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

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

Here you will find Chapter 3 Pair of Linear Equations in Two Variables Class 10 Maths NCERT Solutions is best way to understand the concepts given in the chapter. By practicing, you will get to know about the important points. It will prove useful if you want to complete your homework in time. NCERT Class 10 Maths textbook is also useful for the state boards students such as Bihar Board, UP Board, JAC and many others.

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


Exercise 3.1

Page No. 44

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.

Solution

Let present age of Aftab be x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab = -7
Age of daughter = y-7
According to the question,
(- 7)  = 7 (– 7 )
– 7 = 7 – 49
x- 7= - 49 + 7
– 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-707
y567

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

x036
y-2-10

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

Graphical representation


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.

Solution

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

x39001300-1300
y-130001300

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

x39001300-1300
y-130001300

Algebraic representation
3+ 6y = 3900 … (i)
+ 2= 1300 … (ii)

Graphical representation,


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.
Solution

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
= 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

x80400
y080160

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2= 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
= 150 – 2x
Putting x = 0 , 50 , 100 we get
= 150 – 2 × 0 = 150
= 150 – 2 ×  50 = 50
= 150 – 2 × (100) = -50

x050100
y15050-50

Algebraic representation,
2y = 160 … (i)
4x + 2y = 300 … (ii)

Graphical representation,

Graph 3

Exercise 3.2

Page No. 49

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.

Solution

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

x105
y05

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

x-404
y048

Graphical representation

Graph Exercise 3.2 1.1

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

x3-4-11
y51015

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

x024
y9.26.43.6

Graphical representation

Graph Exercise 3.2 1.2

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

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.

Solution

(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
a1/a2 = 5/7,
b1/b2 = -4/6 and
c1/c2 = 8/-9
Hence, a1/a2 ≠ b1/b2

Therefore, both are intersecting lines at one point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + c2= 0
We get
a1 = 9, b1 = 3, and c1 = 12
a2 = 18, b2 = 6 and c2 = 24
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b= c1/c2

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
a1/a2 = 6/2 = 3/1
b1/b2 = -3/-1 = 3/1 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b c1/c2

Therefore, both lines are parallel.

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 ; 9– 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12

Solution

(i) 3x + 2y = 5 ; 2x – 3y = 7
a1/a2 = 3/2
b1/b2 = -2/3 and
c1/c2 = 5/7
Hence, a1/a2 ≠ b1/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/a2 = 2/4 = 1/2
b1/b2 = -3/-6 = 1/2 and
c1/c2 = 8/9
Hence, a1/a2 = b1/b c1/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) 3/2x + 5/3y = 7 ; 9– 10y = 14
a1/a2 = 3/2/9 = 1/6
b1/b2 = 5/3/-10 = -1/6 and
c1/c2 = 7/14 = 1/2
Hence, a1/a2 ≠ b1/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/a2 = 5/-10 = -1/2
b1/b2 = -3/6 = -1/2 and
c1/c2 = 11/-22 = -1/2
Hence, a1/a2 = b1/b= c1/c2

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.

(v) 4/3x + 2y =8 ; 2x + 3y = 12
a1/a2 = 4/3/2 = 2/3
b1/b2 = /3 and
c1/c2 = 8/12 = 2/3
Hence, a1/a2 = b1/b= c1/c2

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.

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

Solution


(i) x + y = 5; 2x + 2y = 10
a1/a2 = 1/2
b1/b2 = 1/2 and
c1/c2 = 5/10 = 1/2
Hence, a1/a2 = b1/b= c1/c2
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

x432
y123

And, 2x + 2y = 10
x = 10-2y/2

x432
y123

Graphical representation

Exercise3.2-4-1-Class10-Maths

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
a1/a2 = 1/3
b1/b2 = -1/-3 = 1/3 and
c1/c2 = 8/16 = 1/2
Hence, a1/a2 = b1/b c1/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/a2 = 2/4 = 1/2
b1/b2 = -1/2 and
c1/c2 = -6/-4 = 3/2
Hence, a1/a2 ≠ b1/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


x012
y642

And, 4x - 2y -4 = 0
y = 4x - 4/2

x123
y024

Graphical representation

Exercise 3.2 4 3 Class10 Maths

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
a1/a2 = 2/4 = 1/2
b1/b2 = -2/-4 = 1/2 and
c1/c2 = 2/5
Hence, a1/a2 = b1/b c1/c2

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

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.

Solution

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

x0812
y41216

y + x = 36

x03616
y36020

Graphical representation

Exercise 3.2 4 4 Class10 Maths


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.

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

Solution

(i) Intersecting lines:
For this condition,
a1/a2 ≠ b1/b2
The second line such that it is intersecting the given line is
2x + 4y - 6 = 0 as
a1/a2 = 2/2 = 1
b1/b2 = 3/4 and
a1/a2 ≠ b1/b2

(ii) Parallel lines
For this condition,
a1/a2 = b1/b c1/c2
Hence, the second line can be
4x + 6y - 8 = 0 as
a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = -8/-8 = 1
and a1/a2 = b1/b c1/c2

(iii) Coincident lines
For coincident lines,
a1/a2 = b1/b= c1/c2
Hence, the second line can be
6x + 9y - 24 = 0 as
a1/a2 = 2/6 = 1/3
b1/b2 = 3/9 = 1/3 and
c1/c2 = -8/-24 = 1/3
and a1/a2 = b1/b= c1/c2

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.

