NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 3
Chapter Name : Pair Of Linear Equations In Two Variables

Exercise 3.1

Q1 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 this interesting?)
Represent this situation algebraically and graphically.

Answer. Let the present age of Aftab be x.
And, present age of his daughter = y
Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7
According to the question,
( x - 7 ) = 7( y - 7 )
x - 7 = 7y - 49
x - 7y = - 42 ………… (1)
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the question,
( x + 3 ) = 3( y + 3 )
x + 3 = 3y + 9
x - 3y = 6 ………… (2)
Therefore, the algebraic representation is
x - 7y = - 42
x - 3y = 6
For, x - 7y = - 42
x = - 42 + 7y
The solution table is

For x - 3y = 6
x = 6 + 3y
The solution table is

The graphical representation is as follows:

Page : 44 , Block Name : Exercise 3.1

Q2 The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat
and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and
geometrically.

Answer. Let the cost of a bat be Rs x.
And, cost of a ball Rs y
According to the question, the algebraic representation is
3x + 6y = 3900
x + 2y - 1300
For 3x + 6y = 3900,
3900−6y
x =
3

The solution table is

For x + 2y = 1300
x = 1300 - 2y
The solution table is

The graphical representation is as follows:

Page : 44 , Block Name : Exercise 3.1

Q3 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month,
the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and
geometrically.

Answer. Let the cost of 1 kg of apples be Rs x.
And, cost of 1 kg of grapes = Rs y
According to the question, the algebraic representation is

2x + y= 160
4x + 2y = 300
For 2x + y = 160,
y = 160 - 2x
The solution table is

For 4x + 2y = 300
300−4x
y =
2

The solution table is

The graphical representation is as follows:

Page : 44 , Block Name : Exercise 3.1

Exercise 3.2

Q1 Form the pair of linear equations in the following problems, and nd 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, nd the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46.
Find the cost of one pencil and that of one pen.

Answer. (i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x-y=4
For x + y = 10
x = 10 - y

For x - y = 4,
x=4+y

Hence, the graphic representation is as follows:

From the gure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys n the class a e 7 and 3 respectively.
(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y - 50

7x + 5y = 46
For 5x + 7y = 50,
50−7y
x =
5

7x + 5y = 46
46−5y
x =
7

Hence, the graphic representation is as follows:

From the gure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Page : 49 , Block Name : Exercise 3.2

Q2 On comparing the ratios nd out whether the lines representing the
a1 b1 c1
, and
a2 b2 c2

following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Answer. (i) 5x - 4y + 8 = 0
7x + 6y - 9 = 0
Comparing these equations with a x + b y + c 1 1 1 = 0

and a x + b y + c = 0, we obtain
2 2 2

a1 = 5, b1 = −4, c1 = 8

a2 = 7, b2 = 6, c2 = −9
a1 5
=
a2 7
b1 −4 −2
= =
b2 6 3
a b
1 1
≠
a b
since 2 2

Hence, the lines representing the given pair of equations have a unique solution and the pair
of lines intersects at exactly one point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with a x + b y + c 1 1 1 = 0

and a x + b y + c = 0,, we obtain
2 2 2

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24
a1 9 1
= =
a2 18 2
b1 3 1
= =
b2 6 2
c1 12 1
= =
c2 24 2

Since,
a1 b1 c1
= =
a2 b2 c2

Hence, the lines representing the given pair of equations are coincident and there are in nite
possible solutions for the given pair of equations

(iii) 6x - 3y + 10 = 0
2x - y + 9 = 0
Comparing these equations with a x + b y + c 1 1 1 = 0

and a x + b y + c = 0, we obtain
2 2 2

a1 = 6, b1 = −3, c1 = 10

a2 = 2, b2 = −1, c2 = 9
a1 6 3
= =
a2 2 1
b1 −3 3
= =
b2 −1 1
c1 10
=
c2 9

Since
a1 b1 c1
= ≠
a2 b2 c2

Hence, the lines representing the given pair of equations are parallel to each other and hence,
these lines will never intersect each other at any point or there is no possible solution for the
given pair of equations.

Page : 49 , Block Name : Exercise 3.2

Q3 On comparing the ratios , nd out whether the following pair of linear
a1 b1 c1
, and
a2 b2 c2

equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) x + y = 7; 9x − 10y = 14
3

2
5

3

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) x + 2y = 8; 2x + 3y = 12
4

3

Answer. (i) 3x + 2y = 5
2x - 3y = 7
a1 3 b1 −2 c1 5
= , = , =
a2 2 b2 3 c2 7

a1 b1
≠
a2 b2

These linear equations are intersecting each other at one point and thus have only one
possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x - 3y = 8
4x - 6y = 9
a1 2 1 b1 −3 1 c1 8
= = , = = , =
a2 4 2 b2 −6 2 c2 9

b1
Since
a1 c1
= ≠
a2 b2 c2

Therefore, these linear equations are parallel to each other and thus have no possible solution.
Hence, the pair of linear equations is inconsistent.

(iii)
3 5
x + y = 7
2 3

9x - 10y = 14
3 5
a1 2 1 b1 3 −1 c1 7 1
= = , = = , = =
a2 9 6 b2 −10 6 c2 14 2

Since,
a1 b1
≠
a2 b2

Therefore, these linear equations are intersecting each other at one point and thus have only
one possible solution. Hence, the pair of linear equations is consistent

(iv) 5x - 3y = 11
- 10x + 6y = -22
a1 5 −1 b1 −3 −1 c1 11 −1
= = , = = , = =
a2 −10 2 b2 6 2 c2 −22 2

Since,
a1 b1 c1
= =
a2 b2 c2

Therefore, these linear equations are coincident pair of lines and thus have in nite number of
possible solutions. Hence, the pair of linear equations is consistent.

Page : 49 , Block Name : Exercise 3.2

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent,
obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x - y = 8, 3x - 3y = 16
(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0
(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

Answer. (i) x + y = 5
2x + 2y = 10
a1 1 b1 1 c1 5 1
= , = , = =
a2 2 b2 2 c2 10 2

From the gure, it can be observed that these lines are overlapping each other. Therefore,
in nite solutions are possible for the given pair of equations.
(ii) x - y = 8
3x - 3y = 16
b1
Since,
a1 c1
= ≠
a2 b2 c2

Therefore, these linear equations are parallel to each other and thus have no possible solution.
Hence, the pair of linear equations is inconsistent.

