Official Solution:De Bruyne and Mbappe start racing from the bottom of a 100-meter hill to the top and then back to the bottom by the same route. Each of them runs up at his constant rate and back down at twice that rate. What would be the distance between the two when the fastest one finishes the entire 200-meter race?(1) Mbappe and De Bruyne cross each other when Mbappe has run 140 meters.
This implies they cross each other when Mbappe has run 100 meters uphill and then 40 meters downhill, while De Bruyne has covered 100 - 40 = 60 meters uphill. Assuming the uphill rate of Mbappe is \(m\), making his downhill rate \(2m\), and the uphill rate of De Bruyne is \(b\), we'd get:
\(\frac{100}{m} + \frac{40}{2m} = \frac{60}{b}\)
\(\frac{240}{2m} = \frac{60}{b}\)
\(240b = 120m\)
\(m=2b\)
Hence, Mbappe's uphill rate is twice that of De Bruyne's, which implies that Mbappe's downhill rate is four times that of De Bruyne's. Thus, Mbappe will cover the remaining 60 meters of the race downhill, in the same time, De Bruyne will cover an additional 15 meters to the 60 meters he already has covered. Therefore, the distance between the two when Mbappe finishes the entire 200-meter race will be 60 + 15 = 75 meters. Sufficient.
(2) Mbappe reaches the top of the hill 10 seconds before De Bruyne.
The above implies that \(\frac{100}{m} +10 = \frac{100}{b}\). Unlike the previous statement, this does not give the ratio of \(m\) to \(b\), hence we cannot answer the question asked. Not sufficient.
Answer: A