Solution

x - y + 1 = 0
x = y - 1

x012
y123

3x + 2y - 12 = 0
x = 12 - 2y/3

x420
y036

Graphical representation



\Exercise 3.2 5 Class10 Maths

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 3.3

1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14, x - y = 4   
(ii) s - t = 3, s/3 + t/2 = 6
(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
(i) 3x/2 - 5y/3 = -2, x/3 + y/2 = 13/6

Solution

(i) x + y = 14   ...(i)
x - y = 4    ...(ii)
From the equation (i), we get
x = (14 - y)
Substituting this value of x in (ii), we have:
(14 - y) - y = 4
⇒ 14 - 2y = 4
⇒- 2y = 4 - 14
⇒ - 2y = - 10 or y = 10/2
i.e., y = 5
Now, substituting y = 5 in (i), we have:
x + 5 = 14 ⇒ x = 14 - 5 ⇒ x = 9
Thus, x = 9
y = 5

(ii) s - t = 3   ...(i)
s/3 + t/2 = -6   ...(ii)
From (i), we have
s = (3 + t)    ...(iii)
substituting this value of s in (ii) we get:
(3+t)/3 + t/2 = 6 = 6
⇒ 2 (3 + t) + 3 (t) = 6 × 6 [multiplying throughout by 6]
⇒ 6 + 2t + 3t = 36
⇒ 5t = 36 - 6
⇒ 5t = 30 or t = 30/5 = 6
Substituting t = 6 in (iii),
s = 3 + 6 = 9
Thus, s = 9
t = 6

(iii) 3x - y = 3   ...(i)
9x - 3y = 9   ...(ii)
From the equation (i),
y = (3x - 3)
Substituting this value of y in (ii),
9x - 3 (3x - 3) = 9
⇒ 9x - 9x + 9 = 9
⇒ 9 = 9
which is true,
∴ The equations (i) and (ii) have unlimited solutions.

(iv) 0.2x + 0.3y = 1.3     ...(i)
0.4x + 0.5y = 2.3     ...(ii)
From the equations (i), we get
0.3y = (1.3 - 0.2x)
⇒ y = 1.3 - 0.2x/0.3  ...(iii)
From (2) and (3), we have:
0.4x + 0.5 [1.3 - 0.2x/0.3] = 2.3
⇒ 0.4x + [0.65 - 0.1x/0.3] = 2.3
⇒ 0.3 × 0.4x + 0.65 - 0.1x = 0.3 × 2.3
⇒ 0.12x + 0.65 - 0.1x = 0.69
⇒ 0.12x - 0.1x = 0.69 - 0.65
⇒ 0.02x = 0.04
⇒ x = 0.04/0.02 = 2
From (iii),
y = 1.3 - 0.2(2)/0.3
⇒ y = 1.3 - 0.4 / 0.3 = 0.9/0.3 = 3
Thus, x = 2 and y = 3


From (iii) and (i), we have:
Substituting y = 0 in (iii), we have:
x = √8/√3 (0) = 0
Thus, x = 0 and y = 0.

(vi) 3x/2 - 5y/3 = -2 ... (i)
x/3 + y/2 = 13/6 ... (ii)
From (ii), we have
x/3 = 13/6 - y/2
⇒ x = 3 × 13/6 - 3/2y
⇒ x = [13/2 - 3/2y] ...(iii)
Substituting the value of x in (i),

⇒ 117 - 47y = -24
⇒ -47y = -24 - 117 = -141
⇒ y = -141/-47 = 3
Now, substituting y = 3 in (i), we have:
x = 13/2 - 3/2(3)
Thus, x = 2 and y = 3.

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

Solution

We have 2x + 3y = 11 ...(i)
2x - 4y = - 24 ...(ii)
From (i), we have:
2x = 11 - 3y
⇒ x = [11-3y/2] ...(iii)
Substituting this value of x in (ii), we have
2[11-3y/2] - 4y = -24
⇒ 11 − 3y − 4y = − 24
⇒ 11 − 7y = − 24 ⇒− 7y = − 24 − 11 = − 35
⇒ y = -35/-7 = 5
Substituting y = 5 in (iii), we get
x = 11-3(5)/2
⇒ x = 11-15/2
⇒ x = 11-15/2
⇒ x = -4/2 = -2
∴ x = 2 and y = 5
Now, y = mx + 3
⇒ 5 = m (− 2) + 3
⇒ − 2 m = 5 − 3
⇒ − 2 m = 2
or m = -2/2 = -1
Thus, m = -1

3. Form the pair of linear equations for the following problems and find 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 fixed 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 fixed 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 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.
(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?

