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 2 of 11.
Mamatha said:
1 decade ago
Please can any one of you tell me the problem in easy method ?
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.
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?
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.
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.
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?
Yogesh said:
1 decade ago
Multiply The eq1 by 10, IT will become 90y-10x=50. Now subtract eq1 from from 2.
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?
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.
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.
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