NCERT Solutions Class 8 Maths Chapter 2

NCERT Solutions for Class 8 Mathematics Chapter 2- Linear equations in one variable

Mathematics is a difficult yet interesting subject. Once you learn how to solve numerical problems in Mathematics, you will start building interest in  this subject. It is used in various day-to-day life applications, making it the most demanding subject.

The Chapter linear equations Class 8 include topics like understanding and solving mathematical  expressions and equations. . It is interlinked with different Chapters of NCERT Class 9 Mathematics and NCERT Class 10 Mathematics making it one of the most crucial chapters.

Our NCERT Solutions for Class 8 Mathematics Chapter 2 includes  all the topics students need to focus on during their preparation. You will find solutions to all the questions covered in NCERT textbook as per updated CBSE syllabus. The step by step methods of solving problems in the NCERT Solutions facilitate students in  solving NCERT questions. Thus, enhancing their analytical and problem solving skills .

You can find the NCERT Solutions for Class 8 Mathematics Chapter 2 on the Extramarks’ website. At the same time, you can also avail NCERT study material, NCERT revision notes, NCERT Exemplar and mock tests on our website.

Visit the Extramarks’ website to keep yourself updated about the CBSE syllabus, NCERT Solutions and exam patterns. You can also search for NCERT Solutions Class 9 to keep yourself a step ahead in your preparation.

Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 2

Linear equations in two variables are entirely based on algebraic expressions covered in primary classes. This chapter is also interlinked with the chapter quadratic equations of the secondary classes. This suggests that it has immense importance in Mathematics. 

Extramarks NCERT Solutions for Class 8 Mathematics Chapter 2 has solved key to the textual questions which includes all the topics explained in detail. The chapter briefs about the ways to solve various sets of equations which covers a wide range of Mathematics Algebra Syllabus. 

This chapter will require students to use their logical as well as analytical thinking. They will learn to find solutions to problems in a more calculated and smarter way.


An equation is made up of a variable and a constant marked by their equalities. When an equation contains only linear polynomials, it is called linear equations.

If a linear equation has only one variable, it is called a linear equation in one variable, whereas if a linear equation has two variables, it is called a linear equation in two variables.

In this section, we will briefly talk about linear equations in one variable and all the applications and terminologies related to it.

Solving equations which have linear expressions on one side and a number on the other side.

There are different cases of writing an equation. One such case is linear equations on one side and a number on the other side. We will learn more about solving such types of equations in this section. 

First, let us see what a linear equation on one side and a number on the other side look like.

For example: 

  • 3z-4=8
  • 7y+2=12

Steps to solve equations which have linear expressions on one side and a number on the other side

  1. Write the equation
  2. Bring all the numbers on the right-hand side
  3. Add or subtract them according to the given linear equation
  4. Divide the obtained number with the coefficient of the constant
  5. Calculate the result and interpret it.

For instance, taking the case of above mentioned linear equation, we get






Some applications

In this section, you will learn to solve some word problems based on the types of equations we have completed in the previous section.

You can find tips on how to form linear equations based on word problems and how to calculate results based on them in our NCERT Solutions for Class 8 Mathematics Chapter 2.

Solving equations having the variables on both sides

The other set of equations include variables on both sides. These sets of equations are used in applications of various problems in higher classes.

The examples of such types of equations are as follows:

  • 4y-5= 7y+ 8
  • 10x+4= 7x-3.

Following are the steps to solve such types of equations

  1. Write the equation
  2. Bring the same variables on one side and the other group of variables on the other side
  3. Change the signs while shifting from RHS to LHS and vice versa
  4. Perform addition, subtraction, multiplication or division accordingly
  5. Solve further as linear equations on one side and a number on the other side
  6. Calculate and interpret the results.

For instance, taking the examples mentioned above, we get


4y-7y= 8+5

-3y= 13

y= 13/-3

y= -13/3

Some more applications

In this section, you will solve detailed word problems on equations with variables on both sides. You can find solutions to these problems in our NCERT Solutions for Class 8 Mathematics Chapter 2.

Reducing equations to a simpler form

You may come across various complex sets of equations. It becomes very difficult to solve equations in their complex form. Hence, it is necessary to reduce them to their simpler form.

This section gives steps to reduce an equation to its simpler form. You can find the sequential written steps along with complete explanation and illustrations in our NCERT Solutions for Class 8 Mathematics Chapter 2.

Equations reducible to linear form

Two equations in fractional form can be converted to linear form by cross-multiplying their variables.