(iii) 2x + y - 6 = 0
4x - 2y - 4 = 0
a1 2 1 b1 −1 c1 −6 3
= = , = , = =
a2 4 2 b2 2 c2 −4 2

Since
Therefore, these linear equations are coincident pair of lines and thus have in nite number of
possible solutions. Hence, the pair of linear equations is consistent. And, 2K + 2y - 10 10-2y 2

32 Hence, the graph c representation is as follows.

Page : 49 , Block Name : Exercise 3.2

Q5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36
m. Find the dimensions of the garden.

Answer. Let the width of the ga den bex and length be y.
Since
a1 b1
≠
a2 b2

Therefore, these linear equations are intersecting each other at one point and thus
have only one possible solution. Hence, the pair of linear equations is consistent.
2x + y - 6 =0
Y = 6 - 2x
x 0 1 2

y 6 4 2

And 4x − 2y − 4 = 0

4x−4
y =
2

x 1 2 3

y 0 2 4

Hence, the graphlc representation is as follows.
From the gure, it can be observed that these lines are intersecting each other at
only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m
and 16 m respectively.

Page : 50 , Block Name : Exercise 3.2

Q6 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

Answer.
(i) Intersectina lines:

For this condition,

a1 b1
≠
a2 b2

The second line such that it is intersecting the given line is

a1 2 b1 3 a1 b1
2x + 4y − 6 = 0 as = = 1, = and ≠
a2 2 b2 4 a2 b2

(ii) parallel lines:
For this condition,

a1 b1 c1
= ≠
a2 b2 c2

Hence, the second line can be
4x + 6y − 8 = 0

a1 2 1 b1 3 1 c1 −8
as = = , = = , = = 1
a2 4 2 b2 6 2 c2 −8

a1 b1 c1
= ≠
a2 b2 c2

(iii)Coincident lines:
For coincident lines,
According to the question,
y − x = 4(1)

y + x = 36(2)

y − x = 4

y = x + 4

x 0 8 12

y 4 12 16

y + x = 36

x 0 36 16

y 36 0 20

Hence, the graphic representation is as follows.

a1 b1 c1
= =
a2 b2 c2

Hence, the second line can be
6x + 9y - 24 = 0
a1 2 1 b1 3 1 c1 −8 1
as = = , = = , = =
a2 6 3 b2 9 3 c2 −24 3

a1 b1 c1
And clearly, = =
a2 b2 c2

Page : 50 , Block Name : Exercise 3.2

Q7 Draw the graphs of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0. Determine the
coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the
triangular region.

Answer.
x − y + 1 = 0

x = y − 1

x 0 1

y 1 2 3

3x + 2y − 12 = 0
12−2y
x =
3

x 4 2 0

y 0 3 6

Hence, the graphic representation is as follows.

From the gure, it can be observed that these lines are intersecting each other at
point (2, 3) and x-axis at (-1, O) and (4, O). Therefore, the vertices of the triangle
are (2, 3), (-1, 0), and (4, 0).

Page : 50 , Block Name : Exercise 3.2

Exercise 3.3

Q1 Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x-y=4
(ii) s – t = 3
s t
+ = 6
3 2

(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2x + √3y = 0
√3x − √8y = 0

3x 5y
− = −2
(vi) 2

y
3
?
x 13
+ =
3 2 6

Answer.
(i)x + y = 14(1)

x − y = 4(2)

From (1), we obtain

x = 14 − y(3)

Substituting this value in equation (2), we obtain

(14 − y) − y = 4

14 − 2y = 4

10 = 2y

y = 5 … … . (4)

Substituting this in equation (3), we obtain

x = 9

∴ x = 9, y = 5

(ii) s − t = 3 … … . (1)
s r
+ = 6 … … . (2)
3 2

From (1), we obtain

s = t + 3 … … . . (3)

Substituting this value in equation (2), we obtain

t+3 t
+ = 6
3 2

2t + 6 + 3t = 36

5t = 30

t = 6 … … . . (4)

Substituting in equation (3), we obtain

s = 9

∴ S = 9, t = 6

( iii )3x − y = 3(1)

9x − 3y = 9(2)

y = 3x − 3(3)

Substituting this value in equation (2), we obtain

9x − 3(3x − 3) = 9

9x − 9x + 9 = 9

9 = 9

This is always true.
Hence, the given pair of equations has in nite possible solutions and the relation
between these variables can be given by
y = 3x − 3

Therefore, one of its possible solutions is x = 1, y = 0 .

0.2x + 0.3y = 1.3 … . . (1)

0.4x + 0.5y = 2.3 … … . (2)

From equation (1), we obtain

1.3−0.3y
x = . . (3)
0.2

Substituting this value in equation (2), we obtain

1.3−0.3y
0.4 ( ) + 0.5y = 2.3
0.2

2.6 − 0.6y + 0.5y = 2.3

2.6 − 2.3 = 0.1y

0.3 = 0.1y

y = 3 … … (4)

Substituting this value in equation (3), we obtain
1.3−0.3×3
x =
0.2

1.3−0.9 0.4
= = = 2
0.2 0.2

∴ x = 2, y = 3

(v) √2x + √3y = 0

√3x − √8y = 0

From equation (1), we obtain

−√3y
x = (3)
√2

Substituting this value in equation (2), we obtain

√3y
√3 (− ) − √8y = 0
√2

3y
− − 2√2y = 0
√2

3
y (− − 2√2) = 0
√2

y = 0 (4)

Substituting this value in equation (3), we obtain
x = 0

∴ x = 0, y = 0
3 5
(vi) x − y = −2
2 3

x y 13
+ =
3 2 6

From equation ( 1), we obtain

9x − 10y = −12

−12+10y
x = … (3)
9

Substituting this value in equation (2), we obtain
−12+10y

+
9 y 13
=
3 2 6
−12+10y y 13
+ =
27 2 6

47y = 117 + 24

47y = 141

y = 3 … (4)

Substituting this value in equation (3), we obtain
−12+10×3 18
x = = = 2
9 9

Hence, x = 2, y = 3

Page : 53 , Block Name : Exercise 3.3

Q2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence nd the value of ‘m’ for which
y = mx + 3?

