GreenlightTestPrep wrote:
Today Dave drove from home to work at twice his regular speed and arrived at work 16 minutes earlier than he regularly arrives. If Dave regularly drives at a constant speed of 40 miles per hour, what is the distance, in miles, from Dave’s home to work?
(A) 16/3
(B) 32/3
(C) 40/3
(D) 64/3
(E) 80/3
Dave's regular driving speed =
40 mph
Dave's driving speed TODAY =
80 mph
Let d = the distance from Dave's home to work
Dave arrived at work 16 minutes earlier than he regularly arrives.We can write:
regular travel time in hours - 16/60 hours = today's travel time in hourstime = distance/rate, which means we can write:
d/40 - 16/60 hours = d/80Rewrite each fraction with denominator 240 to get:
6d/240 - 64/240 hours = 3d/240Multiply both sides of the equation by 240 to get:
6d - 64 = 3dSubtract 6d from both sides:
-64 = -3dDivide both sides by -3 to get:
64/3 = dAnswer: D