Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 27)
27.
A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
Answer: Option
Explanation:
| 2 kmph = |  |
2 x | 5 |  |
m/sec = | 5 | m/sec. |
| 18 | 9 |
| 4 kmph = | |
4 x | 5 | |
m/sec = | 10 | m/sec. |
| 18 | 9 |
Let the length of the train be x metres and its speed by y m/sec.
| Then, |  |
x |  |
= 9 and | |
x | |
= 10. |
|
|
9y - 5 = x and 10(9y - 10) = 9x
9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
Length of the train is 50 m.
Discussion:
106 comments Page 8 of 11.
Dayana said:
8 years ago
Can anyone explain this x/y-5/9, why do use this?, which formulae did you applied, please explain.
Sovaki said:
8 years ago
Can someone explain to me how you get (5/18)?
Naresh said:
8 years ago
The right answer is 50m. You will get the speed of the train is 22 kmph.So, Length is.
22-2 (velocity of first person) = length/9.
20*5/18 = length/9.
So, the length of the train is 50 m.
22-2 (velocity of first person) = length/9.
20*5/18 = length/9.
So, the length of the train is 50 m.
IDPRASAD said:
8 years ago
4-2=2 speed diff (same direction).
Kmph to mph.
2 * 5/18 = 5/9.
D=s * t
5/9 * 9 * 10 = 50.
Kmph to mph.
2 * 5/18 = 5/9.
D=s * t
5/9 * 9 * 10 = 50.
Shilpa Rana said:
8 years ago
Thank @Mancy.
Anurag Sinha said:
8 years ago
The simple ans just use this formula:-
Length of the train=(diff in speed *t1*t2)/diff in time,
Length of the train=(4*5*9*10/18*1)=50 m).
Length of the train=(diff in speed *t1*t2)/diff in time,
Length of the train=(4*5*9*10/18*1)=50 m).
Soumen said:
8 years ago
Lenth=(diff between speed * t1 * t2)/diff in time.
Lenth=(2 * 5 * 9 * 10)/(1 * 18),
= 50m.
Lenth=(2 * 5 * 9 * 10)/(1 * 18),
= 50m.
Rajesh nande said:
8 years ago
Best solution @Idprasad.
Jaimin said:
8 years ago
Can anyone explain the relation of time in detail: (t1*t2)/(t1-t2).
Farooqui mehtab said:
8 years ago
let x be the speed of train;
(x-2) *(5/18) *9 = (x-4) *(5/18) *10
9x-18 = 10x-40
x = 22 m/sec.
Now, substitute this in,
Length of train = (x-2) *(5/18) *9,
= (22-2) *(5/18) *9 = 20*5/2,
= 50 mins.
(x-2) *(5/18) *9 = (x-4) *(5/18) *10
9x-18 = 10x-40
x = 22 m/sec.
Now, substitute this in,
Length of train = (x-2) *(5/18) *9,
= (22-2) *(5/18) *9 = 20*5/2,
= 50 mins.
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