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A train met with an accident 60km away from station A. It co [#permalink]

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18 Oct 2011, 07:37

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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

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

let the original speed be 6x. A/q, traveling 60 km at 5/6th of original speed costs him 12 minutes etc

My solution: s=original spped, t=time taken in minutes to cover journey usually (t+72-60/s)(5/6s)+60=(t+60-120/s)(5/6s)+120...........as total distance covered is same in both cases solving, s=1 km/min=60km/hr hence ans A I know its too long.....but taking shortcuts was messing me up....

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

The logical approach here is this:

The accident took place 60 km from A. The train was delayed by 1hr 12 min. If it had instead taken place 120 km from A, it would have been only 1 hr late. What does this 12 min gap in time signify? It is the delay caused by reducing the speed by (1/6)th over a distance of 60 km. The ratio of the speed over this 60 km stretch is 5:6 (speed reduced by a sixth : original speed) Time taken 6:5 (Time taken with reduced speed : original time taken) We know this difference of 1 in time taken is given by 12 min. So original time taken is 12*5 = 60 min = 1 hr Time taken is 1 hr to cover a distance of 60 km so speed = 60 km/hr
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In case 2, the 60 kms further which the train traveled was with speed v. In case 1, these 60 kms (which in case 2 the train traveled further where accident happened) was traveled at a speed of (5v)/6. And this caused the extra 12 mins delay in case 1.

Re: A train met with an accident 60km away from station A. It co [#permalink]

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14 Feb 2014, 02:36

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Can someone please help me think through this problem?

Given: It arrives 1 hr 12 min late if the accident took 60 km further from A It arrives 1 hr late if the accident took 120 km further from A

It implies: the two "accident spots" are separated by 12 min and 60 km

Hence: R = (60)*(60/12) = 300 kmph

You are assuming that the train covered 60 km in 12 mins. That gives you the speed as 300 kmph. But is that really the case? 12 mins is the difference in the time taken to cover 60 kms at two different speeds. It is the delay caused by reducing the speed by (1/6)th over a distance of 60 km. The ratio of the speed over this 60 km stretch is 5:6 (speed reduced by a sixth : original speed)

Time taken 6:5 (Time taken with reduced speed : original time taken)

We know this difference of 1 in time taken is given by 12 min. So original time taken is 12*5 = 60 min = 1 hr

Time taken is 1 hr to cover a distance of 60 km so speed = 60 km/hr
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