1

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 present age of Aftab be x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab = x -7
Age of daughter = y-7
According to the question,
(x - 7) = 7 (y – 7 )
x – 7 = 7 y – 49
x- 7y = - 49 + 7
x – 7y = - 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 - 42 = 35 - 42 = - 7
x = 7 × 6 - 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7

x	-7	0	7
y	5	6	7

Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x -3y = 9-3
x -3y = 6 …(ii)
x = 3y + 6
Putting, y = -2, -1 and 0, we get
x = 3 ×(-2) + 6 = -6 + 6 =0
x = 3 ×(-1) + 6 = -3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6

x	0	3	6
y	-2	-1	0

Algebraic representation
From equation (i) and (ii)
x – 7y = – 42
x - 3y = 6

Graphical representation

Exercise 3.1 Page Number 44

2

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 situation algebraically and geometrically.

Answer

Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3x + 6y = 3900 … (i)
Dividing equation by 3, we get
x + 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = -1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
x = 1300 -2(0) = 1300 - 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = - 1300

x	3900	1300	-1300
y	-1300	0	1300

Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2y = 1300 … (ii)
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = - 1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
x = 1300 – 2 (0) = 1300 - 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = -1300

x	3900	1300	-1300
y	-1300	0	1300

Algebraic representation
3x + 6y = 3900
x + 2y = 1300

Graphical representation,

Exercise 3.1 Page Number 44

3

The cost of 2 kg of apples and 1kg 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.

Answer

Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
2 x + y = 160 … (i)
2x = 160 - y
x = (160 – y)/2
Let y = 0 , 80 and 160, we get
x = (160 – ( 0 )/2 = 80
x = (160- 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0

x	80	40	0
y	0	80	160

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2y = 300 … (ii)
Dividing by 2 we get

2x + y = 150
Subtracting 2x both side, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 × 50 = 50
y = 150 – 2 × (100) = -50

x	0	50	100
y	150	50	-50

Algebraic representation,
2x + y = 160
4x + 2y = 300

Graphical representation,

Exercise 3.1 Page Number 44

1

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.

Answer

(i) Let number of boys = x

Number of girls = y
Given that total number of student is 10 so that
x + y = 10
Subtract y both side we get
x = 10 – y
Putting y = 0 , 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0

x	10	5
y	0	5

Given that If the number of girls is 4 more than the number of boys
So that,
y = x + 4
Putting x = -4, 0, 4, and we get
y = - 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8

x	-4	0	4
y	0	4	8

Graphical representation

Therefore, number of boys = 3 and number of girls = 7

(ii) Let cost of pencil = Rs x

Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 - 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = - 4

x = 10 – 7 × 15/5 = 10 – 21 = - 11

x	3	-4	-11
y	5	10	15

Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 - 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6

x	0	2	4
y	9.2	6.4	3.6

Graphical representation

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

Exercise 3.2 Page Number 49

2

On comparing the ratios a₁/a₂ , b₁/b₂ and c₁/c₂, find out whether the lines representing the 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 equation with
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂= 0
We get,
a₁ = 5, b₁ = -4, and c₁ = 8
a₂ =7, b₂ = 6 and c₂ = -9

Therefore, both are intersecting lines at one point.

9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with

Therefore, both lines are coincident.

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0
Comparing these equations with
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂= 0
We get
a₁ = 6, b₁ = -3, and c₁ = 10
a₂ = 2, b₂ = -1 and c₂ = 9

Therefore, both lines are parallel.

Exercise 3.2 Page Number 49

3

On comparing the ratios a₁/a₂ , b₁/b₂ and c₁/c₂ find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12

Answer

(i) 3x + 2y = 5 ; 2x – 3y = 7

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

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

(iii)

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

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

Exercise 3.2 Page Number 49

4

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

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
x + y = 5
x = 5 - y

x	4	3	2
y	1	2	3

And, 2x + 2y = 10
x = 10−

x	4	3	2
y	1	2	3

Graphical representation

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) x – y = 8, 3x – 3y = 16

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

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
y = 4x

x	1	2	3
y	0	2	4

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

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

Exercise 3.2 Page Number 49

5

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 length of rectangle = x m
Width of the rectangle = y m
According to the question,
y - x = 4 ...(i)
y + x = 36 ... (ii)
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

Graphical representation

From the figure, 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.

