On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)
A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes
[spoiler]OA=A[/spoiler].
I don't know how to solve this PS question? May someone gives me some help? Thanks.
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On his morning jog, Charles runs through a park in the
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rate = distance/time.Gmat_mission wrote:On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)
A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes
Let the diameter = 7 meters, implying that the circumference of the semicircle = (Ï€d)/2 = [(22/7)(7)]/2 = 11 meters.
Let t = the time to travel the diameter, implying that t+4 = the time to travel the semicircle.
Since the rate to travel the diameter is the same as the rate to travel the semicircle, we get:
7/t = 11/(t+4)
7t + 28 = 11t
28 = 4t
7 = t.
The correct answer is A.
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Jeff@TargetTestPrep
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We can let the diameter = d; thus, the semicircular arc = ½ x π x d = ½ x 22/7 x d = 11d/7.Gmat_mission wrote:On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)
A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes
Furthermore, we can let Charles' rate = r. Thus, his time running along the diameter is d/r and his time running along the semicircular arc is (11d/7)/r = 11d/(7r).
We are given that his time running along the diameter is 4 minutes less than his time running along the semicircular arc. Thus, we can create the equation:
11d/(7r) - d/r = 4
11d/(7r) - 7d/(7r) = 4
4d/(7r) = 4
d/(7r) = 1
d/r = 7
Notice that d/r is the time running along the diameter, so Charles completes this part of his run in 7 minutes.
Answer: A
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