# Ex.3.6 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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## Question

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream $$20\,\rm{ km}$$ in $$2$$ hours, and upstream $$4 \,\rm{km}$$ 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 $$300 \,\rm{km}$$ to her home partly by train and partly by bus. She takes $$4$$ hours if she travels $$60 \,\rm{km}$$ by train and remaining by bus. If she travels $$100\,\rm{ km}$$ by train and the remaining by bus, she takes $$10$$ minutes longer. Find the speed of the train and the bus separately.

Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.6 | Question 2

## Text Solution

Reasoning:

Steps:

(i)

Let the Ritu’s speed of rowing in still water and the speed of stream be $$x\,{\rm{ km/h}}$$and $$y\,{\rm{km/h }}$$ respectively.

Ritu’s speed of rowing;

Upstream $$= \left( {x – y} \right) \,\rm{km/h}$$

Downstream $$= \left( {x + y} \right)\,\rm{km/h}$$

According to question,

Ritu can row downstream $$20 \,\rm{km}$$ in $$2$$ hours,

\begin{align}2\left( {x + y} \right) &= 20\\x + y& = 10 \qquad\left( 1 \right)\end{align}

Ritu can row upstream $$4\,\rm{ km}$$ in $$2$$ hours,

\begin{align}2\left( {x - y} \right) &= 4\\x - y &= 2 \qquad \left( 2 \right)\end{align} Adding equation (1) and (2), we obtain

\begin{align} 2x&=12 \\ x&=6 \\\end{align}

Putting $$x = 6$$ in equation (1), we obtain

\begin{align}6 + y &= 10\\y& = 4\end{align}

Hence, Ritu’s speed of rowing in still water is $$6 \,\rm{km/h}$$ and the speed of the current is $$4 \,\rm{km/h}.$$

(ii)

Let the number of days taken by a woman and a man to finish the work be $$x$$ and $$y$$ respectively.

Therefore, work done by a woman in $$1$$ day $$= \frac{1}{x}$$

and work done by a man in 1 day $$= \frac{1}{y}$$

According to the question,

$$2$$ women and $$5$$ men can together finish an embroidery work in $$4$$ days; $\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \qquad \left( 1 \right)$

$$3$$ women and $$6$$ men can finish it in $$3$$ days $\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \qquad \left( 2 \right)$

Substituting \begin{align}\frac{1}{x} = p \end{align} and \begin{align}\frac{1}{y} = q \end{align}in equations (1) and (2), we obtain \begin{align} \frac{2}{x} + \frac{5}{y} &= \frac{1}{4} \Rightarrow 2p + 5q = \frac{1}{4} \Rightarrow 8p + 20q - 1 = 0 \quad & & \left( 3 \right)\\ \frac{3}{x} + \frac{6}{y} &= \frac{1}{3} \Rightarrow 3p + 6q = \frac{1}{3} \Rightarrow 9p + 18q - 1 = 0 \quad & & \left( 4 \right) \end{align}

By cross-multiplication, we obtain

\begin{align}\frac{p}{{ - 20 - ( - 18)}} &= \frac{q}{{ - 9 - ( - 8)}} = \frac{1}{{144 - 180}}\\ \frac{p}{{ - 2}} &= \frac{q}{{ - 1}} = \frac{1}{{ - 36}}\\\frac{p}{{ - 2}}& = \frac{1}{{ - 36}}{\text{ and }}\frac{q}{{ - 1}} = \frac{1}{{ - 36}}\\p &= \frac{1}{{18}}\quad {\text{ and }} \quad q = \frac{1}{{36}}\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{x} = \frac{1}{{18}}\\ & \Rightarrow x = 18\\{\text{and, }}q &= \frac{1}{y} = \frac{1}{{36}}\\& \Rightarrow y = 36\end{align}

Hence, number of days taken by a woman is $$18$$ and by a man is $$36.$$

(iii)

Let the speed of train and bus be $$u \,\rm{km/h}$$ and $$v\,\rm{ km/h}$$ respectively.

According to the given information,

Roohi travels $$300\,\rm{ km}$$ and takes $$4$$ hours if she travels $$60 \,\rm{km}$$ by train and the remaining by bus $\frac{{60}}{u} + \frac{{240}}{v} = 4 \qquad \left( 1 \right)$

If she travels $$100\,\rm{ km}$$ by train and the remaining by bus, she takes $$10$$ minutes longer

$\frac{{100}}{u} + \frac{{200}}{v} = \frac{{25}}{6} \qquad \left( 2 \right)$

Substituting \begin{align}\frac{1}{u}=p\end{align} and \begin{align}\frac{1}{v}=q \end{align} in equations (1) and (2), we obtain

\begin{align}\frac{{60}}{{u}} + \frac{{240}}{v} &= 4{\rm{ }} \Rightarrow 60p + 240q = 4 \qquad \left( 3 \right)\\\frac{{100}}{u} + \frac{{200}}{v} &= \frac{{25}}{6}{\rm{ }} \Rightarrow 100p + 200q = \frac{{25}}{6}{\rm{ }} \Rightarrow 600p + 1200q = 25 \qquad \left( 4 \right)\end{align}

Multiplying equation (3) by 10, we obtain

$600p + 2400q = 40 \quad \left( 5 \right)$

Subtracting equation (4) from (5), we obtain

\begin{align}1200q &= 15\\q &= \frac{{15}}{{1200}}\\q& = \frac{1}{{80}}\end{align}

Substituting \begin{align}q = \frac{1}{{80}} \end{align} in equation (3), we obtain

\begin{align}60p + 240 \times \frac{1}{{80}} &= 4\\60p &= 4 - 3\\p &= \frac{1}{{60}}\end{align}

\begin{align}{\text{Therefore, }}p &= \frac{1}{u} = \frac{1}{{60}}\\& \Rightarrow u = 60\\{\text{and, }}q &= \frac{1}{v} = \frac{1}{{80}}\\& \Rightarrow v = 80\end{align}

Hence, speed of the train $$= 60\,{\rm{ km/h}}$$

And speed of the bus $$= 80\,{\rm{ km/h}}$$

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