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Problems on Trains
1.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}=\left(60×\frac{5}{18}\right)\text{m/sec}=\left(\frac{50}{3}\right)\text{m/sec}\\ & \text{Length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}=\left(\text{Speed}×\text{Time}\right)\\ & \therefore \text{Length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}\\ & =\left(\frac{50}{3}×9\right)m=150m\end{array}$

2.A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{relative}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{man}\\ & =\left(\frac{125}{10}\right)\text{m/sec}\\ & =\left(\frac{25}{2}\right)\text{m/sec}\\ & =\left(\frac{25}{2}×\frac{18}{5}\right)\text{km/hr}\\ & =45\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}.\\ & Then,\phantom{\rule{thinmathspace}{0ex}}relative\phantom{\rule{thinmathspace}{0ex}}speed=\left(x-5\right)\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & \therefore x-5=45⇒x=50\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\end{array}$

3.The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}=\left(45×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left(\frac{25}{2}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \text{Time}=30\phantom{\rule{thinmathspace}{0ex}}\text{sec}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{bridge}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{metres}\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\frac{130+x}{30}=\frac{25}{2}\\ & ⇒2\left(130+x\right)=750\\ & ⇒x=245\phantom{\rule{thinmathspace}{0ex}}m\end{array}$

4.A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}=\left(54×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}=15\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \text{Length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}=\left(15×20\right)\text{m}=300\phantom{\rule{thinmathspace}{0ex}}\text{m}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{platform}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{metres}\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\frac{x+300}{36}=15\\ & ⇒x+300=540\\ & ⇒x=240\phantom{\rule{thinmathspace}{0ex}}\text{m}\end{array}$

5.A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

Explanation:

Solution:

6.Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{each}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{metres}.\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\text{distance}\phantom{\rule{thinmathspace}{0ex}}\text{covered}=2x\phantom{\rule{thinmathspace}{0ex}}\text{metres}.\\ & \text{Relative}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\\ & =\left(46-36\right)\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & =\left(10×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & =\left(\frac{25}{9}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \therefore \frac{2x}{36}=\frac{25}{9}\\ & ⇒2x=100\\ & ⇒x=50\end{array}$

7.A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

Explanation:

Solution:
$\begin{array}{rl}& \text{Formula}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\text{converting}\phantom{\rule{thinmathspace}{0ex}}\text{from}\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{m/s:}\\ & X\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}=\left(X×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/s}\\ & \text{Therefore,}\phantom{\rule{thinmathspace}{0ex}}\text{Speed}\\ & =\left(45×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}=\frac{25}{2}\text{m/sec}\\ & \text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{distance}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\text{covered}\\ & =\left(360+140\right)m=500\phantom{\rule{thinmathspace}{0ex}}m\\ & \text{Formula}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\text{finding}\phantom{\rule{thinmathspace}{0ex}}\text{Time}\\ & =\left(\frac{\text{Distance}}{\text{Speed}}\right)\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{time}\\ & =\left(\frac{500×2}{25}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{sec}\\ & =40\phantom{\rule{thinmathspace}{0ex}}\mathrm{sec}.\end{array}$

8.A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{relative}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{jogger}\\ & =\left(45-9\right)\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & =36\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & \left(36×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & =10\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \text{Distance}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\text{covered}\\ & =\left(240+120\right)\phantom{\rule{thinmathspace}{0ex}}m\\ & =360\phantom{\rule{thinmathspace}{0ex}}m\\ & \therefore \text{Time}\phantom{\rule{thinmathspace}{0ex}}\text{taken}\\ & =\left(\frac{360}{10}\right)\phantom{\rule{thinmathspace}{0ex}}\text{sec}\\ & =36\phantom{\rule{thinmathspace}{0ex}}\text{sec}\end{array}$

9.A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

Explanation:

Solution:
$\begin{array}{rl}& \text{Relative}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\\ & =\left(120+80\right)\phantom{\rule{thinmathspace}{0ex}}\text{km/hr}\\ & =\left(200×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & =\left(\frac{500}{9}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{other}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\text{x}\phantom{\rule{thinmathspace}{0ex}}\text{metres}\text{.}\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\frac{x+270}{9}=\frac{500}{9}\\ & ⇒x+270=500\\ & ⇒x=230\end{array}$

10.A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

$\begin{array}{rl}& \text{Speed}=\left(72×\frac{5}{18}\right)\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=20\phantom{\rule{thinmathspace}{0ex}}\text{m/sec}\\ & \text{Time}=26\phantom{\rule{thinmathspace}{0ex}}\text{sec}\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{length}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{train}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{metres}\text{.}\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\frac{x+250}{26}=20\\ & ⇒x+250=520\\ & ⇒x=270\end{array}$