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05 Apr 2025, 23:28
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42%
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A certain train left City X at 7:00 am and arrived at City Y the same morning. Immediately after arriving at City Y, this train was held on the track for 2 hours, after which it returned to City X along the same route. If the train arrived at City X at 2:00 pm on the same day, by what percent did the average speed of the train on its way back exceed its average speed on the way from City X to City Y?
(1) The average speed of the train on its way from City X to City Y was 40 miles per hour.
(2) The return trip took the train one hour less than the trip from City X to City Y.
(1) The average speed of the train on its way from City X to City Y was 40 miles per hour.
(2) The return trip took the train one hour less than the trip from City X to City Y.
08 Apr 2025, 11:02
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(1) The average speed of the train on its way from City X to City Y was 40 miles per hour.
You don't have the distance between X and Y or the time the train took to get from X to Y or back so this is insufficient.
If you need to visualize this, imagine the train took 4 hours the first time (7:00AM to 11:00AM), waited 2 hours (11:00AM to 1:00PM), then went back from (1:00PM to 2:00PM). The train would be 300% faster. Now imagine the train took 3 hours the first time (7:00AM to 10:00AM), waited 2 hours (10:00AM to 12:00PM), then went back from (12:00PM to 2:00PM). The train would only be 50% faster.
(2) The return trip took the train one hour less than the trip from City X to City Y.
There is only one possibility that fits this condition which makes this statement sufficient.
The total time is 7 hours since the difference between 7:00AM and 2:00PM is 7 hours (7:00-12:00 = 5hrs, 12:00 - 2:00 is 2hrs). We can put this int a equation as follows:
7 = 2x + 1 + 2
x is the time that the return trip took (Y to X). The extra 2 that we have is the wait time that the train had to take immediately after reaching Y
Solving, we get x = 2, meaning that the return trip took 2 hours. Since we know that the trip from X to Y was an extra hour, the trip was 3 hours long.
Since we have the time it took for each trip, we can calculate the average speed difference by percentage.
3/2 = 1.5 <- the return trip had an average speed of 150% of the trip from X to Y
This means that the average speed increase was 50%, thus making this statement sufficient and statement 1 insufficient.
You don't have the distance between X and Y or the time the train took to get from X to Y or back so this is insufficient.
If you need to visualize this, imagine the train took 4 hours the first time (7:00AM to 11:00AM), waited 2 hours (11:00AM to 1:00PM), then went back from (1:00PM to 2:00PM). The train would be 300% faster. Now imagine the train took 3 hours the first time (7:00AM to 10:00AM), waited 2 hours (10:00AM to 12:00PM), then went back from (12:00PM to 2:00PM). The train would only be 50% faster.
(2) The return trip took the train one hour less than the trip from City X to City Y.
There is only one possibility that fits this condition which makes this statement sufficient.
The total time is 7 hours since the difference between 7:00AM and 2:00PM is 7 hours (7:00-12:00 = 5hrs, 12:00 - 2:00 is 2hrs). We can put this int a equation as follows:
7 = 2x + 1 + 2
x is the time that the return trip took (Y to X). The extra 2 that we have is the wait time that the train had to take immediately after reaching Y
Solving, we get x = 2, meaning that the return trip took 2 hours. Since we know that the trip from X to Y was an extra hour, the trip was 3 hours long.
Since we have the time it took for each trip, we can calculate the average speed difference by percentage.
3/2 = 1.5 <- the return trip had an average speed of 150% of the trip from X to Y
This means that the average speed increase was 50%, thus making this statement sufficient and statement 1 insufficient.
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