NCERT Solutions Class 10 Maths Chapter 3

NCERT Class 10 Mathematics Chapter 3 defines and explains the line and how the students can plot on a graph. It also talks about geometric representations of the pair of linear equations in two variables, parallel, intersection, and coinciding lines, along with other relevant concepts.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 3 that have answers to all the exercise questions given at the end of NCERT Chapter 3. Students can refer to these solutions to cross-check if they have derived the right answers or understand how to solve a question accurately.

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

Access NCERT Solutions for Class 10 Mathematics Chapter 3 – Pair of Linear Equations in Two Variables

Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables NCERT Solutions Exercises PDF Download

The chapter discusses how a linear equation in two variables can form a straight line. NCERT Solutions Class 10 Mathematics Chapter 3 covers all the exercise questions given in respective chapters. The subject-matter experts have prepared the solutions based on the latest CBSE syllabus.

NCERT Solutions for Class 10 Mathematics Chapter 3 – Free Download

Students can refer to NCERT Solutions for Class 10 Mathematics Chapter 3 on Extramarks to get help in preparing for board exams. The answers in the solutions are explained in an elaborate manner while ensuring they are highly accurate and meet the guidelines of CBSE.

Topics Covered Under Mathematics Chapter 10 NCERT Solutions Chapter 3

Let us look at the topics included in the chapter.

Section 3.1: The students learn that a linear equation in two variables has plenty of solutions. The section helps the students remember the concepts in a linear equation in two variables.

Section 3.2: The section discusses the topic in detail. The exercise includes the Geometric representations of the pair of linear equations in two variables.

Section 3.3: In this section, the Class 10 students learn how to draw the pair of linear equations in two variables in the form of lines. The lines can be parallel, intersecting, or coinciding.

Section 3.4: The section teaches the students: how to find solutions to the two equations with graphical representations. The topic has been explained in a step-by-step manner. Examples have been provided for elimination, cross-multiplication, and substitution.

Section 3.5: The students learn: how to find solutions to the linear equations in two variables. Students must practise this section thoroughly.

Section 3.6: The section presents the gist of all the points associated with the chapter.

Importance of NCERT Solutions Mathematics Class 10 Chapter 3

NCERT Solutions aid the Class 10 students’ learning process comprehensively. The benefits of NCERT Solutions of Class 10 Mathematics Chapter 3 include:

Subject-matter experts have prepared solutions to benefit the students in their learning process.
The solutions have answers to all the questions given at the end of Chapter 3.
NCERT Solutions follow the CBSE syllabus and guidelines.
The solutions are written in simple language making them easy to comprehend.

Related Question

Q1. How do you solve the system 5x−2y=3 and y=2x?

Ans. We are given the two equations 5x−2y=3 and y=2x

We know the steps required to solve a system of equations in two variables. Let’s take the second equation, we get

⇒y=2x

Substituting this in the equation 5x−2y=3, we get

⇒5x−2(2x)=3

Simplifying the above equation, we get

⇒x=3

Substituting this value in the second equation to find the value of y, we get

⇒y=2×3

Multiplying 2 and 3 we get 6, substituting this above

⇒y=6

Hence, the solution values for the system of equations are

x=3 and y=6

Q.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.

Ans. Let the present age of Aftab and his daughter be x years and y years respectively.
So, seven years ago,
Aftab’s age = (x–7) years and his daughter’s age = (y–7) years
According to the question,
x–7 = 7(y–7)
or x – 7y + 42 = 0
Three years hence,
Age of Aftab = (x + 3) years
Age of daughter = (y + 3) years
According to the question,
x + 3 = 3(y + 3)
or x – 3y – 6 = 0
Hence, the given information is represented algebraically by the two equations below.
x – 7y + 42 = 0
x – 3y – 6 = 0
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
x – 7y + 42 = 0
or x = 7y – 42

x	0	–7	7
y	6	5	7

Also,
x – 3y – 6 = 0
or x = 3y + 6

x	0	–3	3
y	–2	–3	–1

The graphical representation is given below.

Q.2 The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

Ans.

Let the price of a bat and a ball be ₹ x and ₹ y respectively.
According to the question,
3x + 6y = 3900 …(1)
or 3 (x + 2y) = 3900
or x + 2y = 3900/3 =1300 …(2)
Equation (1) represents the total price of 3 bats and 6 balls whereas equation (2) represents the total price of one bat and 2 balls.
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
3x + 6y = 3900
or y = (3900 – 3x)/6

x	100	300
y	600	500

(i)
Also,
x + 2y =1300
or y = (1300 – x)/2

x	500	900
y	400	200

(ii)
The below given graphical representation shows that
graphs of both the equations coincide.

Q.3 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.

Ans. Let the price of 1 kg of apple and 1 kg of grapes be ₹ x and ₹ y respectively.
According to the question,
2x + y = 160 …(1)
And
4x + 2y = 300
or 2x + y = 150 …(2)
Equations (1) and (2) represent the given situation algebraically.
To represent the given situation graphically, we need at least two solutions for each equation. We write these solutions in table.
2x + y = 160 …(1)
or y = 160 – 2x

x	60	40
y = 160 – 2x	40	80

(i)
Also,
2x + y = 150 …(2)
or y = 150 – 2x

x	60	40
y = 150 – 2x	30	70

(ii)
The graphical representation of the situation is given below. The two lines never intersect each other, i.e., the two lines are parallel.

Q.4 Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in 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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans.

(i) Let the number of girls and the number of boys be x and y respectively.
According to question,
x + y = 10
x – y = 4
For x + y = 10,
y = 10 – x

x	4	6	8
y = 10 – x	6	4	2

For x – y = 4,
y = x – 4

x	4	6	8
y = x – 4	0	2	4

The graphs of equations are drawn below which shows that the two lines intersect at (7, 3).
Hence the number of boys and the number of girls are 7 and 3 respectively.

(ii)
Let the cost of 1 pencil and the cost of 1 pen be ₹ x and ₹ y respectively.
According to the question,
5x + 7y = 50
and
7x + 5y = 46
For 5x + 7y = 50,
y = (50 – 5x)/7

x	3	-4	10
y = (50 – 5x)/7	5	10	0

For 7x + 5y = 46,
y = (46 – 7x)/5

x	8	3	–2
y = (46 – 7x)/5	–2	5	12

The graphs of equations are drawn below which shows that the two lines intersect at (3, 5).
Hence the cost of 1 pencil and the cost of 1 pen are ₹ 3 and ₹ 5 respectively.

Q.5 On comparing the ratios a1a2,b1b2, c1c2, 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

Ans.

$\begin{array}{l} (i) 5x-4y+8 = 0 \\ 7x+6y-9 = 0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} = 0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {=5, b}_{1} {=-4, c}_{1} =8 \\ a_{2} {=7, b}_{2} {=6, c}_{2} =-9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{5}{7}, \frac{b_{1}}{b_{2}} = \frac{-4}{6} = \frac{-2}{3} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pairs of linear equations intersect at a point. \end{array}$ $\begin{array}{l} (ii) 9x+3y+12=0 \\ 18x+6y+24=0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} = 0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {=9, b}_{1} {=3, c}_{1} =12 \\ a_{2} {=18, b}_{2} {=6, c}_{2} =24 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given pairs of linear equations are coincident. \end{array}$ $\begin{array}{l} (iii) 6x-3y+10 = 0 \\ 2x-y+9 = 0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} =0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {= 6, b}_{1} {= -3, c}_{1} = 10 \\ a_{2} {= 2, b}_{2} {= -1, c}_{2} = 9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{6}{2} = 3, \frac{b_{1}}{b_{2}} = \frac{-3}{-1} = 3, \frac{c_{1}}{c_{2}} = \frac{10}{9} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given pairs of linear equations are parallel. \end{array}$

Q.6

\begin{array}{l} On comparing the ratios \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}, \frac{c_{1}}{c_{2}}, find out whether \\ the following pair of linear equations are consistent, \\ or inconsistent. \\ (i) 3 x + 2 y = 5; 2 x - 3 y = 7 \\ (ii) 2 x - 3 y = 8; 4 x - 6 y = 9 \\ (iii) \frac{3}{2} x + \frac{5}{3} y = 7; 9 x - 10 y = 14 \\ (iv) 5 x - 3 y = 11; - 10 x + 6 y = - 22 \\ (v) \frac{4}{3} x + 2 y = 8; 2 x + 3 y = 12 \end{array}

Ans.