Answer.
2x + 3y = 11 … … . (1)

2x − 4y = −24 … . . (2)

From equation (1), we obtain

11−3y
x = . . . (2)
2

Substituting this value 'n equation (2), we obtain

Substituting this value in equation (2), we obtain

11−3y
2( ) − 4y = −24
2

11 − 3y − 4y = −24

−7y = −35

y = 5 … . . (4)

Putting this value in equation (3), we obtain
11−3×5 4
x = = − = −2
2 2

Hence, x = −2, y = 5

Also,

y = mx + 3

5 = −2m + 3

−2m = 2

m = −1

Page : 53 , Block Name : Exercise 3.3

Q3 Form the pair of linear equations for the following problems and nd their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find
them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5
balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a xed charge together with the charge for the distance
covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the
charge paid is ₹ 155. What are the xed charges and the charge per km? How much does a
person have to pay for travelling a distance of 25 km?
(v) A fraction becomes , if 2 is added to both the numerator and the denominator. If, 3 is
9

11

added to both the numerator and the denominator it becomes 5 6 . Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s
age was seven times that of his son. What are their present ages?

Answer.
(1) Let the first number be x and the other number be y such that y > x .

According to the given information,

y = 3x

y − x = 26

On substituting the value of y from equation (1) into equation (2), we obtain

3x − x = 26

x = 13

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

Hence, the numbers are 13 and 39.

(ii) Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always
180 .
∘

According to the given information,
∘
x + y = 180 . … . (1)

∘
x − y = 18 … … (2)

From (1), we obtain

x = 1800 − y(3)

Substituting this in equation ( 2), we obtain

∘ ∘
180 − y − y = 18

∘
162 = 2y

∘
81 = y … . . (4)

Putting this in equation (3), we obtain
x = 1800 − 810
∘
= 99

Hence, the angles are 99 and 81 .
∘ ∘

(iii) Let the cost of a bat and a ball be x and y respectively.
According to the given information,
7x + 6y = 3800 … … (1)

3x + 5y = 1750 … … (2)

From (1), we obtain

3800−7x
y = . . . . . . (3)
6

Substituting this value in equation (2), we obtain
3800−7x
3x + 5 ( ) = 1750
6

9500 35x
3x + − = 1750
3 6

35x 9500
3x − = 1750 −
6 3

18x−35x 5250−9500
=
6 3

17x −4250
− =
6 3

−17x = −8500

x = 500

Substituting this in equation ( 3), we obtain

3800−7×500
y =
6

300
= = 50
6

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

(iv) Let the xed charge be Rsx and per km charge be Rs y.
According to the given information,
x + 10y = 105 … … (1)

x + 15y = 155 … … . (2)

From (3), we obtain

x = 105 − 10y … … . (3)

From (3), we obtain

x = 105 − 10y … . (3)

Substituting this in equation (2), we obtain

105 − 10y + 15y = 155

5y = 50

y = 10 … . . (4)

Putting this in equation (3), we obtain
x = 105 - 10 x 10
X=5
Hence, xed charge = Rs 5
And per km charge = Rs 10
Charge for 25 km = x + 25y
= 5 + 250 = Rs 255
(x + 5) = 3(y + 5)

x − 3y = 10 … . . (1)

(x − 5) = 7(y − 5)

x − 7y = −30 … … (2)

From (1), we obtain

x = 3y + 10 … … (3)

Substituting this value in equation ( 2 ), we obtain

3y + 10 − 7y = −30

−4y = −40

y = 10 … … (4)

Substituting this value in equation ( 3), we obtain

x = 3 x 10 + 10

= 40
Hence, the present age of Jacob is 40 years whereas the present age of his son is 10
years.

(v) Let numerator be x
and denominator be y
So, fraction is
x

y

Given that
if 2 is added to both the numerator and the denominator,
9
fraction becomes
11

Numerator+2 9
=
Denominator +2 11

x+2 9
=
y+2 11

11(x + 2) = 9(y + 2)

11x + 22 = 9y + 18

11x − 9y + 22 − 18 = 0

11x − 9y + 4 = 0

Also,
given that if 3 is added to both the numerator and the
5
denominator, fraction becomes .
6

Numerator +3 5
=
Denominator +3 6

x+3 5
=
y+3 6

6(x + 3) = 5(y + 3)

6(x + 3) = 5(y + 3)

6x + 18 = 5y + 15

6x − 5y + 18 − 15 = 0

6x − 5y + 3 = 0

Hence, our equations are
11X—9y +4
6x-5Y+3=O
From (1)

11x − 9y + 4 = 0

11x = 9y − 4

9y−4
x = ( )
11

Putting x in (2)

6x − 5y + 3 = 0

9y−4
6( ) − 5y + 3 = 0
11

Multiply both sides by 11
6(9y−4)
11 × − 5y × 11 + 3 × 11 = 0 × 11
11

6(9y − 4) − 55y + 33 = 0

6(9y) − 6(4) − 55y + 33 = 0

54y − 24 − 55y + 33 = 0

−y + 9 = 0

−y = −9

y = 9

Putting y = 9 in (1)
11x − 9y + 4 = 0

11x − 9(9) + 4 = 0

11x − 81 + 4 = 0

11x − 77 = 0

11x − 77 = 0

11x = 77

x = 7

Therefore x = 7, y = 9

So,

Numerator = x = 7

Denominator = y = 9

Numerator x 7
Hence, original fraction = = =
Denominator y 9

(vi) Let present age of Jacob = x years
& present age of Jacob's son = y years
Five years hence(later), Jacob's Age = x + 5
Jacob son's Age = y + 5 Age of
Jacob will be three times of his son.
Age of Jacob will be three times of his son.
x + 5 = 3(y + 5)

x + 5 = 3y + 15

x − 3y + 5 − 15 = 0

x − 3y − 10 = 0

Also,
Five years ago,
Jacob's Age = x— 5
Jacob son's Age = y — 5
Age of Jacob was seven times of his son.

x − 5 = 7(y − 5)

x − 5 = 7y − 7(5)

x − 5 = 7y − 35

x − 7y − 5 + 35 = 0

x − 7y − 30 = 0

Our equations are
x − 3y − 10 = 0

x − 7y + 30 = 0

From (I)
x- 3y-10=0
x - 3Y+ 10
Putting x in (2)

x − 7y + 30 = 0

(3y + 10) − 7y + 30 = 0

3y + 10 − 7y + 30 = 0

−4y + 40 = 0

−4y = −40
−40
y =
−4

y = 10

Putting y = 10 in (1)

x − 3y − 10 = 0

x − 3(10) − 10 = 0

x − 30 − 10 = 0

x − 40 = 0

x = 40

So, x = 40, y = 10

Thus
Present age of Jacob = x = 40 years
Present age of Jacob's son = y 10 years

Page : 53 , Block Name : Exercise 3.3

Exercise 3.4

Q1 Solve the following pair of linear equations by the elimination method and the substitution
method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv)
x 2y y
+ = −1 and x − = 3
2 3 3