Exercise 3.2 Page Number 50

6

Given the linear equation 2x + 3y - 8 = 0, write another linear equations 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) Intersecting lines:

Equation of first line: 2x + 3y – 8 = 0

Let the equation of 2^nd line be ax + by + c = 0

Condition for the lines to be intersecting,

Here,

a₁ = 2, b₁ = 3, c₁ = -8
a₂ = a, b₁ = b, c₁ = c
The equation for second line is 2x + 4y - 6 = 0 which will follow the given condition as,

(ii) Parallel lines
Equation of first line: 2x + 3y – 8 = 0

Let the equation of 2^nd line be ax + by + c = 0

Condition for the lines to be parallel,

Here,

a₁ = 2, b₁ = 3, c₁ = -8
a₂ = a, b₁ = b, c₁ = c
Hence, the equation for second line can be 4x + 6y - 8 = 0 as

(iii) Coincident lines
Equation of first line: 2x + 3y – 8 = 0

Let the equation of 2^nd line be ax + by + c = 0

Condition for the lines to be coincident,

Hence, the equation of second line can be 6x + 9y - 24 = 0 as,

Exercise 3.2 Page Number 50

7

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	2
y	1	2	3

3x + 2y - 12 = 0
⇒ x = 12 -

x	4	2	0
y	0	3	6

Graphical representation

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

Exercise 3.2 Page Number 50

1

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

(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)

(vi)

Answer

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

x – y = 4 ... (2)
From equation (i), we get

x = 14 - y ... (3)
Putting this value in equation (2), we get

(14 - y) - y = 4

14 - 2y = 4

10 = 2y

y = 5 ... (4)

Putting this in equation (3), we get

x = 9

∴ x = 9 and y = 5

(ii) s – t = 3 ... (1)

(iii) 3x - y = 3 ... (1)
9x - 3y = 9 ... (2)
From equation (1), we get
y = 3x - 3 ... (3)
Putting this value in equation (2), we get
9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9
This is always true.
Hence, the given pair of equations has infinite 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.

(iv) 0.2x + 0.3y = 1.3 ... (1)
0.4x + 0.5y = 2.3 ... (2)
0.2x + 0.3y = 1.3
Solving equation (1), we get
0.2x = 1.3 – 0.3y
Dividing by 0.2, we get

(vi) Given of pair of equations,

Exercise 3.3 Page Number 53

2

Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which y =mx + 3.

Answer

2x + 3y = 11 ... (1)

Subtracting 3y both side we get

2x = 11 – 3y … (2)

Putting this value in equation second we get

2x – 4y = – 24 … (3)

11- 3y – 4y = - 24

7y = - 24 – 11

-7y = - 35

y = -35/-7

y = 5

Putting this value in equation (3) we get

2x = 11 – 3 × 5

2x = 11- 15

2x = - 4

Dividing by 2 we get

x = - 2

Putting the value of x and y

y = mx + 3.

5 = -2m +3

2m = 3 – 5

m = -2/2

m = -1

Exercise 3.3 Page Number 53

3(i)

Form the pair of linear equations for the following problems and find their solution by substitution method

The difference between two numbers is 26 and one number is three times the other. Find them.

Answer

Let larger number = x
Smaller number = y
The difference between two numbers is 26
x – y = 26
x = 26 + y --- (1)
Given that one number is three times the other
So, x = 3y --- (2)
Putting the value of x from (1) we get
26y = 3y
⇒ -2y = - 26
⇒ y = 13
So value of x = 3y
Putting value of y, we get
x = 3 × 13 = 39
Hence the numbers are 13 and 39.

Exercise 3.3 Page Number 53

3(ii)

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer

Let first angle = x

And second number = y
As both angles are supplementary so that sum will 180°.
x + y = 180°
⇒ x = 180° - y ...(1)
Difference is 18° so,
x – y = 18
Putting the value of x we get
180° – y – y = 18°
- 2y = -162°
⇒ y = -162/-2
⇒ y = 81°
Putting the value in equation (1), we get
x = 180° – 81° = 99°

Hence, the angles are 99° and 81°.

