# NCERT Solutions for Class 10 Maths Chapter 3 Free PDF Download

## NCERT Solutions for Class 10 Maths Chapter 3 â€“ Pair of Linear Equations in Two Variable

CBSE prescribes NCERT textbooks for class 10. We have compiled ncert solutions for class 10 maths chapter 3 for you.Â Mathematics requires a clear understanding of the concepts, logic and a lot of practice. Hence, we are providing you ncert solutions for class 10 maths chapter 3.Â Our NCERT solutions for class 10 maths chapter 3Â cover all the basic concepts provided in the NCERT textbook.

These ncert solutions for class 10 maths chapter 3 will help you to effectively prepare for your board examination. ncert solutions for class 10 maths chapter 3 are prepared by subject experts. In ncert solutions for class 10 maths chapter 3 the concepts are explained in details and all the doubts are instantly resolved by our subject matter experts during live doubt solving sessions. You will also be able to understand all the difficult concepts by referring to ncert solutions for class 10 maths chapter 3 and score good marks in your exams.

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### CBSE Class 10 Maths Chapter 3 NCERT Solutions â€“ Pair of Linear Equations in Two Variable

NCERT solutions for class 10 maths chapter 3 Â explains the concept of pair of linear equations in two variables, graphical method of solution of a pair of linear equations in two variable,Â  algebraic methods of solving a pair of linear equations, substitution,Â  elimination, cross-multiplication method and equations reducible to a pair of linear equations in two variable, different forms of equations which can be converted into a pair of linear equations.Â The chapter concludes with summarizing points that help you to revise the chapter and its concepts quickly.

### NCERT Solutions for Class 10 Maths Chapter 3

Q.1Â Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.Â

SolutionÂ

The given points are collinear then the area of triangle formed by these points will be 0 âˆ´ 12 [ x1(y2 âˆ’ y1) + x2(y3 âˆ’ y1) + x3(y1 âˆ’ y2) ]Â

= 0Â

Here (x1, y1) = (x, y)Â

(x2, y2) = (1, 2)Â

(x3, y3) = (7, 0) âŸ¹ 21[x(2 âˆ’ 0) + 1(0 âˆ’ y) + 7(y âˆ’ 2)] = 0 âŸ¹ 21[2x âˆ’ y + 7y âˆ’ 14] = 0 âŸ¹ 21[2x + 6y âˆ’ 14] = 0Â

âŸ¹ 2x + 6y âˆ’ 14 = 0Â

âŸ¹ x + 3y âˆ’ 7 = 0Â

Hence this is the required relation between x and y.Â

Q.2 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t thisÂ interesting?) Represent this situation algebraically and graphically.Â

SolutionÂ

Consider Aftab’s age as x and his daughter’s age as y.Â

Then , seven years ago,Â

Aftab’s age = x âˆ’ 7Â

His daughter’s age = y âˆ’ 7Â

According to the question,Â

x âˆ’ 7 = 7(y âˆ’ 7)Â

x âˆ’ 7 = 7y âˆ’ 49Â

x âˆ’ 7y = âˆ’ 49 + 7Â

x âˆ’ 7y = âˆ’ 42 …(i)Â

After three years,Â

Aftab’s age = x + 3Â

His daughter’s age = y + 3Â

According to the question,Â

x + 3 = 3(y + 3)Â

x + 3 = 3y + 9Â

x âˆ’ 3y = 9 âˆ’ 3Â

x âˆ’ 3y = 6 …(ii)Â

Representing equation (i) and (ii) geometrically, we plot these equations by finding points on the lines representing these two equationsÂ

x âˆ’ 7y = âˆ’ 42 âŸ¹ x = 7y âˆ’ 42Â

x = 7y âˆ’ 42 42 35 49 y 12 11 13Â

x âˆ’ 3y = 6 âŸ¹ x = 3y + 6Â

x = 3y + 6 42 36 48 y 12 10 14Â

From the graph we can see that two lines will intersect at a point.Â

x âˆ’ 7y = âˆ’ 42Â

x âˆ’ 3y = 6Â

On subtracting the two equations, we getÂ

4y = 48Â

or y = 12Â

Substituting value of y in (2),Â

x âˆ’ 36 = 6Â

âˆ´ x = 42

Q. 3Â Â The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situationÂ algebraically and geometrically.Â

SolutionÂ

Let the cost of a bat be Rs. x and the cost of a ball be Rs. y.Â

Then algebraically representation of the question is –Â

3x + 6y = 3900 â‡’ x + 2y = 1300……(1)Â

x + 3y = 1300……..(2)Â

To represent it geometrically, we plot the two equationsÂ

x + 3y = 1300 â‡’ y = 1300 3Â

âˆ’ xÂ

x 100 1000 1300 y = 1300 3 âˆ’ xÂ

400 100 0Â

x + 2y = 1300 â‡’ y = 1300 2Â

âˆ’ xÂ

x 100 500 1300 y = 1300 2 âˆ’ xÂ

600 400 0Â

As we can see, the lines intersect at (1300, 0) which is the solution for this question.

