Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 13)
13.
Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:
Answer: Option
Explanation:
Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
 |
(100 + 100) | = 3x |
| 8 |
24x = 200
| x = |
25 | . |
| 3 |
| So, speed of the faster train = | 50 | m/sec |
| 3 |
| = |  |
50 | x | 18 |  km/hr |
| 3 | 5 |
= 60 km/hr.
Discussion:
74 comments Page 2 of 8.
Sarang said:
7 years ago
Distance =speed *time,
100+100=speed*8.
200=speed *8,
Speed=200/8,
Speed =25m/s,
Converting m/s into km/h.
25*18/5=90km/h,
As we want twice speed so first find single speed.
1+2 one is the speed of first train and 2 twice a second so combing we get 3 now,
Speed =90/3=30.
Now twice speed is 60km/h.
100+100=speed*8.
200=speed *8,
Speed=200/8,
Speed =25m/s,
Converting m/s into km/h.
25*18/5=90km/h,
As we want twice speed so first find single speed.
1+2 one is the speed of first train and 2 twice a second so combing we get 3 now,
Speed =90/3=30.
Now twice speed is 60km/h.
(1)
Indraneel Mal said:
2 years ago
Let speed be x and 2x kmph.
The total relative speed of the two trains = 3xkmph.
3600secs = 3000x.
1sec = 3000x/3600,
8secs=3000x/3600×8,
= 20x/3.
Btp;
20x/3 = 200,
x = 200×3/20,
=30.
Speed of slower train = 30kmph.
The faster train = 30 × 2 = 60kmph.
The total relative speed of the two trains = 3xkmph.
3600secs = 3000x.
1sec = 3000x/3600,
8secs=3000x/3600×8,
= 20x/3.
Btp;
20x/3 = 200,
x = 200×3/20,
=30.
Speed of slower train = 30kmph.
The faster train = 30 × 2 = 60kmph.
(9)
Mini said:
1 decade ago
Hi @jelli, I think this is correct.
length of each train=100 m
let speed of one train=x km/h
other is moving twice as fast the other i.e=2*x km/h
speed=distance/time
(x+2x)*5/18=200/8
x=30m
they are asking the speed of faster train which is 2*x
here we land up with x=30
So 2*30=60m
length of each train=100 m
let speed of one train=x km/h
other is moving twice as fast the other i.e=2*x km/h
speed=distance/time
(x+2x)*5/18=200/8
x=30m
they are asking the speed of faster train which is 2*x
here we land up with x=30
So 2*30=60m
Sravan said:
1 decade ago
Actually,
if speed of the slower train be x m/sec.
Then, speed of the faster train y = 2x m/sec=> x=y/2
Relative speed =(x+y)= (y/2+ y) m/sec = 3y/2 m/sec.
(100 + 100)/8 = 3y/2
3y=50
y=50/3 m/s
y=(50/3)*(18/5)kmph
speed of the faster train = 60 kmph
if speed of the slower train be x m/sec.
Then, speed of the faster train y = 2x m/sec=> x=y/2
Relative speed =(x+y)= (y/2+ y) m/sec = 3y/2 m/sec.
(100 + 100)/8 = 3y/2
3y=50
y=50/3 m/s
y=(50/3)*(18/5)kmph
speed of the faster train = 60 kmph
Praveen Sharma said:
10 years ago
Here (100+100)/8 = 3x.
200/8 = 3x.
25 = 3x.
So x = 25/3 m/s it is speed of slow train.
We want speed of fast train, it is 2x where x = 25/2 m/s.
So speed of fast train = 2*25/3 m/s.
So 50/3 m/s.
And we convert in km/s using 18/5.
So (50*18)/(3*5) = 900/15 = 60 km/h.
200/8 = 3x.
25 = 3x.
So x = 25/3 m/s it is speed of slow train.
We want speed of fast train, it is 2x where x = 25/2 m/s.
So speed of fast train = 2*25/3 m/s.
So 50/3 m/s.
And we convert in km/s using 18/5.
So (50*18)/(3*5) = 900/15 = 60 km/h.
(1)
R.Rakesh KUmar said:
8 years ago
@Navin.
In question they have mentioned to calculate the speed of faster train i.e to calculate 2X
X=25/3 is the speed of the slower train and ( " If one is moving twice as fast the other" given in question).
2 * X = 2 * 25/3 => 50/3.
In question they have mentioned to calculate the speed of faster train i.e to calculate 2X
X=25/3 is the speed of the slower train and ( " If one is moving twice as fast the other" given in question).
2 * X = 2 * 25/3 => 50/3.
IKram said:
1 decade ago
Hey Jelli...
1) if 2 things moves in same direction:
time they meet t=(x+y)/(u-v)
where x,y-distance of trains or smthng
u,v-speed's in mps.
2) if 2 things moves in opposite direction:
time they meet t=(x+y)/(u+v)
1) if 2 things moves in same direction:
time they meet t=(x+y)/(u-v)
where x,y-distance of trains or smthng
u,v-speed's in mps.
2) if 2 things moves in opposite direction:
time they meet t=(x+y)/(u+v)
Aashu said:
4 years ago
@Fabarah Israr.
It's not 25/6 it's 25/3.
And it is asked to calculate the speed of a faster train.
x= 25/3 (speed of the slower train)
whereas the speed of the faster train is 2x, therefore 2*25/3 gives 50/3.
It's not 25/6 it's 25/3.
And it is asked to calculate the speed of a faster train.
x= 25/3 (speed of the slower train)
whereas the speed of the faster train is 2x, therefore 2*25/3 gives 50/3.
(1)
Arifur Rahman said:
6 years ago
Let,
The speed of slower train be x m/sec.
The speed of faster train be 2x m/s.
Now,
(x+2x)÷2(two trains) = 100/8.
or,x=25/3.
The Speed of faster train = 2x = 2 * 25/3 = 50/3 m/sec * 18/5 = 60km/hr.
The speed of slower train be x m/sec.
The speed of faster train be 2x m/s.
Now,
(x+2x)÷2(two trains) = 100/8.
or,x=25/3.
The Speed of faster train = 2x = 2 * 25/3 = 50/3 m/sec * 18/5 = 60km/hr.
Jelli said:
1 decade ago
It's very simple i think
WKT t=(a+b)/(u-v)
here
t=8s,a=b=100m,u=x,v=2x(twice fast of other)
8=(100+100)/(x+2x)*5/18
x=30m
then in the problem they asked length of the faster train is 2x=60m.
WKT t=(a+b)/(u-v)
here
t=8s,a=b=100m,u=x,v=2x(twice fast of other)
8=(100+100)/(x+2x)*5/18
x=30m
then in the problem they asked length of the faster train is 2x=60m.
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