GMAT Club World Cup 2022 (DAY 5): De Bruyne and Mbappe start racing

Question

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.

(2) Mbappe reaches the top of the hill 10 seconds before De Bruyne.

Bunuel · Accepted Answer

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

soudipsengupta · Answer

Please refer to the attachment for solution

Dwiti88 · Answer

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 with twice that constant 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.
(2) Mbappe reaches the top of the hill 10 seconds before De Bruyne does.

Suppose the speed of mbappe =x
and speed of de bruyne =y
1) As per statement 1 mbappe and De Bruyne cross each other when Mbappe has run 140 meters.
Thus Time taken by mbappe to go uphill 100 meters + down hill 40 meters = time taken by de bruyne to go uphill 60 meters
100/x+40/2x=60/y
240/2x=60/y
120/x=60/y
x=2y
Now the total time taken by Mbappe= (100/x +100/2x)=300/2x=150/x .And Time taken by De Bruyne to complete up hill 100 meters=100/y =100/x/2=200/x
as 200/x>150/x i.e by the time mbappe completed 200 meters de bruyne was goin up hill
Distance covered by De Bruyne in 150/x seconds= y*150/x=x/2*150/x=75 meters
Thus 1 is sufficient.B,C,E out
2)As per statement 2 100/x=100/y +10
100/x=(100+10y)/y
x=y*100/(100+10y)
We cannot solve x and y from here.Thus 2 is insufficient.
Thus A is the answer

manish1708 · Answer

Answer: A

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 with twice that constant rate. What would be the distance between the two, when the fastest one finishes the entire 200-meter race ?

Let the speed of De Bruyne(D) be Sb uphill then his speed downhill will be 2Sb
Let the speed of Mbappe(M) be Sm uphill then his speed downhill will be 2Sm

Time taken by D to run uphill = 100/Sb
Time taken by D to run downhill = 100/2Sb

Time taken by M to run uphill = 100/Sm
Time taken by M to run downhill = 100/2Sm

(1) Mbappe and De Bruyne cross each other when Mbappe has run 140 meters.

=> Mbappe is the faster runner.
=> De Bruyne has run only 60 meters till this time.
Time taken by both is same to run 60 and 140 meters respectively.
=> 60/Sb = 100/Sm + 40/2Sm
=> 3/Sb = 5/Sm + 1/Sm
=> Sm = 2Sb
Since we know the ratio of speed of the two, we can do the required calculations.

=> speed of M is twice the speed of B
=> By the time M will complete 100 meters uphill, B would have completed 50 meters. 
Downhill speed of M becomes twice => M's speed now would be 4 time speed of B
=> By the time M will complete 100 meters downhill, B would have completed another 25 meters.
So the distance between the two would be 75.

Sufficient .

(2) Mbappe reaches the top of the hill 10 seconds before De Bruyne does.
 
=> 100/Sm = 100/Sb - 10
=> 10Sb = 10Sm - Sb*Sm
Since we can't find the values of Sm or Sb or their ratio, we can't do the required calculations.

Insufficient .

rs24 · Answer

Note: Solution below is not the conventional way to solve the question, but it avoids most of the calculations.

Statement A:

We know Mbappe (MB) and De Bruyne (DB) meet each other when MB has covered 140: 100 uphill (at speed x) and 40 downhill (at speed 2x). If MB is going down... DB should be going uphill (if we consider DB moving downhill then both are travelling towards the same direction and GIVEN THEY STARTED TOGETHER, they cannot meet each other when running in the same direction - if confused, use yourself & a friend and then think about it from a real life perspective). If DB is uphill, he should have covered 60 mtrs (total distance 100 - MB covered 40 from towards down, DB should cover 60 towards up).

We know MB has travelled 100 m uphill (at speed x) and 40 m downhill (at speed 2x) and DB has travelled 60 m uphill (at speed y).

If MB has travelled 40 m at speed 2x, he would have travelled 20 m at speed x (half speed and same time = half distance). Now we know that the at speed x, MB would have covered 120 m (100 up and 20 down) and at speed y, DB has covered 60 m (all up). Hence, x:y = 2:1 (y is half of x)

From x:y, we can say MB is faster and will be the first one to complete the uphill and downhill journey.

During uphill, MB travels 100 m (at speed x) - DB will travel 50 m uphill (because his speed, y, is half of MB's speed, x). During downhill MB travels 100 m (at speed 2x) - DB will travel 25 m uphill (because his speed, y, is one-fourth of MB's new speed, 2x).

Total distance of DB = 75 m uphill.

We can conclude statement A is sufficient.

Statement B:

We know that MB reached the top 10 seconds before DB.

Question is about DB's distance travelled, can we establish anything about DB's distance? No, it could be 1 m as well as 99 m, we do not know anything.

Statement B is clearly not sufficient.

bumpbot · Answer

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GMAT Club World Cup 2022 (DAY 5): De Bruyne and Mbappe start racing

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