RD Sharma Solutions for CBSE Class 10 Mathematics chapter 3 - Pairs of Linear Equations in Two Variables

Now scoring more marks in Mathematics is easy in CBSE Class 10. TopperLearning provides study materials for CBSE Class 10 Mathematics which include a comprehensive set of materials which can help you to score a perfect 100 in the board exams. Our study materials for CBSE Class 10 Mathematics include video lessons, revision notes, question banks, sample papers and past year papers.

Our study materials for CBSE Class 10 Mathematics are created by subject experts who teach students Maths. Our materials will help you to score high marks in your Mathematics examination as we focus on providing solutions which simplify the complex fundamentals of a subject. At TopperLearning, we believe in providing quality solutions at a low cost, and we strictly follow the latest CBSE syllabus; our study materials are revised from time to time. Our TopperLearning packages involve all the study resources for CBSE Class 10 such as solved question papers, video lessons and revision notes which will help you to score high marks. We also provide free NCERT and RD Sharma textbook solutions which provide students with step-by-step solutions.

Page / Exercise

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.1

Question 1

Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.

Solution 1

Question 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' this interesting?) Represent this situation algebraically and graphically.

Solution 2

Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7

According to the given condition,

Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3

According to the given condition,

Thus, the given conditions can be algebraically represented as:
x - 7y = -42
x - 3y = 6

Three solutions of this equation can be written in a table as follows:

x	-7	0	7
y	5	6	7

Three solutions of this equation can be written in a table as follows:

x	6	3	0
y	0	-1	-2

The graphical representation is as follows:

Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

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 situation algebraically and geometrically.

Solution 11

Let the cost of 1 kg of apples and 1 kg grapes be Rsx and Rsy.
The given conditions can be algebraically represented as:

Three solutions of this equation can be written in a table as follows:

x	50	60	70
y	60	40	20

Three solutions of this equation can be written in a table as follows:

x	70	80	75
y	10	-10	0

The graphical representation is as follows:

Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Since, the graph of the two lines coincide, the given system of equations have infinitely many solutions.

Question 13

Solution 13

Question 14

Solution 14

Question 15

Show graphically that each one of the following systems of equations is in-coinsistent (i.e. has no solution):

3x - 5y = 20

6x - 10y = - 40

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

2x + y - 11 = 0

x - y - 1 = 0

Solution 26

Question 27

Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

x + 2y - 7 = 0

2x - y - 4 = 0

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Draw the graphs of the equations 5x - y = 5 and 3x - y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis. Calculate the area of the triangle so formed.

Solution 44

Three solutions of this equation can be written in a table as follows:

x	0	1	2
y	-5	0	5

x	0	1	2
y	-3	0	3

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

$Area space of space Triangle space increment ABC space equals space 1 half cross times BC cross times AO equals 1 half cross times 2 cross times 1 equals 1 space sq. space unit$

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

Question 45

Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

(iii) Champa went to a 'Sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased". Help her friends to find how many pants and skirts Champa a bought.

Solution 45

(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x - y = 4
x + y = 10 x = 10 - y
Three solutions of this equation can be written in a table as follows:

x	4	5	6
y	6	5	4

x - y = 4 x = 4 + y
Three solutions of this equation can be written in a table as follows:

x	5	4	3
y	1	0	-1

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x	3	10	-4
y	5	0	10

Three solutions of this equation can be written in a table as follows:

x	8	3	-2
y	-2	5	12

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.

(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x - 2 ...(1)

and y = 4x - 4 ...(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x	2	0
y = 2x - 2	2	-2

x	0	1
y = 4x - 4	-4	0

Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) parallel lines

(iii) Coincident lines

Solution 49

(i) For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y - 16 = 0

(ii) For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii) For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y - 32 = 0,

Concept insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.

Question 50

Solution 50

Question 51

Solution 51

Question 52

Graphically, solve the following pair of equations:

2x + y = 6

2x - y + 2 = 0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Solution 52

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

From R, draw RM ⊥ X-axis and RN ⊥ Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

Question 53

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8.

Solution 53

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).

Question 54

Draw the graph of the equations x = 3, x = 5 and 2x - y - 4 = 0. Also, find the area of the quadrilateral formed by the lines and the x-axis.

Solution 54

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

Question 55

Draw the graphs of the lines x = -2, and y = 3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure.

