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61. Two train 100 meters and 95 meters long respectively pass each other in 27 seconds, when they run in the same direction and in 9 seconds when they run in opposite directions. Speed of the two trains are?

A. 44 km/hr, 22 km/hr

B. 52 km/hr, 26 km/hr

C. 36 km/hr, 18 km/hr

D. 40 km/hr, 20 km/hr

Discuss

Solution:

\begin{aligned} Let the speed of first train be \\ S_{1} km/hr and speed of second train \\ is S_{2} km/hr \\ As we know, \\ Time \\ = \frac{total distance}{relative speed in same/opposite direction} \\ In the same direction \\ \Rightarrow 27 sec  = \frac{(100 + 95)}{(S_{1} - S_{2}) \times \frac{5}{18}} \\ \Rightarrow 27 = \frac{195 \times 18}{(S_{1} - S_{2}) \times 5} \\ \Rightarrow S_{1} - S_{2} = 26....................... (i) \\ In the opposite direction, \\ \Rightarrow 9 = \frac{(100 + 95)}{(S_{1} + S_{2}) \times \frac{5}{18}} \\ \Rightarrow 9 = \frac{195 \times 18}{(S_{1} + S_{2}) \times 5} \\ \Rightarrow S_{1} + S_{2} = 39 \times 2 \\ \Rightarrow S_{1} + S_{2} = 78 \\ From equation (i) and (ii) \\ \Rightarrow S_{1} - S_{2} = 26 \\ \Rightarrow S_{1} + S_{2} = 78 \\ \Rightarrow S_{1} = \frac{26 + 78}{2} \\ \Rightarrow S_{1} = \frac{104}{2} \\ \Rightarrow S_{1} =  52 km/hr and \\ S_{2} =  26 km/hr \end{aligned}

62. A train running at the speed of 84 km/hr passes a man walking in opposite direction at the speed of 6 km/hr in 4 seconds. What is the length of train (in meter)?

A. 150 m

B. 120 m

C. 100 m

D. 90 m

Discuss

Solution:

\begin{aligned} Let length of train \\ = l metre \\ \Rightarrow Time \\ = \frac{total distance}{relative speed in opposite direction} \\ \Rightarrow 4 \sec = \frac{l + 0}{(84 + 6) \times \frac{5}{18} m/s} \\ \Rightarrow 4 = \frac{l}{90 \times \frac{5}{18}} \\ \Rightarrow l = 100 m \\ ∴ length of the train  =  100 m \end{aligned}

63. A train passes two bridges of length 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is?

A. 152 m

B. 125 m

C. 250 m

D. 120 m

Discuss

Solution:

\begin{aligned} Let the length of train x m \\ Speed of train \\ = \frac{(Length of train  +  length of bridge)}{Time taken in crossing} \\ According to information we get \\ \Rightarrow \frac{x + 500}{100} = \frac{x + 250}{60} \\ \Rightarrow 60 (x + 500) = 100 (x + 250) \\ \Rightarrow 3 (x + 500) = 5 (x + 250) \\ \Rightarrow 5 x + 1250 = 3 x + 1500 \\ \Rightarrow 5 x - 3 x = 1500 - 1250 \\ \Rightarrow 2 x = 250 \\ \Rightarrow x = \frac{250}{2} = 125 m \end{aligned}

64. Train A passes a lamp post in 3 seconds and 900 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)

A. 24 seconds

B. 37 seconds

C. 33 seconds

D. 27 seconds

Discuss

Solution:

\begin{aligned} Let the length of train be x m \\ When a train crosses a light \\ post in 3 second the distance covered \\ =  length of train \\ \Rightarrow speed of train  = \frac{x}{3} \\ Distance covered in crossing a \\ 900 meter platfrom in 30 seconds \\ =  Length of platfrom  +  length of train \\ Speed of train  = \frac{x + 900}{30} \\ \Rightarrow \frac{x}{3} = \frac{x + 900}{30} [∵ Speed  = \frac{Distance}{Time}] \\ \Rightarrow \frac{x}{1} = \frac{x + 900}{10} \\ \Rightarrow 10 x = x + 900 \\ \Rightarrow 10 x - x = 900 \\ \Rightarrow 9 x = 900 \\ \Rightarrow x = \frac{900}{9} = 100 m \\ When the length of the platform be 800m, \\ then time T be taken by train to cross 800m \\ long platfrom \\ \frac{x}{3} = \frac{x + 800}{T} \\ \Rightarrow T x = 3 x + 2400 \\ \Rightarrow 100 T = 300 + 2400 \\ \Rightarrow 100 T = 2700 \\ \Rightarrow T = \frac{2700}{100} = 27 seconds \end{aligned}

65. A train cover a distance of 3584 km in 2 days 8 hours. If it covers 1440 km on the first day and 1608 km on the second day, by how much does the average speed of the train for the remaining part of the journey differ from that for the entire journey?

