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:
231 comments Page 1 of 24.
SAURABH PANDEY said:
2 weeks ago
Very well done. Thanks.
(1)
Samiddho said:
3 weeks ago
Time taken by train 1 = 27s.
Time taken by train 2 = 17s.
Time taken for them to cross each other = 23s
Let the velocity of train 1 be v1 and that of train 2 is v2, and their lengths be l1 and l2 resp.
Net velocity when they cross each other v = v1 + v2.
Thus;
(l1+l2) = (v1+v2)x23---> i
and
l1 = v1 * 27 and l2 = v2 * 17---> ii
By Replacing ii in i.
27v1 + 17v2 = 23(v1+v2)
4v1 = 6v2
v1:v2 = 3:2.
Time taken by train 2 = 17s.
Time taken for them to cross each other = 23s
Let the velocity of train 1 be v1 and that of train 2 is v2, and their lengths be l1 and l2 resp.
Net velocity when they cross each other v = v1 + v2.
Thus;
(l1+l2) = (v1+v2)x23---> i
and
l1 = v1 * 27 and l2 = v2 * 17---> ii
By Replacing ii in i.
27v1 + 17v2 = 23(v1+v2)
4v1 = 6v2
v1:v2 = 3:2.
(3)
Shadab alam said:
6 months ago
@All.
If a train T1 takes 27 sec to cross the man which is more than the time taken by train T2 (17 sec) then how can the speed of Train T1 is greater than the speed of train T2?
option 2 is only correct when the length of train T1 is much greater than the length of train T2.
If a train T1 takes 27 sec to cross the man which is more than the time taken by train T2 (17 sec) then how can the speed of Train T1 is greater than the speed of train T2?
option 2 is only correct when the length of train T1 is much greater than the length of train T2.
(35)
Sourabh kumar said:
10 months ago
Let 1st train be x = 27.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
(23)
Jason valentine daniel said:
11 months ago
Let the speed of the two trains be x m/sec and y m/sec.
Then, the length of the first train = 27x metres,
And length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
So, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
Then, the length of the first train = 27x metres,
And length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
So, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
(44)
Adesh Katiya said:
11 months ago
u=27s
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
(10)
Dhinesh said:
1 year ago
Well, I have something different, 1st train is 27 sec so take this time as a total as 100% minus the 2nd train which is 17 sec so 62. 96 % is the percent of 1st train slower than the second train or vice versa.
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
(23)
Hemanth Jetti said:
1 year ago
Let 1st train be x = 27.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
27x - 23x = 23y - 17y.
4x = 6y.
x = 6y/4.
x/y = 3/2.
= 3:2.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
27x - 23x = 23y - 17y.
4x = 6y.
x = 6y/4.
x/y = 3/2.
= 3:2.
(140)
Anubhav Bharti said:
1 year ago
Let the speed of the two trains be x m/sec. and y m/sec.
Then, the length of the first train = 27x metres,
and length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
so, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
Then, the length of the first train = 27x metres,
and length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
so, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
(32)
Prem said:
2 years ago
Let's assume the distance is constant.
1st train has 27 sec time and X speed
2nd train has 17 sec time and Y speed.
Know that, two trains cross a platform in 23 sec.
The relative seed in opposite directions is (s1+s2).
s1--------x.
s2----------y.
Distance = speed * time.
total distance = Distance 1 + distance 2.
D = d1 + d2
(x+y) 23 = [x * 27] + [y * 17]
23x + 23y = x27 + y17
x/y = 3/2.
1st train has 27 sec time and X speed
2nd train has 17 sec time and Y speed.
Know that, two trains cross a platform in 23 sec.
The relative seed in opposite directions is (s1+s2).
s1--------x.
s2----------y.
Distance = speed * time.
total distance = Distance 1 + distance 2.
D = d1 + d2
(x+y) 23 = [x * 27] + [y * 17]
23x + 23y = x27 + y17
x/y = 3/2.
(45)
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