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

## 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.

## 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 \cdots \left( 3 \right)\\ \\

&\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \\ \Rightarrow \, & 3p + 6q = \frac{1}{3} \\ \Rightarrow \, & 9p + 18q - 1 = 0 \cdots \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 }} \\ 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

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

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 \(= 80\,{\rm{ km/h}}\)