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An emergency vehicle travels 10 miles at a speed of 50 miles
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09 Mar 2020, 02:47
09 Mar 2020, 02:47
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An emergency vehicle travels 10 miles at a speed of 50 miles per hour. How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
Kudos for the right answer and explanation
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
Kudos for the right answer and explanation
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Joined: 10 Apr 2015
Posts: 6218
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Re: An emergency vehicle travels 10 miles at a speed of 50 miles
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09 Mar 2020, 07:19
09 Mar 2020, 07:19
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Carcass wrote:
An emergency vehicle travels 10 miles at a speed of 50 miles per hour. How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
An emergency vehicle travels 10 miles at a speed of 50 miles per hour
time = distance/rate
So, time = 10/50 = 1/5 hours = 12 minutes
How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?
The vehicle already travelled for 12 minutes on the outbound trip, so the 10-mile inbound trip must take 8 minutes
8 minutes = 8/60 HOURS = 2/15 HOURS
Speed = distance/time
So, inbound speed = 10/(2/15)
= (10)(15/2)
= 150/2
= 75 mph
Answer: E
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Brent
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Re: An emergency vehicle travels 10 miles at a speed of 50 miles
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09 Mar 2020, 07:49
09 Mar 2020, 07:49
Carcass wrote:
An emergency vehicle travels 10 miles at a speed of 50 miles per hour. How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
Kudos for the right answer and explanation
(A) 55 mph
(B) 60 mph
(C) 65 mph
(D) 70 mph
(E) 75 mph
Kudos for the right answer and explanation
For the onward journey, time taken = 10/50 = 1/5 hours = 12 minutes
Thus, the return journey must be made in 20 - 12 = 8 minutes
=> Time taken for return journey = 8/12 = 2/3 times of the time taken for the onward journey
=> Speed during return journey = \frac{1}{(2/3)} = 3/2 times the speed during the onward journey = 3/2 * 50 = 75 mph
Answer E
