The posted problem is virtually the same as the one below:
Two cars painted Black and Blue are racing against each other on a track of length 4000 meters. The race ends in a tie as both cars finish the race in exactly the same time. However at first when the race starts the Black car moves 50% faster than the Blue car. The Blue car then quickened its pace and for the remaining distance moved 50% faster than the Black car. When the Blue car quickened its pace what distance had it already covered?
(A) 1200 meters
(B) 1600 meters
(C) 2400 meters
(D) 2800 meters
(E) 3000 meters
Before the blue car speeds up:
Here, the black car travels 3/2 as fast as the blue car.
Thus, the black car travels 3/2 the distance of the blue car.
After the blue car speeds up:
Here, the blue car travels 3/2 as fast as the black car.
Thus, the blue car travels 3/2 the distance of the black car.
We can plug in the answers, which represent the distance traveled by the blue car before it speeds up.
Answer choice C: 2400m
Before the blue car speeds up:
Distance traveled by the blue car = 2400m.
Distance traveled by the black car = (3/2) * 2400 = 3600m.
After the blue car speeds up:
Remaining distance traveled by the black car = 4000-3600 = 400m.
Distance traveled by the blue car = (3/2) * 400 = 600m.
Total distance traveled by the blue car = 2400+600 = 3000m.
Doesn't work: the total distance traveled by the blue car must be 4000m.
To INCREASE the blue car's total distance to 4000m, the blue car must speed up EARLIER.
Thus, the distance traveled by the blue car before it speeds up must DECREASE.
Eliminate C, D and E.
Answer choice B: 1600m
Before the blue car speeds up:
Distance traveled by the blue car = 1600m.
Distance traveled by the black car = (3/2) * 1600 = 2400m.
After the blue car speeds up:
Remaining distance traveled by the black car = 4000-2400 = 1600m.
Distance traveled by the blue car = (3/2) * 1600 = 2400m.
Total distance traveled by the blue car = 1600+2400 = 4000m.
Success!
The correct answer is
B.
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