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 3 of 6.
Ravanan said:
8 years ago
Consider, train length is x now we find out the speed s=d/t here train length is d so s=x/15
x/15=(100+x)/25,
25x=(100+x)*15,
25x=1500+15x,
25x-15x=1500,
10x=1500,
x=1500/10,
x=150m.
x/15=(100+x)/25,
25x=(100+x)*15,
25x=1500+15x,
25x-15x=1500,
10x=1500,
x=1500/10,
x=150m.
(1)
SK Lodhi said:
1 decade ago
T = 15 Sec.
T+100 = 25 Sec.
--------------
T = 100/10 = 10 m/s where 10 sec. = 25-10(difference).
T = Length of the train
T = 15 Sec.*10 m/s=150 m.
Length of the train = 150 m.
T+100 = 25 Sec.
--------------
T = 100/10 = 10 m/s where 10 sec. = 25-10(difference).
T = Length of the train
T = 15 Sec.*10 m/s=150 m.
Length of the train = 150 m.
Priya said:
2 decades ago
@pavan
Since we have declared x-distance,y-speed
According to the below given formula,
Time=Distance/speed
Here Train speed time is specified as 15 sec so,
15 =x/y
Since we have declared x-distance,y-speed
According to the below given formula,
Time=Distance/speed
Here Train speed time is specified as 15 sec so,
15 =x/y
Mr.N.P. said:
7 years ago
Time Difference of Platform and Pole = 25-15= 10s.
Platform is 100m long.
speed of Train is = (D/T)=(100/10) =10m/s.
Length of pole(Train) = S * T = 10 * 15 = 150m.
Platform is 100m long.
speed of Train is = (D/T)=(100/10) =10m/s.
Length of pole(Train) = S * T = 10 * 15 = 150m.
(1)
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)
Shirish said:
9 years ago
Why are you adding the length of the platform and length of the train X+100?
In the question, the distance between pole and platform is nowhere mentioned.
In the question, the distance between pole and platform is nowhere mentioned.
James said:
1 decade ago
Let Speed = x m/s.
Distance = 15x m = 15*10 = 150.
Time = 15 secs.
Using d = st,
15x + 100) = 25x.
x = 10.
Therefore - distance = 15x m = 15*10 = 150.
Distance = 15x m = 15*10 = 150.
Time = 15 secs.
Using d = st,
15x + 100) = 25x.
x = 10.
Therefore - distance = 15x m = 15*10 = 150.
Neha said:
1 decade ago
Please let me know if we can take time as 25-15=10sec.
And solve it as,
Speed=100/10=10sec.
So D/T=S
It means x+100/25=10
x+100=10*25
x=250-100
x=150m.
And solve it as,
Speed=100/10=10sec.
So D/T=S
It means x+100/25=10
x+100=10*25
x=250-100
x=150m.
Swap said:
9 years ago
Suppose the length of the train is X.
Therefore speed X/15,
As given 25 * X/15 = 100 + X,
25X = 1500 + 15X,
25X - 15X = 1500,
10 X = 1500,
X = 150.
Therefore speed X/15,
As given 25 * X/15 = 100 + X,
25X = 1500 + 15X,
25X - 15X = 1500,
10 X = 1500,
X = 150.
Sia Siam Changsan said:
6 years ago
Let x meter be the length of the train.
We have, speed = distance/time.
S=x/15 ( to pole),
Again, s=d/t,
x/15= x+100/25(to platform),
x=150m.
We have, speed = distance/time.
S=x/15 ( to pole),
Again, s=d/t,
x/15= x+100/25(to platform),
x=150m.
(3)
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