Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 16)
16.
A train travelling at a speed of 75 mph enters a tunnel 3
miles long. The train is
mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?


Answer: Option
Explanation:
Total distance covered |
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= 3 min. |
Discussion:
129 comments Page 9 of 13.
Nivin said:
9 years ago
I agree with you @Sanjay Singh.
Abi said:
9 years ago
Why finally it convert into hours? Please explain.
Raju Ahmed said:
9 years ago
Total distance covered = 7/2(tunnel) + 1/4(length of train) + 1/4(length of train since the rear part is to be considered emerging) = 4 miles.
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is [C].
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is [C].
Appa rao said:
9 years ago
The answer should be option C.
Total distance covered = 7/2(tunnel) + 1/4(length of train) + 1/4(length of train since the rear part is to be considered emerging) = 4 miles.
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is option C.
Total distance covered = 7/2(tunnel) + 1/4(length of train) + 1/4(length of train since the rear part is to be considered emerging) = 4 miles.
So total time taken = (4 miles/75mph ) * 60 mins.
= 3.2 mins.
So the answer is option C.
Ajinath khedkar said:
9 years ago
Yes, Option C is the correct answer.
Sathish said:
9 years ago
Why this problem isn't convert the miles to meters or K.meters?
Priya said:
9 years ago
Please explain how come 60?
Priya said:
8 years ago
Can anyone explain 7/2+1/4=15/4 clearly?
Varinder said:
8 years ago
The total distance should be 7/2(tunnel)+1/4(while front entering in the tunnel)+1/4(while the rear is out of the tunnel) = 4 miles.
Prudhvi (palakol) said:
8 years ago
Simple 3.5 + 0.25 = 3.75,
3.75/75 = 0.05,
0.05 * 60 = 3.
3.75/75 = 0.05,
0.05 * 60 = 3.
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