Solution

 (i) Let the two numbers be x and y such that x > y
∵ Difference of the numbers = 26
∴  x - y = 26    ...(i)
Again
One number = 3 [the other number]
⇒  x = 3y  ...(2)
Substituting x = 3y in (i), we get
3y - y = 26
⇒ 2y = 26 ∴ y = 26/2 = 13
Now, substituting y = 13 in (ii), we have:
x = 3 (13)
⇒ x = 39
Thus, the required solution is:
x = 39  and  y = 13

(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18°
∴ x = y + 18°   ...(i)
Also, sum of two supplementary angles = 180°
∴ x + y = 180°   ...(ii)
From (i), x = (18° + y)
Substituting this value of x in (ii),
(18° + y) + y = 180°
⇒ 2y = 180° - 18° = 162°
⇒y =162º/2 = 81º
Substituting y = 81 in (i), we get
x = 18° + 81° = 99°
Thus x = 99°  and  y = 81°

(iii) Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800
⇒ 7x + 6y = 3800  ...(i)
Again
[cost of 3 bats] + [cost of 5 balls] = ₹ 1750
⇒ 3x + 5y = 1750   ..(ii)
From (2), we have:
y = [1750 - 3x/5]
Substituting this value of y in (i), we have:
7x + 6 [1750-3x/5 = 3800
⇒ 35x + 6 (1750 - 3x) = 5 × 3800
⇒ 35x + 10500 - 18x = 19000
⇒ 17x = 19000 - 10500
⇒ x = 8500/17 = 500
Substituting x = 500 in (iii), we have:
y = 1750 -3 (500)/
⇒ y = 1750 - 1500 / 5 = 250/5 = 50
y = 1750-3(500)/5
⇒ y = 1750-1500/5 - 250/5 - 50
Thus, x = 500  and  y = 50
⇒ Cost of a bat = ₹ 500
Cost of a ball = ₹ 50

(iv) Let fixed charges be ₹ x
And charges per km be ₹ y
∵ Charges for the journey of 10 km = ₹ 105
∴  x + 10y = 105    ...(i)
Again, charge for the journey of 15 km is ₹ 155.
∴ x + 15y = 155   ...(ii)
From (1), x = (105 - 10y)   ...(iii)
From (2) and (3),
(105 - 10y) + 15y = 155
⇒ 5y = 155 - 105 = 50
⇒ y = 50/5 = 10
Substituting y = 10 in (iii), we get
x = 105 - 10 (10)
⇒ x = 105 - 100 = 5
Thus, x = 5  and  y = 10
⇒ Fixed charges = ₹ 5
Charges per km = ₹ 10
Now, Charges for 25 km
= x + 25y
= 5 + 25 (10)
= 5 + 250 = 255
∴The charges for 25 km journey = ₹ 255

(v) Let the numerator = x
And the denominator = y
⇒ Fraction = x/y

Case I:
x+2/y+2 = 9/11
⇒ 11 (x + 2) = 9 (y + 2)
11x + 22 = 9y + 18
⇒ 11x - 9y = 18 - 22
⇒11x - 9y + 4 = 0   ...(1)

Case II:
x+3/y+3 = 5/6
⇒ 6 (x + 3) = 5 (y + 3)
⇒ 6x + 18 = 5y + 15
⇒ 6x - 5y + 18 - 15 = 0
⇒ 6x - 5y + 3 = 0   ...(ii)
Now, from (ii),
x = [5y-3/6] ...(iii)
Substituting this value of x in (i), we get:
11 [5y - 3/6] -9y + 4 = 0
⇒ 11 × 5y - 11 × 3 - 6 × 9y + 6 × 4 = 0
55y - 33 - 54y + 24 = 0
y - 9 = 0
y = 9
Now, substituting y = 9 in (iii), we get
x = 5(9) -3/6
⇒ x = 45-/6, 42/6, = 7
∴ x = 7 and y = 9
⇒ Fraction = 7/9·

(vi) Let the present age of Jacob = x years.
And the present age of his son = y years.
∴ 5 years hence
Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
Condition:
[Age of Jacob] = 3 [Age of his son]
∴ x + 5 = 3 (y + 5)
x + 5 = 3y + 15
x - 3y = 15 - 5
x - 3y - 10 = 0  ...(i)
5 years ago:
Age of Jacob = (x - 5) years
Age of his son = (y - 5) years
Condition:
[Age of Jacob] = 7 [Age of his son]
∴ (x - 5) = 7 (y - 5)
x - 5 = 7y - 35
x - 7y = - 35 + 5 = - 30
x - 7y + 30 = 0     ...(ii)
From (1), x = [10 + 3y]   ...(iii)
Substituting this value of x in (ii),
(10 + 3y) - 7y + 30 = 0
⇒ 10 + 3y - 7y + 30 = 0
⇒ - 4y = - 40
⇒ y = -40/40 = 10
Now substituting y = 10 in (iii), we get
x = 10 + 3 (10)
⇒ x = 10 + 30 = 40
Thus, x = 40  and  y = 10
⇒ Present age of Jacob = 40 years
⇒ Present age of his son = 10 years.

