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 4 of 11.
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)
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)
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?
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.
Ronie said:
7 years ago
@Mancy
Check your equation.
You wrote : 9y = 5+x.
y = 5+x/9. This is not possible you have divided x by 9 but haven't divided 5 by 9.
It would be y=5/9+ x/9.
Check your equation.
You wrote : 9y = 5+x.
y = 5+x/9. This is not possible you have divided x by 9 but haven't divided 5 by 9.
It would be y=5/9+ x/9.
Surbhi said:
7 years ago
Let the speed of train is x.
Then (x-2)9=(x-4)10.
9x-18=10x-40,
X=22.
Now let the length of the train is L.
L/(22-2)5/18=9,
So, L=50.
Then (x-2)9=(x-4)10.
9x-18=10x-40,
X=22.
Now let the length of the train is L.
L/(22-2)5/18=9,
So, L=50.
Vivek Tiwari said:
4 weeks ago
@All.
Just, let the length of train = x, then find the speed of the train after that equals the relative speed of person and the relative sped of the train.
Just, let the length of train = x, then find the speed of the train after that equals the relative speed of person and the relative sped of the train.
Aadi said:
1 decade ago
In the practical: if the length of train is 50m then why train crossing two mens in 9 and 10 sec. When they are walking 2kmph and 4kmph constantly.
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.
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).
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