$\begin{array}{l} (i) 3 x + 2 y = 5; 2 x - 3 y = 7 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 3, b_{1} = 2, c_{1} = - 5 \\ a_{2} = 2, b_{2} = - 3, c_{2} = - 7 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{3}{2}, \frac{b_{1}}{b_{2}} = - \frac{2}{3}, \frac{c_{1}}{c_{2}} = \frac{5}{7} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pair of linear equations has a unique \\ solution and hence the two linear equations are consistent. \end{array}$ $\begin{array}{l} (ii) 2 x - 3 y = 8; 4 x - 6 y = 9 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = - 3, c_{1} = - 8 \\ a_{2} = 4, b_{2} = - 6, c_{2} = - 9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{8}{9} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel and \\ hence the two linear equations are inconsistent . \end{array}$ $\begin{array}{l} (iii) \frac{3}{2} x + \frac{5}{3} y = 7; 9 x - 10 y = 14 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = \frac{3}{2}, b_{1} = \frac{5}{3}, c_{1} = - 7 \\ a_{2} = 9, b_{2} = - 10, c_{2} = - 14 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \frac{b_{1}}{b_{2}} = \frac{\frac{5}{3}}{- 10} = - \frac{1}{6}, \frac{c_{1}}{c_{2}} = \frac{- 7}{- 14} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pair of linear equations has a unique \\ solution and hence the two linear equations are consistent. \end{array}$ $\begin{array}{l} (iv) 5 x - 3 y = 11; - 10 x + 6 y = - 22 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 5, b_{1} = - 3, c_{1} = - 11 \\ a_{2} = - 10, b_{2} = 6, c_{2} = 22 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{5}{- 10} = - \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{6} = - \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 11}{22} = - \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they consistent. \end{array}$ $\begin{array}{l} (v) \frac{4}{3} x + 2 y = 8; 2 x + 3 y = 12 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = \frac{4}{3}, b_{1} = 2, c_{1} = - 8 \\ a_{2} = 2, b_{2} = 3, c_{2} = - 12 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{2}{3}, \frac{c_{1}}{c_{2}} = \frac{- 8}{- 12} = \frac{2}{3} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they are consistent. \end{array}$

Q.7

\begin{array}{l} 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 \end{array}

Ans.

$\begin{array}{l} (i) x + y = 5, 2 x + 2 y = 10 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 1, b_{1} = 1, c_{1} = - 5 \\ a_{2} = 2, b_{2} = 2, c_{2} = - 10 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 5}{- 10} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they are consistent. \\ Graphs of the two equations coincide and hence each and \\ every point on this graph is a solution of these equations. \end{array}$

$\begin{array}{l} (ii) x - y = 8, 3 x - 3 y = 16 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 1, b_{1} = - 1, c_{1} = - 8 \\ a_{2} = 3, b_{2} = - 3, c_{2} = - 16 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{3}, \frac{c_{1}}{c_{2}} = \frac{- 8}{- 16} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel \\ and hence they are inconsistent. \end{array}$ $\begin{array}{l} (iii) 2 x + y - 6 = 0, 4 x - 2 y - 4 = 0 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = 1, c_{1} = - 6 \\ a_{2} = 4, b_{2} = - 2, c_{2} = - 4 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = - \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 6}{- 4} = \frac{3}{2} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given linear equations are consistent. \\ Now, \\ 2 x + y - 6 = 0 \Rightarrow y = 6 - 2 x \\ Some points which satisfy this equation are written \\ in the following table. \\ \begin{matrix} x & 0 & 2 & 3 \\ y = 6 - 2 x & 6 & 2 & 0 \end{matrix} \\ Again, \\ 4 x - 2 y - 4 = 0 \Rightarrow y = 2 x - 2 \\ Some points which satisfy this equation are written \\ in the following table. \\ \begin{matrix} x & 0 & 2 & 3 \\ y = 2 x - 2 & - 2 & 2 & 4 \end{matrix} \\ We get the following graphs of the given equations and \\ find that they intersect at (2, 2). Hence, x = 2 and y = 2. \end{array}$

$\begin{array}{l} (iv) 2 x - 2 y - 2 = 0, 4 x - 4 y - 5 = 0 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = - 2, c_{1} = - 2 \\ a_{2} = 4, b_{2} = - 4, c_{2} = - 5 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{1}{3} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 2}{- 5} = \frac{2}{5} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel \\ and hence they are inconsistent. \end{array}$

Q.8 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.

Ans. Let the width of the garden be x and length be y.
According to the question, length is 4 m more than its width. Therefore,
y − x = 4 …(1)
Also, half the perimeter of a rectangular garden is 36 m.
Therefore,
x + y = 36 …(2)
Now,
y − x = 4 …(1)
or y = x + 4

x	–4	0
y = x + 4	0	4

Similarly,
x + y = 36 …(2)
or y = 36 – x

x	28	20
y = 36 – x	8	16

Graphs of the equations (1) and (2) are drawn below and from there we observe that they intersect at (16, 20).
Therefore, x = 16 and y = 20. Hence, width of the garden is 16 m and length of the garden is 20 m.

Q.9 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

Ans.

$\begin{array}{l} (i) Intersecting lines: \\ A line which will intersect the given line 2 x + 3 y - 8 = 0 \\ is 3 x + 2 y - 8 = 0 as \\ \frac{a_{1}}{a_{2}} = \frac{2}{3} and \frac{b_{1}}{b_{2}} = \frac{3}{2} impllies that \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} . \\ (ii) P lines: \\ A line which is paralel to the given line 2 x + 3 y - 8 = 0 \\ is 4 x + 6 y - 7 = 0 as \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2} and \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} impllies that \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} . \\ (iii) : \\ A line which is to the given line 2 x + 3 y - 8 = 0 \\ is 4 x + 6 y - 16 = 0 because \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{1}{2} . \end{array}$

Q.10 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.

Ans. We have,
x – y + 1 = 0
or y = x + 1
We have the following table.

x	0	–1
y = x + 1	1	0

Again,
3x + 2y – 12 = 0
or y = (12–3x)/2
and so we have the following table.

x	0	4
y = (12–3x)/2	6	0

Graphs of the given equations are drawn below and from there we find that coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0); (4, 0) and (2, 3).

Q.11

\begin{array}{l} Solve the following pair of linear equations by the \\ substitution method. \\ (i) x + y = 14 (ii) s - t = 3 \\ x - y = 4 \frac{s}{3} + \frac{t}{2} = 6 \\ (iii) 3 x - y = 3 (iv) 0 .2 x + 0 .3 y = 1 .3 \\ 9 x - 3 y = 9 0 .4 x + 0 .5 y = 2 .3 \\ (v) \sqrt{2} x + \sqrt{3} y = 0 (vi) \frac{3 x}{2} - \frac{5 y}{3} = - 2 \\ \sqrt{3} x - \sqrt{8} y = 0 \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \end{array}

Ans.