Q. 4 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situationÂ algebraically and geometrically.

SolutionÂ

x 65 55 45 35 y 20 40 60 80Â

Let the cost of 1 kg of apples be x and that of 1 kg be yÂ

So the algebraic representation can be as follows:Â

2x + y = 160Â

4x + 2y = 300 â‡’ 2x + y = 150Â

The situation can be represented graphically by plotting these two equations.Â

2x + y = 160 â‡’ y = 160 âˆ’ 2xÂ

x 70 60 50 40 y = 160 âˆ’ 2x 20 40 60 80Â

2x + y = 150 â‡’ y = 150 âˆ’ 2xÂ

x 65 55 45 35 y = 150 âˆ’ 2x 20 40 60 80Â

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.Â

### Solved Questions for You

Question 1: The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as oldÂ as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differÂ by 30 years. Find the ages of Ani and Biju.

Answer: Let the age of Ani and Biju be x and y respectively.

Then according to the question-

xâˆ’y=3……….(i)

Dharam is twice the age of Ani

Hence, age of DharamÂ =2x

Age of Cathy is half the age of Biju

Then age of CathyÂ =y/2â€‹

Then according to the question

2xâˆ’y/2â€‹=30

â‡’4xâˆ’y=60……(ii)

Subtracting (ii) from (i)

â‡’xâˆ’yâˆ’4x+y=3âˆ’60

â‡’âˆ’3x=âˆ’57

â‡’x=âˆ’57/-3â€‹=19

Substitute the vale of x in (i)

â‡’19âˆ’y=3

â‡’âˆ’y=âˆ’16

â‡’y=16

âˆ´Â Age of AniÂ =19Â years and age of BijuÂ =16Â years.

Question 2: One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

Answer: Let the initial amount with them be Rs x and Rs y respectively.

Then according to the question-

â‡’x+100=2(yâˆ’100)

â‡’x+100=2yâˆ’200

â‡’xâˆ’2y=âˆ’300…….. (i)

And,

â‡’6(xâˆ’10)=(y+10)

â‡’6xâˆ’60=y+10

â‡’6xâˆ’y=70…… (ii)

Multiplying equation (ii) by 2, we obtain

12xâˆ’2y=140……. (iii)

Subtracting equation (i) from equation (iii), we gets

â‡’11x=140+300

â‡’11x=440

â‡’x=40

Substitute the vale of x in Â equation (i), we gets

â‡’40âˆ’2y=âˆ’300

â‡’40+300=2y

â‡’2y=340

y=340/2â€‹=170

âˆ´Â The Â friends had Rs 40 and Rs 170 with them respectively.

Question 3: A train covered a certain distance at a uniform speed. If the train would have beenÂ 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer: Let the speed of the train beÂ xkm/hÂ and the time taken by train to travel the given distance beÂ tÂ hours and the distance to travel beÂ dkm. We know that,

â‡’Speed=Distanceâ€‹/Time

â‡’x=tdâ€‹

âˆ´d=xt…….(i)

Case 1

â‡’(x+10)Ã—(tâˆ’2)=d

â‡’xt+10tâˆ’2xâˆ’20=d

â‡’d+10tâˆ’2xâˆ’20=d

â‡’âˆ’2x+10t=20…… (ii)

Case 2

â‡’(xâˆ’10)Ã—(t+3)=d

â‡’xtâˆ’10t+3xâˆ’30=d

â‡’dâˆ’10t+3xâˆ’30=d

â‡’3xâˆ’10t=30……… (iii)

Adding equations (ii) and (iii), we gets

â‡’x=50

Substitute the value of x in (ii) we gets

â‡’(âˆ’2)Ã—(50)+10t=20

â‡’âˆ’100+10t=20

â‡’10t=120

â‡’t=12Â hours

Substitue the value ofÂ tÂ  andÂ xÂ in Â equation (i), we gets

Distance to travelÂ =d=xt

â‡’d=12Ã—50=600Km

Hence, the distance covered by the train isÂ 600km.

Question 4: The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find theÂ number of students in the class.

Answer: Â Let the number of rows be x and number of students in a row be y.

Total students of the class= Number of rowsÂ Ã—Â Number of students in a row

=xÃ—y=xy

Case 1

Total number of studentsÂ =(xâˆ’1)(y+3)

â‡’xy=(xâˆ’1)(y+3)=xyâˆ’y+3xâˆ’3

â‡’3xâˆ’yâˆ’3=0

â‡’3xâˆ’y=3….. (i)

Case 2

Total number of students=(x+2)(yâˆ’3)

â‡’xy=xy+2yâˆ’3xâˆ’6

â‡’3xâˆ’2y=âˆ’6….. (ii)

Subtracting equation (ii) from (i),

â‡’(3xâˆ’y)âˆ’(3xâˆ’2y)=3âˆ’(âˆ’6)

â‡’âˆ’y+2y=3+6

â‡’y=9

By substituting value of y in (i), we get

â‡’3xâˆ’9=3

â‡’3x=9+3=12

â‡’x=4

Number of rowsÂ =x=4

Number of students in a rowÂ =y=9

Number of total students in a classÂ =xÃ—y=4Ã—9=36

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