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.
Ronie said:
7 years ago
You first solve for y which will give y=55/9m/s now convert it into km/h which will be 22km/h
But the actual speed of the train is:
(22-2)km/h=20km/h
2km/h is the speed of the 1st person
20km/h is 50/9 m/s when you convert
Now as we know,
Length of train = speed * time.
So, L=(50/9)*9,
or, L=50m.
Go through the formula how to convert kmph to mps. Because when it is given in seconds you have to convert it to mph.
But the actual speed of the train is:
(22-2)km/h=20km/h
2km/h is the speed of the 1st person
20km/h is 50/9 m/s when you convert
Now as we know,
Length of train = speed * time.
So, L=(50/9)*9,
or, L=50m.
Go through the formula how to convert kmph to mps. Because when it is given in seconds you have to convert it to mph.
Ronie said:
7 years ago
@Vamc.
You need to convert the 2kmph and 4kmph to mps. So on converting it gives 5/9 m/s and 10/9m/s.
According to the formula for the same direction movement: (speed of train -speed of man).
As the speed of man is slow.
You need to convert the 2kmph and 4kmph to mps. So on converting it gives 5/9 m/s and 10/9m/s.
According to the formula for the same direction movement: (speed of train -speed of man).
As the speed of man is slow.
Charan said:
7 years ago
Well said. Thanks all.
Divya Bisht said:
7 years ago
Can anyone tell me that why we take 2kmph nd 4kmph separate? Why we not done 4kmph+2kmph?
Meenu said:
7 years ago
Sir, how can you take 9y-5=x?
If we multiply by 9 all over the equation the that is 9(9y-5)=x then their answer will be 81y-45=x. Please explain.
If we multiply by 9 all over the equation the that is 9(9y-5)=x then their answer will be 81y-45=x. Please explain.
Ramya said:
6 years ago
Relative speed respect to 1 person and train= (x-2).
Relative speed respect to 2nd person and train= (x-4).
Dist. = speed * time.
9(x-2) = 10(x-4).
x = 22.
So, 22-2 = dist/9.
dist = 180*(5/18).
> 50.
Relative speed respect to 2nd person and train= (x-4).
Dist. = speed * time.
9(x-2) = 10(x-4).
x = 22.
So, 22-2 = dist/9.
dist = 180*(5/18).
> 50.
Kubstar said:
6 years ago
@Ramya.
Why are you subtract "2" from 22?
Why are you subtract "2" from 22?
Muttu said:
6 years ago
@Amith is well solved.
Mihir gandhi said:
5 years ago
Assume speed for s and find length means (distance)
D1=D2
(S-2)*T1=(s-4)*T2.
(S-2)*9/3600=(s-4)*10/3600.
(S-2)*9=(s-4)*10.
9s-18=10s-40.
10s-9s=40-18.
S=22.
So find length
L=D=(s-2)*T1.
=(22-2)*9/3600,
=20*9/3600,
=2*9/360,
=18/360,
=1/20,
=0.05km.
=50 meter.
D1=D2
(S-2)*T1=(s-4)*T2.
(S-2)*9/3600=(s-4)*10/3600.
(S-2)*9=(s-4)*10.
9s-18=10s-40.
10s-9s=40-18.
S=22.
So find length
L=D=(s-2)*T1.
=(22-2)*9/3600,
=20*9/3600,
=2*9/360,
=18/360,
=1/20,
=0.05km.
=50 meter.
Vishal k. Ramteke said:
5 years ago
Its pretty simple just follow the short trick i.e we know that,
Distance= Time*speed.
So,
Distance=9*10*2*5/18 just solve this you will get the answer in a minute i.e 50 m.
Distance= Time*speed.
So,
Distance=9*10*2*5/18 just solve this you will get the answer in a minute i.e 50 m.
(1)
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