# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7

Go back to  'Pair of Linear Equations in Two Variables'

## Question 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 differs by $$30$$ years. Find the ages of Ani and Biju.

### Solution

Reasoning:

The difference between the ages of Biju and Ani is $$3$$ years. Either Biju is $$3$$ years older than Ani or Ani is $$3$$ years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be $$30$$ years more than that of Cathy’s age.

Steps:

Let the age of Ani and Biju be $$x$$ and $$y$$ years respectively.

Therefore, age of Ani’s father, Dharam be $$2x$$ years

And age of Biju’s sister Cathy be $$\frac{y}{2}$$ years

Case (I) When Ani is older than Biju

The ages of Ani and Biju differ by $$3$$ years,

$x - y = 3 \qquad \left( 1 \right)$

The ages of Cathy and Dharam differs by $$30$$ years,

\begin{align}2x - \frac{y}{2} &= 30\\4x - y &= 60 \qquad \left( 2 \right)\end{align}

Subtracting $$(1)$$ from $$(2),$$ we obtain

\begin{align}3x &= 57\\x &= 19\end{align}

Substituting $$x = 19$$ in equation $$(1),$$ we obtain

\begin{align}19 - y &= 3\\y& = 16\end{align}

Therefore, Ani is $$19$$ years old and Biju is $$16$$ years old

Case (II) When Biju is older than Ani.

The ages of Ani and Biju differ by $$3$$ years,

\begin{align}y - x &= 3\\ - x + y &= 3 \qquad \left( 1 \right)\end{align}

The ages of Cathy and Dharam differs by $$30$$ years,

\begin{align}2x - \frac{y}{2} &= 30\\4x - y &= 60 \qquad (2)\end{align}

Adding $$(1)$$ and $$(2),$$ we obtain

\begin{align}3x &= 63\\x &= 21\end{align}

Substituting $$x = 21$$ in equation $$(1),$$ we obtain

\begin{align} - 21 + y &= 3\\y &= 24\end{align}

Therefore, Ani is $$21$$ years old and Biju is $$24$$ years old.

Hence, Ani is $$19$$ years old and Biju is $$16$$ years old or Ani is $$21$$ years old and Biju is $$24$$ years old.

## Question 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]

$$\begin{bmatrix} \textbf{Hint:} \, x + 100 = 2 ({y - 100}),\\ y + 10 = 6 (x - 10) \end{bmatrix}$$

### Solution

Reasoning:

Assume the friends have ₹ $$x$$ and ₹ $$y$$ with them. Then based on given conditions, two linear equations can be formed which can be easily solved.

Steps:

Let the first friend has ₹ $$x$$

And second friend has ₹ $$y$$

Using the information given in the question,

When second friend gives ₹ $$100$$ to first friend;

\begin{align}x + 100 &= 2\left( {y - 100} \right)\\x + 100 &= 2y - 200\\x - 2y &= - 300 \qquad \qquad \left( 1 \right)\end{align}

When first friend gives ₹ $$10$$ to second friend;

\begin{align}y + 10 &= 6\left( {x - 10} \right)\\y + 10 &= 6x - 60\\6x - y &= 70 \qquad \qquad \left( 2 \right)\end{align}

Multiplying equation $$(2)$$ by $$2,$$ we obtain

$12x - 2y = 140 \qquad {\rm{ }}\left( 3 \right)$

Subtracting equation $$(1)$$ from equation $$(3),$$ we obtain

\begin{align}11x &= 440\\x &= \frac{{440}}{{11}}\\x &= 40\end{align}

Substituting $$x = 40$$ in equation $$(1)$$, we obtain

\begin{align}40 - 2y &= - 300\\2y &= 40 + 300\\y &= \frac{{340}}{2}\\y &= 170\end{align}

Therefore, first friend has ₹ $$40$$ and second friend has ₹ $$170$$ with them.

## Question 3

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

### Solution

What is Known?

Changes in speed of the train as well in the time.