Exercise 3.3 Page Number 53

3(iii)

The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer

Let cost of each bat = Rs x

Cost of each ball = Rs y
Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
7x + 6y = 3800
⇒ 6y = 3800 – 7x
Dividing by 6, we get

Given that she buys 3 bats and 5 balls for Rs 1750 later.
3x + 5y = 1750
Putting the value of y

Multiplying by 6, we get

18x + 19000 – 35x = 10500

⇒ -17x =10500 - 19000

⇒ -17x = -8500

⇒ x = -8500/-17

⇒ x = 500

Putting this value in equation (1) we get

⇒ y = 300/6
⇒ y = 50

Hence cost of each bat = Rs 500 and cost of each balls = Rs 50.

Exercise 3.3 Page Number 53

3(iv)

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Answer

Let the fixed charge for taxi = Rs x

And variable cost per km = Rs y
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10y = 105 … (1)
⇒ x = 105 – 10y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15y = 155
Putting the value of x we get
105 – 10y + 15y = 155
⇒ 5y = 155 – 105
⇒ 5y = 50
Dividing by 5, we get
y = 50/5 = 10
Putting this value in equation (1) we get
x = 105 – (10 × 10)
⇒ x = 5
People have to pay for traveling a distance of 25 km
= x + 25y
= 5 + 25 × 10
= 5 + 250
= 255

A person has to pay Rs 255 for 25 Km.

Exercise 3.3 Page Number 54

3(v)

A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction.

Answer

Let Numerator = x
Denominator = y
Fraction will = x/y

A fraction becomes 9/11, if 2 is added to both the numerator and the denominator.

By Cross multiplication, we get
11x + 22 = 9y + 18
Subtracting 22 both side, we get
11x = 9y – 4
Dividing by 11, we get

Given that 3 is added to both the numerator and the denominator it becomes 5/6 .
If 3 is added to both the numerator and the denominator it becomes 5/6

Exercise 3.3 Page Number 54

3(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

Let present age of Jacob = x year
And present Age of his son is = y year
Five years hence,
Age of Jacob will = x + 5 year
Age of his son will = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5 = 3(y + 5)
Adding 5 both side, we get
x = 3y + 15 - 5
⇒ x = 3y + 10 … (1)
Five years ago,
Age of Jacob will = x - 5 year
Age of his son will = y - 5 year
Jacob’s age was seven times that of his son
x – 5 = 7(y -5)
Putting the value of x from equation (1) we get
3y + 10 – 5 = 7y – 35
⇒ 3y + 5 = 7y – 35
⇒ 3y – 7y = -35 – 5
⇒ -4y = - 40
⇒ y = -40/-4
y = 10 years
Putting the value of y in equation first we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years

Exercise 3.3 Page Number 54

1(i)

Solve the following pair of linear equations by the elimination method and the substitution method:
x + y =5 and 2x –3y = 4

Answer

x + y =5 and 2x –3y = 4
By elimination method
x + y =5 ... (1)
2x –3y = 4 ... (2)
Multiplying equation (1) by (2), we get
2x + 2y = 10 ... (3)
2x –3y = 4 ... (2)
Subtracting equation (2) from equation (3), we get
5y = 6

By substitution method

x+ y = 5 ... (1)
Subtracting y both side, we get
x = 5 - y ... (4)
Putting the value of x in equation (2) we get
2(5 – y) – 3y = 4
⇒ -5y = - 6

Exercise 3.4 Page Number 56

1(ii)

3x + 4y = 10 and 2x – 2y = 2

Answer

3x + 4y = 10 and 2x – 2y = 2
By elimination method
3x + 4y = 10 ....(1)
2x – 2y = 2 ....(2)
Multiplying equation (2) by 2, we get
4x – 4y = 4 ....(3)
3x + 4y = 10 ....(1)
Adding equation (1) and (3), we get
7x + 0 = 14
Dividing both side by 7, we get

Hence, answer is x = 2, y = 1

By substitution method
3x + 4y = 10 ... (1)
Subtract 3x both side, we get
4y = 10 – 3x
Divide by 4 we get

Hence, answer is x = 2, y = 1 again.

