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Re: Excluding stoppages, the speed of a train is 40 km/h and including sto [#permalink]
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Bunuel wrote:
Excluding stoppages, the speed of a train is 40 km/hand including stoppage, it is 32 km/h. For how many minutes does the train stop per hour ?

(A) 10
(B) 12
(C) 15
(D) 18
(E) 20


I got the point of the above-mentioned methods, but I offer an alternative from my point of view much more intuitive.

LCM for a smart distance ----> 160km
At 40 km/h this distance is covered in 4 hours
At 32 km/h is covered in 5 hours
Good 1 hour of difference, hence (1/5) * 60 min is time per min spent in stoppages ---> 12 minutes
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Re: Excluding stoppages, the speed of a train is 40 km/h and including sto [#permalink]
­Please guide me where am I getting this incorrectly? Bunuel

I assumed the total distance travelled by the trains be x.
Now the time taken to traverse the distance without stoppages be t
Lets say the time for stoppage be T.
Thus the total time taken including stoppages is t+T.

Now since the distance traversed is x so, x= 40*t
And considering the stoppages we get x = 32(t+T)

Thus 40*t = 32*(t+T)
Thus 8t=32T
thus, T=(1/4)*t in hrs.

to convert it into mins we multiply by 60, Thus T = 15t

Now t=1, as we need the stoppage time per hour. 
Hence T=15*1 = 15mins.

Answer C­
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Re: Excluding stoppages, the speed of a train is 40 km/h and including sto [#permalink]
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Budhaditya_Saha wrote:
­Please guide me where am I getting this incorrectly? Bunuel

I assumed the total distance travelled by the trains be x.
Now the time taken to traverse the distance without stoppages be t
Lets say the time for stoppage be T.
Thus the total time taken including stoppages is t+T.

Now since the distance traversed is x so, x= 40*t
And considering the stoppages we get x = 32(t+T)

Thus 40*t = 32*(t+T)
Thus 8t=32T
thus, T=(1/4)*t in hrs.

to convert it into mins we multiply by 60, Thus T = 15t

Now t=1, as we need the stoppage time per hour. 
Hence T=15*1 = 15mins.

Answer C­

­t + T should be equal 1 hour, not t. This way you get T = 1/5 = 12 minutes.

Similar questions to prctice:
https://gmatclub.com/forum/excluding-st ... 37716.html
https://gmatclub.com/forum/excluding-st ... 47257.html
https://gmatclub.com/forum/without-stop ... 92836.html
https://gmatclub.com/forum/without-stop ... 47638.html
https://gmatclub.com/forum/excluding-st ... 55143.html
https://gmatclub.com/forum/a-train-trav ... 87511.html 
https://gmatclub.com/forum/without-stop ... 15585.html
https://gmatclub.com/forum/the-average- ... 13861.html
https://gmatclub.com/forum/without-stop ... 47253.html

Hope this helps.

 
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Excluding stoppages, the speed of a train is 40 km/h and including sto [#permalink]
Bunuel wrote:
Budhaditya_Saha wrote:
­Please guide me where am I getting this incorrectly? Bunuel

I assumed the total distance travelled by the trains be x.
Now the time taken to traverse the distance without stoppages be t
Lets say the time for stoppage be T.
Thus the total time taken including stoppages is t+T.

Now since the distance traversed is x so, x= 40*t
And considering the stoppages we get x = 32(t+T)

Thus 40*t = 32*(t+T)
Thus 8t=32T
thus, T=(1/4)*t in hrs.

to convert it into mins we multiply by 60, Thus T = 15t

Now t=1, as we need the stoppage time per hour. 
Hence T=15*1 = 15mins.

Answer C­

­t + T should be equal 1 hour, not t. This way you get T = 1/5 = 12 minutes.

Similar questions to prctice:
https://gmatclub.com/forum/excluding-st ... 37716.html
https://gmatclub.com/forum/excluding-st ... 47257.html
https://gmatclub.com/forum/without-stop ... 92836.html
https://gmatclub.com/forum/without-stop ... 47638.html
https://gmatclub.com/forum/excluding-st ... 55143.html
https://gmatclub.com/forum/a-train-trav ... 87511.html 
https://gmatclub.com/forum/without-stop ... 15585.html
https://gmatclub.com/forum/the-average- ... 13861.html
https://gmatclub.com/forum/without-stop ... 47253.html

Hope this helps.

 


Thank You! Bunuel

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Excluding stoppages, the speed of a train is 40 km/h and including sto [#permalink]
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