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13. A train 300m long crossed a platform 900 m long in 1 minute 12 seconds. The speed of the train in km/hr was :
a. 45
b. 50
c. 54
d. 60

Answer: D

Explanation:
Distance covered in 72 seconds =(300 + 900)m
Speed = $\left( {\displaystyle\frac{{1200}}{{72}}} \right)$ m/sec = $\left( {\displaystyle\frac{{50}}{3}} \right)$ m/sec = $\left( {\displaystyle\frac{{50}}{3} \times \displaystyle\frac{{18}}{5}} \right)$ km/hr = 60 km/hr

14. A train crosses a platform 100 metres long in 60 seconds at a speed of 45 km per hour. The time taken by the train to cross an electric pole is :
a. 8 seconds
b. 1 minute
c. 52 seconds
d. Date inadequate.

Answer: C

Explanation:
Let the length of train = x metres
Speed = 45 km/h = $\left( {45 \times \displaystyle\frac{5}{{18}}} \right)$ m/sec = $\left( {\displaystyle\frac{{25}}{2}} \right)$ m/sec
Distance covered in crossing the platform = (x+100) m
$ \Rightarrow (x + 100) = \dfrac{{25}}{2} \times 60$
$ \Rightarrow (x + 100) = 750$
$ \Rightarrow $ x = 650
Now, time taken to cross the pole = $\dfrac{D}{s} = \dfrac{{650}}{{\left( {\dfrac{{25}}{2}} \right)}}$ sec = 52 sec

15. A train moving at the rate of 36 km per hour crosses a standing man in 10 seconds. It will cross a platform 55 metres long in :
a. ${5\displaystyle\frac{1}{2}}$ seconds
b. 6 seconds
c. ${7\displaystyle\frac{1}{2}}$ seconds
d. ${15\displaystyle\frac{1}{2}}$ seconds

Answer: D

Explanation:
Speed = $\left( {36 \times \displaystyle\frac{5}{{18}}} \right)$ m/sec = 10 m/sec
Let the length of the train be x metres
Then, $\displaystyle\frac{x}{{10}} = 10 \Rightarrow x = 100$ m.
Time taken to cross the platform = $\left( {\displaystyle\frac{{100 + 55}}{{10}}} \right)$ sec = $15\displaystyle\frac{1}{2}$ sec

16. A train of length 150 metres take 10 seconds to pass over another train 100 metres long coming from the opposite direction. If the speed of the first train is 30 kmph, the speed of the second train is :
a. 54 kmph
b. 60 kmph
c. 72 kmph
d. 36 kmph

Answer: B

Explanation:
Relative speed = $\dfrac{{150 + 100}}{{10}} = 25$ m/s = $\left( {25 \times \displaystyle\frac{{18}}{5}} \right)$ km/hr = 90 km/hr
Speed of second train = (90 - 30) km/hr = 60 km/hr

17. A 150 meter long train crosses a man walking at the speed of 6 kmph in the opposite direction in 6 seconds. The speed of the train in km/hr is :
a. 66
b. 84
c. 96
d. 106

Answer: B

Explanation:
Let the speed of the train be x km/hr
Relative speed = (x + 6) km/hr = $\left[ {(x + 6) \times \displaystyle\frac{5}{{18}}} \right]$ m/sec
$\dfrac{{150}}{6} = \dfrac{{(x + 6) \times 5}}{{18}}$
$\Rightarrow $ 5x + 30 = 450 (or) x = 84 km/hr

18. A train speeds past a pole in 15 seconds and speeds past a platform 100 meters long in 25 seconds. Its length in meters is :
a. 200
b. 150
c. 50
d. Data inadequate

Answer: B

Explanation:
Let the length of the train be $x$ metres and its speed be $y$ meters/sec.
Then, $\displaystyle\frac{x}{y} = 15 \Rightarrow y = \displaystyle\frac{x}{{15}}$
Now, $\displaystyle\frac{{x + 100}}{{25}} = \displaystyle\frac{x}{{15}} \Rightarrow x = 150$ m

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