For example: If the equation is of the form


Then we can convert it into a linear form by cross-multiplying as follows


One can achieve mastery in this section with a lot of practice and comprehension of the concepts. To sum up, this is one of the root chapters as it is laying the foundation for Algebra of NCERT Class 9 and NCERT Class 10 Mathematics. Hence, students are advised to cover this chapter with complete conceptual understanding and try to make it easier for themselves. 

NCERT Solutions for Class 8 Mathematics Chapter 2 Exercise & Solutions

After solving every question, it is necessary that you have a proper reference solution for cross-checking the answers. Hence, Extramarks’ website provides NCERT Solutions for Class 8 Mathematics Chapter 2 which helps students to check whether their answers are right and up to the examination standards or not. This will help them to understand all the topics in a better way and motivate them to solve different kinds of problems with greater confidence. Click on the below links to view exercise-specific questions and solutions for NCERT Solutions for Class 8 Mathematics Chapter 2:

Chapter 2 – Linear Equations in One Variable Exercises
Exercise 2.1 Questions & Solutions
Exercise 2.2 Questions & Solutions
Exercise 2.3 Questions & Solutions
Exercise 2.4 Questions & Solutions
Exercise 2.5 Questions & Solutions
Exercise 2.6 Questions & Solutions

Along with Class 8 Mathematics Solutions, you can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

NCERT Exemplar Class 8 Mathematics 

NCERT Exemplar book is a great source of knowledge for every Class 8 Mathematics student. The questions cover every corner of NCERT chapters so that students don’t miss out on any of the concepts. It covers most of the comprehension-based questions thus requiring students to have a strong conceptual understanding of every topic they have learned.

The book is specially made for students preparing for various competitive examinations. The questions are updated year by year to ensure that the solutions are up-to-date. It aids in developing logical, analytical, and critical thinking which plays a great role in the skill development of the students in various disciplines of their lives. Another key skill that they will acquire is conveying their answer clearly and concisely.

Students find themselves fully prepared and ready to solve any question after referring to the NCERT Solutions and NCERT Exemplar. Hence, they are able to solve all the advanced-level questions with ease.

Key Features of NCERT Solutions for Class 8 Mathematics Chapter 2

Students with quality study material always excel with good grades. Hence, NCERT Solutions for Class 8 Mathematics Chapter 2 offers a complete solution guide to all the NCERT questions. The key features are provided: 

  • Students can find a detailed solution to each exercise and every question which enables them to test themselves easily during exam preparations. 
  • Students will learn to compose an answer that meets the question’s requirements After completing the NCERT Solutions for Class 8 Mathematics Chapter 2, students will start seeing Mathematics as an interesting and scoring subject
  • This will help students to boost confidence in their preparation.

Q.1 Solve the following equations.

1. x2=72. y+3=103.6=z+2



10.14y8=1311.17+6p=912.  x3+1=715


1.​x2=7Transposing 2 to R.H.S,we obtainx=7+2x=92.y+3=10Transposing 3 to R.H.S,we obtainy=103y=73.6=z+2Transposing 2 to L.H.S,we obtain62=zz=44.37+x=177Transposing 37 to R.H.S,we obtainx=17737x=1737x=147x=25.6x=12Dividing both sides by 6,we get6x6=126x=26.t5=10Multiplying both the sides by 5,we gett5×5=10×5t=507. ​2x3=18Multiplying both the sides by32,we get2x3×32=18×32x=278.1.6=y1.5Multiplying both the sides by 1.5, we get1.6×1.5=y1.5×1.5y=2.49.7x9=16 Transposing 9 to R.H.S,we get7x=16+97x=25Dividing both the sides by 77x7=257x=25710.14y8=13Transposing 8 to R.H.S,we get14y=13+814y=21Dividing both the sides by 1414y14=2114y=3211.17+6p=9 Transposing 17 to R.H.S,we get6p=9176p=8Dividing both sides by 66p6=86p=4312.x3+1=715Transposing1 to R.H.S,we getx3=7151x3=71515x3=815Multiplying both the sides by 3x3×3=8×315x=85


If you subtract 12 from a number and multiply the resultby 12, you get 18. What is the number?


