the number of fifty paise coins be x

The number of twenty-five paise coins be 2x

Amount due to fifty paise coins = (50×x)/100 = 0.50x

Amount due to twenty five paise coins = (25×2x)/100 = 0.50x

So the total amount is Rs 9

0.50x + 0.50x = 9

∴ The number of fifty paise coins is x = 9

Number of twenty-five paise coins, 2x = 2×9 = 18

11. Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?

Let the present age of Ashima be ‘x’ years

The present age of Sunita is 2x years

Ashima’s new age = (x – 6) years

Sunita’s new age = (2x + 4) years

So, (2x + 4) = 4 (x – 6)

2x + 4 = 4x – 24

2x – 4x = -24 – 4

∴ The age of Ashima is x years = 14 years

Age of Sunita is 2x years = 2(14) = 28 years

Two years ago, age of Ashima is 14 – 2 = 12 years, age of Sunita = 28 – 2 = 26 years

12. The ages of Sonu and Monu are in the ratio 7:5 ten years hence, the ratio of their ages will be 9:7. Find their present ages.

Let the present age of Sonu be 7x years

The present age of Monu is 5x years

Sonu’s age after 10 years = (7x + 10) years

Monu’s age after 10 years = (5x + 10) years

(7x + 10) / (5x + 10) = 9/7

by using cross-multiplication, we get,

7(7x + 10) = 9(5x + 10)

49x + 70 = 45x + 90

49x – 45x = 90 – 70

∴ Present age of Sonu is 7x = 7(5) = 35years

Present age of Monu is 5x = 5(5) = 25years

13. Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Let the age of the son five years ago be x years

The age of man five years ago be 7x years

After five years, the son’s age is x + 5 years

After five years father’s age is 7x + 5 years

So, since five years, the relation in their ages are

7x + 5 + 5 = 3(x + 5 + 5)

7x + 10 = 3x + 15 + 15

7x + 10 = 3x + 30

7x – 3x = 30 – 10

∴ Present father’s age is 7x + 5 = 7(5) + 5 = 35 + 5 = 40years

Present son’s age is x + 5 = 5 + 5 = 10years

14. I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?

Let the present son’s age be x years

Present father’s age be 5x years

Son’s age after 6 years = (x + 6) years

Fathers’ age after 6 years = (5x + 6) years

5x + 6 = 3(x + 6)

5x + 6 = 3x + 18

5x – 3x = 18 – 6

∴ present son’s age is x = 6years

Present father’s age is 5x = 5(6) = 30years

15. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Let the number of five rupee notes be x

The number of ten rupee notes be (x + 10)

Amount due to five rupee notes = 5 × x = 5x

Amount due to ten rupee notes = 10 (x + 10) = 10x + 100

The total amount = Rs 1000

5x + 10x +100 = 1000

∴ the number of five rupee notes is x = 60

The number of ten rupee notes is x + 10 = 60+10 = 70

16. At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two-fifths drank fruit juice, and just three did not drink anything. How many guests were in all?

Let the number of guests be x

The given details are the number of guests who drank colas are x/4

The number of guests who drank squash is x/3

The number of guests who drank fruit juice is 2x/5

The number of guests who did not drink anything was 3

x/4 + x/3 + 2x/5 + 3 = x

By taking LCM for 4, 3 and 5, we get 60

(15x+20x+24x-60x)/60 = -3

(15x+20x+24x-60x) = -3(60)

∴ The total number of guests in all was 180

17. There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

Let the number of correct answers be x

The number of questions answered wrong is (180 – x)

Total score when answered right = 4x

Marks deducted when answered wrong = 1(180 – x) = 180 – x

4x – (180 – x) = 450

4x – 180 + x = 450

5x = 450 + 180

∴ 126 questions he answered correctly.

18. A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works, and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, how many days he remained absent?

Let us consider the number of absent days as x

So, the number of present days is (20 – x)

The wage for one day of work = Rs 60

Fine for absent day = Rs 5

60(20 – x) – 5x = 745

1200 – 60x – 5x = 744

-65x = 744-1200

-65x = -456

x = -456/-65

∴ For 7 days, the labourer was absent.

19. Ravish has three boxes whose total weight is 60 ½ Kg. Box B weighs 3 ½ kg more than box A, and box C weighs 5 1/3 kg more than box B. Find the weight of box A.