$\begin{array}{l} (i) The given pair of linear equations are: \\ x + y = 14 . .. (1) \\ x - y = 4 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = 14 - y \\ We substitute this value of x in equation (2) to get \\ 14 - y - y = 4 \\ i.e., - 2 y = 4 - 14 = - 10 \\ i.e., y = 5 \\ Putting this value of y in equation (2), we get \\ x - 5 = 4 \\ i.e., x = 9 \\ (ii) The given pair of linear equations are: \\ s - t = 3 . .. (1) \\ \frac{s}{3} + \frac{t}{2} = 6 . .. (2) \\ We express s in terms of t from equation (1) to get \\ s = t + 3 \\ We substitute this value of s in equation (2) to get \\ \frac{t + 3}{3} + \frac{t}{2} = 6 \\ i.e., \frac{5 t + 6}{6} = 6 \\ i.e., 5 t + 6 = 36 \\ i.e., t = 6 \\ Putting this value of t in equation (1), we get \\ s - 6 = 3 \\ i.e., s = 9 \end{array}$ $\begin{array}{l} (iii) The given pair of linear equations are: \\ 3 x - y = 3 . .. (1) \\ 9 x - 3 y = 9 . .. (2) \\ We express y in terms of x from equation (1) to get \\ y = 3 x - 3 \\ We substitute this value of y in equation (2) to get \\ 9 x - 9 x + 9 = 9 \\ i.e., 9 = 9 \\ This statement is true for all values of x and so we can not \\ obtain a specific value of x. We observe that both the given \\ equations are the same as one is derived from another. \\ Therefore, given equations have infinitely many solutions. \\ (iv) The given pair of linear equations are: \\ 0 .2 x + 0 .3 y = 1 .3 . .. (1) \\ 0 .4 x + 0 .5 y = 2 .3 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = (1 .3 - 0 .3 y) / (0 .2) = (13 - 3 y) / 2 \\ We substitute this value of x in equation (2) to get \\ 0.2 (13 - 3 y) + 0 .5 y = 2 .3 \\ i.e., 2 .6 - 0 .6 y + 0 .5 y = 2 .3 \\ i.e., - 0.1 y = 2 .3 - 2 .6 = - 0 .3 \\ i.e., y = 3 \\ Putting this value of y in equation (2), we get \\ 0 .4 x + 0 .5 \times 3 = 2 .3 \\ i.e., x = (2 .3 - 1 .5) / 0 .4 = 8 / 4 = 2 \end{array}$ $\begin{array}{l} (v) The given pair of linear equations are : \\ \sqrt{2} x + \sqrt{3} y = 0 . .. (1) \\ \sqrt{3} x - \sqrt{8} y = 0 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = \frac{\sqrt{3}}{\sqrt{2}} y \\ We substitute this value of x in equation (2) to get \\ \sqrt{3} \frac{\sqrt{3}}{\sqrt{2}} y - \sqrt{8} y = 0 \\ i . e ., \frac{3 - 4}{\sqrt{2}} y = 0 \\ i . e ., y = 0 \\ Putting this value of y in equation (2), we get \\ \sqrt{3} x = 0 \\ i . e ., x = 0 \\ (vi) The given pair of linear equations are : \\ \frac{3 x}{2} - \frac{5 y}{3} = - 2 \\ or 9 x - 10 y = - 12 . .. (1) \\ and \\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \\ or 2 x + 3 y = 13 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = (- 12 + 10 y) / 9 \\ We substitute this value of x in equation (2) to get \\ 2(\frac{- 12 + 10 y}{9})+ 3 y = 13 \\ i . e ., - 24 + 20 y + 27 y = 117 \\ i . e ., 47 y = 117 + 24 = 141 \\ i . e ., y = \frac{141}{47} = 3 \\ Putting this value of y in equation (2), we get \\ 2 x + 9 = 13 \\ i . e ., x = \frac{4}{2} = 2 \end{array}$

Q.12 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Ans.

\begin{array}{l} The given linear equations are: \\ 2 x + 3 y = 11 . .. (1) \\ 2 x - 4 y = - 24 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = \frac{11 - 3 y}{2} \\ We substitute this value of x in equation (2) to get \\ 11 - 3 y - 4 y = - 24 \\ i.e., 11 - 7 y = - 24 \\ i.e., 7 y = 24 + 11 = 35 \\ i.e., y = 5 \\ Putting this value of y in equation (1), we get \\ 2 x + 3 \times 5 = 11 \\ i.e., 2 x = 11 - 15 = - 4 \\ i.e., x = - 2 \\ Now, \\ y = mx + 3 \\ or 5 = - 2 m + 3 \\ or m = - 1 \end{array}

Q.13

\begin{array}{l} Form the pair of linear equations for the following \\ problems and find their solution by substitution method. \\ (i) The difference betrween 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 a 7 bats and 6 balls \\ for ₹ 3800. Later she buys the 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 \frac{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 \frac{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? \end{array}

Ans.

(i)
Let the larger number is y and the smaller number is x.
According to question,
y = 3x …(1)
and
y – x = 26 …(2)
We substitute the value of y from equation (1) in equation (2) and get
3x – x = 26
or 2x = 26
or x = 13
Putting this value of x in equation (1), we get
y = 39
Hence, the numbers are 13 and 39.

(ii)
Let the larger angle is y and the smaller angle is x.
According to question,
y – x = 18° …(1)
and
y + x = 180° …(2)
We write y in terms of x from (1) to get
y = 18° + x
Putting this value of y in equation (2), we get
18° + x + x = 180°
or 2x = 180° – 18° = 162°
or x = 81°
Putting this value of x in equation (1), we get
y = 18° + 81° = 99°
Hence, the two supplementary angles are 81° and 99°.

(iii)
Let the cost of one bat and one ball be ₹ x and ₹ y respectively.
According to question,
7x + 6y = ₹ 3800 …(1)
and
3x + 5y = ₹ 1750 …(2)
We write y in terms of x from (1) to get
y= (3800 – 7x)/6
Putting this value of y in equation (2), we get
3x + 5(3800 – 7x)/6 = 1750
or 18x – 35x = 1750 × 6 – 19000 = –8500
or x = 8500/17 = 500
Putting this value of x in equation (1), we get
7×500 + 6y = 3800
or y = (3800 – 3500)/6 = 50
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

(iv)
Let the fixed charge be ₹ x and charge per km be ₹ y.
According to question,
x + 10y = ₹ 105 …(1)
and
x + 15y = ₹ 155 …(2)
We write x in terms of y from (1) to get
x = 105 – 10y
Putting this value of x in equation (2), we get
105 – 10y + 15y = 155
or 5y = 155 – 105 = 50
or y = 10
Putting this value of y in equation (1), we get
x + 10×10 = 105
or x = 105 – 100 = 5
Hence, the fixed charge is ₹ 5 and charge per km is ₹ 10.
Charge for 25 km = 5 + 25 ×10 = ₹ 255

$\begin{array}{l} (v) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x + 2}{y + 2} = \frac{9}{11} \Rightarrow 11 x - 9 y = - 4 . .. (1) \\ and \\ \frac{x + 3}{y + 3} = \frac{5}{6} \Rightarrow 6 x - 5 y = - 3 . .. (2) \\ From equation (2), we get \\ x = \frac{5 y - 3}{6} \\ We put this value of x in equation (1) and get \\ 11 \frac{5 y - 3}{6} - 9 y = - 4 \\ \Rightarrow 55 y - 33 - 54 y = - 24 \\ \Rightarrow y = - 24 + 33 = 9 \\ Putting this value of y in equation (1), we get \\ 11 x - 9 \times 9 = - 4 \\ \Rightarrow 11 x = 81 - 4 = 77 \\ \Rightarrow x = 7 \\ Hence the fraction is \frac{7}{9} . \end{array}$

(vi)
Let the age of Jacob be x and the age of his son be y.
According to question,
x + 5 = 3(y + 5)
or x – 3y = 10 …(1)
Also,
x – 5 = 7(y – 5)
or x – 7y = –30 …(2)
From equation (1), we find
x = 10 + 3y
We substitute this value of x in equation (2) and get
10 + 3y – 7y = –30
or –4y = –40
or y = 10
Putting this value of y in (1), we get
x – 3×10 = 10
or x = 40
Hence the present age of Jacob is 40 years and that of his son is 10 years.

Q.14

$\begin{array}{l} Solve the following pair of linear equations by the \\ elimination method and the substitution method: \\ (i) x + y = 5 and 2x – 3y = 4 \\ (ii) 3x + 4y = 10 and 2x – 2y = 2 \\ (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 \\ (iv) \frac{x}{2} + \frac{2y}{3} = –1 and x – \frac{y}{3} = 3 \end{array}$

Ans.