Answer. (i) By elimination method
2x - 3y = 5 ... (1)
2x - 3y = 4 ... (2)
Multiplying equation (1) by 2, we obtain
2x+2y=10 …. (3)
Subtracting equation (2) from equation (3), we obtain
5y = 6
y =
6

5
(4)
Substituting the value in equation (1), we obtain
6 19
x = 5 − =
5 5

19 6
∴ x = ,y =
5 5

By substitution method
From equation (1), we obtain
X = 5 - y (5)
Putting this value in equation (2), we obtain
2(5 - y) - 3y = 4
-5y = -6
6
y =
5

Substituting the value in equation (5), we obtain
6 19
x = 5 − =
5 5

19 6
∴ x = ,y =
5 5

(ii) By elimination method
5x + 4y = 10(1)

2x − 2y = 2 (2)

Multiplying equation (2) by 2, we obtain
4x - 4y = 4 (3)
Adding equation (1) and (3), we obtain
7x = 14
X = 2 (4)
Substituting in equation (1), we obtain
x = 1 + y(5)

butting this value in equation (1), we obtain

3(1 + y) + 4y = 10

7y = 7

y = 1

Substituting the value in equation (5), we obtain
x= 1 +1 = 2
∴ x = 2, y = 1

(iii) By elimination method
3x − 5y − 4 = 0

9x = 2y + 7

9x − 2y − 7 = 0

Multiplying equation (1) by 3, we obtain
9x-15y-12=0 (3)
Subtracting equation (3) from equatia-l (2), we obtain
13y = −5

−5
y = . . . . . (4)
13

Substituting in equation (1), we obtain
25
3x + − 4 = 0
13

27
3x =
13

9
x =
13

9 −5
∴ x = ,y =
13 13

By substitution method
From equation (1), we obtain
5y+4
x = … . (5)
3

Putting this value in equation (2), we obtain
5y+4
9( ) − 2y − 7 = 0
3

13y = −5

5
y = −
13

Substituting the value in equation (5), we obtain
−5
5( )+4
13

x =
3

9
x =
13

9 −5
∴ x = ,y =
13 13

(iv) By elimination method
x 2y
+ = −1
2 3

3x + 4y = −6 (1)
y
x − = 3
3

3x − y = 9 … (2)

Subtracting equation (2) from equation (1), we obtain
5y = −15

y = −3 … . (3)

Substituting this value in equation (1), we obtain

3x − 12 = −6

3x = 6

x = 2

Hence, x = 2, y = −3

By substitution method
From equation (2), we obtain
y+9
x =
3

Putting this value in equation (1), we obtain
y+9
3( ) + 4y = −6
3

5y = −15

y = −3

Substituting this value in equation (1), we obtain
−3+9
x = = 2
3

∴ x = 2, y = −3

Page : 56 , Block Name : Exercise 3.4

Q2 Form the pair of linear equations in the following problems, and nd their solutions (if
they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1.
It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as
Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the
number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100
notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a xed charge for the rst three days and an additional charge for
each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for
the book she kept for ve days. Find the xed charge and the charge for each extra day.

Answer. (i)Let the fraction be x

y

According to the given Information,
x+1
= 1 ⇒ x − y = −2 … . . (1)
y−1

x 1
= ⇒ 2x − y = 1 … … (2)
y+1 2

Subtracting equation (1) from equation (2), we obtain
x= 3 …... (3)
Substituting this value in equation (1), we obtain
3 − y = −2

−y = −5

y = 5

Hence, the fraction is 3

5

(ii)Let present age of Nuri x
and present age of Sonu = y
According to the given information,
(x − 5) = 3(y − 5)

x − 3y = −10 … … (1)

(x + 10) = 2(y + 10)

x − 2y = 10 … … (2)

Subtracting equation (1) from equation (2), we obtain
y = 20(3)

Substituting it in equation (1), we obtain
x − 60 = −10

x = 50

Hence, age ot Nuri = 50 years
And, age of Sonu = 20 years
(iii)Let the unit digit and tens digits of the number be x and V respectively. The
Number = 10y + x
Number after reversing the digits = l0x + y
According to the given information,
x + y = 9(1)

9(10y + x) = 2(10x + y)

88y − 11x = 0

−x + 8y = 0 … . . . (2)

Adding equation (1) and (2), we obtain
9y = 9

y = 1 … . (3)

Substituting the value in equation (1), we obtain
Hence, the number is 10y+ x=10 x 1+ 18 = 18
(iv) The number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the given information,
x + y = 25 … (1)

50x + 100y = 2000 … . . (2)

Multiplying equation (1) by 50, we obtain
50x + 50y = 1250

Subtracting equation (3) from equation (2), we obtain
50x + 50y = 1250 ... (3)
Y = 15
Substituting in equation (1), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100
(v)Let the xed charge for rst three days and each day charge thereafter be Rs x
and Rs y respectively.
According to the given information,

x + 4y = 27 … . . (1)

x + 2y = 21 … … . . (2)

Subtracting equation (2) from equation (1), we obtain
2y = 6

y = 3 … . (3)

Substituting in equation (1), we obtain
x+12 = 27
X = 15
Hence, xed charge = Rs 15
And Charge per day = Rs 3

Page : 57 , Block Name : Exercise 3.4

Exercise 3.5

Q1 Which of the following pairs of linear equations has unique solution, no solution, or
in nitely many solutions. In case there is a unique solution, nd it by using cross
multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
(ii) 2x + y = 5
3x + 2y = 8
(iii) 3x – 5y = 20
6x – 10y = 40
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0

Answer. (i)
x − 3y − 3 = 0

3x − 9y − 2 = 0

a1 1 b1 −3 1 c1 −3 3
= , = = , = =
a2 3 b2 −9 3 c2 −2 2

a1 b1 c1
= ≠
a2 b2 c2

Therefore, the given sets of lines are parallel to each other. Therefore, they will not
intersect each other and thus, there will not be any solution for these equations.
(ii)
2x + y = 5

