lheiannie07 wrote:Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?
A) 45
B) 48
C) 50
D) 55
E) 58
Let t = the number of minutes allotted for Mr. Ben to arrive at work on time..
When he averages 40 miles per hour, he arrives at his workplace three minutes late.
(40 miles/1 hour)(1 hour/60 seconds) = 40/60 mile per minute = 2/3 mile per minute.
Since Mr. Ben arrives 3 minutes LATE, he travels for 3 minutes MORE than the allotted number of minutes:
t+3.
At a rate of 2/3 mile per minutes, the distance traveled in t+3 minutes = rt = (2/3)(t+3) =
(2/3)t + 2.
When he averages 60 miles per hour, he arrives three minutes early.
(60 miles/1 hour)(1 hour/60 seconds) = 60/60 mile per minute = 1 mile per minute.
Since Mr. Ben arrives 3 minutes EARLY, he travels for 3 minutes LESS than the allotted number of minutes:
t-3.
At a rate of 1 mile per minute, the distance traveled in t-3 minutes = rt = (1)(t-3) =
t-3.
Since the distance traveled in each case is THE SAME, the expressions in blue are equal:
(2/3)t + 2 = t-3
5 = (1/3)t
t = 15.
Since Mr. Ben travels for t-3 minutes at a rate of 1 mile per minute, the distance to work = (r)(t-3) = (1)(15-3) = 12 miles.
To travel the 12 miles to work in the allotted number of minutes -- the red value above -- the required rate = d/t = 12/15 = 4/5 mile per minute.
(4/5 miles/1 minute)(60 minutes/1 hour) = 48 miles per hour.
The correct answer is
B.
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