The given details are the total weight of three boxes is 60 ½ kg = 121/2 kg

Let the weight of box A be x kg

Weight of box B be x + 7/2 kg

Weight of box C be x + 7/2 + 16/3 kg

x + x + 7/2 + x + 7/2 + 16/3 = 121/2

3x = 121/2 – 7/2 – 7/2 – 16/3

3x = (363 – 21 – 21 – 32)/6

∴ The weight of box A is 289/18 kg

20. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.

Le the denominator be x and the numerator be (x – 3)

By using the formula

Fraction = numerator/denominator

= (x – 3)/x

So, when the numerator is increased by 2 and Denominator is increased by 5, then the fraction is ½

(x – 3 + 2)/(x + 5) = 1/2

(x – 1)/(x + 5) = 1/2

By using cross-multiplication, we get

2(x – 1) = x + 5

2x – 2 = x + 5

2x – x = 2 + 5

∴ Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4

And the fraction = numerator/denominator = 4/7

21. In a rational number, twice the numerator is 2 more than the denominator, if 3 is added to each, the numerator and the denominator. The new fraction is 2/3. Find the original number.

Le the numerator be x and the denominator be (2x – 2)

= x / (2x – 2)

So, the numerator and denominator are increased by 3, then the fraction is 2/3

(x + 3)/(2x – 2 + 3) = 2/3

(x + 3)/(2x + 1) = 2/3

3(x + 3) = 2(2x + 1)

3x + 9 = 4x + 2

3x – 4x = 2 – 9

∴ The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12

And the fraction is numerator/denominator = 7/12

22. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

Let the speed of one train be x km/hr.

The speed of the other train be (x + 5) km/hr.

The total distance between the two stations = 340 km

Distance = speed × time

So, the distance covered by one train in 2 hrs. Will be x×2 = 2x km

Distance covered by the other train in 2 hrs. Will be 2(x + 5) = (2x + 10) km

The distance between the trains is 30 km

2x + 2x + 10 + 30 = 340

4x + 40 = 340

4x = 340 – 40

∴ The speed of one train is x = 75 km/hr.

Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.

23. A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr., find the speed of the steamer in still water and the distance between the ports.

Let the speed of the steamer be x km/hr.

Speed of stream = 1 km/hr.

Downstream speed = (x + 1) km/hr.

Upstream speed = (x – 1) km/hr.

= (x + 1) × 9 and

= (x – 1) × 10

9x + 9 = 10x – 10

9x – 10x = -10 -9

x = 19 km/hr.

∴ The speed of the steamer in still water is 19 km/hr.

Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.

24. Bhagwanti inherited Rs 12000.00. She invested part of it at 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?

At a rate of 10%, let the investment be Rs x

At the rate of 12%, the investment will be Rs (12000 – x)

At 10% of rate the annual income will be x × (10/100) = 10x/100

At 12% of rate, the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100

Total investment = 1280

So, 10x/100 + (144000 – 12x)/100 = 1280

(10x + 144000 – 12x)/100 = 1280

(144000 – 2x)/100 = 1280

144000 – 2x = 1280(100)

-2x = 128000 – 144000

-2x = -16000

x = -16000/-2

∴ At 10% of rate, she invested Rs 8000, and at 12% of the rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000

25. The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm 2 more than that of the given rectangle. Find the length and breadth of the given rectangle.

Let the breadth of the rectangle be x meter

Length of the rectangle be (x + 9) meter

Area of the rectangle length×breadth = x(x +9) m 2

When length and breadth increased by 3cm, then,

New length = x + 9 + 3 = x + 12

New breadth = x + 3

So, the area is

(x + 12) (x + 3) = x (x + 9) + 84

x 2 + 15x + 36 = x 2 + 9x + 84

15x – 9x = 84 – 36

∴ The length of the rectangle (x + 9) = (8 + 9) = 17cm, and the breadth of the rectangle is 8cm.

26. The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?

Let the age of Anup be x years

So the age of Anup’s father will be (100 – x) years

The age of Anuj is (100-x)/5 years

So, When Anup is as old as his father after (100 – 2x) years,

Then Anuj’s age = present age of his father (Anup) + 8

Present age of Anuj + 100 – 2x = Present age of Anup + 8

(100 – x)/5 + (100 – 2x) = x + 8

(100-x)/5 – 3x = 8 – 100

(100 – x – 15x)/5 = -92

100 – 16x = -460

-16x = -460 – 100

-16x = -560

x = -560/-16

∴ The present age of Anup is 35 years then, the age of Anup’s father will be (100-x) = 100-35 = 65 years

The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years

27. A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?

Let the amount lady had be Rs x

Amount spent for hankies and given to beggar is x/2 + 1

Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2

Amount spent for lunch (x-2)/2×1/2 = (x-2)/4

The amount given as a tip is Rs 2

Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4

Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8

The bus fare is Rs 3

Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8

So from the question, we know that the amount left = Rs 1

(x-34)/8 = 1

x – 34 = 8

∴ the lady started with Rs. 42

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## CBSE Class 10 Maths Case Study Questions for Chapter 3 - Pair of Linear Equations in Two Variables (Published by CBSE)

Cbse's question bank on case study for class 10 maths chapter 3 is available here. these questions will be very helpful to prepare for the cbse class 10 maths exam 2022..