What is Unknown?

Distance covered by the train.

Reasoning:

Assuming uniform speed of the train be $$x\,{\rm{km/h }}$$ and time taken to travel a given distance be $$t$$ hours. Then distance can be calculated by;

${\text{Distance}} = {\rm{ Speed }} \times {\rm{ Time}}$

Steps:

Let the uniform speed of the train be $$x\,{\rm{ km/h}}$$ and the scheduled time to travel the given distance be $$t$$ hours

Then the distance be $$xt\,{\rm{ km}}$$

When the train would have been $$10\,{\rm{ km/h}}$$ faster, it would have taken $$2$$ hours less than the scheduled time;

\begin{align}\left( {x + 10} \right)\left( {t - 2} \right)& = xt\\xt - 2x + 10t - 20 &= xt\\ - 2x + 10t &= 20 \qquad \quad \left( 1 \right)\end{align}

When the train were slower by $$10\,{\rm{ km/h,}}$$ it would have taken $$2$$ hours\] more than the scheduled time;

\begin{align}\left( {x - 10} \right)\left( {t + 3} \right) &= xt\\xt + 3x - 10t - 30 &= xt\\3x - 10t &= 30 \qquad \quad \left( 2 \right)\end{align}

Adding equations $$(1)$$ and $$(2),$$ we obtain

$x = 50$

Substituting $$x = 50$$ in equation $$(1),$$ we obtain

\begin{align} - 2 \times 50 + 10t &= 20\\- 100 + 10t &= 20\\10t &= 120\\t &= \frac{{120}}{{10}}\\t &= 12 \end{align}

Therefore, distance,

$$xt = 50 \times 12 = 600$$

Hence, the distance covered by the train is $$600 \,\rm{km.}$$

## Question 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

What is Known?

Changes in number of students in a row and number of rows.

What is Unknown?

Number of students in the class.

Reasoning:

Assume number of rows equal to $$x$$ and number of students in each row be $$y.$$ Then the total number of students in the class can be calculated by;

Total number of students $$=$$  Number of rows $$\times$$ Number of students in each row

Steps:

Let the number of rows be $$x$$

And number of students in each row be $$y$$

Then the number of students in the class be $$xy$$

Using the information given in the question,

Condition 1 If $$3$$ students are extra in a row, there would be $$1$$ row less

\begin{align}\left( {x - 1} \right)\left( {y + 3} \right) &= xy\\xy + 3x - y - 3 &= xy\\3x - y &= 3 \qquad \quad \left( 1 \right)\end{align}

Condition 2 If $$3$$ students are less in a row, there would be $$2$$ rows more

\begin{align}\left( {x + 2} \right)\left( {y - 3} \right) &= xy\\xy - 3x + 2y - 6 &= xy\\ - 3x + 2y &= 6 \qquad \quad \left( 2 \right)\end{align}

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

Substituting $$y = 9$$ in equation $$(1),$$ we obtain

\begin{align}3x - 9 &= 3\\3x &= 12\\x &= 4\end{align}

Hence, number of students in the class, $$xy = 4 \times 9 = 36$$

## Question 5

In $$\Delta ABC$$ ,

$$\angle {C} = 3\angle B = 2(\angle A + \angle B)$$.

Find the three angles.

### Solution

What is Known?

Relation between the angles of the triangle.

What is Unknown?

Measurement of each angles of the triangle.