Exercise 3.4

1. 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/2 + 2y/3 = - 1 and x – y/3 = 3


Solution

(i) x + y =5 and 2x –3y = 4
By elimination method
x + y =5 ... (i)
2x –3y = 4 ... (ii)
Multiplying equation (i) by (ii), we get
2x + 2y = 10 ... (iii)
2x –3y = 4 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 - (6/5) = 19/5
Hence, x = 19/5 and y = 6/5

By substitution methodx + y = 5 ... (i)
Subtracting y both side, we get
x = 5 - y ... (iv)
Putting the value of x in equation (ii) we get
2(5 – y) – 3y = 4
-5y = - 6
y = -6/-5 = 6/5
Putting the value of y in equation (iv) we get
x = 5 – 6/5
x = 19/5
Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2
By elimination method
3x + 4y = 10 .... (i)
2x – 2y = 2 ... (ii)
Multiplying equation (ii) by 2, we get
4x – 4y = 4 ... (iii)
3x + 4y = 10 ... (i)
Adding equation (i) and (iii), we get
7x + 0 = 14
Dividing both side by 7, we get
x = 14/7 = 2
Putting in equation (i), we get
3x + 4y = 10
3(2) + 4y = 10
6 + 4y = 10
4y = 10 – 6
4y = 4
y = 4/4 = 1
Hence, answer is x = 2, y = 1

By substitution method
3x + 4y = 10 ... (i)
Subtract 3x both side, we get
4y = 10 – 3x
Divide by 4 we get
y = (10 - 3x )/4
Putting this value in equation (ii), we get
2x – 2y = 2 ... (i)
2x – 2(10 - 3x )/4) = 2
Multiply by 4 we get
8x - 2(10 – 3x) = 8
8x - 20 + 6x = 8
14x = 28
x = 28/14 = 2
= (10 - 3x)/4
= 4/4 = 1
Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By elimination method
3x – 5y – 4 = 0
3x – 5y = 4 ...(i)
9x = 2y + 7
9x – 2= 7 ... (ii)
Multiplying equation (i) by 3, we get
9 x – 15 y = 11 ... (iii)
9x – 2y = 7 ... (ii)
Subtracting equation (ii) from equation (iii), we get
-13y = 5
y = -5/13
Putting value in equation (i), we get
3x – 5y = 4 ... (i)
3x - 5(-5/13) = 4
Multiplying by 13 we get
39x + 25 = 52
39x = 27
x =27/39 = 9/13
Hence our answer is x = 9/13 and y = - 5/13

By substitution method
3x – 5y = 4 ... (i)
Adding 5y both side we get
3x = 4 + 5y
Dividing by 3 we get
x = (4 + 5y )/3 ... (iv)
Putting this value in equation (ii) we get
9x – 2y = 7 ... (ii)
9 ((4 + 5)/3) – 2y = 7
Solve it we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = - 5
y = -5/13
Hence we get x = 9/13 and y = - 5/13 again.

(iv) x/2 + 2y/3 = - 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = -1 ... (i)
x – y/3 = 3 ... (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = - 2 ... (iii)
x – y/3 = 3 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = -5
Dividing by 5 and multiplying by 3, we get
y = -15/5
y = - 3
Putting this value in equation (ii), we get
x – y/3 = 3 ... (ii)
x – (-3)/3 = 3
x + 1 = 3
x = 2
Hence our answer is x = 2 and y = −3.

By substitution method
x – y/3 = 3 ... (ii)
Add y/3 both side, we get
x = 3 + y/3 ... (iv)
Putting this value in equation (i) we get
x/2 + 2y/3 = - 1 ... (i)
(3+ y/3)/2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = - 1
Multiplying by 6, we get
9 + y + 4y = - 6
5y = -15
y = - 3
Hence our answer is x = 2 and y = −3.

2. Form the pair of linear equations in the following problems, and find 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 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.

(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.

Solution

(i) Let the fraction be x/y
According to the question,x + 1/y - 1 = 1
⇒ x - y = -2 ... (i)x/y+1 = 1/2
⇒ 2x - y = 1 ... (ii)
Subtracting equation (i) from equation (ii), we get
x = 3 ... (iii)
Putting this value in equation (i), we get
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 ... (i)
(x + 10y) = 2(y + 10)
x - 2y = 10 ... (ii)
Subtracting equation (i) from equation (ii), we get
y = 20 ... (iii)
Putting this value in equation (i), we get
x - 60 = -10
x = 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years.

(iii) Let the unit digit and tens digits of the number be x and y respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 ... (i)
9(10y + x) = 2(10x + y)
88y - 11x = 0
- x + 8y =0 ... (ii)
Adding equation (i) and (ii), we get
9y = 9
y = 1 ... (iii)
Putting the value in equation (i), we get
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.

(iv) Let the number of ₹ 50 notes and Rs 100 notes be x and y respectively.
According to the question,
x + y = 25 ... (i)
50x + 100y = 2000 ... (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 ... (iii)
Subtracting equation (iii) from equation (ii), we get
50y = 750
y = 15
Putting this value in equation (i), we have x = 10
Hence, Meena has 10 notes of ₹ 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be ₹ x and ₹ y respectively.
According to the question,
x + 4y = 27 ... (i)
x + 2y = 21 ... (ii)
Subtracting equation (ii) from equation (i), we get
2y = 6
y = 3 ... (iii)
Putting in equation (i), we get
x + 12 =27
x = 15
Hence, fixed charge = ₹ 15 and Charge per day = ₹ 3.