3x + 2y = 8

a1 2 b1 1 c1 −5
= , = , =
a2 3 b2 2 c2 −8

a1 b1
≠
a2 b2

Therefore, they will intersect each other at a unique point and thus, there will be
unique solution for these equations.
By cross-multiplication method,
x y 1
= =
bc2 −b2 c1 c1 a2 −c2 a1 a1 b2 −a2 b1
x y 1
= =
−8−(−10) −15+16 4−3

x y
= = 1
2 1
x y
= 1, = 1
2 1

x = 2, y = 1

∴ x = 2, y = 1

(iii)
3x − 5y = 20

6x − 10y = 40

a1 3 1 b1 −5 1 c1 −20 1
= = , = = , = =
a2 6 2 b2 −10 2 c2 −40 2

a1 b1 c1
= =
a2 b2 c2

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be
coincident to each other and thus, there are in nite solutions possible for these
equations.

(iv)
x − 3y − 7 = 0

3x − 3y − 15 = 0

a1 1 b1 −3 c1 −7 7
= , = = 1, = =
a2 3 b2 −3 c2 −15 15

a1 b1
≠
a2 b2

Therefore, they will intersect each other at a unique point and thus, there will be a
unique solution for these equations.
By cross-multiplication,
x y 1
= =
45−(21) −21−(−15) −3−(−9)

x y 1
= =
24 −6 6

x 1 y 1
= and =
24 6 −6 6

x = 4 and y = −1

∴ x = 4, y = −1

Page : 62 , Block Name : Exercise 3.5

Q2 (i) For which values of a and b does the following pair of linear equations have an in nite
number of solutions?
2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1

Answer. (i)
2x + 3y − 7 = 0

(a − b)x + (a + b)y − (3a + b − 2) = 0

a1 2 b1 3 c1 −7 7
= , = , = =
a2 a−b b2 a+b c2 −(3a+b−2) (3a+b−2)

For in nitely many solutions,
a1 b1 c1
= =
a2 b2 c2

2 7
=
a−b 3a+b−2

6a + 2b − 4 = 7a − 7b

a − 9b = −4 … . (1)

2 3
=
a−b a+b

2a + 2b = 3a − 3b

a − 5b = 0. . … (2)

Subtracting (1) from (2), we obtain
4b = 4

b = 1

Substituting this in equation (2), we obtain
a − 5 × 1 = 0

a = 5

Hence, a= 5 and b=1 are the values for which the given equations give in nitely
many solutions.

(ii) 3x + y − 1 = 0

(2k − 1)x + (k − 1)y − 2k − 1 = 0

a1 3 b1 1 c1 −1 1
= , = , = =
a2 2k−1 b2 k−1 c2 −2k−1 2k+1

For no solution,
a1 b1 c1
= ≠
a2 b2 c2

3 1 1
= ≠
2k−1 k−1 2k+1

3 1
=
2k−1 k−1

3k − 3 = 2k − 1

k=2
Hence, for k = 2, the given equation has no solution.

Page : 62 , Block Name : Exercise 3.5

Q3 Solve the following pair of linear equations by the substitution and cross- multiplication
methods : 8x + 5y = 9
3x + 2y = 4

Answer.
8x + 5y = 9 (i)

3x + 2y = 4 (ii)

From equation (ii), we obtain
4−2y
x = (iii)
3

Substituting this value in equation (i), we obtain
4−2y
8( ) + 5y = 9
3

32 − 16y + 15y = 27

−y = −5

y = 5

Substituting this value in equation (ii), we obtain
3x + 10 = 4

x = −2

Hence, x = −2, y = 5

Again, by cross-multiplication method, we obtain
8x + 5y − 9 = 0

3x + 2y − 4 = 0

x y 1
= =
−20−(−18) −27−(−32) 16−15

x y 1
= =
−2 5 1

x y
= 1 and = 1
−2 5

x = −2 and y = 5

Page : 62 , Block Name : Exercise 3.5

Q4 Form the pair of linear equations in the following problems and nd their solutions (if they
exist) by any algebraic method :
(i) A part of monthly hostel charges is xed and the remaining depends on the number of days
one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000
as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel
charges. Find the xed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when
8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for
each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been
deducted for each incorrect answer, then Yash would have scored 50 marks. How many

questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at
the same time. If the cars travel in the same direction at different speeds, they meet in 5
hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two
cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and
breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units,
the area increases by 67 square units. Find the dimensions of the rectangle.

Answer. (i) Let x be the xed charge of the food and V be the charge for food per day.
According to the given information,
x + 20y = 1000

x + 26y = 1180

Subtracting equation (1) from equatica (2), we obtain
6y = 180

y = 30

Substituting this value in equation (1), we obtain
x + 20 × 30 = 1000

x = 1000 − 600

x = 400

Hence, xed charge= Rs 400
And charge per day = Rs 30

(ii) Let the fraction be x

y

According to the given information,
x−1 1
= ⇒ 3x − y = 3 … (1)
y 3

x 1
= ⇒ 4x − y = 8 … . . (2)
y+8 4

Subtracting equation (1) from equation (2), we obtain
x = 5 … (3)

Putting this value in equation (1), we obtain
15 − y = 3

y = 12

Hence, the fractlon is 5

12

(iii) Let the number of right answers and wrong answers be x and y
According to the given information,
3x − y = 40 … . . (1)

4x − 2y = 50

⇒ 2x − y = 25 … . (2)

Subtracting equation (2) frorn equation (1), we obtain

x = 15 … . (3)

Substituting this in equation (2), we obtain
30-y = 25
Y=5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20

(iv) Let the speed of 1 car and 2
st nd
car be u km/h and vkm/h

Respective speed of both cars While they are travelling in same direction = (u - v) km/h
Respective speed of both cars while they are travelling in opposite directions i.e.,
travelling towards each other = ( u +v) km/h
According to the given information,
5(u − v) = 100

⇒ u − v = 20 … (1)

1(u + v) = 100 … (2)

Adding both the equations, we obtain
2u = 120

u = 60km/h (3)

Substituting this value in equation (2), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
(v) Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
(x − 5)(y + 3) = xy − 9

⇒ 3x − 5y − 6 = 0 … . (1)

(x + 3)(y + 2) = xy + 67

⇒ 2x + 3y − 61 = 0 … . (2)

By cross multiplication method, we obtain
x y 1
= =
305−(−18) −12−(−183) 9−(−10)

x y 1
= =
323 171 19

x = 17, y = 9

Hence, the length and breadth of the rectangle are 17 units and g units respectively.

