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 7 of 24.
Jasminder said:
1 decade ago
27x+17y total distance.
X+y=total speed.
Total distance/total speed=total time.
Total time= 23 given.
Solve it.
X+y=total speed.
Total distance/total speed=total time.
Total time= 23 given.
Solve it.
Maha said:
1 decade ago
Distance=speed*time.
1st train time =27sec. 2nd train time =17sec.
two trains can cross the each other in opposite direction is 23 sec.
so total time=(total distance)/totalspeed;
total distance,
1st&2nd train have speed x,y;
total distance=27x+17y; total speed=x+y;
23=(27x+17y)/x+y;
23x+23y=27x+17y;
6y=4x;
6/4=x/y;
3/2=x/y;
3:2
1st train time =27sec. 2nd train time =17sec.
two trains can cross the each other in opposite direction is 23 sec.
so total time=(total distance)/totalspeed;
total distance,
1st&2nd train have speed x,y;
total distance=27x+17y; total speed=x+y;
23=(27x+17y)/x+y;
23x+23y=27x+17y;
6y=4x;
6/4=x/y;
3/2=x/y;
3:2
NICK said:
1 decade ago
Can any one tell me how to calculate mod(23-17) : mod(23-27)
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
Avinash said:
1 decade ago
First you have to know that mod(positive or negative value) = positive value
So mod(23-17)=6
mod(23-27)=mod(-4)=4
Hope you understood.
So mod(23-17)=6
mod(23-27)=mod(-4)=4
Hope you understood.
Rabish pandey said:
1 decade ago
Really it is helpful to get qualify related aptitude exam I thank from my side to indiabix. Com and easy to understand solution also.
Pardhu said:
1 decade ago
Avinash could you please explain clearly.
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 ???
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