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Two trains cross each other in $14 \; \text{seconds}$ when running in opposite directions along parallel tracks. The faster train is $160 \; \text{m}$ long and crosses a lamp post in $12 \; \text{seconds}.$ If the speed of the other train is $6 \; \text{km/hr}$ less than the faster one, its length, in $\text{m},$ is

1. $190$
2. $192$
3. $184$
4. $180$

Given that, two trains cross each other in $14$ seconds.

Let the length of the slower train be $\text{L}$ meter.

The faster train crosses a lamp pole in $12$ seconds.

$S_{\text{faster train}} = \frac{160}{12} \;\text{m/sec}=\frac{40}{3}\;\text{m/sec}$

The speed of slower train $=S _{\text{faster train}} = \frac{40}{3}\;\text{m/sec} – 6 \times \frac{5}{18}\;\text{m/sec}$

$\qquad \qquad = \left(\frac{40}{3} – \frac{5}{3}\right)\;\text{m/sec}$

$\qquad \qquad = \left(\frac{40-5}{3}\right)\;\text{m/sec}$

$\qquad \qquad=\frac{35}{3}\;\text{m/sec}$

We know that,

• When two trains are moving in opposite directions, then their speed will be added.
• when two trains are moving in the same directions, then their speed will be subtracted.

Now,  $\frac{L+160}{14}= \frac{35}{3} + \frac{40}{3}$

$\Rightarrow \frac{L+160}{14}= \frac{75}{3}$

$\Rightarrow \frac{L+160}{14}= 25$

$\Rightarrow L+160= 25 \times 14$

$\Rightarrow L+160= 350$

$\Rightarrow L= 350-160$

$\Rightarrow \boxed{L= 190\;\text{meters}}$

$\therefore$ The length of slower train is $190\;\text{meters.}$

Correct Answer $:\text{A}$

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