Exercise 3.5

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
a1/a2 = 1/3
b1/b2 = -3/-9 = 1/3 and
c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b c1/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/a2 = 2/3
b1/b2 = 1/2 and
c1/c2 = -5/-8 = 5/8
a1/a2 ≠ b1/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 method,
x/b1c2-b2c= y/c1a2-c2a= 1/a1b2-a2b1
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ = 2, = 1.

(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 = 1/2 and
c1/c2 = -20/-40 = 1/2
a1/a2 = b1/b= c1/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 infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a1/a2 = 1/3
b1/b2 = -3/-3 = 1 and
c1/c2 = -7/-15 = 7/15
a1/a2 ≠ b1/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/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
= 4 and = -1
∴ = 4, = -1.

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

Solution

2x + 3y -7 = 0
(a – b)x + (a + b)y - (3a +b –2) = 0
a1/a2 = 2/a-b = 1/2
b1/b2 = -7/a+b and
c1/c2 = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a1/a2 = b1/b2 = c1/c2

2/a-= 7/3a+b-26a + 2b - 4 = 7a - 7b
a - 9b = -4 ... (i)

2/a-= 3/a+b
2a + 2b = 3a - 3b
a - 5b = 0 ... (ii)

Subtracting equation (i) from (ii), we get
4b = 4
b = 1
Putting this value in equation (ii), 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) For which value of k will the following pair of linear equations have no solution?
3x + = 1
(2k –1)x + (k –1)y = 2k + 1

Solution

3x + -1 = 0
(2k –1)x + (k –1)y - (2k + 1) = 0
a1/a2 = 3/2k-1
b1/b2 = 1/k-1 and
c1/c2 = -1/-2k-1 = 1/2k+1
For no solutions,
a1/a2 = b1/b c1/c2
3/2k-1 = 1/k-1 ≠ 1/2k+1
3/2k-1 = 1/k-1
3k - 3 = 2k - 1
k = 2
Hence, for k = 2, the given equation has no solution.

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

Solution

8x +5y = 9 ... (i)
3x +2y = 4 ... (ii)
From equation (ii), we get
x = 4-2y/3 ... (iii)
Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9
32 - 16y +15y = 27
-y = -5
y = 5 ... (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, = 5
By cross multiplication again, we get
8x + 5y -9 = 0
3x + 2y - 4 = 0
x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

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

(i) 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.


Solution

Let be the fixed charge of the food and be the charge for food per day.

According to the question,
x + 20y = 1000 ... (i)
x + 26y = 1180 ... (ii)
Subtracting equation (i) from equation (ii), we get
6y = 180
y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 - 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30

(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.

Solution

Let the fraction be x/y
According to the question,
x-1/y = 1/3
⇒ 3x - y = 3... (i)
x/y+8 = 1/4
⇒ 4x - y = 8 ... (ii)
Subtracting equation (i) from equation (ii), we get
x = 5 ... (iii)
Putting this value in equation (i), we get
15 - = 3
= 12
Hence, the fraction is 5/12.

(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?

Solution

Let the number of right answers and wrong answers be and respectively.

According to the question,
3x - = 40 ... (i)
4x - 2y = 50
⇒ 2x - = 25 ... (ii)
Subtracting equation (ii) from equation (i), we get
= 15 ... (iii)
Putting this value in equation (ii), we get
30 - = 25
= 5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20

(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?

Solution

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 = (v) km/h
According to the question,
5(v) = 100
⇒ u - v = 20 ... (i)
1(u + v) = 100 ... (ii)
Adding both the equations, we get
2u = 120
u = 60 km/h ... (iii)
Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(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.

Solution

Let length and breadth of rectangle be unit and y unit respectively.
Area = xy
According to the question,
(x - 5) (y + 3) = xy - 9
⇒ 3x - 5y - 6 = 0 ... (i)
(x + 3) (y + 2) = xy + 67
⇒ 2x - 3y - 61 = 0 ... (ii)
By cross multiplication, we get
x/305-(-18) = y/-12-(-183) = 1/9-(-10)
x/323 = y/171 = 1/19
= 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Exercise 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2= 13/6

(ii) 2/x +3/y = 2
4/x - 9/y = -1

(iii) 4/x + 3y = 14
3/x - 4y = 23

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 - 3/y-2 = 1

(v) 7x-2y/xy = 5
8x + 7y/xy = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) 10/x+y + 2/x-y = 4
15/x+y - 5/x-y = -2

(viii) 1/3x+y + 1/3x-y = 3/4
1/2(3x-y) - 1/2(3x-y) = -1/8


Solution

(i) 1/2x + 1/3y = 2
1/3x + 1/2= 13/6
Let 1/p and 1/q, then the equations changes as below:
p/2 + q/3 = 2
⇒ 3p + 2q -12 = 0 ... (i)
p/3 + q/2 = 13/6
⇒ 2p + 3q -13 = 0 ... (ii)

