# NCERT Solutions for Class 8 Maths Exercise 2.2 Chapter 2- Linear Equations in One Variable

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**NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.2) Exercise 2.2**

On Extramarks’ website and app, students can find NCERT solutions for Class 8 for every chapter. NCERT Solutions are essential for students to study because they will help them understand the subjects. Students should keep in mind that working through NCERT textbooks can help them perform better on entrance exams like JEE, CUET, and others as well as in board exams. NCERT books can help students improve their fundamental understanding of the chapter as well as how the questions should be answered. Students can obtain assistance from NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, to understand various topics.

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**Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable**

The goal of NCERT exercises is to gauge how well students comprehend the concepts covered in a given chapter. NCERT solutions are crucial for board examinations. If students are having difficulty understanding linear equations in one variable, they should refer to the study materials for Mathematics Chapter 2 Class 8, such as the NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2, which are available on the Extramarks website and application. Many students find this chapter difficult and turn to the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, which offers thorough solutions.

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**What You’ll Learn In NCERT Class 8 Maths Chapter 2 Exercise 2.2**

Through the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, it can be concluded that this chapter begins with a brief introduction before covering a number of concepts, including how to solve equations with linear expressions on one side and numbers on the other, how to apply the aforementioned exercises, and how to solve equations with variables on both sides, equations that can be transformed into a simpler form or a linear form. The second exercise helps students put what they learned in the first exercise into practice. It provides the basis for NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Therefore, it is crucial for students to comprehend the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 in order to achieve a higher grade.

**NCERT Maths Class 8 Chapter 2 Exercise 2.2: Get Detailed Solutions**

Students can understand the concepts in Exercise 2.2 by using the exact solutions and explanations provided in the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Students should remember that every NCERT exercise, including Exercise 2.2, has solutions like the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Students must fully understand the content presented in Exercise 2.2 in order to choose how to respond to various questions. When addressing problems where the query only offers scant information, they can benefit from this. Students could benefit from being aware of the various processes. To score well on exams, students must consequently absorb and retain this information. They could consult the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, if they require additional assistance. Any queries students may have regarding Exercise 13.2 can be answered by reading the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. The ideas in Exercise 2.2 are crucial for students’ exams. As a result, it is crucial for students to comprehend how certain questions should be answered because doing so can improve their academic achievement. If necessary, students must use Extramarks to assist them. For assistance for their exam, they might consult the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Hence, the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is a helpful tool for students that they must use to improve their scores in board exams or year-end exams.

**About Chapter 2 of NCERT Textbook: Linear Equations in One Variable**

Chapter 2 of NCERT Class 8 Mathematics is named Linear Equations in One Variable. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 will provide an overview of the chapter. This chapter contains a brief introduction to the chapter followed by various concepts such as solving equations that have linear expressions on one side and numbers on the other, some applications of the above exercise, and solving equations that have a variable on both sides. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is based on the second exercise that helps students apply the things they have learned in the previous exercise. Hence, the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 is important for students to understand in order to get a good score in the exams.

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**NCERT Solutions Class 8 Maths Chapter 2 All Exercises**

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**NCERT Solutions for Class 8**

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**Q.1 **

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

**Ans**

$\begin{array}{l}\text{Let the number be x}.\\ \mathrm{}\text{According to the question the equation becomes},\\ (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}=\frac{1}{8}\\ \text{On multiplying both the sides by 2,we get}\\ \Rightarrow (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}\times 2=\frac{1}{8}\times 2\\ \Rightarrow \mathrm{x}-\frac{1}{2}=\frac{1}{4}\\ \text{On transposing}\frac{\text{1}}{\text{2}}\text{to R.H.S, we get}\\ \Rightarrow \mathrm{x}=\frac{1}{4}+\frac{1}{2}\\ \Rightarrow \mathrm{x}=\frac{1+2}{4}\\ \Rightarrow \mathrm{x}=\frac{3}{4}\end{array}$

**Q.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?

**Ans**

The perimeter of a rectangular swimming pool is 154 m.

Let the breadth be *x* m. The length will be (2*x* + 2) m.

