Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 5)
5.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Answer: Option
Explanation:
| Speed = |  |
54 x | 5 |  m/sec = 15 m/sec. |
| 18 |
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
| Then, | x + 300 | = 15 |
| 36 |
x + 300 = 540
x = 240 m.
Discussion:
276 comments Page 18 of 28.
Tina said:
1 decade ago
Total time = 36sec - 20sec = 16 sec.
Speed = 54*(5/18) = 15 m/sec.
Length = speed * time = 15 * 16 = 240m.
Speed = 54*(5/18) = 15 m/sec.
Length = speed * time = 15 * 16 = 240m.
R.sai said:
9 years ago
Another method we do that problem.
Train reach time-man standing time = 36-20 = 16.
Length = 15*16 = 240m.
Train reach time-man standing time = 36-20 = 16.
Length = 15*16 = 240m.
Sai said:
9 years ago
Length = Speed * Time.
Speed = 54 * 5/18 = 15 m/sec.
Time = 36 - 20 = 16.
Length = 15 * 16.
= 240m.
Speed = 54 * 5/18 = 15 m/sec.
Time = 36 - 20 = 16.
Length = 15 * 16.
= 240m.
Shelke ganesh mariba said:
6 years ago
According to speed of trai = 15m/s.
And time = 36-20 = 16sec.
Therefore length of platform = 15x16 = 240m.
And time = 36-20 = 16sec.
Therefore length of platform = 15x16 = 240m.
Sravana said:
6 years ago
D = TxS.
As we mentioned here,
54kmph = 54x(5÷18) = 15mps.
D = TxS.
= (36-20)x15.
= 16x15.
= 240mts.
As we mentioned here,
54kmph = 54x(5÷18) = 15mps.
D = TxS.
= (36-20)x15.
= 16x15.
= 240mts.
Varad shinde said:
8 years ago
He guys,
In the problem, there is a common formula used from the examiner is,
Distance = Speed x Time.
In the problem, there is a common formula used from the examiner is,
Distance = Speed x Time.
Sheik said:
2 decades ago
Me also have the same doubt.
Sahil,
Dis= s*t is okay.
Why didn't we take 300=15*36.
Please explain.
Sahil,
Dis= s*t is okay.
Why didn't we take 300=15*36.
Please explain.
Baskar.R said:
1 decade ago
Speed = 54x5m/sec = 15 m/sec
15*36=540
man standing= 15*20=300
ans= platform length=540-200
ans= 240
15*36=540
man standing= 15*20=300
ans= platform length=540-200
ans= 240
Mvk said:
1 decade ago
S (km/hr to m/s) => 54 * 5/18 = 15.
T = (36s - 20s) = 16s.
D = S * T = 15 * 16 = 240m.
T = (36s - 20s) = 16s.
D = S * T = 15 * 16 = 240m.
Manisha dambe said:
7 years ago
Time to pass platform = 16s,
length of platform=t*s,
= 16*54*5/18=245.
The length of platform=245 m.
length of platform=t*s,
= 16*54*5/18=245.
The length of platform=245 m.
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