$\begin{array}{l} (i) \\ By elimination method : \\ Given that \\ x + y = 5 . .. (1) \\ 2 x - 3 y = 4 . .. (2) \\ We multiply equation (1) by 3 and then add it to \\ equation (2) to eliminate the variable y and get \\ the value of x as follows. \\ 3 x + 3 y = 15 \\ \underline{2 x - 3 y = 4} \\ 5 x = 19 \\ or x = \frac{19}{5} \\ Using this value of x in equation (1), we get \\ \frac{19}{5} + y = 5 \\ or y = 5 - \frac{19}{5} = \frac{6}{5} \\ By substitution method : \\ From equation (1), we get \\ y = 5 - x \\ Substituting this in equation (2), we get \\ 2 x - 3 (5 - x) = 4 \\ or 5 x = 4 + 15 = 19 \\ or x = \frac{19}{5} \\ Substituting this value of x in equation (1), we get \\ \frac{19}{5} + y = 5 \\ or y = 5 - \frac{19}{5} = \frac{6}{5} \end{array}$ $\begin{array}{l} (ii) \\ By elimination method : \\ Given that \\ 3 x + 4 y = 10 . .. (1) \\ 2 x - 2 y = 2 . .. (2) \\ We multiply equation (2) by 2 and then add it to \\ equation (1) to eliminate the variable y and get \\ the value of x as follows. \\ 3 x + 4 y = 10 \\ \underline{4 x - 4 y = 4} \\ 7 x = 14 \\ or x = 2 \\ Using this value of x in equation (1), we get \\ 6 + 4 y = 10 \\ or y = \frac{10 - 6}{4} = 1 \\ By substitution method : \\ From equation (2), we get \\ x = 1 + y \\ Substituting this in equation (1), we get \\ 3 (1 + y) + 4 y = 10 \\ or 7 y = 10 - 3 = 7 \\ or y = 1 \\ Substituting this value of y in equation (2), we get \\ 2 x - 2 = 2 \\ or x = \frac{2 + 2}{2} = 2 \end{array}$ $\begin{array}{l} (iii) \\ By elimination method : \\ Given that \\ 3 x - 5 y - 4 = 0 . .. (1) \\ 9 x = 2 y + 7 . .. (2) \\ We multiply equation (1) by 3 and then subtract \\ equation (2) from it to eliminate the variable x and get \\ the value of y as follows. \\ 9 x - 15 y - 12 = 0 \\ \underline{\begin{array}{l} 9 x = 2 y + 7 \\ - - - \end{array}} \\ - 15 y - 12 = - 2 y - 7 \\ or - 15 y + 2 y = - 7 + 12 = 5 \\ or y = - \frac{5}{13} \\ Using this value of y in equation (2), we get \\ 9 x = - \frac{5}{13} \times 2 + 7 \\ or x = \frac{- 10}{117} + \frac{7}{9} = \frac{- 10 + 91}{117} = \frac{81}{117} = \frac{9}{13} \\ By substitution method : \\ From equation (2), we get \\ x = \frac{2 y + 7}{9} \\ Substituting this in equation (1), we get \\ 3 \times \frac{2 y + 7}{9} - 5 y - 4 = 0 \\ or \frac{2 y + 7}{3} - 5 y = 4 \\ or 2 y + 7 - 15 y = 12 \\ or - 13 y = 12 - 7 \\ or y = - \frac{5}{13} \\ Substituting this value of y in equation (2), we get \\ 9 x = - 2 \times \frac{5}{13} + 7 = \frac{- 10 + 91}{13} \\ or x = \frac{81}{13 \times 9} = \frac{9}{13} \end{array}$ $\begin{array}{l} (iv) \\ By elimination method : \\ Given that \\ \frac{x}{2} + \frac{2 y}{3} = - 1 . .. (1) \\ x - \frac{y}{3} = 3 . .. (2) \\ We multiply equation (2) by 2 and then add it to \\ equation (1) to eliminate the variable y and get \\ the value of x as follows. \\ \frac{x}{2} + \frac{2 y}{3} = - 1 \\ \underline{2 x - \frac{2 y}{3} = 6} \\ \frac{x}{2} + 2 x = 5 \\ or \frac{5 x}{2} = 5 \\ or x = 2 \\ Using this value of x in equation (2), we get \\ 2 - \frac{y}{3} = 3 \\ or - y = 3 (3 - 2) = 3 \\ or y = - 3 \\ By substitution method : \\ From equation (2), we get \\ x = 3 + \frac{y}{3} \\ Substituting this in equation (1), we get \\ \frac{1}{2} (3 + \frac{y}{3}) + \frac{2 y}{3} = - 1 \\ or \frac{9 + y}{6} + \frac{2 y}{3} = - 1 \\ or \frac{9 + y + 4 y}{6} = - 1 \\ or 9 + y + 4 y = - 6 \\ or y = \frac{- 15}{5} = - 3 \\ Substituting this value of y in equation (2), we get \\ x = 3 + \frac{- 3}{3} = 2 \end{array}$

Q.15 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 ½ 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 fixed charge for the first 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 five days. Find the fixed charge and the charge for each extra day.

Ans.

$\begin{array}{l} (i) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x + 1}{y - 1} = 1 \\ \Rightarrow x + 1 = y - 1 \\ \Rightarrow x - y = - 2 . .. (1) \\ Also, \\ \frac{x}{y + 1} = \frac{1}{2} \\ \Rightarrow x = \frac{y + 1}{2} \\ \Rightarrow 2 x = y + 1 \\ \Rightarrow 2 x - y = 1 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 3 \\ Substituting this value in equation (1), we get \\ y = 5 \\ Hence, the fraction is \frac{3}{5} . \end{array}$ $\begin{array}{l} (ii) \\ Let the present age of Nuri and Sonu be x years \\ and y years respectively . \\ So, five years ago, \\ Nuri ‘ s age = (x - 5) years \\ and \\ Sonu’s age = (y - 5) years \\ According to question, \\ x - 5 = 3 (y - 5) \\ \Rightarrow x - 5 = 3 y - 15 \\ \Rightarrow x - 3 y = - 10 . .. (1) \\ Again, \\ Ten years hence, \\ Nuri ‘ s age = (x + 10) years \\ and \\ Sonu’s age = (y + 10) years \\ According to the question, \\ x + 10 = 2 (y + 10) \\ or x - 2 y = 10 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ y = 20 \\ Putting this value in (2), we get \\ x = 50 \\ Hence, Nuri and Sonu are 50 years \\ and 20 years old respectively . \end{array}$ $\begin{array}{l} (iii) \\ Let the two digit number be 10 x + y . \\ According to question, \\ x + y = 9 . .. (1) \\ and \\ 9 (10 x + y) = 2 (10 y + x) \\ \Rightarrow 90x + 9 y = 20 y + 2 x \\ \Rightarrow 90 x - 2 x + 9 y - 20 y = 0 \\ \Rightarrow 88 x - 11 y = 0 \\ \Rightarrow 8 x - y = 0 . .. (2) \\ Adding equations (1) and (2), we get \\ x = 1 \\ Putting this value of x in equation (2), we get \\ y = 8 \\ Hence, \\ Number = 10 x + y = 10 + 8 = 18 \end{array}$ $\begin{array}{l} (iv) \\ Let number of notes of ₹ 50 and ₹ 100 bexandyrespectively . \\ According to question, \\ x + y = 25 . .. (1) \\ and \\ 50 x + 100 y = 2000 \\ \Rightarrow 50 (x + 2 y) = 2000 \\ \Rightarrow x + 2 y = 40 . .. (2) \\ Subtracting equation (1) from equation(2), we get \\ x + 2 y = 40 \\ \underline{\begin{array}{l} x + y = 25 \\ - - - \end{array}} \\ y = 15 \\ Putting this value in equation (1), we get \\ x = 10 \\ Hence, \\ Meena received 10 notes of ‘ 50 and 15 notes of ‘ 100 . \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (v) \\ Let the fixed charge for first three day be ₹ x and additional \\ charge for each day thereafter be ₹ y . \\ According to question, \\ x + 4 y = 27 . .. (1) \\ and \\ x + 2 y = 21 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x + 4 y = 27 \\ \underline{\begin{array}{l} x + 2 y = 21 \\ - - - \end{array}} \\ 2 y = 6 \\ or y = 3 \\ Putting this value in equation (1), we get \\ x = 27 - 12 = 15 \\ Hence, fixed charge is ₹ 15 and additional charge for \\ each extra day is ₹ 3 . \end{array}$

Q.16

\begin{array}{l} 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                         (ii) 2x + y = 5 \\ 3x - 9y - 2 = 0                           3x + 2y = 8 \\ (iii) 3x - 5y = 20                         (iv)  x - 3y - 7 = 0 \\ 6x - 10y = 40                             3x - 3y - 15 = 0 \end{array}

Ans.

$\begin{array}{l} (i) x - 3 y - 3 = 0 \\ 3 x - 9 y - 2 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 9} = \frac{1}{3}, \frac{c_{1}}{c_{2}} = \frac{- 3}{- 2} = \frac{3}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have no solution. \\ (ii) 2 x + y = 5 i.e., 2 x + y - 5 = 0 \\ 3 x + 2 y = 8 i.e., 3 x + 2 y - 8 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{5}{8} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 8 - (- 10)} = \frac{y}{- 15 - (- 16)} = \frac{1}{4 - 3} \\ \Rightarrow \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \\ \Rightarrow x = 2 and y = 1 \end{array}$ $\begin{array}{l} (iii) 3 x - 5 y = 20 i.e., 3 x - 5 y - 20 = 0 \\ 6 x - 10 y = 40 i.e., 6 x - 10 y - 40 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 5}{- 10} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 20}{- 40} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have infinitely \\ many solutions. \\ (iv) x - 3 y - 7 = 0 \\ 3 x - 3 y - 15 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 3} = 1, \frac{c_{1}}{c_{2}} = \frac{- 7}{- 15} = \frac{7}{15} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{45 - 21} = \frac{y}{- 21 - (- 15)} = \frac{1}{- 3 - (- 9)} \\ \Rightarrow \frac{x}{24} = \frac{y}{- 6} = \frac{1}{6} \\ \Rightarrow x = 4 and y = - 1 \end{array}$

Q.17

\begin{array}{l} (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 \end{array}