Exercise 3.4 Page Number 56

1(iii)

3x – 5y – 4 = 0 and 9x = 2y + 7

Answer

3x – 5y – 4 = 0 and 9x = 2y + 7
By elimination method
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ...(1)
9x = 2y + 7
⇒ 9x – 2y = 7 ... (2)
Multiplying equation (1) by 3, we get
9x – 15y = 12 ... (3)
Subtracting equation (2) from equation (3), we get
(9x – 15y) – (9x – 2y) = 12 – 7
⇒ 9x – 15y – 9x + 2y = 5
⇒ -13y = 5

By substitution method
3x – 5y = 4 ... (1)
Adding 5y both side we get
3x – 5y + 5y = 4 + 5y
⇒ 3x = 4 + 5y
Dividing by 3 we get

Multiplying by 3 both side, we get
9(4 + 5y) – 6y = 21
⇒ 36 + 45y – 6y = 21
⇒ 39y = 21 – 36
⇒ 39y = – 15

Exercise 3.4 Page Number 56

1(iv)

Answer

Given pair of linear equation is

Exercise 3.4 Page Number 56

2(i)

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

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?

Answer

Let the fraction be x/y
According to the question,

⇒ x+1 = y−1
⇒ x+1−y= −1
⇒ x−y = − 1−1
⇒ x − y = −2 ... (1)
⇒ 2x = y+1
⇒ 2x − y = 1 ... (2)
Subtracting equation (1) from equation (2), we get
(2x−y)− (x−y) = 1−(−2 )
⇒ 2x− y –x + y = 1+2
x = 3 ...(3)
Putting this value in equation (1), we get
x − y = −2
3− y = − 2
−y = −5
y = 5
Hence, the fraction is 3/5

Exercise 3.4 Page Number 57

2(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?

Answer

Let present age of Nuri = x
and present age of Sonu = y

Hence, age of Nuri = 50 years and age of Sonu = 20 years.

Exercise 3.4 Page Number 57

2(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.

Answer

Let the unit digit and tens digits of the number be x and y respectively.
Then, Required number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 ... (1)
9(10y + x) = 2(10x + y) [ given ]
⇒ 90y + 9x = 20x + 2y
⇒ 90y + 9x – 20x - 2y = 0
⇒ 88y - 11x = 0
⇒ 11(8y – x ) = 0
⇒ - x + 8y =0 ... (2)
Adding equation (1) and (2), we get
(x + y) + (-x + 8y) = 9 + 0
⇒ x+ y – x + 8y = 9
⇒ 9y = 9
⇒ y = 1 ... (3)
Putting the value in equation (1), we get
x + y = 9
x + 1 = 9
x = 9 – 1
x = 8
Hence, the number is 10y + x = 10×1 + 8 = 18.

Exercise 3.4 Page Number 57

2(iv)

Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer

Let Meena received x notes of Rs 50 and y notes of Rs 100.

According to question,

x + y = 25 ...(1)

50x + 100y = 2000 ... (2)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 ... (3)
Subtracting equation (3) from equation (2), we get
(50x + 100y ) – (50x + 50y) = 2000 – 1250
⇒ 50x +100y – 50x – 50y = 750
⇒ 50y = 750
⇒ y = 15
Putting this value in equation (1), we get
x + y = 25
⇒ x + 15 = 25
⇒ x = 25 – 15
⇒ x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

Exercise 3.4 Page Number 57

2(v)

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer

Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,
x + 4y = 27 ... (1)
x + 2y = 21 ... (2)
Subtracting equation (2) from equation (1), we get
(x + 4y) – (x +2y) = 27 – 21
x + 4y – x – 2y = 6
2y = 6
y = 3 ... (3)
Putting the value in equation (1), we get
x + 4y = 27
⇒ x + 4×3 = 27
⇒ x + 12 =27
⇒ x = 27 – 12
⇒ x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

Exercise 3.4 Page Number 57

1

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find 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

