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:
239 comments Page 2 of 24.
Anubhav Bharti said:
3 years 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.
(33)
Dhinesh said:
3 years 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?
(31)
Sourabh kumar said:
2 years 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.
(30)
Konatham preetham said:
6 months ago
@All.
You can solve alligation like take right side top 27 left side top take 17 middle of the alligation is 23 (17-23 = 6) and (27-23 = 4) and 2:3 inversely proportion 3:2 So, you can solve like only same like sec or km/hr or litres or kgs same terms you can solve like that.
You can solve alligation like take right side top 27 left side top take 17 middle of the alligation is 23 (17-23 = 6) and (27-23 = 4) and 2:3 inversely proportion 3:2 So, you can solve like only same like sec or km/hr or litres or kgs same terms you can solve like that.
(27)
Thavapriya A said:
1 year ago
How to solve the following step?
27x+17y =23x+23y.
Could anyone tell me please?
27x+17y =23x+23y.
Could anyone tell me please?
(24)
Ishwarya said:
3 years ago
But why we can't use the formula of B:A?
Anyone, please explain me.
Anyone, please explain me.
(20)
Adesh Katiya said:
2 years 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).
(14)
Venkatesh said:
10 months ago
Let speeds be x and y.
Then, l1 = x * 27.
l2 = y * 17.
Time taken to cross each other = 23 seconds.
=> (l1+l2)/(x+y) = 23 seconds.
=> 27x + 17y = 23x + 23y.
=> x : y = 3:2.
Then, l1 = x * 27.
l2 = y * 17.
Time taken to cross each other = 23 seconds.
=> (l1+l2)/(x+y) = 23 seconds.
=> 27x + 17y = 23x + 23y.
=> x : y = 3:2.
(14)
DHANASURIYA M G said:
2 months ago
Step 1: Express lengths in terms of speed and time.
Train 1 crosses a man in 27 seconds:
L1 = 27x.
Train 2 crosses a man in 17 seconds: L2 = 17y.
Step 2: Use the time to cross each other.
When trains move in opposite directions, their relative speed is x+y.
They cross each other in 23 seconds, so:
L1 + L2 = (x+y) × 23.
Substitute L1 and L2: 27x + 17y = 23(x+y)
Step 3: Simplify the equation
27x + 17y = 23x + 23y⟹
4x = 6y⟹x/y = 3/2.
Train 1 crosses a man in 27 seconds:
L1 = 27x.
Train 2 crosses a man in 17 seconds: L2 = 17y.
Step 2: Use the time to cross each other.
When trains move in opposite directions, their relative speed is x+y.
They cross each other in 23 seconds, so:
L1 + L2 = (x+y) × 23.
Substitute L1 and L2: 27x + 17y = 23(x+y)
Step 3: Simplify the equation
27x + 17y = 23x + 23y⟹
4x = 6y⟹x/y = 3/2.
(14)
Anome said:
12 months ago
Very useful. Thanks all.
(12)
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