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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