Case study questions are going to be new for CBSE Class 10 students. These are the competency-based questions that are completely new to class 10 students. To help students understand the format of the questions, CBSE has released a question bank on case study for class 10 Maths. Students must practice with these questions to get familiarised with the concepts and logic used in the case study and understand how to answers them correctly. You may check below the case study questions for CBSE Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables. You can also check the right answer at the end of each question.

Check Case Study Questions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

CASE STUDY-1:

1. If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?

2. How many questions did he guess?

3. If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?

4. If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?

Let the no of questions whose answer is known to the student x and questions attempted by cheating be y

x – 1/4y =90

solving these two

x = 96 and y = 24

1. He answered 96 questions correctly.

2. He attempted 24 questions by guessing.

3. Marks = 80- ¼ 0f 40 =70

4. x – 1/4 of (120 – x) = 95

5x = 500, x = 100

CASE STUDY-2:

Amit is planning to buy a house and the layout is given below. The design and the measurement has been made such that areas of two bedrooms and kitchen together is 95 sq.m.

Based on the above information, answer the following questions:

1. Form the pair of linear equations in two variables from this situation.

2. Find the length of the outer boundary of the layout.

3. Find the area of each bedroom and kitchen in the layout.

4. Find the area of living room in the layout.

5. Find the cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m.

1. Area of two bedrooms= 10x sq.m

Area of kitchen = 5y sq.m

10x + 5y = 95

Also, x + 2+ y = 15

2. Length of outer boundary = 12 + 15 + 12 + 15 = 54m

3. On solving two equation part(i)

x = 6m and y = 7m

area of bedroom = 5 x 6 = 30m

area of kitchen = 5 x 7 = 35m

4. Area of living room = (15 x 7) – 30 = 105 – 30 = 75 sq.m

5. Total cost of laying tiles in the kitchen = Rs50 x 35 = Rs1750

Case study-3 :

It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:

Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110.

Situation 2: In a city B, for a journey of 8km, the charge paid is Rs91 and for a journey of 14km, the charge paid is Rs 145.

Refer situation 1

1. If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing the situation is

a) x + 10y =110, x + 15y = 75

b) x + 10y = 75, x + 15y = 110

c) 10x + y = 110, 15x + y = 75

d) 10x + y = 75, 15x + y = 110

Answer: b) x + 10y = 75, x + 15y = 110

2. A person travels a distance of 50km. The amount he has to pay is

Answer: c) Rs.355

Refer situation 2

3. What will a person have to pay for travelling a distance of 30km?

Answer: b) Rs.289

4. The graph of lines representing the conditions are: (situation 2)

Answer: (iii)

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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Case Study Questions for Class 8 Maths Chapter 2 Linear Equations in One Variable Here we are providing Case Study questions for Class 8 Maths Chapter 2 Linear Equations in One Variable. Maths Class 8 Chapter 2 Linear Equations in One Variable Maths CBSE Class 8 Chapter Covered Class 8 Maths Chapter 2 Topics Solving … Continue reading Case Study Questions for Class 8 Maths Chapter 2 Linear ...

Case Study Questions on Linear Equations in One Variable. Questions. Passage 1: It is common that government revises fares from time to time based on various factors such as taxes, economy and inflation, for various vehicles like auto-rickshaw, taxis and radio cab etc. The auto and Taxi charge in a city comprise of fixed charge and the charge for the distance covered.

@mathscluster5737 Case study based question | Class 8 maths | linear equations in one variable| DAV maths | Case studyFor more videos of HOTS questions for c...

Extra Practice Questions For Class 8 Maths Chapter 2 Linear Equations in One Variable. 1. The tens and unit digits of a number are the same. By adding the number to its reverse, you get the sum 110. Calculate the number. 2. A man sold his bicycle for an amount that is over Rs 988 by half the price that he paid for it. He made a profit of Rs, 300.

