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Vinay and Phil are driving in separate cars to Los Angeles,
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Updated on: 10 Jun 2021, 08:24
Updated on: 10 Jun 2021, 08:24
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Question Stats:
33% (01:46) correct
66% (01:54) wrong based on 24 sessions
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Vinay and Phil are driving in separate cars to Los Angeles, both leaving from the same place and traveling along the same route. If Vinay leaves at 1 a.m. and travels at 40 miles per hour, and Phil leaves at 5 a.m. and travels at 50 miles per hour, at what time does Phil catch up to Vinay?
A. 1 p.m.
B. 5 p.m.
C. 7 p.m.
D. 9 p.m.
E. 11 p.m.
A. 1 p.m.
B. 5 p.m.
C. 7 p.m.
D. 9 p.m.
E. 11 p.m.
Drill 3
Question: 2
Page: 528
Question: 2
Page: 528
Re: Vinay and Phil are driving in separate cars to Los Angeles,
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29 Dec 2017, 14:16
29 Dec 2017, 14:16
Expert Reply
Explanation
Before Phil leaves, Vinay has traveled for 4 hours; the rate formula is distance = rate × time, so Vinay has gone 40 miles per hour × 4 hours = 160 miles.
Upon leaving, Phil is gaining on Vinay at a rate of 10 miles per hour, because he travels 10 more miles per hour than Vinay.
Now your equation is 160 miles = 10 miles per hour × time, so time = 16 hours. Phil left at 5 a.m., so he’ll catch up to Vinay at 9 p.m., so the answer is choice D.
Before Phil leaves, Vinay has traveled for 4 hours; the rate formula is distance = rate × time, so Vinay has gone 40 miles per hour × 4 hours = 160 miles.
Upon leaving, Phil is gaining on Vinay at a rate of 10 miles per hour, because he travels 10 more miles per hour than Vinay.
Now your equation is 160 miles = 10 miles per hour × time, so time = 16 hours. Phil left at 5 a.m., so he’ll catch up to Vinay at 9 p.m., so the answer is choice D.
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Re: Vinay and Phil are driving in separate cars to Los Angeles,
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17 Jan 2020, 10:58
17 Jan 2020, 10:58
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1
sandy wrote:
Vinay and Phil are driving in separate cars to Los Angeles, both leaving from the same place and traveling along the same route. If Vinay leaves at 1 a.m. and travels at 40 miles per hour, and Phil leaves at 5 a.m. and travels at 50 miles per hour, at what time does Phil catch up to Vinay?
A. 1 p.m.
B. 5 p.m.
C. 7 p.m.
D. 9 p.m.
E. 11 p.m.
A. 1 p.m.
B. 5 p.m.
C. 7 p.m.
D. 9 p.m.
E. 11 p.m.
Drill 3
Question: 2
Page: 528
Question: 2
Page: 528
This is a shrinking gap question
When Phil starts driving at 5 a.m., Vinay has already been driving for 4 hours.
Vinay's speed is 40 mph
Distance = (rate)(time)
So, in those 4 hours, Vinay's travel distance = (40)(4) = 160 miles
When Phil eventually catches up to Vinay, the distance between them will have shrunk from 160 miles to 0 miles
In other words, the distance between them will have shrunk 160 miles
Phil's travel speed is 50 mph, and Vinay's travel speed is 40 mph
So, the shrink rate = 50 - 40 = 10 mph
Time = distance/rate
So, the time to shrink the gap 160 miles = 160/10 = 16 hours (from the time that Phil began driving).
Phil began driving at 5 a.m.
So the time when Phil catches up to Vinay = 5 a.m. + 16 hours = 9 p.m.
Answer: D
Cheers,
Brent