According to the question the equation becomes,

$\begin{array}{l}\text{2}\left(\text{2}x+\text{2}+x\right)=\text{154}\\ \Rightarrow \text{2}\left(\text{3}x+\text{2}\right)=\text{154}\\ \text{Dividing both sides by 2, we obtain}\\ \Rightarrow \frac{\text{2}\left(\text{3}x+\text{2}\right)\text{}}{2}=\frac{\text{154}}{2}\\ \Rightarrow x+\text{2}=\text{77}\\ \text{On transposing 2 to R.H.S, we obtain}\\ \Rightarrow \text{3}x=\text{77}-\text{2}\\ \Rightarrow \text{3}x=\text{75}\\ \text{On dividing both sides by 3, we obtain}\\ \frac{\text{3}x}{3}=\frac{\text{75}}{3}\\ x=\text{25}\\ \therefore \text{2}x+\text{2}=\text{2}\times \text{25}+\text{2}=\text{52}\end{array}$

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

**Q.3 **

$\begin{array}{l}\mathrm{The}\mathrm{base}\mathrm{of}\mathrm{an}\mathrm{is}\mathrm{osceles}\mathrm{triangleis}\frac{4}{3}\mathrm{cm}.\mathrm{The}\mathrm{perimeter}\\ \mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}4\frac{2}{15}\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{either}\\ \mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{equal}\mathrm{sides}?\end{array}$

**Ans**

$\begin{array}{l}\text{Let the length of equal sides be x cm.}\\ \text{Perimeter =xcm+xcm+Base=4}\frac{\text{2}}{\text{15}}\text{cm}\\ \Rightarrow 2\mathrm{x}+\frac{4}{3}=\frac{62}{15}\\ \text{On transposing}\frac{\text{4}}{\text{3}}\text{to R.H.S, we obtain}\\ \Rightarrow 2\mathrm{x}=\frac{62}{15}-\frac{4}{3}\\ \Rightarrow 2\mathrm{x}=\frac{62-20}{15}\\ \Rightarrow 2\mathrm{x}=\frac{42}{15}\\ \text{On dividing both sides by 2, we obtain}\\ \Rightarrow \frac{2\mathrm{x}}{2}=\frac{42}{15\times 2}\\ \Rightarrow \mathrm{x}=\frac{7}{5}\\ \text{Therefore, the length of equal sides is}\frac{\text{7}}{\text{5}}\text{cm or 1}\frac{\text{2}}{\text{5}}\text{.}\end{array}$

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

**Ans**

Let one number be* x*.

Therefore, the other number will be *x* + 15.

According to the question,

*x + x* + 15 = 95

2*x* + 15 = 95

On transposing 15 to R.H.S, we obtain

2*x* = 95 − 15

2*x* = 80

On dividing both sides by 2, we get

x = 40

Therefore, x+15 = 40+15 = 55

Hence the numbers are 40 and 55.

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

**Ans**

Let the ratio between these numbers be *x*.

Therefore, the numbers will be 5*x* and 3*x* respectively.

Difference between these numbers = 18

According, to the question the equation becomes

5*x* − 3*x* = 18

2*x* = 18

$\begin{array}{l}\text{Dividing both sides by 2,we get}\\ \frac{2\mathrm{x}}{2}=\frac{18}{2}\\ \Rightarrow \mathrm{x}=9\end{array}$

Therefore, the first number is 5x=5×9=45 and,

The second number is 3x=3×9=27.

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

**Ans**

Let three consecutive integers be* x*, *x* + 1, and* x* + 2.

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

3*x* + 3 = 51

On transposing 3 to R.H.S, we obtain

3*x* = 51 − 3

3*x* = 48

$\begin{array}{l}\mathrm{On}\text{}\mathrm{dividing}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}3,\mathrm{we}\text{}\mathrm{get}\\ \frac{3\mathrm{x}}{3}=\frac{48}{3}\\ \Rightarrow \mathrm{x}=16\\ \therefore \mathrm{x}+1=17\\ \mathrm{x}+2=18\end{array}$

Hence, the integers are 16, 17 and 18.

**Q.7** The sum of three consecutive multiples of 8 is 888. Find the multiples.

**Ans**

Let the three consecutive multiples of 8 be 8*x*, 8(*x* + 1), 8(*x* + 2).

Sum of these numbers* =* 8*x *+ 8(*x* + 1) + 8(*x* + 2) = 888

8(*x* + *x* + 1 +* x *+ 2) = 888

8(3*x* + 3) = 888

\begin{array}{l}\text{On dividing both sides by 8, we get}\\ \Rightarrow \frac{8(3x+3)}{8}=\frac{888}{8}\\ \Rightarrow 3x+3=111\\ \text{On transposing 3 to RHS,we get}\\ \Rightarrow 3x=111-3\\ \Rightarrow 3x=108\\ \text{On dividing both sides by 3,we get}\\ \Rightarrow \frac{3x}{3}=\frac{108}{3}\\ \Rightarrow x=36\end{array}

Therefore, 8x = 8×36 = 288

8(x+1) = 8(36+1) = 8×37 = 296

8(x+2) = 8(36+2) = 8×38 = 304

Hence, the numbers are 288,296 and 304.

