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Train Sum

by shankar.ashwin » Wed Sep 14, 2011 6:29 am
A train after travelling 50km from A meets with an accident and proceeds at 4/5th of the former speed and reaches B, 45 mins late. Had the accident happened 20kms further on, it would have arrived 12 mins sooner. Find the original speed and distance.

Ans is [spoiler]25kms/hr and 125kms [/spoiler]

Could someone explain this?
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by gmatboost » Wed Sep 14, 2011 7:49 am
Let's call the distance from the point of the accident to the end of the trip d.
Let's call the initial rate r.
Let's also remember that time should ALWAYS BE CONVERTED TO HOURS when speed is given in km/hr or mi/hr, which it almost always is.
So, 12 mins= 1/5 hour. 45 mins= 3/4 hour.
A train after travelling 50km from A meets with an accident and proceeds at 4/5th of the former speed and reaches B, 45 mins late.
How much time should the trip have taken?
Total distance = 50 + d. Rate = r. Time = Distance/Rate = (50 + d)/r.

How much time did the trip actually take?
Distance of first piece = 50. Rate of first piece = r. Time = Distance/Rate = 50/r.
Distance of second piece = d. Rate of second piece = (4/5)r. Time = Distance/Rate = d/([4/5]r) = 5d/(4r).

Since this trip was 45 mins longer than it was supposed to be:
(50 + d)/r + 3/4 = 50/r + 5d/(4r)
Multiply through by r:
(50 + d) + 3r/4 = 50 + 5d/4
Multiply through by 4:
200 + 4d + 3r = 200 + 5d
[spoiler]3r = d[/spoiler]
Had the accident happened 20kms further on, it would have arrived 12 mins sooner. Find the original speed and distance.
Similar approach. This 20km would have been covered 12 minutes faster.
How long was it supposed to take?
20/r
How long did it actually take?
20/([4/5]r) = 100/(4r) = 25/r

20/r + 1/5 = 25/r
Multiply through by r:
20 + r/5 = 25
100 + r = 125
[spoiler]r = 25[/spoiler]

[spoiler]Since 3r = d, d = 75. The total distance was 50 + d, so it was 50 + 75 = 125.[/spoiler]
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by GMATGuruNY » Wed Sep 14, 2011 1:07 pm
shankar.ashwin wrote:A train after travelling 50km from A meets with an accident and proceeds at 4/5th of the former speed and reaches B, 45 mins late. Had the accident happened 20kms further on, it would have arrived 12 mins sooner. Find the original speed.

Ans is 25kms.

Could someone explain this?
Had the accident happened 20 kilometers further on -- implying that the train would be traveling at the original speed for 20 more kilometers -- it would have arrived 1/5 of an hour sooner.
Thus, the time to travel 20 kilometers at the slower speed must be 1/5 of an hour more than the time to travel 20 kilometers at the original speed.
On the GMAT we could plug in the answers, which would represent the original speed.

Answer choice: 25 kph
Time to travel 20 kilometers = d/r = 20/25 = 4/5 of an hour.
Slower speed = (4/5)*25 = 20 kph.
Time to travel 20 kilometers at the slower speed = 20/20 = 1 hour.
Difference in time = 1 - 4/5 = 1/5 of an hour.
Success!

To determine the distance, we also could plug in the answer choices.
Even without answer choices, we can quickly reason our way to the correct distance.

After the accident, the train arrives 3/4 of an hour late.
Thus, the time to finish the trip at the slower speed must be 3/4 of an hour more than the time to finish the trip at the original speed.

Let the remaining distance after the accident = 100 kilometers.
Time at the original speed = d/r = 100/25 = 4 hours.
Time at the slower speed = d/r = 100/20 = 5 hours.
Difference = 5-4 = 1 hour.
The difference is too great.
The remaining distance must be smaller.

Let the remaining distance after the accident = 75 kilometers.
Time at the original speed = d/r = 75/25 = 3 hours.
Time at the slower speed = d/r = 75/20 = 15/4 hours.
Difference = 15/4 - 3 = 3/4 of an hour.
Success!

Since the distance before the accident is 50 kilometers, the total distance = 50+75 = 125 kilometers.
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