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 5 of 6.
Yakesh said:
1 decade ago
Thank you! gowthami.
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.
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
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
Vinay rawat said:
1 decade ago
How can we assume that the distance is equal ?
Gururaj said:
1 decade ago
But the two man is also moving then there is change in distance, so how then (x-1.25)x8.4=(x-1.5)x8.5 both are moving so both the distance are varying.
Can some one please explain this?
Can some one please explain this?
Vinay said:
1 decade ago
But as manasa said why to equal both the relative speed's.
I am not understanding. Can any one please help me.
I am not understanding. Can any one please help me.
Swetha said:
1 decade ago
Good explanation. Thank you manasa.
Manasa said:
1 decade ago
There is no need of confusion regarding this problem.
in this problem...train and the two persons are moving in the same direction...
Consider the speed of the train be X km/hr.
**(note:the SPEED and LENGTH remains same in both the cases)****
----------------------------------------------------------------
now let us consider relative speed of the train w.r.to the first person..we get...(X-1.25)
in similar manner,relative speed w.r.to the second train...we get...(X-1.5)
(NOTE:why i'm doing subtraction here ,since they are moving in the same direction)
now we know,distance=speed*time
(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 * (5/18)=81km/hr
in this problem...train and the two persons are moving in the same direction...
Consider the speed of the train be X km/hr.
**(note:the SPEED and LENGTH remains same in both the cases)****
----------------------------------------------------------------
now let us consider relative speed of the train w.r.to the first person..we get...(X-1.25)
in similar manner,relative speed w.r.to the second train...we get...(X-1.5)
(NOTE:why i'm doing subtraction here ,since they are moving in the same direction)
now we know,distance=speed*time
(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 * (5/18)=81km/hr
Gowthami said:
1 decade ago
@Gowthami
4.5km/hr=4.5x5m/sec =5m/sec=1.25m/sec
5.4km/hr=5.4x5m/sec=3m/sec=1.5m/sec
then speed of the train=x
distance=speed*time
so distance=(x-1.25)*8.4-------------1
distance=(x-1.5)*8.5----------------2
(1)=(2)
(x - 1.25) x 8.4 = (x - 1.5) x 8.5
4.5km/hr=4.5x5m/sec =5m/sec=1.25m/sec
5.4km/hr=5.4x5m/sec=3m/sec=1.5m/sec
then speed of the train=x
distance=speed*time
so distance=(x-1.25)*8.4-------------1
distance=(x-1.5)*8.5----------------2
(1)=(2)
(x - 1.25) x 8.4 = (x - 1.5) x 8.5
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