Reasoning:

Sum of the measures of all angles of a triangle is $$180^\circ.$$

Steps:

Let the measurement of $$\angle A = {x^{\rm{\circ}}}$$

And the measurement of $$\angle B = {y^{\rm{\circ}}}$$

Using the information given in the question,

$$\angle C = 3\angle B = 2\left( {\angle A \!+\! \angle B} \right)$$

\begin{align} \Rightarrow 3\angle B &= 2\left( {\angle A + \angle B} \right)\\ \Rightarrow 3y &= 2\left( {x + y} \right)\\ \Rightarrow 3y &= 2x + 2y\\ \Rightarrow \;2x - y &= 0 \qquad \qquad \qquad \left( 1 \right)\end{align}

We know that the sum of the measures of all angles of a triangle is $$180^\circ.$$

Therefore,

\begin{align}\angle A + \angle B + \angle C &= {180^{\circ}}\\\angle A + \angle B + 3\angle B &= {180^{\circ}} \\&\begin{bmatrix} \because \angle C = 3\angle B \end{bmatrix} \\\angle A + 4\angle B &= {180^{\circ}}\\ x + 4y &= 180 \qquad \quad \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$4,$$ we obtain

$8x - 4y = 0 \qquad \left( 3 \right)$

Adding equations $$(2)$$ and $$(3),$$ we obtain

\begin{align}9x &= 180\\x &= 20\end{align}

Substituting $$x = 20$$ in equation $$(1)$$, we obtain

\begin{align}2 \times 20 - y &= 0\\y &= 40\end{align}

Therefore,

\begin{align}\angle A &= {x^{\circ}} = {20^{\circ}}\\\angle B &= {y^{\circ}} = {40^{\circ}}\\\angle C &= 3\;\angle B = 3 \times {40^{\circ}} = {120^{\circ}}\end{align}

## Question 6

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

### Solution

Steps:

\begin{align}5x - y &= 5\\\Rightarrow \quad y &= 5x - 5\end{align}

The solution table will be as follows.

 $$x$$ $$0$$ $$2$$ $$y$$ $$-5$$ $$5$$

\begin{align}3x - y &= 3\\ \Rightarrow \quad y &= 3x - 3\end{align}

The solution table will be as follows.

 $$x$$ $$0$$ $$2$$ $$y$$ $$-3$$ $$3$$

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is $$ABC$$ formed by these lines and $$y-$$axis.

The coordinates of vertices are $$A( 1, 0),\; B (0, - 3),\; C (0, - 5)$$.

## Question 7

Solve the following pair of linear equations.

(i) \begin{align} px + qy &= p - q\\qx - py &= p + q\end{align}

(ii) \begin{align} ax + by &= c\\bx + ay &= 1 + c\end{align}

(iii) \begin{align} \frac{x}{a} - \frac{y}{b} &= 0\\ax + by &= {a^2} + {b^2}\end{align}

(iv) \begin{align} \begin{bmatrix}(a - b)x + \\(a + b)y \end{bmatrix}\!&= \! {a^2} \! -\! 2ab \! -\! {b^2} \\(a + b)(x + y) &= {a^2} + {b^2}\end{align}

(v) \begin{align} 152x - 378y &= - 74\\- 378x + 152y &= - 604\end{align}

### Solution

Steps:

(i)

\begin{align} px + qy &= p - q \qquad \ldots \left( 1 \right)\\qx - py &= p + q \qquad \ldots \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$p$$ and equation $$(2)$$ by $$q,$$ we obtain

\begin{align}{p^2}x + pqy &= {p^2} - pq \qquad \ldots \left( 3 \right)\\{q^2}x - pqy &= pq + {q^2} \qquad \ldots \left( 4 \right)\end{align}

Adding equations $$(3)$$ and $$(4),$$ we obtain

\begin{align}{p^2}x + {q^2}x &= {p^2} + {q^2}\\\left( {{p^2} + {q^2}} \right)x &= {p^2} + {q^2}\\ x &= \frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}}\\x &= 1\end{align}

Substituting $$x = 1$$ in equation $$(1),$$ we obtain

\begin{align}p \times 1 + qy &= p – q\\qy &= - q\\y &= - 1\end{align}

Therefore, $$x = 1$$ and $$y = - 1$$

(ii)

\begin{align}ax + by &= c \qquad \quad\;\; \ldots \left( 1 \right)\\ bx + ay &= 1 + c \qquad \ldots \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$a$$ and equation $$(2)$$ by $$b$$, we obtain

\begin{align}{a^2}x + aby &= ac \qquad \quad\;\; \ldots \left( 3 \right)\\{b^2}x + aby &= b + bc \qquad \ldots \left( 4 \right)\end{align}

