[Math Revolution GMAT math practice question]
A car and a bicycle traveled in the same direction along the equal route at their constant speed rates of 40 miles per hour and 30 miles per hour, respectively. After 15 minutes the car passed the bicycle, the car reaches a waiting point, how long it takes the bicycle to reach the waiting point?
A. 15 min
B. 20 min
C. 25 min
D. 30 min
E. 35 min
A car and a bicycle traveled in the same direction along the
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- Max@Math Revolution
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Since the car travelled for 15 minutes with the speed 40 mph, it moved 10 miles. It take 20 minutes the bicycle to travel 10 miles with the speed 30 mph.
Therefore, the answer is B.
Answer: B
Since the car travelled for 15 minutes with the speed 40 mph, it moved 10 miles. It take 20 minutes the bicycle to travel 10 miles with the speed 30 mph.
Therefore, the answer is B.
Answer: B
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speed rate of car = 40 mph
speed rate of bicycle = 30 mph
15 minutes to an hour = 1/4 hour
In 15 minutes, the car ha travelled $$40\cdot\frac{1}{4}=10miles$$
and the bicycle travelled $$30\cdot\frac{1}{4}=\frac{30}{4}miles$$
At constant speed, the bicycle will have to travel additional miles to reach the position of the car
$$Therefore\ \frac{10}{1}-\frac{30}{4}=\frac{40-30}{4}=\frac{10}{4}=\frac{5}{2}miles$$
i.e bicycle will travel additional 5/2 miles to reach the car.
if the bicycle travels 30 miles in 60 minutes
it will travel 5/2 miles in x meter
$$30=60$$
$$\frac{5}{2}=x$$
by cross multiplying
$$30\cdot x=60\cdot\frac{5}{2}$$
$$\frac{30x}{30}=\frac{150}{30}$$
$$x=5miles$$
Bearing in mind that the bicycle has already travelled 15 minutes
Total time-taken to reach waiting point = 15+5 mile = 20 miles
$$Option\ B$$
speed rate of bicycle = 30 mph
15 minutes to an hour = 1/4 hour
In 15 minutes, the car ha travelled $$40\cdot\frac{1}{4}=10miles$$
and the bicycle travelled $$30\cdot\frac{1}{4}=\frac{30}{4}miles$$
At constant speed, the bicycle will have to travel additional miles to reach the position of the car
$$Therefore\ \frac{10}{1}-\frac{30}{4}=\frac{40-30}{4}=\frac{10}{4}=\frac{5}{2}miles$$
i.e bicycle will travel additional 5/2 miles to reach the car.
if the bicycle travels 30 miles in 60 minutes
it will travel 5/2 miles in x meter
$$30=60$$
$$\frac{5}{2}=x$$
by cross multiplying
$$30\cdot x=60\cdot\frac{5}{2}$$
$$\frac{30x}{30}=\frac{150}{30}$$
$$x=5miles$$
Bearing in mind that the bicycle has already travelled 15 minutes
Total time-taken to reach waiting point = 15+5 mile = 20 miles
$$Option\ B$$
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In 15 minutes, or 1/4 of an hour, the car travels 40/4 miles and the bicycle travels 30/4 miles.Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
A car and a bicycle traveled in the same direction along the equal route at their constant speed rates of 40 miles per hour and 30 miles per hour, respectively. After 15 minutes the car passed the bicycle, the car reaches a waiting point, how long it takes the bicycle to reach the waiting point?
A. 15 min
B. 20 min
C. 25 min
D. 30 min
E. 35 min
Thus, the bicycle needs to travel 40/4 - 30/4 = 10/4 = 5/2 miles to reach the car.
The bicycle will travel 5/2 miles in (5/2)/30 = 5/60 = 1/12 hour, which is 60 x 1/12 = 5 minutes.
Including the 15 minutes that the bike has already traveled, it takes 15 + 5 = 20 minutes for the bike to reach the waiting point.
Answer: B
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