Ans.

\begin{array}{l} (i) \\ Given pair of linear equations are: \\ 2 x + 3 y = 7 \\ (a - b) x + (a + b) y = 3 a + b - 2 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{a - b}, \frac{b_{1}}{b_{2}} = \frac{3}{a + b}, \frac{c_{1}}{c_{2}} = \frac{7}{3 a + b - 2} \\ For a pair of linear equations to have infinitely many solutions: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ So, we need \\ \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3 a + b - 2} \\ or \frac{2}{a - b} = \frac{7}{3 a + b - 2} and \frac{2}{a - b} = \frac{3}{a + b} \\ or 2 (3 a + b - 2) = 7 (a - b) and 2 (a + b) = 3 (a - b) \\ or 6 a + 2 b - 4 = 7 a - 7 b and 2 a - 3 a = - 3 b - 2 b \\ or - a + 9 b - 4 = 0 and a = 5 b \\ Now, we put a = 5 b in - a + 9 b - 4 = 0 . \\ Thus, \\ - 5 b + 9 b = 4 \\ or 4 b = 4 \\ or             b = 1 \\ Also, \\ a = 5 b = 5 \times 1 = 5 \end{array}

\begin{array}{l} (ii) \\ Given pair of linear equations are: \\ 3 x + y = 1 \\ (2 k - 1) x + (k - 1) y = 2 k + 1 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{2 k - 1}, \frac{b_{1}}{b_{2}} = \frac{1}{k - 1}, \frac{c_{1}}{c_{2}} = \frac{1}{2 k + 1} \\ For a pair of linear equations to have no solution: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ So, the given pair of linear equations have no solution when \\ \frac{3}{2 k - 1} = \frac{1}{k - 1} \\ or 3 k - 3 = 2 k - 1 \\ or 3 k - 2 k = - 1 + 3 \\ or k = 2 \end{array}

Q.18

\begin{array}{l} Solve the following pair of linear equations by the \\ substitution and cross-multiplication methods: \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \end{array}

Ans.

$\begin{array}{l} 8 x + 5 y = 9 . .. (1) \\ 3 x + 2 y = 4 . .. (2) \\ By substitution mwthod: \\ From equation (2), we get \\ y = \frac{4 - 3 x}{2} \\ Substituting this value of y in equation (1), we get \\ 8 x + 5 y = 9 \\ or 8 x + 5 \times \frac{4 - 3 x}{2} = 9 \\ or 16 x + 20 - 15 x = 18 \\ or x = 18 - 20 = - 2 \\ Putting this value of x in equation (2), we get \\ 3 \times (- 2) + 2 y = 4 \\ or - 6 + 2 y = 4 \\ or 2y = 4 + 6 = 10 \\ or y = 5 \end{array}$ $\begin{array}{l} By cross-multiplication method : \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \\ Or \\ 8 x + 5 y - 9 = 0 . .. (1) \\ 3 x + 2 y - 4 = 0 . .. (2) \\ {Here, a}_{1} = 8, b_{1} {= 5, c}_{1} = - 9 \\ a_{2} = 3, b_{2} {= 2, c}_{2} = - 4 \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 20 + 18} = \frac{y}{- 27 + 32} = \frac{1}{16 - 15} \\ \Rightarrow \frac{x}{- 2} = \frac{y}{5} = \frac{1}{1} \\ \Rightarrow x = - 2 and y = 5 \end{array}$

Q.19

\begin{array}{l} 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 ₹ 1000 \\ as hostel charges whereas a student B, who \\ takes food for 26 days, pays ₹ 1180 as hostel \\ charges. Find the fixed charges and the cost of \\ food per day. \\ (ii) A fraction becomes \frac{1}{3} when 1 is subtracted from \\ the numerator and it becomes \frac{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? \end{array}

$\begin{array}{l} (iv) Places A and B are 100 km apart on a highway. One \\ car starts from A and another from B at the same \\ time. If the cars travel in the same direction at \\ different speeds, they meet in 5 hours. If they \\ travel towards each other, they meet in 1 hour. \\ What are the speeds of the two cars? \\ (v) The area of a rectangle gets reduced by 9 square \\ units, if its length is reduced by 5 units and \\ breadth is increased by 3 units. If we increase \\ the length by 3 units and the breadth by 2 units, \\ the area increases by 67 square units . Find the \\ dimensions of the rectangle. \end{array}$

Ans.

$\begin{array}{l} (i) \\ Let the fixed charge be ₹ x and the cost of food per day be ₹ y . \\ According to question, \\ x + 20 y = 1000 . .. (1) \\ and \\ x + 26 y = 1180 . .. (2) \\ Subtracting equation (1) from equation (2), we get \\ \underline{\begin{array}{l} x + 26 y = 1180 \\ x + 20 y = 1000 \\ - - - \end{array}} \\ 6 y = 180 \\ or y = 30 \\ Putting this value of y in equation (1), we get \\ x + 20 \times 30 = 1000 \\ or x = 1000 - 600 = 400 \\ Hence, fixed charge is ₹ 400 and the cost of food per \\ day is ₹ 30 . \end{array}$ $\begin{array}{l} (ii) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x - 1}{y} = \frac{1}{3} \\ \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 x - y = 3 . .. (1) \\ Also, \\ \frac{x}{y + 8} = \frac{1}{4} \\ \Rightarrow x = \frac{y + 8}{4} \\ \Rightarrow 4 x = y + 8 \\ \Rightarrow 4 x - y = 8 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 5 \\ Substituting this value in equation (1), we get \\ 3 \times 5 - y = 3 \\ or - y = 3 - 15 \\ or y = 12 \\ Therefore, \\ \frac{x}{y} = \frac{5}{12} \\ Hence, the fraction is \frac{5}{12} . \end{array}$ $\begin{array}{l} (iii) \\ Let the number of right and the number of wrong \\ answers be x and y respectively. \\ According to questions, \\ 3 x - y = 40 . .. (1) \\ and \\ 4 x - 2 y = 50 \\ \Rightarrow 2 (2 x - y) = 2 \times 25 \\ \Rightarrow 2 x - y = 25 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 15 \\ Putting this value of x in equation (2), we get \\ 2 \times 15 - y = 25 \\ \Rightarrow - y = 25 - 30 = - 5 \\ \Rightarrow y = 5 \\ Thus, \\ Number of right answers = x = 15 \\ Number of wrong answers = y = 5 \\ and \\ Number of questions = x + y = 15 + 5 = 20 \end{array}$ $\begin{array}{l} (iv) \\ {Let the speeds of I}^{st} {and 2}^{nd} cars be x km/h and y km/h \\ respectively. \\ According to question, \\ x + y = 100 . .. (1) \\ and \\ 5 x - 100 = 5 y \\ \Rightarrow 5 (x - 20) = 5 y \\ \Rightarrow x - 20 = y \\ \Rightarrow x - y = 20 . .. (2) \\ Adding equations (1) and (2), we get \\ 2 x = 120 \Rightarrow x = 60 \\ Putting this value of x in equation (1), we get \\ 60 + y = 100 \\ \Rightarrow y = 100 - 60 = 40 \\ Thus, \\ {Speed of I}^{st} car = x km/h = 60 km/h \\ {Speed of 2}^{nd} car = y km/h = 40 km/h \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (v) \\ Let length and breadth of the rectangle be x units and y units \\ respectively. \\ According to question, \\ (x - 5) (y + 3) = xy - 9 \\ \Rightarrow xy - 5 y + 3 x - 15 = xy - 9 \\ \Rightarrow - 5 y + 3 x = 15 - 9 = 6 \\ \Rightarrow 3 x - 5 y - 6 = 0 . .. (1) \\ Also, \\ (x + 3) (y + 2) = xy + 67 \\ \Rightarrow xy + 3 y + 2 x + 6 = xy + 67 \\ \Rightarrow 3 y + 2 x = 67 - 6 = 61 \\ \Rightarrow 2 x + 3 y - 61 = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{305 - (- 18)} = \frac{y}{- 12 - (- 183)} = \frac{1}{9 - (- 10)} \\ \Rightarrow x = \frac{323}{19} = 17 and y = \frac{171}{19} = 9 \\ Hence, length and breadth of the rectangle are 17 units and \\ 9 units respectively. \end{array}$

Q.20

$\begin{array}{l} Solve the following pair of equations by reducing them to a pair of linear equations: \\ (i) \frac{1}{2x} + \frac{1}{3y} =2 (ii) \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} =–1 \\ (iii) \frac{4}{x} + 3y = 14 (iv) \frac{5}{x – 1} + \frac{1}{y – 2} = 2 \\ \frac{3}{x} - 4y = 23 \frac{6}{x – 1} - \frac{3}{y – 2} = 1 \\ (v) \frac{7x – 2y}{xy} = 5 (vi) 6x + 3y = 6xy \\ \frac{8x + 7y}{xy} = 15 2x + 4y = 5xy \\ (vii) \frac{10}{x + y} + \frac{2}{x – y} = 4 (viii) \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \\ \frac{15}{x + y} - \frac{5}{x – y} = –2 \frac{1}{2(3x + y)} - \frac{1}{2(3x – y)} = \frac{–1}{8} \end{array}$

Ans.

