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PS : TRAIN (m09q09)

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Re: PS : TRAIN (m09q09) [#permalink] New post 04 Sep 2013, 16:15
length of train = 20*6 + 5 = 125m

length of tunnel = 1000m

Total distance train has to cover = 1125m @ 60km/h

time = (1125 m/ 60km/h) * (60 min / 1000m) = 1.125 mins

Ans is A
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Re: PS : TRAIN (m09q09) [#permalink] New post 01 Apr 2014, 04:51
A)

Length of the train = 20*6 + 5*1 (gaps) = 125m

Rate 60km/h => 1 km per min

125m = 0.125km => 1/8 of min

Answer => 1 1/8 min
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Re: PS : TRAIN (m09q09) [#permalink] New post 02 Apr 2014, 00:07
given train speed is 60km/hr---> 1km/min
total length of train--> (60*20)+5(gaps)
-->125 mtrs

therefore, total length to travel--> 1km+125 mtrs

clearly, it takes exactly 1 min for train to traverse the length of the tunnel but it has to travel its full length i.e., 125 mtrs to come outside of the tunnel completely.

Now since train is travelling at 1km/min-->1000mtrs/min-->1000mtrs/60 seconds
it takes 7.5 seconds to travel 125mtrs, 7.5 seconds-->1/8th of a minute

so train takes--> 1+1/8 minutes to cross tunnel completely
Re: PS : TRAIN (m09q09)   [#permalink] 02 Apr 2014, 00:07
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