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

**Ans**

Let three consecutive integers be *x*, *x* + 1, *x* + 2. According to the question,

2*x *+ 3(*x* + 1) + 4(*x* + 2) = 74

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

9*x* + 11 = 74

On transposing 11 to R.H.S, we obtain

9*x* = 74 − 11

9*x* = 63

\begin{array}{l}\text{On dividing both sides by 9,we get}\\ \Rightarrow \frac{9x}{9}=\frac{63}{9}\\ \Rightarrow x=7\\ \therefore x+1=8\\ x+2=9\\ \text{Hence,the numbers are 7, 8 and 9}\text{.}\end{array}

**Q.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 the ratio between Rahul’s age and Haroon’s age be *x*.

Therefore, the age of Rahul and Haroon will be 5*x* years and 7*x* years.

Four years later, the age of Rahul and Haroon will be (5*x *+ 4) years and (7*x *+ 4) years.

According to the given question, the equation becomes,

(5*x *+ 4 + 7*x *+ 4) = 56

12*x *+ 8 = 56

On transposing 8 to R.H.S, we obtain

12*x* = 56 − 8

12*x* = 48

$\begin{array}{l}\mathrm{On}\text{}\mathrm{dividing}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}12,\mathrm{we}\text{}\mathrm{get}\\ \Rightarrow \frac{12\mathrm{x}}{12}=\frac{48}{12}\\ \Rightarrow \mathrm{x}=4\\ \therefore \mathrm{Rahul}\u2018\mathrm{s}\text{}\mathrm{age}=5\mathrm{x}=5\times 4=20\mathrm{years}\phantom{\rule{0ex}{0ex}}\mathrm{Haroon}\u2018\mathrm{s}\text{}\mathrm{age}\text{}\mathrm{is}=7\mathrm{x}\text{}\mathrm{years}=7\times 4=28\text{}\mathrm{years}\end{array}$

**Q.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 ratio between the number of boys and numbers of girls be *x*.

Then, number of boys = 7*x*

Number of girls = 5*x*

According to the given question,

Number of boys = Number of girls + 8

7*x* = 5*x* + 8

On transposing 5*x* to L.H.S, we obtain

7*x* − 5*x* = 8

2*x* = 8

$\begin{array}{l}\mathrm{On}\text{}\mathrm{dividing}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}2,\mathrm{we}\text{}\mathrm{get}\\ \Rightarrow \frac{2x}{2}=\frac{8}{2}\\ \Rightarrow x=4\\ \therefore \mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{boys}=7\mathrm{x}=7\times 4=28\\ \text{}\mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{girls}=\mathrm{x}=5\times 4=20\text{}\\ \mathrm{Hence},\mathrm{the}\text{}\mathrm{total}\text{}\mathrm{strength}\text{}\mathrm{of}\text{}\mathrm{class}=\text{}28+20=48\text{}\mathrm{students}\end{array}$

**Q.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?

**Ans**

Let Baichung’s father’s age be *x* years.

Therefore, Baichung’s age will be (*x* − 29) years and Baichung’s grandfather’s age will be (*x* + 26) years.

According to the given question, the equation becomes:

*x + x* − 29 + *x* + 26 = 135

3*x* − 3 = 135

On transposing 3 to R.H.S, we obtain

3*x* = 135 + 3

3*x* = 138

$\begin{array}{l}\mathrm{On}\text{}\mathrm{dividing}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}3,\mathrm{we}\text{}\mathrm{get}\\ \Rightarrow \frac{3\mathrm{x}}{3}=\frac{138}{3}\\ \Rightarrow \mathrm{x}=46\\ \\ \therefore \text{}\mathrm{Baichung}\u2018\mathrm{s}\text{}\mathrm{father}\u2018\mathrm{s}\text{}\mathrm{age}\text{}=\mathrm{x}\text{}\mathrm{years}=46\text{}\mathrm{years}\\ \text{}\mathrm{Baichung}\u2018\mathrm{s}\text{}\mathrm{age}\text{}=(\mathrm{x}-29)\text{}\mathrm{years}=(46-29)=17\text{}\mathrm{years}\\ \text{}\mathrm{Baichung}\u2018\mathrm{s}\text{}\mathrm{grandfather}\u2018\mathrm{s}\text{}\mathrm{age}\text{}=(\mathrm{x}+26)\text{}\mathrm{years}\\ =(46+26)\mathrm{years}\phantom{\rule{0ex}{0ex}}=72\text{}\mathrm{years}\end{array}$

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

**Ans**

Let Ravi’s present age be *x* years.

