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A train leaves Pune at $7: 30 \mathrm{am}$ and reaches Mumbai at $11: 30 \mathrm{am}$. Another train leaves Mumbai at $9: 30 \mathrm{am}$ and reaches Pune at $1: 00 \mathrm{pm}$. Assuming that the two trains travels at constant speeds, at what time do the two trains cross each other -
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The correct answer is:
$11: 30 \mathrm{am}$
First train from Pune to Mumbai takes $4 \mathrm{hrs}$,
Second train from Mumbai to Pune takes $3.5 \mathrm{hrs}$.
Speed of train from Pune to Mumbai is $V_{1}=\frac{d}{4}$
Speed of train from Mumbai to Pune is $\mathrm{V}_{2}=\frac{\mathrm{d}}{3.5}$
Distance traveled by first train till $9: 30$ is $\mathrm{x}=\frac{\mathrm{d}}{4} \times 2=\frac{\mathrm{d}}{2}$
Say now both trains meet after time from $9: 30$
$\frac{x}{d / 4}=\frac{\frac{d}{2}-x}{\frac{d}{3.5}} \Rightarrow \frac{4 x}{d}=\frac{3.5}{2}-\frac{3.5 x}{d} \Rightarrow \frac{7.5 x}{d}=\frac{3.5}{2} \Rightarrow 15 x=3.5 \mathrm{~d} \Rightarrow 30 x=7 d$
time taken by first train from 9: 30 is $\frac{4 \mathrm{x}}{\mathrm{d}}=4\left(\frac{7}{30}\right)$
$\begin{array}{l}
=\frac{14}{15} \\
=56 \mathrm{~min}
\end{array}$
Then both train meet at $9.30+56 \mathrm{~min}=10: 26 \mathrm{am}$
Second train from Mumbai to Pune takes $3.5 \mathrm{hrs}$.
Speed of train from Pune to Mumbai is $V_{1}=\frac{d}{4}$
Speed of train from Mumbai to Pune is $\mathrm{V}_{2}=\frac{\mathrm{d}}{3.5}$
Distance traveled by first train till $9: 30$ is $\mathrm{x}=\frac{\mathrm{d}}{4} \times 2=\frac{\mathrm{d}}{2}$
Say now both trains meet after time from $9: 30$
$\frac{x}{d / 4}=\frac{\frac{d}{2}-x}{\frac{d}{3.5}} \Rightarrow \frac{4 x}{d}=\frac{3.5}{2}-\frac{3.5 x}{d} \Rightarrow \frac{7.5 x}{d}=\frac{3.5}{2} \Rightarrow 15 x=3.5 \mathrm{~d} \Rightarrow 30 x=7 d$
time taken by first train from 9: 30 is $\frac{4 \mathrm{x}}{\mathrm{d}}=4\left(\frac{7}{30}\right)$
$\begin{array}{l}
=\frac{14}{15} \\
=56 \mathrm{~min}
\end{array}$
Then both train meet at $9.30+56 \mathrm{~min}=10: 26 \mathrm{am}$
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