Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 4)
4.
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
Answer: Option
Explanation:
Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
 |
27x + 17y | = 23 |
| x+ y |
27x + 17y = 23x + 23y
4x = 6y
| |
x | = | 3 | . |
| y | 2 |
Discussion:
233 comments Page 5 of 24.
Ajit said:
1 decade ago
Dear rohit,
Your calculation right but this furmula use in every cases or limited case?
Your calculation right but this furmula use in every cases or limited case?
Jagadeesh said:
1 decade ago
Hi priya u can do like this take speed of train A is x and speed of B is Y
x(take difference between speed of train A and crossing time)=y(take difference between train B and crossing time) i.e
x(27-23)=y(23-17)
4x=6y
x/y=3/2
x(take difference between speed of train A and crossing time)=y(take difference between train B and crossing time) i.e
x(27-23)=y(23-17)
4x=6y
x/y=3/2
Riyas said:
1 decade ago
My explain,
You know 2train x and y
crossing the time =27then17 so 27x 17y
Given
Then each other train crossing 23 second.
so 23x 23y
27x+17y=23x+23y
27x-23x=23y-17y
Tf 4x=6y
x=6y/4
x/y=6/4
=3/2
You know 2train x and y
crossing the time =27then17 so 27x 17y
Given
Then each other train crossing 23 second.
so 23x 23y
27x+17y=23x+23y
27x-23x=23y-17y
Tf 4x=6y
x=6y/4
x/y=6/4
=3/2
BhargaV said:
1 decade ago
Why km/hr to m/sec conversion takes 5/18 ???
Nisha said:
1 decade ago
Hi friends .....
1 km =1000m
1 hr=3600s
So, km/hr = 1000/3600 = 5/18
It's clear.
1 km =1000m
1 hr=3600s
So, km/hr = 1000/3600 = 5/18
It's clear.
Dhiraj said:
1 decade ago
(la+lb)/(Sa+Sb)=23
la/sa=27
la/sb=17
sa/sb=?
sa/sb=3:2
la/sa=27
la/sb=17
sa/sb=?
sa/sb=3:2
Swetha said:
1 decade ago
It can also be solved like this using the methods of allegation and mixture 27 17.
23.
6 4.
Thus ratio is 6:4 3:2.
23.
6 4.
Thus ratio is 6:4 3:2.
Sachin kamble said:
1 decade ago
According to my opinion the first train crosses the man after crossing the other train i.e. 4 sec late and second train will crosses the man 6 sec earliar before it cross the other train .
Just take the ratio of 6:4 we get 3:2 i.e speed of seocnd train:speed of first train.
Just take the ratio of 6:4 we get 3:2 i.e speed of seocnd train:speed of first train.
Rajkumar said:
1 decade ago
Let the speed of the two trains be a=x m/sec and b=y m/sec,
crossing time=23sec
formula :distance=speed*time
for 1st train,distance,u=x*27=27x
for 2nd train,distance,v=y*17=17x
suppose in opposite direction the formula is time=u+v/a+b
therefore,27x+17x/x+y=23
27x+17y=23(x+y)
27x+17y=23x+23y
27x-23x=23y-17y
4x=6y
x/y=6/4
x/y=3/2
x:y=3:2
crossing time=23sec
formula :distance=speed*time
for 1st train,distance,u=x*27=27x
for 2nd train,distance,v=y*17=17x
suppose in opposite direction the formula is time=u+v/a+b
therefore,27x+17x/x+y=23
27x+17y=23(x+y)
27x+17y=23x+23y
27x-23x=23y-17y
4x=6y
x/y=6/4
x/y=3/2
x:y=3:2
Jyotirmoy said:
1 decade ago
Let the speed of the two trains be a=x m/sec and b=y m/sec
Length of a=27x and b=17y
So (27x+17y)=23(x+y).
So 4x=6y
So x/y=3/2
[In First sight I thought as the 1st train is taking more time to cross the man , so it will be the slower train, but as the lenght of the 1st train longer so it has taken longer time althought it is the faster train]
Length of a=27x and b=17y
So (27x+17y)=23(x+y).
So 4x=6y
So x/y=3/2
[In First sight I thought as the 1st train is taking more time to cross the man , so it will be the slower train, but as the lenght of the 1st train longer so it has taken longer time althought it is the faster train]
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