Subtracting equation $$(4)$$ from equation $$(3),$$

\begin{align}\left( {{a^2} – {b^2}} \right)x &= ac – bc – b\\x &= \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}{\rm{ }}\end{align}

Substituting \begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}} \end{align} in equation $$(1),$$ we obtain

\begin{align}ax + by &= c \\ a\left[ {\frac{{c(a – b) – b}}{{{a^2} – {b^2}}}} \right] + by &= c\\\frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}} + by &= c\\ \end{align}

\begin{align} by &= c - \frac{{ac(a – b) – ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{{a^2}c – {b^2}c – {a^2}c + abc + ab}}{{{a^2} – {b^2}}}\\by &= \frac{{abc – {b^2}c + ab}}{{{a^2} – {b^2}}}\\ by &= \frac{{bc(a – b) + ab}}{{{a^2} – {b^2}}}\\by& = \frac{{b\left[ {c(a – b) + a} \right]}}{{{a^2} – {b^2}}}\\ y &= \frac{{c(a – b) + a}}{{{a^2} – {b^2}}}\end{align}

Therefore,

\begin{align}x = \frac{{c(a – b) – b}}{{{a^2} – {b^2}}}\end{align}

and

\begin{align}y=\frac{c(a-b)+a}{{{a}^{2}}-{{b}^{2}}}\end{align}

(iii)

\begin{align} \frac{x}{a} - \frac{y}{b} &= 0 \qquad \qquad\;\; \ldots \left( 1 \right)\\ax + by &= {a^2} + {b^2} \qquad \ldots \left( 2 \right)\end{align}

By solving equation $$(1),$$ we obtain

\begin{align}\frac{x}{a} - \frac{y}{b} &= 0\\x &= \frac{{ay}}{b} \qquad \ldots \left( 3 \right)\end{align}

Substituting \begin{align}x = \frac{{ay}}{b} \end{align} in equation $$(2),$$ we obtain

\begin{align}a \times \left( {\frac{{ay}}{b}} \right) + by &= {a^2} + {b^2}\\ \frac{{{a^2}y + {b^2}y}}{b} &= {a^2} + {b^2}\\\left( {{a^2} + {b^2}} \right)y &= b\left( {{a^2} + {b^2}} \right)\\y &= b\end{align}

Substituting $$y = b$$ in equation $$(3),$$ we obtain

\begin{align}x &= \frac{{a \times b}}{b}\\x &= a\end{align}

Therefore, $$x = a$$ and $$y = b$$

(iv)

\begin{align} &\begin{bmatrix} (a - b)x +\\ (a + b)y \end{bmatrix} \! = \! {a^2} \! - \! 2ab \! - \! {b^2} \ldots \left( 1 \right) \\ \\ & (a + b)(x + y) \! = \! {a^2} \! + \! {b^2}\ldots \left( 2 \right)\end{align}

By solving equation $$(2),$$ we obtain

\begin{align}(a + b)(x + y) &= {a^2} + {b^2}\\ \begin{bmatrix}(a + b)x +\\ (a + b)y \end{bmatrix}&= {a^2} + {b^2} \ldots \left( 3 \right)\end{align}

Subtracting equation $$(3)$$ from $$(1),$$ we obtain

\begin{align}\!\begin{bmatrix}(a – b)x –\\ (a + b)x \end{bmatrix}&=\! \begin{bmatrix} \left( {{a^2} – 2ab – {b^2}} \right) \!-\! \\\left( {{a^2} + {b^2}} \right)\end{bmatrix} \\ \begin{bmatrix} {(a – b) – }\\{(a + b)} \end{bmatrix}x \!& =\! \begin{bmatrix}{a^2} – 2ab – {b^2} – \\{a^2} – {b^2}\end{bmatrix} \\\left[ {a – b – a – b} \right]x \!&=\! - 2ab – 2{b^2}\\ - 2bx \!&= \!- 2b\left( {a + b} \right)\\x\!& = \!\left( {a + b} \right)\end{align}