By cross-multiplication method, we get
p/-26-(-36) = q/-24-(-39) = 1/9-4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
= 2 and = 3
1/= 2 and 1/= 3
Hence, = 1/2 and = 1/3

(ii) 2/x +3/y = 2
4/x - 9/y = -1
Let 1/p and 1/y = q, then the equations changes as below:
2p + 3q = 2 ... (i)
4p - 9q = -1 ... (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 ... (iii)
Adding equation (ii) and (iii), we get
10p = 5
p = 1/2 ... (iv)
Putting in equation (i), we get
2 × 1/2 + 3q = 2
3q = 1
q = 1/3

= 1/x = 1/2
x = 2
x = 4
and
= 1/y = 1/3
y = 3
y = 9
Hence, x = 4, y = 9

(iii) 4/x + 3y = 14
3/x - 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
⇒ 4p + 3y - 14 = 0
3p - 4y = 23
⇒ 3p - 4y -23 = 0
By cross-multiplication, we get
p/-69-56 = y/-42-(-92) = 1/-16-9
⇒ -p/125 = y/50 = -1/25
Now,
-p/125 = -1/25 and y/50 = -1/25
⇒ p = 5 and y = -2
Also, p = 1/x = 5
⇒ x = 1/5
So, x = 1/5 and y = -2 is the solution.

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 - 3/y-2 = 1
Putting 1/x-1 = p and 1/y-2 = q in the given equations, we obtain
5p + q = 2 ... (i)
6p - 3q = 1 ... (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 ... (iii)
Now, adding equation (ii) and (iii)
21p = 7
⇒ p = 1/3
Putting this value in equation (ii) we get,
 6×1/3 - 3q =1
 ⇒ 2-3q = 1
 ⇒ -3q = 1-2
 ⇒ -3q = -1
 ⇒ q = 1/3
Now,
p = 1/x-1 = 1/3
 ⇒1/x-1 = 1/3
 ⇒ 3 = - 1
 ⇒ x = 4
Also,
q = 1/y-2 = 1/3
 ⇒ 1/y-2 = 1/3
 ⇒ 3 = y-2
 ⇒ y = 5
Hence, x = 4 and y = 5 is the solution.

(v) 7x-2y/xy = 5
 ⇒ 7x/xy - 2y/xy = 5
 ⇒ 7/y - 2/x = 5 ... (i)
8x+7y/xy = 15
 ⇒ 8x/xy + 7y/xy = 15
 ⇒ 8/y + 7/x = 15 ... (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
7q - 2p = 5 ... (iii)
8q + 7p = 15 ... (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q - 14p = 35 ... (v)
16q + 14p = 30 ... (vi)
Now, adding equation (v) and (vi) we get,
49q - 14p + 16q + 14p = 35 + 30
⇒ 65q = 65
⇒ q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
⇒ 7p = 7
⇒ p = 1
Now,
p = 1/x = 1
⇒ 1/x = 1
⇒ x = 1
also, q = 1 = 1/y
⇒ 1/y = 1
⇒ = 1
Hence, =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy
⇒ 6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6 ... (i)
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5 ... (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p - 6 = 0
2q + 4p - 5 = 0
By cross multiplication method, we get
p/-30-(-12) = q/-24-(-15) = 1/6-24
p/-18 = q/-9 = 1/-18
p/-18 = 1/-18 and q/-9 = 1/-18
= 1 and q = 1/2
= 1/x = 1 and q = 1/y = 1/2
= 1, y = 2
Hence, = 1 and = 2

(vii) 10/x+y + 2/x-y = 4
15/x+y - 5/x-y = -2
Putting 1/x+y = p and 1/x-y = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
⇒ 15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/x-y = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3x-y = 3/4
1/2(3x-y) - 1/2(3x-y) = -1/8
Putting 1/3x+y = p and 1/3x-y = q in the given equations, we get
p + q = 3/4 ... (i)
p/2 - q/2 = -1/8
= -1/4 ... (ii)
Adding (i) and (ii), we get
2p = 3/4 - 1/4
2p = 1/2
= 1/4
Putting the value in equation (ii), we get
1/4 - q = -1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x+y = 1/4
3x + y = 4 ... (iii)
q = 1/3x-y = 1/2
3x - y = 2 ... (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find 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.


Solution

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 ... (i)
2(x - y) = 4
⇒ x - y = 2 ... (ii)
Adding equation (i) and (ii), we get
Putting this equation in (i), we get
y = 4
Hence, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Solution

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
According to the question,
4(2/x + 5/y) = 1
2/x + 5/y = 1/4
3(3/x + 6/y) = 1
3/x + 6/y = 1/3
Putting 1/x = p and 1/q in these equations, we get
2p + 5q = 1/4
By cross multiplication, we get
p/-20-(-18) = q/-9-(-18) = 1/144-180
p/-2 = q/-1 = 1/-36
p/-2 = -1/36 and q/-1 = 1/-36
= 1/18 and q = 1/36
p = 1/x = 1/18 and q = 1/y = 1/36
= 18 and = 36
Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(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 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.