Page : 62 , Block Name : Exercise 3.5

Exercise 3.6

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i)
1 1
+ = 2
2x 3y

1 1 13
+ =
3x 2y 6

(ii) 2
+
3
= 2
√x √y

4 9
− = −1
√x √y

(iii)
4
+ 3y = 14
x

3
− 4y = 23
x

(iv)
5 1
+ = 2
x−1 y−2

6 3
− = 1
x−1 y−2

(v)
7x−2y
= 5
xy

8x+7y
= 15
xy

(vi)
6x + 3y = 6xy

2x + 4y = 5xy

(vii)
10 2
+ = 4
x+y x−y

15 5
− = −2
x+y x−y

(viii)
1 1 3
+ =
3x+y 3x−y 4

1 1 −1
− =
2(3x+y) 2(3x−y) 8

Answer. (i)
1 1
+ = 2
2x 3y

1 1 13
+ =
3x 2y 6

Let 1

x
= p
and
1 1

y
= q then the equations change as follows.
p q
+ = 2 ⇒ 3p + 2q − 12 = 0 … … . (1)
2 3

p q 13
+ = ⇒ 2p + 3q − 13 = 0 … … . (2)
3 2 6

Using cross-multiplication method, we obtain
p q 1
= =
−26−(−36) −24−(−39) 9−4

p q 1
= =
10 15 5

p 1 q 1
= and =
10 5 15 5

p = 2 and q = 3

1 1
= 2 and = 3
x y

1 1
x = and y =
2 3

(ii)
2 3
+ = 2
√x √y

4 9
− = −1
√x √y

Putting 1
= p and
1
= q
√x √y

in the given equations, we obtain
2p + 3q = 2

4p − 9q = −1

Multiplying equation (1) by 3, we obtain
6p + 9q = 6 ….(3)
Adding equation (2) and (3), we obtain
10p = 5

1
p = … (4)
2

Putting in equation (1), we obtain
1
2 × + 3q = 2
2

3q = 1

1
q =
3

1 1
p = =
√x 2

√x = 2

x = 4
1 1
and q = =
√y 3

√y = 3

y = 9

Hence, x = 4, y = 9

(iii)
4
+ 3y = 14
x

3
− 4y = 23
x

Substituting 1

x
= p in the given equat ons, we obtain
4p + 3y = 14 ⇒ 4p + 3y − 14 = 0

3p − 4y = 23 ⇒ 3p − 4y − 23 = 0

By cross-multiplication, we obtain

p y 1
= =
−69−56 −42−(−92) −16−9

p y −1
= =
−125 50 25

p −1 y −1
= and =
−125 25 50 25

p = 5 and y = −2

1
p = = 5
x

1
x =
5

y = −2

(iv)
5 1
+ = 2
x−1 y−2

6 3
− = 1
x−1 y−2

Putting x−1
1
= p and
y−2
1
= q in the given equation, we obtain
5p + q = 2 (1)

6p − 3q = 1 (2)

Multiplying equation (1) by 3, we obtain
1
5 × + q = 2
3

5 1
q = 2 − =
3 3

1 1
p = =
x−1 3

⇒ x − 1 = 3

⇒ x = 4
1 1
q = =
y−2 3

y − 2 = 3

y = 5

∴ x = 4, y = 5

(v)
7x−2y
= 5
xy

7 2
− = 5. . . (1)
y x

8x+7y
= 15
xy

8 7
+ = 15 … . (2)
y x

Putting 1

x
= p and
1

y
= q in the given equation, we obtain
−2p + 7q = 5 ⇒ −2p + 7q − 5 = 0 … . . (3)

7p + 8q = 15 ⇒ 7p + 8q − 15 = 0 … . (4)

By cross-multiplication method, we obtain

p q 1
= =
−105−(−40) −35−30 −16−49

p q 1
= =
−65 −65 −65

P 1 q 1
= and =
−65 −65 −65 −65

p = 1 and q = 1
1 1
p = = 1 q = = 1
x y

x = 1 y = 1

(vi)
6x + 3y = 6xy

6 3
⇒ + = 6 … . (1)
y x

2x + 4y = 5xy

2 4
+ = 5 … (2)
y x

and Y
Putting 1

x
= p and
1

y
= q in these equations, we obtain
3p + 6q − 6 = 0

4p + 2q − 5 = 0

By cross-multiplication method, we obtain
p q 1
= =
−30−(−12) −24−(−15) 6−24

p q 1
= =
−18 −9 −18
p 1 q 1
= and =
−18 −18 −9 −18

1
p = 1 and q =
2

1 1 1
p = = 1 q = =
x y 2

x = 1 y = 2

Hence ,x = 1, y = 2

(vii)
10 2
+ = 4
x+y x−y

15 5
− = −2
x+y x−y

Putting x+y
1
= p
1

x−y
= q in the given equation a
, we obtain
10p + 2q = 4 ⇒ 10p + 2q − 4 = 0 … . (1)

15p − 5q = −2 ⇒ 15p − 5q + 2 = 0 … … (2)

Using cross-multiplication method, we obtain

p q 1
= =
4−20 −60−(20) −50−30

p q 1
= =
−16 −80 −80

P 1 q 1
= and =
−16 −80 −80 −80

1
p = and q = 1
5
1 1 1
p = = and q = = 1
x+y 5 x−y

x + y = 5 … . (3)

and x − y = 1. . . (4)

Adding equation (3) and (4), we obtain
2x = 6
X = 3 …. (5)
Substituting in equation (3), we obtain
y = 2

Hence, x = 3, y = 2

(viii)
1 1 3
+ =
3x+y 3x−y 4

1 1 −1
− =
2(3x+y) 2(3x−y) 8

Putting 3x+y
1
= p and
3x−y
1
= q in these equations ,we obtain
3
p + q = … … (1)
4

p q −1
− =
2 2 8

−1
p − q = … . . (2)
4

Adding (1) and (2), we obtain
3 1
2p = −
4 4

1
2p =
2

1
p =
4

Substituting in (2), we obtain
1 −1
− q =
4 4

1 1 1
q = + =
4 4 2

1 1
p = =
3x+y 4

3x + y = 4 … . (3)

1 1
q = =
3x−y 2

3x − y = 2 … (4)

Adding equations (3) and (4), we obtain
6x = 6

x = 1 … . (5)

Substituting in (3), we obtain

3(1) + y = 4

y = 1

Hence, x = 1, y = 1

Page : 67 , Block Name : Exercise 3.6

Q2 Formulate the following problems as a pair of equations, and hence nd their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of
rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together nish an embroidery work in 4 days, while 3 women and
6 men can nish it in 3 days. Find the time taken by 1 woman alone to nish the work, and
also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if
she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the
remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus
separately.