Let the number be x. According to the question the equation becomes,(x12)×12=18On multiplying both the sides by 2,we get(x12)×12×2=18×2x12=14On transposing 12 to R.H.S, we getx=14+12x=1+24x=34

Q.3 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 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 (2x + 2) m.
According to the question the equation becomes,

22x+2+x= 15423x+2= 154Dividing both sides by 2, we obtain23x+2 2=1542x+2=77On transposing 2 to R.H.S, we obtain 3x=7723x=75On dividing both sides by 3, we obtain3x3=753x=252x+2=2×25+2=52

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


The base of an is osceles triangleis 43 cm.The perimeter of the triangle is 4215cm.What is the length of eitherof the remaining equal sides?


Let the length of equal sides be x cm.Perimeter =xcm+xcm+Base=4215cm2x+43=6215On transposing 43 to R.H.S, we obtain2x=6215432x=6220152x=4215On dividing both sides by 2, we obtain2x2=4215×2x=75Therefore, the length of equal sides is 75cm or 125.

Q.5 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

2x + 15 = 95

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

2x = 95 − 15

2x = 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.6 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 5x and 3x respectively.
Difference between these numbers = 18
According, to the question the equation becomes
5x − 3x = 18
2x = 18

Dividing both sides by 2,we get2x2=182x=9

Therefore, the first number is 5x=5×9=45 and,
The second number is 3x=3×9=27.

Q.7 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

3x + 3 = 51

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

3x = 51 − 3

3x = 48

On dividing both sides by 3,we get3x3=483x=16x+1=17x+2=18

Hence, the integers are 16, 17 and 18.

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

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

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

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

8(3x + 3) = 888

On dividing both sides by 8, we get 8(3x+3) 8 = 888 8 3x+3=111 On transposing 3 to RHS,we get 3x=1113 3x=108 On dividing both sides by 3,we get 3x 3 = 108 3 x=36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C2DB@

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.9 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,

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

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

9x + 11 = 74

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

9x = 74 − 11

9x = 63

On dividing both sides by 9,we get 9x 9 = 63 9 x=7 x+1=8 x+2=9 Hence,the numbers are 7, 8 and 9. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8932@

Q.10 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 5x years and 7x years.

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

According to the given question, the equation becomes,

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

12x + 8 = 56

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

12x = 56 − 8

12x = 48

On dividing both sides by 12,we get12x12=4812x=4Rahuls age=5x=5×4=20yearsHaroons age is=7x years=7×4=28 years

Q.11 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 = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

7x = 5x + 8

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

7x − 5x = 8

2x = 8

On dividing both sides by 2,we get2x2=82x=4Number of boys=7x=7×4=28 Number of girls=x=5×4=20 Hence,the total strength of class= 28+20=48 students

Q.12 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

3x − 3 = 135

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

3x = 135 + 3

3x = 138

On dividing both sides by 3,we get3x3=1383x=46 Baichungs fathers age =x years=46 years Baichungs age =(x29) years=(4629)=17 years Baichungs grandfathers age =(x+26) years=(46+26)years=72 years

Q.13 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 = 4x

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

15 = 4xx

15 = 3x

On dividing both sides by 3,we get153=3x3x=5 Ravis present age =5 years


A rational number is such that when you multiply it by 52and add 23 to the product, you get 712. What is the number ?


Let the number be x.According to the question,the equation becomes52x+23=712On transposing 23 to R.H.S, we get52x=7122352x=781252x=1512On multiplying both sides by25, we get5x2×25=1512×25x=12Hence, the number is 12.

Q.15 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.


200x + 150x + 50x = 400000

⇒ 400x = 400000

On dividing both sides by 400, we obtain

x = 1000


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.16 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 = 3x

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

= 160 − (3x + x) = 160 − 4x

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

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

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

Total amount is ₹300.

Therefore, 160 − 4x + 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.17 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) = ₹ 100x

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

= ₹ (1575 − 25x)

According to the given question,

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 obtain75x75=142575x=19Hence, the number of winners=19


Solve the following equations and check your results.1.3x=2x+182.5t3=3t53.5x+9=5+3x4.4z+3=6+2z5.2x1=14x6.8x+4=3x1+77. x=45x+108.2x3+1=7x15+39.2y+53=263y10.3m=5m85


1. 3x=2x+18On transposing 2x to L.H.S, we get3x2x=18x=18L.H.S=3x=3×18=54R.H.S=2x+18=2×18+18=36+18=54 L.H.S=R.H.S

2. 5t3=3t5On transposing 3t to L.H.S and 3 to R.H.S, we get 5t3t=352t=2t=1L.H.S=5t3=5×(1)3=53=8R.H.S=3t5=3×(1)5=35=8L.H.S=R.H.S

