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:
234 comments Page 9 of 24.
Dileep kumar said:
1 decade ago
27x + 17y/(x+y) = 23;
27x + 17y = 23x + 23y;
4x = 5y;
x/y = 5/4.
27x + 17y = 23x + 23y;
4x = 5y;
x/y = 5/4.
Vipul said:
1 decade ago
First we are doing KM Hr to MS.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Carlo said:
1 decade ago
Let Train A distance = 27x Train A speed = x.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Askar said:
1 decade ago
:) The simple formula use for this problem is S = v*t.
Arunkumar said:
1 decade ago
Hi,
Nice explanation for why we need to.
1. Add the both trains speed which were running in opposite Directions with a reference point.
2. Subtract the both trains speed which were running in Same direction with a reference point.
http://en. Wikipedia. Org/wiki/Relative_velocity.
Nice explanation for why we need to.
1. Add the both trains speed which were running in opposite Directions with a reference point.
2. Subtract the both trains speed which were running in Same direction with a reference point.
http://en. Wikipedia. Org/wiki/Relative_velocity.
R.Abiraman said:
1 decade ago
Relative speed of train = x+y;
Train 1 dis = 27x;
Train 2 dis = 17y;
Formula(speed = distance/time).
So x+y = (27x+17y)/23.
23x+23y = 27x+17y;
-4x+6y = 0;
6y = 4x;
x/y = 6/4.
So 3:2.
Train 1 dis = 27x;
Train 2 dis = 17y;
Formula(speed = distance/time).
So x+y = (27x+17y)/23.
23x+23y = 27x+17y;
-4x+6y = 0;
6y = 4x;
x/y = 6/4.
So 3:2.
Anil vattamwar said:
1 decade ago
Train1 time is = 27.
Train2 time is = 17.
Divide them = (27/17) => (3/2).
Train2 time is = 17.
Divide them = (27/17) => (3/2).
Chacko said:
1 decade ago
Can anyone say why is 27X+17Y = 23X+23Y....?
Ideally if we cross multiply, it should come as 27x+17Y = 23X-23Y.
Ideally if we cross multiply, it should come as 27x+17Y = 23X-23Y.
Kriti said:
1 decade ago
How can the time be taken as the ratio for length?
I mean why 27x and 17y. 27 and 17 are time, then how can they be taken as ratios of length?
I mean why 27x and 17y. 27 and 17 are time, then how can they be taken as ratios of length?
Anil said:
1 decade ago
If, 4x=6y.
Then how to come ratio 3:2?
How it possible?
Then how to come ratio 3:2?
How it possible?
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