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anybody has an easy way to solve this PS, please

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by diebeatsthegmat » Sat Oct 29, 2011 9:09 am
A train met with an accident 60km away from station A. It completed the remaining journey at 5/6th of the original speed and reached station B 1hr 12mins late. Had the accident taken place 60km further, it would have been only 1hr late. what was the original speed of the train?

a. 60 km/hr
b. 55 km/hr
c. 65 km/hr
d. 70 km/hr
e. 48 km/hr
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by parul9 » Sat Oct 29, 2011 10:37 am
Not sure if I am correct, but will try to attempt this.

Case 1: Train meets an accident 60 kms from A
A------60kms---------(Acc1)---------------------------------B =>train is 1hr 12 mins late
xkm/s 5/6x km/s

Case 2: Train meets an accident 120 kms from A.

A------60kms---------(Acc1)-----60kms------(Acc2)------------B =>train is 1hr late
xkm/s 5/6km/s

Comparing the 2 cases, we can see that there is a 12 minute difference when the accident site is moved by 60 kms.
So, covering these 60 kms at normal speed is 12 mins faster than at 5/6th normal speed.

Time taken to cover 60 kms at normal speed = 60/x
Time taken to cover at 5/6th the speed = 6*60/5x

60*6/5x - 60/x = 12/60

Solving this, x = 60km/hr.

So, I guess the answer is A.

Please share the OA.
Also the source of this qsn. It somehow dsnt seem to GMAT like!
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by shankar.ashwin » Sat Oct 29, 2011 11:00 am
One of those sums you could get lost in algebra;

Let the speeds be 5X and 6X,

Had the train travelled in 6X speed for 60 kms longer, it would reach 12 mins earlier, translates,

60/5X = 60/6X + 12/60 (Everything is km/hr)

X=10. Org speed = 6X = 60 A
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