CBSE Maths Value Based Questions for Class 8th download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

Linear Equations in One Variable Class 8 Extra Questions Very Short Answer Type. Question 1. Identify the algebraic linear equations from the given expressions. (a) x 2 + x = 2 is not a linear equation. (b) 3x + 5 = 11 is a linear equation. (c) 5 + 7 = 12 is not a linear equation as it does not contain variable.

With proper practice of these class 8 maths chapter 2 important questions, students can prepare well for their exams. Below are some of the important questions from this chapter. 1. Solve the following linear equations: (i) x - 11 =7. (ii) z + 8 = 9. (iii) 11x = 121. 2. The sum of three consecutive multiples of 8 is 888.

Case Study Class 8 Maths | Cambridge I Did It Mathematics | Ch 7 Linear equation in one variable._____...

Class 8 Maths Chapter 2 | Linear Equations in One Variable | Case Study QuestionIn this video, I have solved case study question of class 8 maths chapter 2 L...

Class 8. 12 units · 55 skills. Unit 1. Rational Numbers. Unit 2. Linear Equations in one Variable. ... This lesson covers skills from the following lessons of the NCERT Math Textbook: (i) 2.2 Equations with variable on both sides. Learn. ... Solving equations reducible to linear form Get 3 of 4 questions to level up!

This section of Maths NCERT Solutions Class 8 Chapter 2 starts by recalling the method for solving equations like the one shown below: 5x - 3 = 8. We solve this by adding 3 on both sides so we get. 5x - 3 + 3 = 8 + 3. 5x = 11, hence x = 2.2. The above example is a linear expression where the highest power of a variable is only 1.

Updated for new NCERT - 2023-24 Book. Get NCERT Solutions of all Exercise Questions and Examples of Chapter 2 Class 8 Linear Equations in One Variable free at Teachoo. Answers to each and every question with detailed explanation for your understanding. Best answers guaranteed! We studied Equations and Expressions in Algebra ( Chapter 9 Class 8 ...

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations In One Variable are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT ...

NCERT Solutions Class 8 Maths Chapter 2 - Free PDF Download. NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable are provided here in PDF format, which can be downloaded for free. The NCERT Solutions for the Chapter Linear Equations in One Variable have been designed by Mathematics experts at BYJU'S accurately. All the solved questions of Linear Equations in One ...

Math; Class 8 (Old) Unit 2: Linear equations in one variable. 800 possible mastery points. Mastered. ... Word problems linear equations (basic) Get 3 of 4 questions to level up! Sums of consecutive integers Get 3 of 4 questions to level up! Linear equations word problems (advanced) Get 3 of 4 questions to level up! Up next for you: Unit test.

MCQs on Class 8 Linear Equations in One Variable. Multiple choice questions (MCQs) are available for Class 8 Linear Equations in One Variable with each problem consisting of four options, out of which one is the correct answer. Students have to solve the problem and select the correct answer. Verify your solution with the answers provided here. 1.

Find these multiples. 5. Solve 3x/4 - 7/4 = 5x + 12. 6. Perimeter of a rectangle is 13cm. if its width is 11/4 cm, find its length. 7. The present of Sita's father is three times the present age of Sita. After six years sum of their ages will be 69 years. Find their present ages.

Benefits Of Solving Important Questions Class 8 Maths Chapter 2 The subject of Maths requires a lot of practice. The Maths of Classes 8, 9, and 10 are the milestones of fundamental knowledge. We suggest students access our platform Extramarks to access linear equations class 8 extra questions. By systematically solving questions and going ...

Solution. (a) If ax = b, then x = b a. Since, a and b are positive integers. So, b a is also positive integer, Hence, the solution of the given equation has to be always positive. Question. 8 Linear equation in one variable has. (a) only one variable with any power. (b) only one term with a variable.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1. Ex 2.1 Class 8 Maths Question 1. Solve the equation: x - 2 = 7. Solution: Given: x - 2 = 7. ⇒ x - 2 + 2 = 7 + 2 (adding 2 on both sides) ⇒ x = 9 (Required solution) Ex 2.1 Class 8 Maths Question 2. Solve the equation: y + 3 = 10.

Find the length and breadth of the given rectangle. Solution: Let the breadth of the rectangle be x meter. Length of the rectangle be (x + 9) meter. Area of the rectangle length×breadth = x (x +9) m 2. When length and breadth increased by 3cm, then, New length = x + 9 + 3 = x + 12. New breadth = x + 3.

Case study questions for CBSE Class 10 Maths Chapter 3 are available here. CBSE has published these questions on its website cbse.nic.in.