Solution

Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
60/u + 240/= 4 ... (i)
100/u + 200/= 25/6 ... (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 ... (iii)
100p + 200q = 25/6
600p + 1200q = 25 ... (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 .... (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 ... (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.

Exercise 3.7 (Optional)

1. 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.

Solution

Let the age of Ani = x years
And the age of Biju = y years
Case I:
y > x
According to 1st condition:
y - x = 3     ...(i)
∵ Age of Ani’s father = 2 Age of Ani
∴ = 2x years
Also, Age of Biju’s sister = 1/2 Age of Biju
= 1/2y
According to IInd condition:
2x - 1/2y = 30
⇒ 4x − y = 60      ...(ii)
Adding (i) and (ii),
⇒ x = 63/3 = 21
From (i), y - 21 = 3  ⇒  y = 3 + 21 = 24
∴ Age of Ani = 21 years
Age of Biju = 24 years
Case II:
x > y
∴ x - y = 3     ...(i)
According to the condition:
2x - 1/2y = 30
⇒ 4x − y = 60     ...(ii)
Subtracting (i) from (ii),
⇒ x = 57/3 = 19
From (1), 19 - y = 3
⇒ - y = 3 - 19
⇒ - y = - 16  ⇒  y = 16
⇒ Age of Ani = 19 years
Age of Biju = 16 years.

2. One says, ‘‘Give me a hundred, friend! I shall then become twice as rich as you’’. The other replies, ‘‘If you give me ten, I shall be six times as rich as you’’. Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2 (y - 100), y + 10 = 6 (x - 10)].

Solution

Let the capital of 1st friend = ₹ x
And capital of 2nd friend = ₹ y
According to the condition,
x + 100 = 2 (y - 100)
⇒ x + 100 - 2y + 200 = 0
⇒ x - 2y + 300 = 0     ...(1)
And 6 (x - 10) = y + 10
⇒ 6x - 60 - y - 10 = 0
⇒ 6x - y - 70 = 0    ...(2)
From (1), x = - 300 + 2y
From (2), 6x - y - 70 = 0
⇒ 6 [- 300 + 2y] - y - 70 = 0
⇒ - 1800 + 12y - y - 70 = 0
⇒ - 1870 + 11y = 0
⇒ y = 1870/11= 170
⇒ y = 1870/11 = 170
Now, x = - 300 + 2y
= - 300 + 2 (170)
= - 300 + 340
= 40
Thus, 1st friend has ₹ 40 and the 2nd friend has ₹ 170.

3. 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.

Solution

Let the actual speed of the train be x km/hr 
Actual time taken = у/hг.
Distance = Speed × Time 

1st Condition:
(x + 10) × (y-2) = xy
⇒ xy - 2x + 10y - 20 = xy
⇒ 2x - 10y + 20 = 0 ...(i)

2nd Condition:
(x - 10) × (y + 3) = xy
⇒ xy + 3x - 10y - 30 = xy
⇒ 3x - 10y - 30 = 0 ... (ii)
Using cross multiplication for solving (1) and (2):
a1 = 2,  b1 = - 10,  c1 = 20
a2 = 3,  b2 = - 10,  c2 = - 30
⇒ x/300+20 = y/60+6 = 1/-20+30
⇒ x/500 = y/120 = 1/50
⇒ x = 1/(10) × 500 = 50
y = 1/(10) × (120 = 12
Thus, the distance covered by the train = 50 × 12 km = 600 km.

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

Solution

Let the number of students = x
And the number of rows = y
∴ Number of students in each row = Number of Students/Number of rows = x/y

1st Condition:
= (x/y +3) × (y-1) = x
[Number of students in a row × Number of rows = Number of students]
⇒ x - x/y + 3y -3 = x
⇒ x/y - 3y + 3 = 0 ...(i)

2nd Condition:
= (x/y -3) × (y+2) = x
⇒ x + 2x/y - 3y -6 = x
⇒ 2x/y - 3y -6 = 0 ...(ii)

Let x/y = p
∴ Equations (i) and (ii) can be expressed as
p - 3y + 3 = 0 ...(iii)
2p - 3y - 6 = 0 ...(iv)
Subtracting (iii) from (iv), we get
p = 9
From (iii), we get
9 − 3y + 3 = 0 ⇒− 3y = − 12
⇒ y = -12/-3 = 4
Since, x /y = 9   [∵ p = 9]
x/4 = 9 ⇒ x = 4 × 9 = 36
Thus, number of students in the class
= xy
= 36 × 4 = 144.

5. In a Δ ABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

Solution

∵ Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° ...(i)
∵ ∠C = 3 ∠B = 2 (∠A + ∠B) ...(ii)
From (1) and (2), we have:
∠A + ∠B + 2 (∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2 ∠A + 2 ∠B = 180°
⇒ 3 ∠A + 3 ∠B = 180°
⇒ ∠A + ∠B = 60 ...(iii)
Also,
∠A + ∠B + 3 ∠B = 180°
⇒ ∠A + 4 ∠B = 180°   ...(iv)
Subtracting (iii) from (iv),
⇒ ∠B = 120/3 = 40°
From (4),
∠A + 4 (40) = 180°  ⇒  ∠A = 180° - (4 × 40)
⇒ ∠A = 180° - 160 = 20°
Again ∠C = 3 ∠B = 3 × 40° = 120°
Thus, ∠A = 20°,  ∠B = 40°  and  ∠C = 120°.