$\begin{array}{l} (i) \\ Let \frac{1}{x} = p and \frac{1}{y} = q \\ then given equations can be written as: \\ \frac{p}{2} + \frac{q}{3} = 2 \Rightarrow 3 p + 2 q - 12 = 0 . .. (1) \\ \frac{p}{3} + \frac{q}{2} = \frac{13}{6} \Rightarrow 2 p + 3 q - 13 = 0 . .. (2) \\ Using cross-multiplication method, we get \\ \frac{p}{- 26 - (- 36)} = \frac{q}{- 24 - (- 39)} = \frac{1}{9 - 4} \\ \Rightarrow \frac{p}{10} = \frac{q}{15} = \frac{1}{5} \\ \Rightarrow \frac{p}{10} = \frac{1}{5} and \frac{q}{15} = \frac{1}{5} \\ \Rightarrow p = 2 and q=3 \\ \Rightarrow \frac{1}{x} = 2 and \frac{1}{y} = 3 \\ \Rightarrow x = \frac{1}{2} and y= \frac{1}{3} \\ (ii) \\ Given that \\ \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ and \\ \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1 \\ Let \frac{1}{\sqrt{x}} = p and \frac{1}{\sqrt{y}} = q, then we get \\ 2 p + 3 q = 2 . .. (1) \\ 4 p - 9 q = - 1 . .. (2) \\ Multiplying equation (1) by 3, we get \\ 6p + 9q = 6 \\ Adding this to equation (2), we get \\ 10p = 5 \\ \Rightarrow p= \frac{1}{2} \\ Putting value of p in equation (1), we get \\ 2 \times \frac{1}{2} + 3 q = 2 \\ \Rightarrow 3 q = 1 \\ \Rightarrow q = \frac{1}{3} \end{array}$ $\begin{array}{l} Now, \\ p = \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4 \\ q = \frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9 \\ Hence, x = 4 and y = 9 . \\ (iii) \\ Given that \\ \frac{4}{x} + 3 y = 14 \\ \frac{3}{x} - 4 y = 23 \\ Substituting \frac{1}{x} = p in above equations, we get \\ 4p + 3 y - 14 = 0 . .. (1) \\ 3 p - 4 y - 23 = 0 . .. (2) \\ By cross-multiplication method, we get \\ \frac{p}{- 69 - 56} = \frac{y}{- 42 - (- 92)} = \frac{1}{- 16 - 9} \\ \Rightarrow \frac{p}{- 125} = \frac{y}{50} = \frac{1}{- 25} \\ \Rightarrow p = 5 and y = - 2 \\ \Rightarrow p = \frac{1}{x} = 5 \Rightarrow x = \frac{1}{5} and y = - 2 \end{array}$

$\begin{array}{l} (iv) \\ Given that \\ \frac{5}{x - 1} + \frac{1}{y - 2} = 2 \\ and \\ \frac{6}{x - 1} - \frac{3}{y - 2} = 1 \\ Putting \frac{1}{x - 1} = p and \frac{1}{y - 2} = q, we get \\ 5p + q = 2 . .. (1) \\ 6p - 3 q = 1 . .. (2) \\ multiplying equation (1) by 3, we get \\ 15p + 3 q = 6 . .. (3) \\ Adding (2) and (3), we get \\ 21 p = 7 \Rightarrow p = \frac{1}{3} \\ Putting value of p in equation (1), we get \\ 5 \times \frac{1}{3} + q = 2 \Rightarrow q = 2 - \frac{5}{3} = \frac{1}{3} \\ Now, \\ p = \frac{1}{x - 1} = \frac{1}{3} \Rightarrow x - 1 = 3 \Rightarrow x = 4 \\ and q = \frac{1}{y - 2} = \frac{1}{3} \Rightarrow y - 2 = 3 \Rightarrow y = 5 \\ ∴ x = 4 and y = 5 . \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (v) \\ We have \\ \frac{7 x - 2 y}{xy} = 5 \Rightarrow \frac{7}{y} - \frac{2}{x} = 5 . .. (1) \\ \frac{8 x + 7 y}{xy} = 15 \Rightarrow \frac{8}{y} + \frac{7}{x} = 15 . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in equation (1) and (2), we get \\ - 2 p + 7 q - 5 = 0 . .. (3) \\ 7p + 8 q - 15 = 0 . .. (4) \\ By cross-multiplication method, we get \\ \frac{p}{- 105 - (- 40)} = \frac{q}{- 35 - 30} = \frac{1}{- 16 - 49} \\ \Rightarrow \frac{p}{- 65} = \frac{q}{- 65} = \frac{1}{- 65} \\ \Rightarrow \frac{p}{- 65} = \frac{1}{- 65} and \frac{q}{- 65} = \frac{1}{- 65} \\ \Rightarrow p = 1 and q = 1 \\ \Rightarrow p = \frac{1}{x} = 1 and q = \frac{1}{y} = 1 \\ \Rightarrow x = 1 and y = 1 \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (vi) \\ We have \\ 6 x + 3 y = 6 xy \Rightarrow \frac{6}{y} + \frac{3}{x} = 6 . .. (1) \\ and \\ 2 x + 4 y = 5 xy \Rightarrow \frac{2}{y} + \frac{4}{x} = 5 . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in above equations, we get \\ 3p + 6 q - 6 = 0 . .. (3) \\ 4p + 2 q - 5 = 0 . .. (4) \\ By cross-multiplication method, we have \\ \frac{p}{- 30 - (- 12)} = \frac{q}{- 24 - (- 15)} = \frac{1}{6 - 24} \\ \Rightarrow \frac{p}{- 18} = \frac{q}{- 9} = \frac{1}{- 18} \\ \Rightarrow \frac{p}{- 18} = \frac{1}{- 18} and \frac{q}{- 9} = \frac{1}{- 18} \\ \Rightarrow p = 1 and q = \frac{1}{2} \\ \Rightarrow p = \frac{1}{x} = 1 and q = \frac{1}{y} = \frac{1}{2} \\ \Rightarrow x = 1 and y = 2 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (vii) We have \\ \frac{10}{x + y} + \frac{2}{x - y} = 4 \\ and \frac{15}{x + y} - \frac{5}{x - y} = - 2 \\ Putting \frac{1}{x + y} = p and \frac{1}{x - y} = q in above equations, we get \\ 10p + 2 q - 4 = 0 . .. (3) \\ and \\ 15p - 5 q + 2 = 0 . .. (4) \\ By cross-multiplication method, we have \\ \frac{p}{4 - 20} = \frac{q}{- 60 - 20} = \frac{1}{- 50 - 30} \\ \Rightarrow \frac{p}{- 16} = \frac{q}{- 80} = \frac{1}{- 80} \\ \Rightarrow \frac{p}{- 16} = \frac{1}{- 80} and \frac{q}{- 80} = \frac{1}{- 80} \\ \Rightarrow p = \frac{1}{5} and q = 1 \\ \Rightarrow p = \frac{1}{x + y} = \frac{1}{5} and q = \frac{1}{x - y} = 1 \\ Thus we have \\ x + y = 5 . .. (5) \\ and x - y = 1 . .. (6) \\ Adding equations (5) and (6), we get \\ 2x = 6 \Rightarrow x = 3 \end{array}$ $\begin{array}{l} Subtracting equation (6) from equation (5), we get \\ y = 2 \\ Hence, x = 3 and y = 2 . \\ (viii) \\ We have \\ \frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4} \\ and \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x + y)} = \frac{- 1}{8} \\ Putting \frac{1}{3 x + y} = p and \frac{1}{3 x - y} = q in above equations, we get \\ p + q = \frac{3}{4} . .. (3) \\ and \frac{p}{2} - \frac{q}{2} = \frac{- 1}{8} \\ \Rightarrow p - q = \frac{- 1}{4} . .. (4) \\ Adding equations (3) and (4), we get \\ 2p = \frac{3}{4} - \frac{1}{4} \Rightarrow p = \frac{1}{4} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Now, \\ p = \frac{1}{3 x + y} = \frac{1}{4} \\ \Rightarrow 3 x + y = 4 . .. (5) \\ Also, \\ q = \frac{1}{3 x - y} = \frac{1}{2} \\ \Rightarrow 3 x - y = 2 . .. (6) \\ Adding equations (5) and (6), we get \\ 6x = 6 \Rightarrow x = 1 \\ Substitute x = 1 in equation (5), we get \\ 3 + y = 4 \Rightarrow y = 1 \\ Hence, x = 1 and y = 1 . \end{array}$

Q.21 Formulate the following problem as a pair of linear equations and hence find their solution:
(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 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.

