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.
Drill 3
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 milesWhen 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/rateSo, 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
_________________
Brent Hanneson - founder of Greenlight Test Prep