3. 5x+9=5+3xOn transposing 3x to L.H.S and 9 to R.H.S, we get 5x3x=592x=4On dividing both sides by 2,we get2x2=42x=2L.H.S=5x+9=5×(2)+9=(10)+9=1R.H.S=5+3x=5+3×(2)=5+(6)=1 L.H.S=R.H.S

4. 4z+3=6+2zOn transposing 2z to L.H.S and 3 to R.H.S, we get 4z2z=632z=3On dividing both sides by 2, we get2z2=32z=32L.H.S=4z+3=4×(32)+3=6+3=9R.H.S=6+2z=6+2×32=6+3=9L.H.S=R.H.S

5. 2x1=14xOn transposing x to L.H.S and 1 to R.H.S,we get2x+x=14+13x=15On dividing both sides by 3,we get3x3=153x=5L.H.S=2x1=2×51=101=9R.H.S=14x=145=9L.H.S=R.H.S

6. 8x+4=3(x1)+78x+4=3x3+78x+4=3x+4On transposing 3x to L.H.S and 4 to R.H.S,we get 8x3x=445x=0x=0L.H.S=8x+4=8×0+4=4R.H.S=3(x1)+7=3(01)+7=3+7=4L.H.S=R.H.S

7. x=45(x+10)Multiplying both sides by 5,we get5x=4(x+10)5x=4x+40Transposing 4x to L.H.S,we get5x4x=40x=40L.H.S=x=40R.H.S=45(x+10)=45(40+10)=45(50)=45×(50)=4×10]=40L.H.S=R.H.S

8. 2x3+1=7x15+3On transposing 7x15 to L.H.S and 1 to R.H.S,we get 2x37x15=31Taking LCM of 3 and 15,we get10x7x15=23x15=23x=30x=303=10L.H.S=2x3+1=2×103+1=203+1=20+33=233


9. 2y+53=263yOn transposing y to L.H.S and 53 to R.H.S,we get 2y+y=263533y=213y=219y=73

L.H.S=2y+53=2×73+53=143+53=14+53=193R.H.S=263y=26373=2673=193 L.H.S=R.H.S

10. 3m=5m85On transposing 5m to L.H.S ,we get 3m5m=852m=85


L.H.S=3m=3×45=125R.H.S=5m85=5×4585=485=2085=125L.H.S=R.H.S MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqGabeGaaa0faGJlbaGaamitaiaac6cacaWGibGaaiOlaiaadofacqGH9aqpieqacaWFGaGaa8hiaiaaiodacaWGTbaabaGaeyypa0JaaG4maiabgEna0oaalaaabaGaaGinaaqaaiaaiwdaaaaabaGaeyypa0ZaaSaaaeaacaaIXaGaaGOmaaqaaiaaiwdaaaaabaaabaGaamOuaiaac6cacaWGibGaaiOlaiaadofacqGH9aqpcaWFGaGaaGynaiaa=bcacaWGTbGaa8hiaiabgkHiTmaalaaabaGaaGioaaqaaiaaiwdaaaaabaGaeyypa0JaaGynaiabgEna0oaalaaabaGaaGinaaqaaiaaiwdaaaGaeyOeI0YaaSaaaeaacaaI4aaabaGaaGynaaaaaeaacqGH9aqpcaaI0aGaeyOeI0YaaSaaaeaacaaI4aaabaGaaGynaaaaaeaacqGH9aqpdaWcaaqaaiaaikdacaaIWaGaeyOeI0IaaGioaaqaaiaaiwdaaaaabaGaeyypa0ZaaSaaaeaacaaIXaGaaGOmaaqaaiaaiwdaaaaabaaabaGaeyinIWLaamitaiaac6cacaWGibGaaiOlaiaadofacqGH9aqpcaWGsbGaaiOlaiaadIeacaGGUaGaam4uaaqaaaaaaa@73D3@


Amina thinks of a number and subtracts 52 from it. Shemultiplies the result by 8. The result now obtained is3 times the same number she thought of.What is thenumber?

Ans. Let the number be x.
According to the given question,


8x − 20 = 3x
Transposing 3x to L.H.S and −20 to R.H.S, we obtain
8x − 3x = 20
5x = 20
On dividing both sides by 5, we get
x = 4
Hence, the number is 4.

Q.20 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? What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans. Let the numbers be x and 5x.