(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.

Ans.

(i)
Let speed of Ritu in still water be x km/h and the speed of stream be y km/h.
Speed of Ritu while rowing upstream = (x– y) km/h
Speed of Ritu while rowing downstream = (x + y) km/h
According to question,
2(x + y) = 20
or x + y = 10 …(1)
Also,
2(x – y) = 4
or x – y = 2 …(2)
Adding equations (1) and (2), we get
2x = 12
or x = 6
Putting this in equation (1), we get
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of current is 4 km/h.

(ii)

$\begin{array}{l} Let the number of days taken by a woman to finish the work \\ is x and the number of days taken by a man to finish the \\ work is y . \\ Therefore, \\ Part of the work finished by a woman in 1 day = \frac{1}{x} \\ and \\ part of the work finished by a man in 1 day = \frac{1}{y} \\ Given that \\ the . \\ Therefore, \\ 4 (\frac{2}{x} + \frac{5}{y}) = 1 \\ \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \end{array}$ $\begin{array}{l} It is also given that \\ the work . \\ Therefore, \\ 3 (\frac{3}{x} + \frac{6}{y}) = 1 \\ \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3} \\ P utting \frac{1}{x} = p and \frac{1}{y} = q in the above equations, we get \\ 2p + 5q = \frac{1}{4} \Rightarrow 8 p + 20 q = 1 \\ 3p + 6q = \frac{1}{3} \Rightarrow 9 p + 18 q = 1 \\ By cross-multiplication method, we get \\ \frac{p}{- 20 - (- 18)} = \frac{q}{- 9 - (- 8)} = \frac{1}{144 - 180} \\ \Rightarrow \frac{p}{- 2} = \frac{q}{- 1} = \frac{1}{- 36} \\ \Rightarrow \frac{p}{- 2} = \frac{1}{- 36} and \frac{q}{- 1} = \frac{1}{- 36} \\ \Rightarrow p = \frac{1}{18} and q = \frac{1}{36} \\ \Rightarrow p = \frac{1}{x} = \frac{1}{18} and q = \frac{1}{y} = \frac{1}{36} \\ \Rightarrow x = 18 and y = 36 \\ Hence, number of days taken by a woman = 18 days \\ and number of days taken by a man = 36 days. \end{array}$ $\begin{array}{l} (iii) \\ Let the speeds of train and bus be x km/h and y km/h respectively. \\ According to question, \\ \frac{60}{x} + \frac{240}{y} = 4 . .. (1) \\ and \\ \frac{100}{x} + \frac{200}{y} = \frac{25}{6} . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in these equations, we get \\ 60 p + 240 q = 4 . .. (3) \\ and 100 p + 200 q = \frac{25}{6} \\ \Rightarrow 600 p + 1200 q = 25 . .. (4) \\ Multiplying equation (3) by 10, we get \\ 600 p + 2400 q = 40 . .. (5) \\ Subtracting equation (4) from (5), we get \\ 1200q = 15 \Rightarrow q = \frac{1}{80} \\ Substituting in equation (3), we get \\ 60p = 1 \Rightarrow p = \frac{1}{60} \\ Now, p = \frac{1}{x} = \frac{1}{60} and q = \frac{1}{y} = \frac{1}{80} \end{array}$

$\begin{array}{l} \Rightarrow x = 60 km/h and y = 80 km/h \\ Hence, speed of train = 60 km/h \\ and speed of bus = 80 km/h \end{array}$

Q.22

$\begin{array}{l} The ages of two friends Ani and Biju differ by 3 years . \\ Ani ‘ s father Dharam is twice as old as Ani and Biju is \\ twice as old as his sister Cathy . The ages of Cathy and \\ Dharam differ by 30 years . Find the ages of Ani and Biju . \end{array}$

Ans.

$\begin{array}{l} Let the ages of Ani and Biju be x and y years respectively . \\ If Ani is older than Biju then \\ x - y = 3 . .. (1) \\ Ani ‘ s father Dharam is twice as old as Ani . So, \\ Dharam ‘ s age = 2 x \\ The ages of Cathy and Dharam differ by 30 years . \\ Therefore, \\ Cathy ‘ s age = 2 x - 30 \\ Biju is twice as old as Cathy . Therefore, \\ y = 2 (2 x - 30) . .. (2) \\ Putting the value of y from equation (2) in equation (1), \\ we get \\ x - 2 (2 x - 30) = 3 \\ \Rightarrow x - 4 x + 60 = 3 \\ \Rightarrow - 3 x = 3 - 60 = - 57 \\ \Rightarrow x = 19 \\ From equation (1), we get \\ y = 19 - 3 = 16 \\ Hence, Ani is 19 years old and Biju is 16 years old . \\ Again, if Biju is older than Ani then \\ y - x = 3 . .. (3) \\ Again, putting the value of y from (2) in (3), we get \\ 2 (2 x - 30) - x = 3 \\ \Rightarrow 4 x - 60 - x = 3 \\ \Rightarrow 3 x = 3 + 60 = 63 \\ \Rightarrow x = 21 \\ Putting this value of x in equation (3), we get \\ y = 21 + 3 = 24 \\ Hence, in this case Ani is of 21 years and Biju is of 24 years . \end{array}$

Q.23

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

Ans.

$\begin{array}{l} Let their respective capitals are ₹ x and ₹ y . \\ According to question, \\ x + 100 = 2 (y - 100) \\ \Rightarrow x - 2 y = - 300 . .. (1) \\ Also, \\ y + 10 = 6 (x - 10) \\ \Rightarrow y - 6 x = - 70 \\ \Rightarrow y = 6 x - 70 . .. (2) \\ Putting this value of y in equation (1), we get \\ x - 2 (6 x - 70) = - 300 \\ \Rightarrow x - 12 x = - 300 - 140 \\ \Rightarrow x = \frac{- 440}{- 11} = 40 \\ Putting this value of x in equation (2), we get \\ y = 6 \times 40 - 70 = 240 - 70 = 170 \\ Hence, their respective capitals are ₹ 40 and ₹ 170 . \end{array}$

Q.24

$\begin{array}{l} A train covered a certain distance at a uniform \\ speed. If the train would have been 10 km/h faster, \\ it would have taken 2 hours less than the scheduled \\ time. And, if the train were slower by 10 km/h; it \\ would have taken 3 hours more than the scheduled \\ time. Find the distance covered by the train. \end{array}$

Ans.

\begin{array}{l} Let the distance covered by the train is x km and the \\ uniform speed of the train is y km/h. Then time taken \\ to travel this distance is \frac{x}{y} hour. \\ According to question, \\ \frac{x}{y} - 2 = \frac{x}{y + 10} \\ \Rightarrow (x - 2 y) (y + 10) = xy \\ \Rightarrow xy + 10 x - 2 y^{2} - 20 y = xy \\ \Rightarrow 10 x - 2 y^{2} - 20 y = 0 . .. (1) \\ Also, \\ \frac{x}{y} + 3 = \frac{x}{y - 10} \\ \Rightarrow (x + 3 y) (y - 10) = xy \\ \Rightarrow xy - 10 x + 3 y^{2} - 30 y = xy \\ \Rightarrow - 10 x + 3 y^{2} - 30 y = 0 . .. (2) \\ Adding equations (1) and (2), we get \\ 10 x - 2 y^{2} - 20 y = 0 \\ \underline{- 10 x + 3 y^{2} - 30 y = 0} \\ y^{2} - 50 y = 0 \\ \Rightarrow y (y - 50) = 0 \\ \Rightarrow y = 0 or y = 50 \end{array}

\begin{array}{l} Here, speed can not be taken as zero unit as then distance \\ travelled would be zero unit. \\ So, y = 50 \\ Putting this value of y in equation (1), we get \\ 10 x - 2 {(50)}^{2} - 20 \times 50 = 0 \\ \Rightarrow 10 x - 5000 - 1000 = 0 \\ \Rightarrow x = 600 \\ Hence, distance covered by the train is 600 km. \end{array}

Q.25 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.