Fifteen years later, Ravi’s age = 4 × His present age

*x* + 15 = 4*x*

On transposing *x* to R.H.S, we obtain

15 = 4*x* − *x*

15 = 3*x*

$\begin{array}{l}\mathrm{On}\text{}\mathrm{dividing}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}3,\mathrm{we}\text{}\mathrm{get}\\ \Rightarrow \frac{15}{3}=\frac{3\mathrm{x}}{3}\\ \Rightarrow \mathrm{x}=5\\ \\ \therefore \text{}\mathrm{Ravi}\u2018\mathrm{s}\mathrm{present}\text{}\mathrm{age}\text{}=5\text{}\mathrm{years}\end{array}$

**Q.13 **

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

**Ans**

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}.\\ \text{According to the question,the equation becomes}\\ \frac{5}{2}\mathrm{x}+\frac{2}{3}=-\frac{7}{12}\\ \text{On transposing}\frac{2}{3}\text{to R.H.S,}\text{we get}\\ \Rightarrow \frac{5}{2}\mathrm{x}=-\frac{7}{12}-\frac{2}{3}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-7-8}{12}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-15}{12}\\ \text{On multiplying both sides by}\frac{2}{5},\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{5\mathrm{x}}{2}\times \frac{2}{5}=\frac{-15}{12}\times \frac{2}{5}\\ \Rightarrow \mathrm{x}=-\frac{1}{2}\\ \text{Hence,}\text{the number is}-\frac{1}{2}.\end{array}$

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

**Ans**

Let the ratio between the numbers of notes of different denominations be x.

Therefore, numbers of ₹ 100 notes, ₹ 50 notes, and ₹ 10 notes will be 2x, 3x, and 5x respectively.

Amount of ₹ 100 notes = ₹ (100×2x) = ₹ 200x

Amount of ₹ 50 notes = ₹ (50×3x) = ₹ 150x

Amount of ₹ 10 notes = ₹ (10×5x) = ₹ 50x

Total cash = ₹ 400000.

Therefore,

200x + 150x + 50x = 400000

⇒ 400x = 400000

On dividing both sides by 400, we obtain

x = 1000

Therefore,

Number of ₹ 100 notes = 2x = 2 × 1000 = 2000 notes

Number of ₹ 50 notes = 3x = 3 × 1000 = 3000 notes

Number of ₹ 10 notes = 5x = 5 × 1000 = 5000 notes

**Q.15 **I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of Rs 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

**Ans**

Let the number of ₹ 5 coins be *x*.

Since the number of ₹ 2 coins is 3 times the number of ₹ 5 coins, so the number of ₹ 2 coins = 3*x*

Therefore, the number of ₹1 coins = 160 − (Number of coins of ₹ 5 and of ₹ 2)

= 160 − (3*x* + *x*) = 160 − 4*x*

Now, Amount of ₹ 1 coins = ₹ [1 × (160 − 4*x*)] = ₹ (160 − 4*x*)

Amount of ₹ 2 coins = ₹ (2 × 3*x*) = ₹ 6*x*

Amount of ₹ 5 coins = ₹ (5 × *x*) = ₹ 5*x*

Total amount is* *₹300.

Therefore, 160 − 4*x* + 6x + 5x =300

160 + 7x = 300

On transposing 160 to R.H.S., we obtain

7x = 300 – 160 = 140

On dividing both sides by 7, we get

x = 20No. of ₹ 1 coins = 160 – 4x = 160 – 4×20 = 80

No. of ₹ 2 coins = 3x = 3×20 = 60

No. of ₹ 5 coins = x = 20

Therefore, number of ₹ 1 coins is 80, ₹ 2 coins is 60 and ₹ 5 coins is 20.

**Q.16** The organisers of an essay competition decide that a winner of the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.

**Ans**

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 = ₹ (100 × *x*) = ₹ 100*x*

Amount given to the participants who did not win = ₹ [25(63 − *x*)]

= ₹ (1575 − 25*x*)

According to the given question,

100*x* + 1575 − 25*x *= 3000

On transposing 1575 to R.H.S, we obtain

75*x *= 3000 − 1575

75*x *= 1425

$\begin{array}{l}\text{On dividing both sides by 75},\text{we obtain}\\ \Rightarrow \frac{75\mathrm{x}}{75}=\frac{1425}{75}\\ \Rightarrow \mathrm{x}=19\\ \\ \mathrm{Hence},\text{the number of winners}=19\end{array}$

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