Substituting $$x = \left( {a + b} \right)$$ in equation (1), we obtain

\begin{align} \begin{bmatrix}(a – b)(a + b) + \\ (a + b)y \end{bmatrix}&= {a^2} \! –\! 2ab \! –\! {b^2} \\ \begin{bmatrix}({a^2} – {b^2}) +\\ (a + b)y \end{bmatrix} &= {a^2} \! – \! 2ab \! – \! {b^2} \end{align}

\begin{align} (a \! + \! b)y & \! = \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! ({a^2} \! – \! {b^2}) \a \! + \! b)y \! &= \! {a^2} \! – \! 2ab \! – \! {b^2} \! – \! {a^2} \! + \! {b^2} \\y \! &= \! \frac{{ - 2ab}}{{(a \! + \! b)}}\end{align} (v) \(\begin{align}152x - 378y &= - 74 \;\; \ldots \left( 1 \right)\\ - 378x + 152y &= - 604 \;\; \ldots \left( 2 \right)\end{align}

Adding equations $$(1)$$ and $$(2),$$ we obtain

\begin{align} - 226x - 226y &= - 678\\ - 226\left( {x + y} \right) &= - 678\\x + y &= 3 \qquad \ldots \left( 3 \right)\end{align}

Subtracting equation $$(2)$$ from $$(1),$$ we obtain

\begin{align}530x - 530y &= 530\\530\left( {x - y} \right) &= 530\\x - y &= 1 \qquad \ldots \left( 4 \right)\end{align}

Adding equations $$(3)$$ and $$(4),$$ we obtain

\begin{align}2x &= 4\\x& = 2\end{align}

Substituting $$x = 2$$ in equation $$(3),$$ we obtain

\begin{align}2 + y &= 3\\y &= 1\end{align}

Therefore, $$x = 2$$ and $$y = 1$$

## Question 8

$$ABCD$$ is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

### Solution

What is Known?

Measurement of the angles of the cyclic quadrilateral in terms of $$x$$ and $$y.$$

What is Unknown?

Measurement of the angles of the cyclic quadrilateral.

Reasoning:

Pairs of opposite angles of a cyclic quadrilateral are supplementary.

Steps:

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is $$180^\circ.$$

Therefore,

\begin{align}\angle A + \angle C &= {180^\circ}\\\left( {4y + 20} \right) + \left( { - 4x} \right) &= 180\\ 4y + 20 - 4x &= 180\\ - 4\left( {x - y} \right)& = 160\\x - y &= - 40 \qquad \left( 1 \right)\end{align}

And

\begin{align}\angle B + \angle D &= {180^\circ}\\\left( {3y - 5} \right) + \left( { - 7x + 5} \right) &= 180\\ 3y - 5 - 7x + 5 &= 180\\ - 7x + 3y &= 180\\7x - 3y &= - 180 \quad \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$3,$$ we obtain

$3x - 3y = - 120\qquad \left( 3 \right)$

Subtracting equation $$(3)$$ from equation $$(2),$$ we obtain

\begin{align}4x &= - 60\\x &= - 15\end{align}

Substituting $$x = - 15$$ in equation $$(1),$$ we obtain

\begin{align} - 15 - y &= - 40\\y &= 25\end{align}

Therefore,

\begin{align}\angle A &= 4 \times 25 + 20 = {120^\circ}\\\angle B &= 3 \times 25 - 5 = {70^\circ}\\ \angle C &= - 4 \times \left( { - 15} \right) = {60^\circ}\\\angle D &= - 7 \times \left( { - 15} \right) + 5 = {110^\circ}\end{align}