Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 28)
28.
A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
Answer: Option
Explanation:
| 4.5 km/hr = |  |
4.5 x | 5 |  |
m/sec = | 5 | m/sec = 1.25 m/sec, and |
| 18 | 4 |
| 5.4 km/hr = | |
5.4 x | 5 | |
m/sec = | 3 | m/sec = 1.5 m/sec. |
| 18 | 2 |
Let the speed of the train be x m/sec.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
8.4x - 10.5 = 8.5x - 12.75
0.1x = 2.25
x = 22.5
 Speed of the train = |
|
22.5 x | 18 | |
km/hr = 81 km/hr. |
| 5 |
Discussion:
54 comments Page 2 of 6.
Krishna said:
1 decade ago
Distance means it,s length of train ---------->(x-1.25)*8.4
(x-1.25)------>is relative speed,
8.4----------->time.
Same as in case of (x-1.5)x8.5.
In both cases gives distence of cover it,s idirectly both are same.
So (x-1.25)x8.4=(x-1.5)x8.5-----> find x from this you can get speed
(x-1.25)------>is relative speed,
8.4----------->time.
Same as in case of (x-1.5)x8.5.
In both cases gives distence of cover it,s idirectly both are same.
So (x-1.25)x8.4=(x-1.5)x8.5-----> find x from this you can get speed
Kishore mech said:
1 decade ago
x is the length of the train; y is the velocity of the train
x/y=t1 ; x/y=t2
x/(y-1.25)=8.4 ; x/(y-1.5)=8.5
[4.5kmph=1.25m/s ;5.4kmph=1.5m/s]
x-8.4y=-10.75..........(1)
x-8.5y=-12.75..........(2)
solving the equ 1 & 2
we get thge solution y=22.5m/s
there for ans;
the speed of the train y=22.5m/s=81kmph
x/y=t1 ; x/y=t2
x/(y-1.25)=8.4 ; x/(y-1.5)=8.5
[4.5kmph=1.25m/s ;5.4kmph=1.5m/s]
x-8.4y=-10.75..........(1)
x-8.5y=-12.75..........(2)
solving the equ 1 & 2
we get thge solution y=22.5m/s
there for ans;
the speed of the train y=22.5m/s=81kmph
Yaswanth said:
1 decade ago
Hi Vinay, the distance here is the length of train.... Consider the human as a point. Then length of train from case 1 is 8.4(x-1.25) and from case 2 is 8.5(x-1.5)...
Since the length of train is constant...
We can equate both.
Since the length of train is constant...
We can equate both.
Yakesh said:
1 decade ago
Thank you! gowthami.
Divya said:
1 decade ago
Thank you manasa. Nice explanation.
Tinku majumder said:
1 decade ago
How 5/18 Came?
Suan said:
1 decade ago
We don't necessarily have to use the conversion of kmph to m/s here since the unnecessary units (sec) eventually cancel out each other on the subsequent steps of solution.
Vaisakh said:
1 decade ago
Just covert seconds to hour by dividing 60x60 = 3600.
No more conversions:
Let length be x.
and Speed be y.
8.4/3600 = x/(y-4.5).
8.4y -37.80 = 3600x ...(eq 1).
8.5/3600 = x/(y-5.4).
8.5y-45.90 = 3600x....(eq 2).
Solving both equations will give you, y= 81kmph.
No more conversions:
Let length be x.
and Speed be y.
8.4/3600 = x/(y-4.5).
8.4y -37.80 = 3600x ...(eq 1).
8.5/3600 = x/(y-5.4).
8.5y-45.90 = 3600x....(eq 2).
Solving both equations will give you, y= 81kmph.
Sanket said:
1 decade ago
In this step:
(x - 1.25) x 8.4 = (x - 1.5) x 8.5.
Why " x-1.25 " taken?
(x - 1.25) x 8.4 = (x - 1.5) x 8.5.
Why " x-1.25 " taken?
Haphyz said:
1 decade ago
Since we are calculating relative speeds, the value of the distance can only imply the length of the train which is constant. Hence the need to equate both distances.
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