According to the question, the equation becomes 21+5x=2(x+21)21+5x=2x+42Transposing 2x to L.H.S and 21 to R.H.S, we obtain5x2x=42213x=21Dividing both sides by 3, we obtainx=75x=5×7=35Hence, the numbers are 7 and 35 respectively.

Q.21 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?

Ans. Let the digits at tens place and ones place be x and 9 − x respectively.
Therefore, original number = 10x + (9 − x) = 9x + 9
On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.

Therefore, new number after interchanging the digits = 10(9 − x) + x
= 90 − 10x + x
= 90 − 9x

According to the given question,
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
54 = 18x

Dividing both sides by 18, we obtain
3 = x and 9 − x = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 9x + 9
= 9 × 3 + 9
= 36

Q.22 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?


Let the digits at tens place and ones place be ‘x and ‘3x.the original number=10x+3x=13xOn interchanging the digits, the digits at ones place will be ‘x’and tens place will be ‘3x.New number=10×3x+x=30x+x=31xAccording to the given question,the equationbecomes 8813x+ 31x= 8844x= 88

Dividing both sides by 44, we obtain44x44=8844x= 2, original number=13x=13×2=26Hence, the twodigit number may be 26 or 62.

Q.23 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Ans. Let Shobo’s age be x years.
His mother’s age will be 6x years.
According to the given question, the equation becomes


Multiplying both sides by 3

(x+5)×3=6x3×33x+15=6xTransposing 3x to R.H.S,we get15=6x3x15=3xDividing both sides by 3, we get153=3x3x=56x=6×5=30

Therefore, the present ages of Shobo and his mother will be 5 years and 30 years.

Q.24 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

Ans. Let the 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)


It is given that the cost of fencing the plot at the rate of ₹ 100 per metre is ₹ 75, 000.

100 × Perimeter = 75000

100 × 30x = 75000

3000x = 75000

Dividing both sides by 3000, we obtain

3000x 3000 = 75000 3000 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaabodacaaIWaGaaGimaiaaicdacaWG4baabaGaaG4maiaaicdacaaIWaGaaGimaaaacqGH9aqpcaqGGaWaaSaaaeaacaqG3aGaaeynaiaaicdacaaIWaGaaGimaaqaaiaaiodacaaIWaGaaGimaiaaicdaaaaaaa@482E@

x = 25

Therefore, Length = 11x m = (11 × 25) m = 275 m

and Breadth = 4x m = (4 × 25) m = 100 m

Hence, the dimensions of the plot are 275 m and 100 m.

Q.25 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

Ans. Let the trouser and shirt material bought by Hasan be 2x and 3x .
The selling price of trouser material is


The selling price of Shirt material is

=( 50+ 50×10 100 ) = 55 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiabg2da9iaaysW7tCvAUfKttLearyGu1bxzLbIrVjxyKLwyUbacgaGae8hyaa2aaeWaaeaacaaI1aGaaGimaiabgUcaRmaalaaabaGaaGynaiaaicdacqGHxdaTcaaIXaGaaGimaaqaaiaaigdacaaIWaGaaGimaaaaaiaawIcacaGLPaaaaeaacqGH9aqpcaaMe8Uae8hyaaMae8hiaaIaaGynaiaaiwdaaaaa@5607@

Total sale =36,600

100.80×(2x)+55×(3x)=36,600201.60x+165x= 36,600366.60x=36600Dividing both sides by 366.60, we obtain366.60x366.60=36600366.60x=99.836x=100Trouser material=2xm =(2×100)m =200 m

Q.26 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.

Ans. Let the number of deer be x.

Number of deer grazing in the field = x / 2

Remaining deer= x – (x / 2)

Therefore, number of deer playing nearby is


The rest 9 are drinking water from the pond, so the equation becomes:



Mutiplying both sides by 8,we getx8×8=9×8x=72

Therefore, the total number of deer in the herd is 72.

Q.27 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Ans. Let the granddaughter’s age be x years.

Therefore, grandfather’s age will be 10x years.

According to the question, the equation becomes:

10x = x + 54

Transposing x to L.H.S, we obtain

10xx = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years

= (10 × 6) years

= 60 years

Therefore, the grandfather’s age is 60 years and granddaughter’s age is 6 years.

Q.28 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Ans. Let Aman’s son’s age be x years.
Therefore, Aman’s age will be 3x years.
Ten years ago, their age was (x − 10) years and (3x − 10) years.
According to the question, the equation is:

3x10=5(x10)3x10=5x50On transposing 5x to L.H.S​ and 10 to R.H.S,we get3x5x=10502x=402x=40

On dividing both the sides by 2,we get2x2=402x=20

Aman’s son’s age is 20 years and
Aman’s age = 3 × 20 years = 60 years.


