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.
Vickey pandy said:
5 years ago
Same direction =>2-4=2.
2*(5/18)= 5/9,
D = S*T.
D = (5/9)*10*9,
D = (5/9)*90,
D = 50.
2*(5/18)= 5/9,
D = S*T.
D = (5/9)*10*9,
D = (5/9)*90,
D = 50.
Mukundan said:
5 years ago
@Mancy.
Good Explanation. Thanks.
Good Explanation. Thanks.
Raksha said:
5 years ago
I did not get it properly. Could someone explain me briefly?
Sumit said:
5 years ago
@Ramya.
Why don't you use x-4 equation?
Why don't you use x-4 equation?
Pooja said:
5 years ago
@All.
Why they do not subtract speed first, then convert into m/sec?
Can anyone explain, please?
Why they do not subtract speed first, then convert into m/sec?
Can anyone explain, please?
Nisma said:
5 years ago
There is a simple formula to do this type of questions.
If a train covers 2 persons with speed S1 and S2 in time T1 and T2, then the length of train is:
(S1-S2 ÷ T1-T2) * T1 * T2.
So applying it in this question;
{(4-2)5/18÷10-9} * 10 * 9 =50m.
Note:(4-2) is multiplied by 5/18 so as to convert km to m.
If a train covers 2 persons with speed S1 and S2 in time T1 and T2, then the length of train is:
(S1-S2 ÷ T1-T2) * T1 * T2.
So applying it in this question;
{(4-2)5/18÷10-9} * 10 * 9 =50m.
Note:(4-2) is multiplied by 5/18 so as to convert km to m.
Karthik said:
5 years ago
Simplify to the easy method :
Distance we don't know:
s = s.
(s-2)*9/60*60=(s-4)*10/60*60---> Equation 1.
9s-18=10 - 40.
s=22--->Equation 2
Substitute equation 2 into 1.
Ans:180
Because we convert into meter & multiply by the 5/18 then we got.
180*5/18 = 50m.
Distance we don't know:
s = s.
(s-2)*9/60*60=(s-4)*10/60*60---> Equation 1.
9s-18=10 - 40.
s=22--->Equation 2
Substitute equation 2 into 1.
Ans:180
Because we convert into meter & multiply by the 5/18 then we got.
180*5/18 = 50m.
(1)
Rakshikasaravanan said:
5 years ago
Speed = 2 - 4 = 2kmps.
S = 2 * 5/18 = 5/9m/s.
L = s * t
L= 5/9 * 9 * 10
L= 50.
S = 2 * 5/18 = 5/9m/s.
L = s * t
L= 5/9 * 9 * 10
L= 50.
(3)
Amigo said:
4 years ago
Instead of taking 2 variables.
Step1 : Simply find the velocity of the train. If velocity is known we can find the distance, since time is already;
Given.
(x - 5/9)*9 = (x - 10/9)*10,
81x - 45 = 90x - 100,
==> x = 55/9.
Step2 : Since we know the time for each man, take anyone say the man with velocity 10/9.
Relative velocity = (55 -10)/9 = 5/9.
Step3 : Distance = relative vel * time.
x = 5*10/9 = 50m.
Thank you.
Step1 : Simply find the velocity of the train. If velocity is known we can find the distance, since time is already;
Given.
(x - 5/9)*9 = (x - 10/9)*10,
81x - 45 = 90x - 100,
==> x = 55/9.
Step2 : Since we know the time for each man, take anyone say the man with velocity 10/9.
Relative velocity = (55 -10)/9 = 5/9.
Step3 : Distance = relative vel * time.
x = 5*10/9 = 50m.
Thank you.
(1)
Partha said:
4 years ago
90y - 5 = x, 90y - 9x = 100.
Putting the value of x in 2nd so 90y - 81y + 45 = 100,
So after solving, y=55/9,
The put the value in any equation then we get x=50m,
Thank you.
Putting the value of x in 2nd so 90y - 81y + 45 = 100,
So after solving, y=55/9,
The put the value in any equation then we get x=50m,
Thank you.
(2)
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