A. 3 km/h

B. 4 km/h

C. 10 km/h

D. 2 km/h

Discuss

Solution:

\begin{aligned} Given , \\ Train cover 3584 kms in 2 days 8 hours \\ (2 days 8 hours  = \frac{7}{3} days) \\ Average speed  = \frac{3584}{\frac{7}{3}} \\ =  1536 km/day  = \frac{1536}{24} = 64 km/h \\ Distance covered in two days \\ =  1440 + 1608 =  3048 km \\ Remaining distance for third day \\ =  3584 - 3048  =  536 km \\ Third day 536 km is covered in \\ 8 hour with speed of \\ = \frac{536}{8} = 67 km/h \\ ( 3rd day total 536 km distance \\ covered by 67 km/hr in 8 hr) \\ ∴ Difference of average speedm \\ =  67 - 64  = 3 km/hr \end{aligned}

66. A train starts from A at 7 a.m. towards B with speed 50 km/h. Another train starts from B at 8 a.m. with speed of 60 km/h towards A. Both of them meet at 10 a.m. at C. The ratio of the distance AC to BC is?

A. 5 : 6

B. 5 : 4

C. 6 : 5

D. 4 : 5

Discuss

Solution:

The speed of train A is 50km/hr and A starts its journey at 7 AM and reaches C at 10 AM. Total Travel time = 3hr
∴ Distance cover by A in 3hr = 50 × 3 = 150KM
Similarly, the speed of train B is 60km/hr and B starts its journey at 8 AM and reaches C at 10 AM. Total Travel time = 2hr
∴ Distance cover by B in 2hr = 60 × 2 = 120KM
The ratio of the distance between AC : BC
= 150 : 120
= 5 : 4

67. Train A passes a lamp post in 9 seconds and 700 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)

A. 32 seconds

B. 31 seconds

C. 33 seconds

D. 30 seconds

Discuss

Solution:

\begin{aligned} Let the length of train be x m \\ When a train crosses a light \\ post in 9 second the distance covered \\ =  length of train \\ \Rightarrow speed of train  = \frac{x}{9} \\ Distance covered in crossing a \\ 700 meter platfrom in 30 seconds \\ =  Length of platfrom  +  length of train \\ Speed of train  = \frac{x + 700}{30} \\ \Rightarrow \frac{x}{9} = \frac{x + 700}{30} [∵ Speed  = \frac{Distance}{Time}] \\ \Rightarrow \frac{x}{3} = \frac{x + 700}{10} \\ \Rightarrow 10 x = 3 x + 2100 \\ \Rightarrow 10 x - 3 x = 2100 \\ \Rightarrow 7 x = 2100 \\ \Rightarrow x = \frac{2100}{7} = 300 m \\ When the length of the platform be 800m, \\ then time T be taken by train to cross 800m \\ long platform \\ \frac{x}{9} = \frac{x + 800}{T} \\ \Rightarrow T x = 9 x + 7200 \\ \Rightarrow 300 T = 2700 + 7200 \\ \Rightarrow 300 T = 9900 \\ \Rightarrow T = \frac{9900}{300} = 33 seconds \end{aligned}

68. Train A traveling at 63 kmph can cross a platform 199.5 m long in 21 seconds. How much would train A take to completely cross (from the moment they meet ) train B, 157 m long and traveling at 54 kmph in opposite direction which train A is traveling? (in seconds)

A. 16

B. 18

C. 12

D. 10

Discuss

Solution:

\begin{aligned} Speed of train A \\ =  63 kmph \\ = (\frac{63 \times 5}{18}) m/sec \\ =  17 .5 m/sec \\ Speed of train B \\ =  54 kmph \\ = (\frac{54 \times 5}{18}) m/sec  =  15 m/sec \\ If the length of train A be x metre, \\ then \\ Speed of train A \\ = \frac{Length of train  +  length of platform}{Time taken in crossing} \\ \Rightarrow 17.5 = \frac{x + 199.5}{21} \\ \Rightarrow 17.5 \times 21 = x + 199.5 \\ \Rightarrow 367.5 = x + 199.5 \\ \Rightarrow x = 367.5 - 199.5 \\ \Rightarrow 168 metres \\ Relative speed \\ =  ( Speed train A +  Speed train B) \\ =  (17 .5 + 15)  m/sec \\ =  32 .5 m/sec \\ Required time \\ = \frac{Length of train A +   Length of train B}{Relative speed} \\ = (\frac{168 + 157}{32.5}) seconds \\ = 10 seconds \end{aligned}

69. A train which is moving at an average speed of 40 km/h reaches its destination on time. When its average speed reduces to 35 km/h, then it reaches its destination 15 minutes late. The distance traveled by the train is?

A. 70 km

B. 80 km

C. 40 km

D. 30 km

Discuss

Solution:

\begin{aligned} Average speed of train \\ =  40 km/hr \\ Reach at its destination at on time \\ New average speed of train \\ =  35 km/h \\ Time  =  15 minutes \\ = \frac{15}{60} hours \\ Then distance travelled \\ = \frac{40 \times 35}{40 - 35} \times \frac{15}{60} \\ = \frac{40 \times 35}{5} \times \frac{15}{60} \\ =  70 km \end{aligned}

70. A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train?

A. 37.50 kmph

B. 36 kmph

C. 48 kmph

D. 30 kmph

Discuss

Solution:

\begin{aligned} Distance  =  Speed \times Time \\ Distance covered by train with the \\ speed of 30 kmph in 12 minutes is \\ =  30 \times \frac{12}{60} = 6 km \\ Distance covered by the same train \\ with the speed of 45 kmph in 8 minutes is \\ =  45 \times \frac{8}{60} = 6 km \\ Average speed \\ = \frac{total distance}{total time} . \\ \Rightarrow \frac{(6 + 6) km}{(12 + 8) min} = \frac{12}{20} \times 60 \\ =  36 kmph \end{aligned}

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