Ans. Let the number of rows be x and the number of students in a row be y.
Total students in the class
= (Number of rows) × (Number of students in a row)
= xy
Given that if 3 students are extra in a row, then there would be 1 row less.
Therefore,
Total number of students = (x – 1)(y + 3)
or xy =(x-1)(y+3)=xy – y + 3x – 3
or 3x – y – 3 = 0 …(1)
It is also given that if 3 students are less in a row, then there would be 2 rows more.
Therefore,
Total number of students = (x+2)(y– 3)
or xy = (x+2)(y– 3) = xy + 2y – 3x – 6
or 3x – 2y + 6 = 0 …(2)
Subtracting equation (2) from equation (1),
– y + 2y = 3 + 6
or y= 9
By using equation (1)
3x – 9 = 3
or 3x = 12
or x = 4
Number of rows = x = 4
Number of students in each row = y = 9
Hence, number of students in the class = 9 × 4 = 36.

Q.26

\begin{array}{l} In a ΔABC, ∠ C=3 ∠ B=2 (∠ A+ ∠ B) . Find the \\ three angles. \end{array}

Ans.

$\begin{array}{l} Given that, \\ ∠ C = 3 ∠ B = 2 (∠ A + ∠ B) \\ \Rightarrow 3 ∠ B = 2 ∠ A + 2 ∠ \\ \Rightarrow ∠ B = 2 ∠ A \\ \Rightarrow 2 ∠ A - ∠ B = 0 . .. (1) \\ We know that the sum of measures of all angles of \\ a triangle is 180 ° . Therefore, \\ ∠ A + ∠ B + ∠ C = 180 ° \\ \Rightarrow ∠ A + ∠ B + 3 ∠ B = 180 ° \\ \Rightarrow ∠ A + 4 ∠ B = 180 ° . .. (2) \\ Multiplying equation (1) by 4, we get \\ 8 ∠ A - 4 ∠ B = 0 . .. (3) \\ Addi ng equations (2) and (3), we get \\ 9 ∠ A = 180° \Rightarrow ∠ A = 20 ° \\ From equation (2), we get \\ 20° + 4 ∠ B = 180 ° \\ \Rightarrow 4 ∠ B = 160 ° \\ \Rightarrow ∠ B = 40 ° \\ and ∠ C = 3 ∠ B \Rightarrow ∠ C = 120 ° . \\ Therefore, ∠ A, ∠ B and ∠ C are respectively \\ 20°, 40° and 120° . \end{array}$

Q.27 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

Ans. The given equations are
5x – y = 5 or y = 5(x – 1) …(1)
3x – y = 3 or y = 3(x – 1) …(2)
We find the value of y when x is zero and the value of x when y is zero for both equations and write the corresponding values in tables as below.

x	0	1
y = 5(x – 1)	–5	0

(1)

x	0	1
y = 3(x – 1)	–3	0

(2)
Now, we draw graphs of the given equations as given below.

From the graph above, we find that the co-ordinates of the vertices of the triangle formed by lines and the y-axis are (1, 0), (0, –3) and (0, –5).

Q.28

\begin{array}{l} Solve the following pair of linear equations: \\ (i)    px+qy = p-q                 (ii) ax+by = c \\ qx-py = p+q                        bx+ay = 1+c \\ (iii) \frac{x}{a} – \frac{y}{b} {= 0                           (iv) (a-b)x+(a+b)y = a}^{2} {-2ab-b}^{2} \\ {ax+by = a}^{2} {+b}^{2} {(a+b)(x+y) = a}^{2} {+b}^{2} \\ (v)  152x-378y = -74 \\ -378x+152y = -604 \end{array}

Ans.

$\begin{array}{l} (i) \\ The given equations can be written as: \\ px + qy - (p - q) = 0 . .. (1) \\ qx - py - (p + q) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- pq - q^{2} - (p^{2} - pq)} = \frac{y}{- pq + q^{2} + p^{2} + pq} = \frac{1}{- p^{2} - q^{2}} \\ \Rightarrow \frac{x}{- q^{2} - p^{2}} = \frac{y}{q^{2} + p^{2}} = \frac{1}{- p^{2} - q^{2}} \\ \Rightarrow x = 1 and y = - 1 \end{array}$ $\begin{array}{l} (ii) \\ The given equations can be written as: \\ ax + by - c = 0 . .. (1) \\ bx + ay - (1 + c) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- b - bc + ac} = \frac{y}{- bc + a + ac} = \frac{1}{a^{2} - b^{2}} \\ \Rightarrow \frac{x}{c (a - b) - b} = \frac{y}{c (a - b) + a} = \frac{1}{a^{2} - b^{2}} \\ \Rightarrow x = \frac{c (a - b) - b}{a^{2} - b^{2}} and y = \frac{c (a - b) + a}{a^{2} - b^{2}} \\ (iii) \\ The given equations can be written as: \\ bx - ay = 0 . .. (1) \\ ax + by - (a^{2} + b^{2}) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{a (a^{2} + b^{2})} = \frac{y}{b (a^{2} + b^{2})} = \frac{1}{a^{2} + b^{2}} \\ \Rightarrow x = a and y = b \end{array}$ $\begin{array}{l} (iv) \\ The given equations can be written as: \\ (a - b) x + (a + b) y - (a^{2} - 2 ab - b^{2}) = 0 . .. (1) \\ (a + b) x + (a + b) y - (a^{2} + b^{2}) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- (a + b) (a^{2} + b^{2}) + (a + b) (a^{2} - 2 ab - b^{2})} \\ = \frac{y}{- (a + b) (a^{2} - 2 ab - b^{2}) + (a - b) (a^{2} + b^{2})} = \frac{1}{a^{2} - b^{2} - {(a + b)}^{2}} \\ \Rightarrow \frac{x}{(a + b) (- a^{2} - b^{2} + a^{2} - 2 ab - b^{2})} \\ = \frac{y}{- a^{3} + 2 a^{2} b + {ab}^{2} - a^{2} b + 2 {ab}^{2} + b^{3} + a^{3} + {ab}^{2} - a^{2} b - b^{3}} \\ = \frac{1}{a^{2} - b^{2} - a^{2} - b^{2} - 2 ab} \\ \Rightarrow \frac{x}{- 2 b {(a + b)}^{2}} = \frac{y}{4 {ab}^{2}} = \frac{1}{- 2 b (a + b)} \\ \Rightarrow x = a + b and y = \frac{- 2 ab}{a + b} \\ (v) \\ 152 x - 378 y = - 74 \\ - 378 x + 152 y = - 604 \\ The given equations can be written as: \\ 76 x - 189 y + 37 = 0 . .. (1) \\ - 189 x + 76 y + 302 = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- 37 \times 76 - 189 \times 302} = \frac{y}{- 189 \times 37 - 76 \times 302} = \frac{1}{76^{2} - 189^{2}} \\ \Rightarrow x = - \frac{37 \times 76 + 189 \times 302}{76^{2} - 189^{2}} = 2 \\ and y = \frac{- 189 \times 37 - 76 \times 302}{76^{2} - 189^{2}} = 1 \end{array}$

Q.29 ABCD is a cyclic quadrilateral. Find its all angles.

Ans.

$\begin{array}{l} We know that sum of the measures of the opposite \\ {angles of a cyclic quadrilateral is 180}^{°} . \\ Therefore, \\ ∠ A + ∠ C = 180 \\ \Rightarrow 4 y + 20 - 4 x = 180 \\ \Rightarrow x - y = - 40 . .. (i) \\ Also, \\ ∠ B + ∠ D = 180 \\ \Rightarrow 3 y - 5 - 7 x + 5 = 180 \\ \Rightarrow - 7 x + 3 y = 180 . .. (ii) \\ Multiplying equation (i) by 3, we get \\ 3 x - 3 y = - 120 . .. (iii) \\ Adding equations (ii) and (iii), we get \\ - 7 x + 3 x = 60 \Rightarrow x = - 15 \\ Putting value of x in equation (i), we get \\ - 15 - y = - 40 \\ \Rightarrow y = 25 \\ Thus, \\ ∠ A = 4 y + 20 = 4 \times 25 + 20 = 120 ° \\ ∠ B = 3 y - 5 = 3 \times 25 - 5 = 70 ° \\ ∠ C = - 4 x = - 4 \times (- 15) = 60 ° \\ ∠ D = - 7 x + 5 = - 7 \times (- 15) + 5 = 110 ° . \end{array}$

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Coordinate Geometry

Introduction to Trigonometry

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NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Quadratic Equations

Arithmetic Progressions

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Statistics

Probability

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