Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 19)
19.
A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
Answer: Option
Explanation:
Let the length of the train be x metres and its speed by y m/sec.
| Then, | x | = 15  y = |
x | . |
| y | 15 |
 |
x + 100 | = | x |
| 25 | 15 |
15(x + 100) = 25x
15x + 1500 = 25x
1500 = 10x
x = 150 m.
Discussion:
55 comments Page 6 of 6.
Shubham lad said:
3 years ago
Can we calculate the speed of the train 100/25=4 m/s and length = 15 * 4 = 60m?
Please explain.
Please explain.
(5)
Bir bdr said:
2 years ago
Let length of train = x m.
From 1st case.
Speed = x/15.
From the Second case;
speed = x + 100/25.
Now x/15 = x + 100/25.
25x = 15x + 1500.
10x = 1500.
x = 150 m.
From 1st case.
Speed = x/15.
From the Second case;
speed = x + 100/25.
Now x/15 = x + 100/25.
25x = 15x + 1500.
10x = 1500.
x = 150 m.
(5)
Arun said:
1 year ago
Every question they used the word " cross the pole" but here they mention " past the pole " (mean we need to reduce the time from the platform).
So time has been 25-15 = 10, T = 10.
S = 100m/10sec = 10m/sec.
S = D/T.
x + D = S*T,
X + 100 = 10* 25,
x+100 = 250,
x = 250-100,
x = 150.
So time has been 25-15 = 10, T = 10.
S = 100m/10sec = 10m/sec.
S = D/T.
x + D = S*T,
X + 100 = 10* 25,
x+100 = 250,
x = 250-100,
x = 150.
(10)
Gautam Kumar Ray said:
8 months ago
We know S = D/T,
for train S = D/15,
Now,
D/15 = (D+100)/25,
3D + 300 = 5D,
D = 150.
for train S = D/15,
Now,
D/15 = (D+100)/25,
3D + 300 = 5D,
D = 150.
(2)
Gautam said:
8 months ago
Let the length of the train = x.
Now,
(X+100)/25 = X/15,
Then, X = 150.
Now,
(X+100)/25 = X/15,
Then, X = 150.
(1)
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