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 10 of 11.
Vinila said:
1 decade ago
In the 1st method.
On solving,
9y - x = 5 and 90y - 9x = 100.
We get: x = 50.
How?
On solving,
9y - x = 5 and 90y - 9x = 100.
We get: x = 50.
How?
Amit said:
1 decade ago
2km/hr = 5/9 m/s,,4km/hr = 10/9 m/s.
let speed=S m/s & lenth = L mtr.
Both man and train moving in same direction so speed of train w.r.to man will be (Speed of train-Speed of man).
case1: S-5/9 = L/9 (Speed=Distance/Time here Length is distance).
Case2: S-10/9 = L/10 (Time--- 9 & 10 sec given).
=>90S-50 = 10L & 90S-100 = 9L.
L = 50 m.
let speed=S m/s & lenth = L mtr.
Both man and train moving in same direction so speed of train w.r.to man will be (Speed of train-Speed of man).
case1: S-5/9 = L/9 (Speed=Distance/Time here Length is distance).
Case2: S-10/9 = L/10 (Time--- 9 & 10 sec given).
=>90S-50 = 10L & 90S-100 = 9L.
L = 50 m.
Nady said:
1 decade ago
(x/(y-5/9)) = 9.
Now cross multiply this.
So we get x = 9*(y-5/9) i.e.,x = 9y-5 and do the same for other one too.
Now cross multiply this.
So we get x = 9*(y-5/9) i.e.,x = 9y-5 and do the same for other one too.
Kuppu said:
1 decade ago
Let the length of the train be x metres and its speed by y m/sec.
Therefore 9y - 5 = x and 10(9y - 10) = 9x.
I can't understand this step. please anyone explain this?
Therefore 9y - 5 = x and 10(9y - 10) = 9x.
I can't understand this step. please anyone explain this?
Kadher said:
1 decade ago
We can take either s-s1 or s-s2.
If we substitute in s-s2.
Then,
(110/18)-(20/18)=(x/10)
(90/18)=x/10
5=x/10
x=50.
If we substitute in s-s2.
Then,
(110/18)-(20/18)=(x/10)
(90/18)=x/10
5=x/10
x=50.
Mamatha said:
1 decade ago
Please can any one of you tell me the problem in easy method ?
Syndhya said:
1 decade ago
@Kapil
Please say why we are taking here why here we taking s-s1 and not s-s2 in your solution.
Please say why we are taking here why here we taking s-s1 and not s-s2 in your solution.
Vinay said:
1 decade ago
Kapil finally it is 55/9 not 55/11 ok, you had solved really great.
Nikita said:
1 decade ago
Let consider the speed of train is x;
then distance is constant.
So
9*(5/9-x)=10*(10/9*x);
by solving
x=55/9
now length of train=
9*(5/9-55/9)
=50
(distance never be -ve)
then distance is constant.
So
9*(5/9-x)=10*(10/9*x);
by solving
x=55/9
now length of train=
9*(5/9-55/9)
=50
(distance never be -ve)
Md Ali Umar said:
1 decade ago
@Kapil
You solved the problem in very easily. I agree with it but finally you got speed 55/9 sec. It is not correct the unit of speed. The unit of speed in SI is m/s.
You solved the problem in very easily. I agree with it but finally you got speed 55/9 sec. It is not correct the unit of speed. The unit of speed in SI is m/s.
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