Answer. (i) Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h
respectively.
Speed of Ritu while rowing
Upstream = (x - y)km/h
Downstream = (x + y)km/h
According to question ,
2(x + y) = 20

⇒ x + y = 10 … . . (1)

2(x − y) = 4

⇒ x − y = 2 … … (2)

Adding equation (1) and (2), we obtain
2x= 12 => x= 6
Putting this in equation (1), we obtain
Y=4
Hence, Ritu’s speed in still is 6 km/h the speed Of the current is 4 km/h.

(ii) Let the number of days taken by a woman and a man be x and y respectively
Therefore, work done by a woman in 1 day =
1

x

Work done by man in 1 day=
1

y

According to the question,
2 5
4( + ) = 1
x y

2 5 1
+ =
x y 4

3 6
3( + ) = 1
x y

3 6 1
+ =
x y 3

Putting x
1
= p and
1

y
= q in these equations,we obtain
1
2p + 5q =
4

⇒ 8p + 20q = 1

1
3p + 6q =
3

⇒ 9p + 18q = 1

By cross-multiplication, we obtain
p q 1
= =
−20−(−18) −9−(−8) 144−180

p q 1
= =
−2 −1 −36

p −1 q 1
= and =
−2 36 −1 −36
p −1 q 1
= and =
−2 36 −1 −36

1 1
p = and q =
18 36

1 1 1 1
p = = and q = =
x 18 y 36

x = 18 y = 36

Hence, number of days taken by a woman =18
Number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
60 240
+ = 4 … … (1)
u v

100 200 25
+ = … . (2)
u v 6

Putting u
1
= p and
1

v
= q in these equations, we obtain
60p + 240q = 4 … (3)

25
100p + 200q =
6

600p + 1200q = 25 … . . (4)

Multiplying equation (3) by 10, we obtain
600p + 2400q = 40 (5)

Subtracting equation (4) from (5), we obtain
1200q = 15

15 1
q = = … . (6)
1200 80

Substituting in equation (3), we obtain
60p+ 3 = 4
60p=1
p= 1

60
1 1 1 1
p = = and q = =
u 60 v 80

u = 60km/h and v = 80km/h

Hence, speed of train = 60 km/h
Speed of bus = 80 km/h

Page : 67 , Block Name : Exercise 3.6

Exercise 3.7

Q1 The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as
Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30
years. Find the ages of Ani and Biju.

Answer. The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older
than Ani or Ani is 3 years older than Biju. However, it is obvious that in both
cases, Ani's father's age will be 30 years more than that of Cathy's age.
Let the age of Ani and Biju be x and y years respectively.
Therefore, age of Ani's father, Dharam = 2 × x = 2x
y
And age of Biju's sister Cathy = years
2

By using the information given in the question,
Case (I) When Ani is older than Bi]u by 3 years,
X - y =3 (i)
y
2x − = 30
2

4x − y = 60 … . (ii)

Subtracting (i) from (ii), we obtain
3K = 60 - 3 = 57
57
x = = 19
3

Therefore, age of Anil =19 years
And age of Biju = 19 - 3 = 16 years
Case (II) When Biju is older than Ani,
y − x = 3(i)
y
2x − = 30
2

4x- y = 60 . ..(ii)
Adding (i) and (ii), we obtain
3x = 63
x - 21
Therefore, age of Ani = 21 years
And age of Biju = 21 + 3 = 24

Page : 68 , Block Name : Exercise 3.7(Optional)

Q2 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]
[Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)]?

Answer. Let those friends were having Rs x and y with them.
using the information given in the question, we obtain
x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300(i)

And, 6(x − 10) = (y + 10)

6x − 60 = y + 10

6x − y = 70(ii)

Multiplying equation (ii) by 2, we obtain
12x - 2y = 140 (iii)
Subtracting equation (i) from equation (iii), we obtain
11x = 140 + 300

11x = 440

x = 40

using this in equation (i), we obtain
40 - 2y = -300
40 + 300 = 2y
2y = 340
Y = 170
Therefore, those friends had Rs 40 and Rs 170 with them respectively.

Page : 68 , Block Name : Exercise 3.7(Optional)

Q3 A train covered a certain distance at a uniform speed. If the train would have been 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 be x km/h and the time taken by train to travel the given
distance be t hours and the distance to travel was d km. We know that,
Distance travelled
Speed =
Time taken to travel that distance

d
x =
t

Or, d = xt(i)

using the information given in the question, we obtain
d
(x + 10) =
(t−2)

(x + 10)(t − 2) = d

xt + 10t − 2x − 20 = d

By using equation (i), we obtain

−2x + 10t = 20(ii)

d
(x − 10) =
(t+3)

(x − 10)(t + 3) = d

xt − 10t + 3x − 30 = d

By using equation (i), we obtain
3x - 10t = 30 (iii)
Adding equations (ii) and (iii), we obtain
x = 50
Using equation (ii), we obtain
(−2) × (50) + 10t = 20

−100 + 10t = 20

10t = 120

t = 12 hours

From equation (i), we obtain
Distance to travel = d = Xt
=50 x 12
= 600 km
Hence, the distance covered by the train is 600 km.

Page : 68 , Block Name : Exercise 3.7(Optional)

Q4 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 the
number of students in the class?

Answer. Let the number of rows be x and number of students in a row be y
Total students of the class
=Number of rows x Number of students in a row
= xy
using the information given in the question,
Condition 1
Total number of students = (x - 1) (y + 3)
xy = (x − 1)(y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3(i)

Condition 2
Total number of students = (x+2)(y-3)
Xy = xy+2y - 3x - 6
3x - 2y =-6 (ii)
Subtracting equation (ii) from (i)
(3x − y) − (3x − 2y) = 3 − (−6)

-y + 2y = 3+6
y= 9

By using equation (i), we obtain
3x - 9 = 3
3x= 9 +3= 12
X=4
Number of rows = x = 4
Number of students in a row = y = 9
Number of total students in a class = xy = 4 x 9 = 36

Page : 68 , Block Name : Exercise 3.7(Optional)

Q5 In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles?

