Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, what was the total time of Marcia's trip?
A) (4D)/v
B) (4v)/D
C) (10D)/(11v)
D) (10v)/(11D)
E) (11D)/(10v)
The OA is A.
Is there a strategic approach to this question? Can any experts help me, please? Thanks!
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GMATGuruNY
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Let v = 1 mph and D = 6 miles.AAPL wrote:Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, what was the total time of Marcia's trip?
A) (4D)/v
B) (4v)/D
C) (10D)/(11v)
D) (10v)/(11D)
E) (11D)/(10v)
Time to travel 5D miles at a speed of 2v mph = d/r = (5*6)/(2*1) = 15 hours.
Time to travel 4D miles at a speed of 3v mph = d/r = (4*6)/(3*1) = 8 hours.
Time to travel D miles at a speed of 6v mph = d/r = 6/(6*1) = 1 hour.
Total time = 15+8+1 = 24 hours.
The target value is 24 hours.
Now plug v=1 and D=6 into the answers to see which yields the target value.
Only A works:
(4D)/v = (4*6)/1 = 24.
The correct answer is A.
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Jeff@TargetTestPrep
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The time for the first leg of her trip is 5D/(2v).AAPL wrote:Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, what was the total time of Marcia's trip?
A) (4D)/v
B) (4v)/D
C) (10D)/(11v)
D) (10v)/(11D)
E) (11D)/(10v)
The time for her second leg is 4D/(3v).
The time for her last leg is: D/(6v).
Thus the total time is:
5D/(2v) + 4D/(3v) + D/(6v)
15D/(6v) + 8D/(6v) + D/(6v)
24D/(6v)
4D/v
Answer: A
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