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03 Jan 2020, 01:32
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (03:03) correct
36%
(02:55)
wrong
based on 259
sessions
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A train met with an accident 150 km from its originating station. It completed the remaining journey at half of the usual speed and reached 1 hour late at the destination station. Had the accident taken place 30 km later, it would have been only half an hour late. Find the distance between the two stations.
A. 180 km
B. 200 km
C. 250 km
D. 400 km
E. None of these
A. 180 km
B. 200 km
C. 250 km
D. 400 km
E. None of these
22 Sep 2024, 08:24
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I solved it using logical reasoning and it helped me get through it a lot quicker, so i figured i'd share my solution:
If when the accident had taken place 30 km later it would have only been half an hour late as opposed to an hour late as it was now, that means that the delay would have been cut exactly in half when it would have happened 30 kilometers later (which means that the speed halving must have happened exactly when half of the remaining distance had yet to be travelled. Therefore, the total remaining distance must have been 30 * 2 and thus, total distance travelled is D = 150 + 60.
Answer: E
If when the accident had taken place 30 km later it would have only been half an hour late as opposed to an hour late as it was now, that means that the delay would have been cut exactly in half when it would have happened 30 kilometers later (which means that the speed halving must have happened exactly when half of the remaining distance had yet to be travelled. Therefore, the total remaining distance must have been 30 * 2 and thus, total distance travelled is D = 150 + 60.
Answer: E
General Discussion
03 Jan 2020, 22:33
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Bunuel
Let the speed be 2x, so after accident speed is x..
If remaining distance is y,
then time taken normally =\( \frac{150+y}{2x}\), but time taken now = \(\frac{150}{2x}+\frac{y}{x}\), so \(\frac{150}{2x}+\frac{y}{x}-\frac{150+y}{2x}=1........\frac{y}{x}-\frac{y}{2x}=1......y=2x\)
Now the remaining distance is y-30,
then time taken normally = \(\frac{180+y-30}{2x}\), but time taken now =\( \frac{180}{2x}+\frac{y-30}{x}\), so \(\frac{180}{2x}+\frac{y-30}{x}-\frac{180+y-30}{2x}=1........\frac{y-30}{x}-\frac{y-30}{2x}=\frac{1}{2}......y-30=x...y=x+30\)
Equating values of y..\(y=2x=x+30...x=30,\) and \(y=2*30=60\)..Total distance =\( 150+60=210..\)
E
Logically too...
By travelling 30 km at original speed rather than half the speed, the train saves half an hour....
So to save the entire 1 hour, the train should have traveled 30*2 or 60 km at original speed.
This means the total distance is 150+60=210
E