Answer. Given that ,
∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2∠A − ∠B = 0 … (i)

We know that the sum of the measures of all angles of a triangle is 1800, Therefore,
∘
∠A + ∠B + ∠C = 180
∘
∠A + ∠B + 3∠B = 180
∘
∠A + 4∠B = 180 … (ii)

Multiplying equation (i) by 4, we obtain
8∠A − 4∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain
∘
9∠A = 180
∘
∠A = 20

∘
9∠A = 180
∘
∠A = 20

∘ ∘
20 + 4∠B = 180
∘
4∠B = 160
∘
∠B = 40

∠C = 3∠B
∘ ∘
= 3 × 40 = 120
∘ ∘ ∘
Therefore, ∠A, ∠B, ∠C are 20 , 40 , and 120 respectively.

Page : 68 , Block Name : Exercise 3.7(Optional)

Q6 Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of
the vertices of the triangle formed by these lines and the y axis?

Answer.
5x − y = 5

Or, y = 5x − 5

The solution table will be as follows.
x 0 1 2

y −5 0 5

3x − y = 3

Or, y = 3x − 3

The solution table will be as follows.
x 0 1

y −3 0 3

The graphical representation of these lines will be as follows

It can be observed that the required triangle is △ABC formed by these lines and y- axis.
The coordinates of vertices are A (1, 0), B(0,-3),C(0, -5).

Page : 68 , Block Name : Exercise 3.7(Optional)

Q7 Solve the following pair of linear equations:
(i) px + qy = p – q
qx - py = p+q
(ii) ax + by = c
bx + ay = 1 + c
(iii) − = 0
x y

a b
2 2
ax + by = a + b

(iv) (a – b)x + (a + b) y = a 2
− 2ab − b
2

(a + b)(x + y) = a + b 2 2

(v) 152x – 378y = – 74
–378x + 152y = – 604

Answer. (i) px + qy = p − q … (1)
qx − py = p + q … (2)

Multiplying equation (1) by p and equation (2) by q, we obtain
2 2
p x + pqy = p − pq … (3)

2 2
q x − pqy = pq + q … (4)

Adding equations (3) and (4), we obtain
2 2 2 2
p x + q x = p + q

2 2 2 2
(p + q )x = p + q

2 2
p +q
x = = 1
2 2
p +q

From equation (1), we obtain
P (1) + qy = p − q

qy = −q

y = −1

(ii)ax + by = c … . (1)

bx + ay = 1 + c … (2)

Multiplying equation (1) by a and equation (2) by b, we obtain
2
a x + aby = ac … (3)

2
b x + aby = b + bc … (4)

Subtracting equation (4) from equation (3),
2 2
(a − b ) x = ac − bc − b

c(a−b)−b
x =
a2 −b2

From equation (1), we obtain
ax + by = c
c(a−b)−b
a{ } + by = c
a2 −b2

ac(a−b)−ab
+ by = c
a2 −b2

ac(a − b) − ab
by = c −
2 2
a − b
2 2 2
a c − b c − a c + abc + ab
by =
2 2
a − b
2
abc − b c + ab
by =
2 2
a − b
bc(a−b)+ab
by =
a2 −b2

c(a−b)+a
y =
a2 −b2

y
(iii)
x
− = 0
a b

or, bx − ay = 0 … (1)

2 2
ax + by = a + b … (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain
2
b x − aby = 0 … (3)

2 3 2
a x + aby = a + ab … (4)

Adding equations (3) and (4), we obtain
2 2 3 2
b x + a x = a + ab

2 2 2 2
x (b + a ) = a (a + b )

x = a

By using (1), we obtain

b(a) − ay = 0

ab − ay = 0

ay = ab

y = b
2 2
( iv )(a − b)x + (a + b)y = a − 2ab − b … (1)

2 2
(a + b)(x + y) = a + b

2 2
(a + b)x + (a + b)y = a + b … (2)

Subtracting equation (2) from (1), we obtain
2 2 2 2
(a − b)x − (a + b)x = (a − 2ab − b ) − (a + b )

2
(a − b − a − b)x = −2ab − 2b

−2bx = −2b(a + b)

x = a + b

Using equation (1), we obtain
2 2
(a − b)(a + b) + (a + b)y = a − 2ab − b

2 2 2 2
a − b + (a + b)y = a − 2ab − b

(a + b)y = −2ab

−2ab
y =
a+b

(v) 152x − 378y = −74
76x − 189y = −37

189y−37
x =
76

−378x + 152y = −604

−189x + 76y = −302 … (2)

Substituting the value of x in equation (2), we obtain
189y−37
−189 ( ) + 76y = −302
76

2 2
−(189) y + 189 × 37 + (76) y = −302 × 76

2 2
189 × 37 + 302 × 76 = (189) y − (76) y

6993 + 22952 = (189 − 76)(189 + 76)y

29945 = (113)(265)y

y = 1

From equation (1), we obtain

189(1)−37
x =
76

189−37 152
x = =
76 76

x = 2

Page : 68 , Block Name : Exercise 3.7(Optional)(Optional)

Q8 ABCD is a cyclic quadrilateral (see Figure). Find the angles of the cyclic quadrilateral.

Answer. We know that the sum of the measures of opposite angles in a cyclic quadrilateral is
∘
180 .

Therefore, ∠A + ∠C = 180

4y + 20 − 4x = 180

−4x + 4y = 160

x − y = −40(i)

Also, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

−7x + 3y = 180(ii)

Multiplying equation (i) by 3, we obtain
3x − 3y = −120(iii)

Adding equations (ii) and (iii), we obtain
-7x + 3x = 180 -120
-4x = 60
X = -15
By using equation (i), we obtain
x − y = −40

−15 − y = −40

y = −15 + 40 = 25
∘
∠A = 4y + 20 = 4(25) + 20 = 120

∘
∠B = 3y − 5 = 3(25) − 5 = 70

∘
∠C = −4x = −4(−15) = 60

∘
∠D = −7x + 5 = −7(−15) + 5 = 110

Page : 68 , Block Name : Exercise 3.7(Optional)