Solution 55

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Question 56

Draw the graphs of the pair of linear equations x - y + 2 = 0 and 4x - y - 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.

Solution 56

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following systems of equation:

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solve the pair of equations:

Solution 39

Question 40

Solution 40

$fraction numerator 10 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 4 fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 9 over denominator straight x minus straight y end fraction equals negative 2 Let space fraction numerator 1 over denominator straight x plus straight y end fraction equals straight p space and space fraction numerator 1 over denominator straight x minus straight y end fraction equals straight q 10 straight p plus 2 straight q minus 4 equals 0....... left parenthesis straight i right parenthesis 15 straight p minus 9 straight q plus 2 equals 0........ left parenthesis ii right parenthesis Using space cross minus multiplication space method comma space we space obtain colon fraction numerator straight p over denominator 4 minus 36 end fraction equals fraction numerator straight q over denominator negative 60 minus 20 end fraction equals fraction numerator 1 over denominator negative 90 minus 30 end fraction fraction numerator straight p over denominator negative 32 end fraction equals fraction numerator straight q over denominator negative 80 end fraction equals fraction numerator 1 over denominator negative 120 end fraction fraction numerator straight p over denominator negative 4 end fraction equals fraction numerator straight q over denominator negative 10 end fraction equals fraction numerator 1 over denominator negative 15 end fraction straight p equals 4 over 15 comma straight q equals 10 over 15 equals 2 over 3 Substituting space the space values space of space straight p space and space straight q comma straight x plus straight y equals 15 over 4....... left parenthesis straight i right parenthesis space straight x minus straight y equals 3 over 2...... left parenthesis ii right parenthesis left parenthesis straight i right parenthesis plus left parenthesis ii right parenthesis rightwards double arrow 2 straight x equals 21 over 4 straight x equals 21 over 8 comma straight y equals 9 over 8$

Question 41

Solution 41

Question 42

Solution 42

Question 43

$152 x space minus space 378 y space equals space minus 74 space space minus 378 x space plus space 152 y space equals space minus 604$

Solution 43

$152 straight x minus 378 straight y equals negative 74....... left parenthesis straight i right parenthesis minus 378 straight x plus 152 straight y equals negative 604....... left parenthesis ii right parenthesis Adding space the space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space obtain colon minus 226 straight x minus 226 straight y equals negative 678 rightwards double arrow straight x plus straight y equals 3........... left parenthesis 3 right parenthesis Subtracting space the space equation space left parenthesis 2 right parenthesis space from space equation space left parenthesis 1 right parenthesis comma space we space obtain colon 530 straight x minus 530 straight y equals 530 rightwards double arrow straight x minus straight y equals 1........... left parenthesis 4 right parenthesis Adding space equations space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space obtain colon straight x equals 2 Substituting space straight x equals 2 space in space equation space left parenthesis 3 right parenthesis comma space we space get straight y equals 1$

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solve the following systems of equation:

21x + 47y = 110

47x + 21y = 162

Solution 48

Question 49

If x + 1 is a factor of 2x³ + ax² + 2bx + 1, the find the values of a and b given that 2a - 3b = 4.

Solution 49

Question 50

Find the solution of the pair of equations and . Hence, find λ, if y = λx + 5.

Solution 50

Question 51

Find the values of x and y in the following rectangle.

Solution 51

Question 52

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x - y = 1. How many such lines can we find?

Solution 52

Question 53

Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?

Solution 53

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solve each of the following systems of equations by the method of cross-multiplication:

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 3 - Pairs of Linear Equations in Two Variables Excercise Ex. 3.5

Question 1

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x - 3y = 3

3x - 9y = 2

Solution 1

Question 2

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

2x + y = 5

4x + 2y = 10

Solution 2

Question 3

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

3x - 5y = 20

6x - 10y = 40

Solution 3

Question 4

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x - 2y = 8

5x - 10y = 10

Solution 4

Question 5

Find the value of k for which the following system of equations has a unique solution:

kx + 2y = 5

3x + y = 1

Solution 5

Question 6

Find the value of k for which the following system of equations has a unique solution:

4x + ky + 8 = 0

2x + 2y + 2 = 0

Solution 6

Question 7

Find the value of k for which the following system of equations has a unique solution:

4x - 5y = k

2x - 3y = 12

Solution 7

Question 8

Find the value of k for which the following system of equations has a unique solution:

x + 2y = 3

5x + ky + 7 = 0

Solution 8