6. 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.

Solution

To draw the graph of 5x - y = 5, we get
5x - y = 5

x 1 2 0
y 0 5 -5
(x,y) (1,0)(2,5) (0, -5)

and for equation 3x - y = 3, we get

x 2 3 0
y 3 6 -3
(x,y) (2,3) (3,6) (0,-3)

Plotting the points (1, 0), (2, 5) and (0, - 5), we get a straight line l1. Plotting the points (2, 3), (3, 6) and (0, - 3) we get a straight line l2.
From the figure, obviously, the vertices of the triangle formed are :
 A (1, 0), B (0, - 5) and C (0, - 3).

Exercise 3.7 2 Class 10 Maths


7. 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) x/a - y/b = 0
ax + by = a2

(iv) (a - b) x + (a + b) y = a2 - 2ab - b2 
(a + b) (x + y) = a2 + b2 

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

Solution

(i) We have:
px + qy = p - q   ...(i)
qx - py = p + q    ...(ii)
Multiply (i) by p and (ii) by q, and adding:
or y =-1
Substitute this value of y in (1), we get
px + 9 (- 1) = p -1
or px - q = p - q
or px = p – q + q
or px = p
or x = 1
Hence, x = 1 and y = - 1.

(ii) We have:
ax + by = c     ...(1)
bx + ay = (1 + c)   ...(2)
By cross multiplication, we have:
A1 = a,  B1 = b,  C1 = - c
A2 = b,  B2 = a,  C2 = - (1 + c)

(iii) We have:
= x/a - y/b = 0 ...(i)
ax + by = a2 + b2  ...(ii)
From (i), we have:
x/a = y/b 
⇒ y = (x/a × b)
Substituting y = (b/a × x) in (ii), we have
ax + b (b/a × x) = a^2+ b^2
⇒ x [a^2 + b^2/a] = a^2+ b^2
⇒ x = a^2+ b^2/a^2+ b^2 × a = a
⇒ x = a
Substituting x = a in y = b/a × x,
y = b/a × a = b
⇒ x = b
Thus, the required solution is
x = a
y = b

(iv) We have:
(a - b) x + (a + b) y = a2 - 2ab - b2   ...(i)
(a + b) (x + y) = a2 + b2    ...(ii)
From (ii),
(a + b) x + (a + b) y = a2 + b2 ...(iii)
Subtracting (iii) from (i), we get
⇒ x [a − b − a − b]= − 2ab − 2b2 
⇒ x (− 2b)= − 2b (a + b)
⇒ x = -2b (a + b)/-2b
⇒ x =(a + b
Substituting x = (a + b) in (1),
(a - b) (a + b) + (a + b) y = a2 - 2ab - b2
⇒ a2 - b2 + (a + b) y = a2 - 2ab - b2
⇒ (a + b) y = a2 - 2ab - b2 - a2 + b2
⇒ (a + b) y = - 2ab
⇒ y = -2ab/a+b
Hence, x =(a + b), y = -2ab/a+b

(v) We have:
152x - 378y = - 74     ...(i)
- 378x + 152y = - 604  ...(ii)
Adding (i) and (ii), we have:
- 226x - 226y = - 678
⇒ x + y = 3     ...(iii)
[Dividing throughout by - 226]
Subtracting (i) from (ii),
⇒− x + y = − 1
⇒ x − y = 1     ...(iv)
Adding (iii) and (iv),
Subtracting (iii) from (iv),
Hence, x = 2  and  y = 1.

8. ABCD is a cyclic quadrilateral (see figure.). Find the angles of the cyclic quadrilateral.

Solution

∵ ABCD is a cyclic quadrilateral.
∴  ∠A + ∠C = 180°  and  ∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180°
⇒ 4y - 4x + 20° - 180° = 0
⇒ 4y - 4x - 160° = 0
⇒ y - x - 40° = 0      ...(i)    
[Dividing throughout by 4]
And
[3y - 5] + [- 7x + 5] = 180°
⇒ 3y - 5 + 5 - 7x - 180° = 0
⇒ 3y - 7x - 180° = 0 ...(ii)
Multiplying (i) by 7 and subtracting from (ii),
Now, substituting y = 25° in (1)
y - x = 40  ⇒  - x = 40 - y
= 40° - 25° = 15°
⇒ x = - 15°
∴ ∠A = 4y + 20° = 4 (25°) + 20°
= 100° + 20° = 120°
∠B = 3y - 5° = 3 (25°) - 5°
= 75° - 5° = 70°
∠C = - 4x = - (- 15°)
= 60°
∠D = - 7x + 5° = - 7 (- 15) + 5°
= 105° + 5° = 120°
Thus, ∠A = 120°,  ∠B = 70°,  ∠C = 60°,  ∠D = 110°.

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