Problems On Trains - Aptitude MCQ Questions and Solutions with Explanations

31

Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

A) 2 : 3 B) 4 : 3 C) 6 : 7 D) 9 : 16

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Let}}\,{\text{us}}\,{\text{name}}\,{\text{the}}\,{\text{trains}}\,{\text{as}}\,{\text{A}}\,{\text{and}}\,{\text{B}}{\text{.}}\,{\text{Then}}, \cr & \left( {{\text{A's}}\,{\text{speed}}} \right):\left( {{\text{B's}}\,{\text{speed}}} \right) \cr & = \sqrt b :\sqrt a \cr & = \sqrt {16} :\sqrt 9 \cr & = 4:3\, \cr} $$

32

A 100 m long train is going at a speed of 60 km/hr. It will cross a 140 m long railway bridge in-

A) 3.6 sec B) 7.2 sec C) 14.4 sec D) 21.6 sec

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Speed }} \cr & {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{3}{\text{ m/sec}} \cr & {\text{Total distance covered}} \cr & {\text{ = (100 + 140) m = 240 m}} \cr & \therefore {\text{Required time}} \cr & {\text{ = }}\left( {240 \times \frac{3}{{50}}} \right){\text{sec}} \cr & {\text{ = }}\frac{{72}}{5}{\text{sec}} \cr & {\text{ = 14}}{\text{.4 sec}} \cr} $$

33

A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train?

A) 70 km/hr B) 72 km/hr C) 79.2 km/hr D) 80 km/hr

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {\frac{{132}}{6}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {22 \times \frac{{18}}{5}} \right){\text{km/sec}} \cr & {\text{ = 79}}{\text{.2 km/hr}} \cr} $$

34

A train running at the speed of 60 kmph crosses a 200 m long platform in 27 seconds. What is the length of the train?

A) 200 meters B) 240 meters C) 250 meters D) 450 meters

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{3}{\text{m/sec}} \cr & {\text{Time = 27 sec}}{\text{.}} \cr & {\text{Let the length of the train be }}x{\text{ metres}}{\text{.}} \cr & {\text{Then,}}\frac{{x + 200}}{{27}}{\text{ = }}\frac{{50}}{3}{\text{ }} \cr & \Leftrightarrow x + 200 = \left( {\frac{{50}}{3} \times 27} \right) = 450 \cr & \Leftrightarrow x = 450 - 200 = 250{\text{ metres}} \cr} $$

35

A train running at a speed of 90 km/hr crosses a platform double its length in 36 seconds. What is the length of the platform in meters?

A) 200 B) 300 C) 450 D) Can not be determined E) None of these

Discuss

Answer & Solution

Answer: Option E

Solution:

$$\eqalign{ & {\text{Let the length of the train be x metres}}{\text{.}} \cr & {\text{Then, length of the platform = (2}}x{\text{) metres}}{\text{.}} \cr & {\text{Speed of the train}} \cr & {\text{ = }}\left( {90 \times \frac{5}{{18}}} \right)m/\sec \cr & = 25m/sec \cr & \therefore \frac{{x + 2x}}{{25}} = 36 \cr & \Rightarrow 3x = 900 \cr & \Rightarrow x = 300 \cr & {\text{Hence, length of platform}} \cr & {\text{ = }}2x = \left( {2 \times 300} \right){\text{m}} = 600{\text{m}} \cr} $$

36

A train of length 150 meters takes 40.5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr?

A) 13.33 B) 26.67 C) 40 D) 66.67

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Speed = }}\left( {\frac{{150 + 300}}{{40.5}}} \right)m/\sec \cr & = \left( {\frac{{450}}{{40.5}} \times \frac{{18}}{5}} \right)km/hr \cr & = 40km/hr \cr} $$

37

A 280 meter long train crosses a platform thrice its length in 50 seconds. What is the speed of the train in km/hr?

A) 60.48 B) 64.86 C) 80.64 D) 82.38 E) None of these

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & \therefore {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & = \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr} $$

38

A train 110 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

A) 5 sec B) 6 sec C) 7 sec D) 10 sec

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Speed of train relative to man}} \cr & {\text{ = }}\left( {60 + 6} \right){\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = \left( {66 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{55}}{3}} \right)m/\sec \cr & \therefore {\text{Time taken to pass the man}} \cr & = \left( {110 \times \frac{3}{{55}}} \right)\sec \cr & = 6\,\sec \cr} $$

39

Two trains A and B start running together from the same point in the same direction, at the speed of 60 kmph and 72 kmph respectively. If the length of each of the trains is 240 meters, how long will it take for B to cross train A?

A) 1 min 12 sec B) 1 min 24 sec C) 2 min 12 sec D) 2 min 24 sec

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (72}} - {\text{60) km/hr}} \cr & {\text{ = 12 km/hr}} \cr & = \left( {12 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{10}}{3}} \right)m/\sec \cr & {\text{Total distance covered}} \cr & {\text{ = Sum of lengths of trains}} \cr & {\text{ = (240 + 240) m}} \cr & {\text{ = 480 m}} \cr & {\text{Time taken}} \cr & {\text{ = }}\left( {480 \times \frac{3}{{10}}} \right)\sec \cr & = 144\sec \cr & = 2\min \,24sec \cr} $$

40

Two trains are moving in opposite directions @60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in second is?

A) 36 B) 45 C) 48 D) 49

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (60 + 90) km/hr}} \cr & {\text{ = }}\left( {150 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{125}}{3}} \right){\text{m/sec}} \cr & {\text{Distance coverd}} \cr & {\text{ = (1}}{\text{.10 + 0}}{\text{.9)km}} \cr & {\text{ = 2 km}} \cr & {\text{ = 2000 m}}{\text{}} \cr & {\text{Required time}} \cr & {\text{ = }}\left( {2000 \times \frac{3}{{125}}} \right)\sec \cr & = 48{\text{ sec}}\cr} $$