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 9 of 11.
Poonam said:
1 decade ago
At last why it is multiplied by 9 not by 10?
Swapnil jagadale said:
1 decade ago
Take Length of Train = X meter.
And Speed of train = Y m/s.
2kmph = 5/9m/s, 4kmph = 10/9 m/s.
For first train X/(Y-5/9) = 9.
That is X-9Y = -5.....eq1.
For Second train X/(Y-10/9) = 10.
That is 9X-90Y = -100.....eq2.
Solving eq1 and eq2.
X = 50 meter.
And speed Y = 22 kmph or 55/9 m/s.
And Speed of train = Y m/s.
2kmph = 5/9m/s, 4kmph = 10/9 m/s.
For first train X/(Y-5/9) = 9.
That is X-9Y = -5.....eq1.
For Second train X/(Y-10/9) = 10.
That is 9X-90Y = -100.....eq2.
Solving eq1 and eq2.
X = 50 meter.
And speed Y = 22 kmph or 55/9 m/s.
K.Anitha said:
1 decade ago
2 kmph = 2*(5/18) = 5/9 (m/s).
4 kmph = 4*(5/18) = 10/9 (m/s).
Let the speed of the train be x.
(x-5/9)9 = (x-10/9)10 (because the distance is same).
On solving we get.
x = 55/9 (m/s).
Train length = (x-5/9)9.
= (55/9-5/9)9.
= (50/9)*9.
= 50 m.
4 kmph = 4*(5/18) = 10/9 (m/s).
Let the speed of the train be x.
(x-5/9)9 = (x-10/9)10 (because the distance is same).
On solving we get.
x = 55/9 (m/s).
Train length = (x-5/9)9.
= (55/9-5/9)9.
= (50/9)*9.
= 50 m.
Jitendra raghuvanshi said:
1 decade ago
What is find out by 5/9 and 2/9.
I mean is it speed of train ?
I mean is it speed of train ?
Jeevitha said:
1 decade ago
@Kapil.
Your method is easy. Thanks for it. But at last step it is:
((55/9)-(5/9))*9=50m.
CNA finish off this way.
Your method is easy. Thanks for it. But at last step it is:
((55/9)-(5/9))*9=50m.
CNA finish off this way.
Angad yadav said:
1 decade ago
Because the man who walking with the velocity greater, the train take maximum time to overtake him than the other guy.
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.
Mancy said:
1 decade ago
@everyone who didn't get how x= 50 came?
Let me explain in easy way.
Here, on solving we get,
9y-x = 5.........eq no.1.
&90y-9x = 100......eq no.2.
Now, take eq no.1.
9y-x = 5.
9y = 5+x.
y = 5+x/9.
Now put the value of y in eq no.2 then,
90y-9x = 100.
90(5+x/9)-9x = 100 ( here y= 5+x/9).
Now in L.H.S 90 is cancelled by 9 and become 10.
So, 10(5+x)-9x = 100.
50+10x-9x = 100.
50 +x = 100.
x = 100-50.
Answer x = 50.
Hope you'll get it now.
Let me explain in easy way.
Here, on solving we get,
9y-x = 5.........eq no.1.
&90y-9x = 100......eq no.2.
Now, take eq no.1.
9y-x = 5.
9y = 5+x.
y = 5+x/9.
Now put the value of y in eq no.2 then,
90y-9x = 100.
90(5+x/9)-9x = 100 ( here y= 5+x/9).
Now in L.H.S 90 is cancelled by 9 and become 10.
So, 10(5+x)-9x = 100.
50+10x-9x = 100.
50 +x = 100.
x = 100-50.
Answer x = 50.
Hope you'll get it now.
Arvind said:
1 decade ago
@Yogesh,
By doing that we are not getting 50.
Got x = 50/8.
Please explain?
By doing that we are not getting 50.
Got x = 50/8.
Please explain?
Yogesh said:
1 decade ago
Multiply The eq1 by 10, IT will become 90y-10x=50. Now subtract eq1 from from 2.
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