Solve the following linear equations.1.x215=x3+142.n23n4+5n6=213.x+78x3=1765x2

4.x53=x355.3t242t+33=23t6.mm12=1m23Simplify and solve the following linear equations.7. 3(t3)=5(2t+1)8. 15(y4)2(y9)+5(y+6)=09. 3(5z7)2(9z11)=4(8z13)1710. 0.25(4f3)=0.05(10f9)


1.x215=x3+14Taking L.C.M on both the sides ,we get 5x210=4x+312Multiplying both the sides by10,we get5x210×10=4x+312×10(5x2)=4x+312×10

Multiplying both the sides by12,we get(5x2)×12=4x+312×10×1260x24=40x+30Transposing 40x to L.H.S and 24 to R.H.S,we get60x40x=30+2420x=54Dividing both sides by 20 ,we get20x20=5420x=2.7

2. n23n4+5n6=21Taking L.C.M of 2,4 and 6,we get6n9n+10n12=217n12=21

Multiplying bith sides by 12, we get7n12×12=21×127n=21×127n=252Dividing both sides by 7, we get7n7=2527n=2527n=36

3. x+78x3=1765x2Transposing 5x2 to L.H.S and 7 to R.H.S, we getx8x3+5x2=1767Taking LCM both the sides, we get

6x16x+15x6=174265x6=256Multiplying both the sides by 65x6×6=256×65x=25Dividing both the sides by 55x5=255x=5

4.x53=x35Multiplying both the sides by 3,we getx53×3=x35×3x5=x35×3

Multiplying both the sides by 5,we get (x5)×5=x35×3×55(x5)=3(x3)5x25=3x95x3x=9+252x=16x=8

5.3t242t+33=23t3(3t2)4(2t+3)12=23t39t68t1212=23t3t1812=23t3Multiplying both the sides by 12,we gett1812×12=23t3×12t18=(23t)4t18=812t

Transposing 12t to L.H.S and 18 to R.H.S ,we gett+12t=8+1813t=26t=2613=2

6.mm12=1m232mm+12=3m+23m+12=5m3Multiplying both the sides by 2,we getm+12×2=5m3×2m+1=5m3×2

Multiplying both the sides by 3,we get ⇒(m+1)×3=5m3×2×33m+3=102m3m+2m=1035m=7m=75


Transposing 10t to L.H.S and 9 to R.H.S,we get3t10t=5+97t=14t=2

8. 15(y4)2(y9)+5(y+6)=015y602y+18+5y+30=015y2y+5y60+18+30=018y12=018y=12y=1218y=23

9. 3(5z7)2(9z11)=4(8z13)1715z2118z+22=32z52173z+1=32z69Transposing 32z to L.H.S and 1 to R.H.S,we get3z32z=69135z=70z=7035=2z=2

10.0.25(4f3)=0.05(10f9)1f0.75=0.5f0.45Transposing 0.5f to L.H.S and 0.75 to R.H.S,we get1f0.5f=0.750.450.5f=0.3f=0.30.5=0.6f=0.6

Q.30 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.

Ans. Let the ratio between their ages be x.

Therefore, Hari’s age and Harry’s age will be 5x years and 7x years

Four years later, their ages will be (5x + 4) years and (7x + 4) years.

According to the question, we get

5x+47x+4=34Multiplying both the sides by (7x+4),we get5x+47x+4×(7x+4)=34×(7x+4)5x+4=34×(7x+4)Multiplying both the sides by 4,we get(5x+4)×4=34×(7x+4)×420x+16=21x+1220x21x=1216x=4x=4Harisage=5x years=20 years and Harry’s age=7x years=28 years.

Q.31 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 3/2.Find the rational number.

Ans. Let the numerator of the rational number be x.

Therefore, its denominator will be x + 8.

The rational number is xx+8.According to the question,we getx+17x+81=32x+17x+7=32Multiplying both the sides by (x+7),wegetx+17x+7×(x+7)=32×(x+7)x+17=32×(x+7)Multiplying both the sides by 2,weget(x+17)×2=32×(x+7)×22x+34=3x+212x3x=2134x=13x=13

Therefore, the numerator is 13 and the denominator is x+8=21.

Hence, the rational number is


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