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

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

∴ x = 2 , y =1

(iii) 3x – 5y = 20
6x – 10y = 40

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 infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0

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 = 4 and y = −1
∴ x =4, y= −1

Exercise 3.5 Page Number 62

2

(i) For which values of a and b does the following pair of linear equations have an infinite 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

6a + 2b – 4 = 7a – 7b
− 4 = 7a – 7b – 6a – 2b
− 4 = a – 9b
a – 9b = − 4 …(1)
Also,

2a + 2b = 3a – 3b
a – 5b = 0 ... (2)
Subtracting equation (1) from (2), we get
a −5b = 0
a – 9b = − 4 …(1)
4b = 4
b = 1
Putting this value in equation (2), we get
a −5 × 1 = 0
a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) 3x + y -1 = 0

(2k –1)x + (k –1)y - (2k + 1) = 0

⇒ 3k – 3 = 2k – 1
⇒ k = 2
Hence, for k = 2, the given equation has no solution.

Exercise 3.5 Page Number 62

3

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

Answer

8x +5y = 9 ... (1)
3x +2y = 4 ... (2)
From equation (2), we get

⇒ 32 – 16y +15y = 27
⇒ − y = 27 – 32
⇒ − y = −5
⇒ y = 5 ... (4)
Putting this value in equation (2), we get
3x +2y = 4 ... (2)
⇒ 3x + 2×5 = 4
⇒ 3x + 10 = 4
⇒ 3x = 4 – 10
⇒ x = -6/3
⇒ x = − 2
Hence, x = -2, y = 5
By cross multiplication again, we get
8x + 5y -9 = 0
3x + 2y - 4 = 0

x = −2 and y = 5

Exercise 3.5 Page Number 62

4(i)

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

A part of monthly hostel charges is fixed 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 Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Answer

Let x be the fixed charge of the food and y be the charge for food per day.
According to the question,

x + 20y = 1000 ... (1)

x + 26y = 1180 ... (2)
Subtracting equation (1) from equation (2), we get

⇒ x + 26y - x - 20y = 1180 - 1100

⇒ 6y = 180

⇒ y = 180/6 = 30
Putting this value in equation (1), we get
x + 20 × 30 = 1000
⇒ x = 1000 − 600
⇒ x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30.

Exercise 3.5 Page Number 62

4(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.

Answer

Let the fraction be x/y

According to the question,

4x − y = 8 ... (2)
Subtracting equation (1) from equation (2), we get
4x − y = 8 ... (2)
3x − y = 3 ... (1)
⇒ 4x − y - 3x + y = 8 - 3

x = 5 ... (3)
Putting this value in equation (1), we get
3x − y = 3 ... (1)
⇒ 3×5 – y = 3
⇒ 15 – y = 3
⇒ y = 12
Hence, the fraction is

Exercise 3.5 Page Number 62

4(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 ?

Answer

Let the number of right answers and wrong answers be x and y respectively.
According to the question,

3x - y = 40 ... (1)

4x - 2y = 50
⇒ 2x - y = 25 ... (2)

Subtracting equation (2) from equation (1), we get
3x - y = 40 ... (1)
2x - y = 25 ... (2)
⇒ 3x - y - 2x + y = 40 - 25

x = 15 ... (3)
Putting this value in equation (2), we get
2x - y = 25 ... (2)
⇒ 2×15 – y = 25
⇒ 30 − y = 25
⇒ y = 5

Therefore, number of right answers = 15
Number of wrong answers = 5
Total number of questions = 20

Exercise 3.5 Page Number 62

4(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 ?

Answer

Let the speed of 1st car and 2nd car be u km/h and v km/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 question,

5(u - v) = 100
u - v = 20 ... (1)

1(u + v) = 100 ... (2)

Adding both the equations, we get

⇒ u - v + u + v = 20 + 100

⇒ 2u = 120

⇒ u = 60 km/h ... (3)

Putting 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

Exercise 3.5 Page Number 62

4(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

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, we get

or, x = 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Exercise 3.5 Page Number 62

1(i)

Solve the following pairs of equations by reducing them to a pair of linear equations:

Answer

Given pair of equations are,

Exercise 3.6 Page Number 67