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John and Amanda stand at opposite ends of a straight road

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by AAPL » Mon Apr 30, 2018 5:42 am
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John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

A. 3/16
B. 5/16
C. 3/8
D. 1/2
E. 13/16

The OA is A.

Is there a strategic approach to solve this question? Can anyone help me, please? Thanks!
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by GMATGuruNY » Mon Apr 30, 2018 6:43 am
AAPL wrote:John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

A. 3/16
B. 5/16
C. 3/8
D. 1/2
E. 13/16[/quote

P = (good outcomes)/(all possible outcomes).

All possible outcomes:
Number of possible rates for John = 4.
Number of possible rates for Amanda = 4.
To combine the number of rate options for John with the number of rate options for Amanda, we multiply:
4*4 = 16.

Good outcomes:
John will travel farther than Amanda if his rate is greater than Amanda's rate, yielding the following options:
John's rate = 5 mph, Amanda's rate = 4 mph
John's rate = 6 mph, Amanda's rate = 4 mph
John's rate = 6 mph, Amanda's rate = 5 mph
Total ways = 3.

Thus:
P(John travels farther than Amanda) = (good outcomes)/(all possible outcomes) = 3/16.

The correct answer is A.
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by Jeff@TargetTestPrep » Tue May 01, 2018 9:26 am
AAPL wrote:John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

A. 3/16
B. 5/16
C. 3/8
D. 1/2
E. 13/16
John will travel farther than Amanda if John is traveling faster than Amanda. That is, John’s rate must be greater than Amanda’s rate. Thus, John must travel either 5 or 6 miles per hour in order to have a chance to be faster than Amanda.

If John travels 5 mph (there is a 1/4 chance he would do that), then Amanda has to travel 4 mph (there is a 1/4 chance she would do that). Thus, the probability is 1/4 x 1/4 = 1/16.

If John travels 6 mph (there is a 1/4 chance he would do that), then Amanda has to travel either 4 or 5 mph (there is a 2/4 = 1/2 chance she would do that). Thus, the probability is 1/4 x 1/2 = 1/8.
Therefore, the overall probability that John will travel farther than Amanda is 1/16 + 1/8 = 3/16.

Alternate Solution:

Since there are 4 possibilities for the rates of John and Amanda each, there are 4 x 4 = 16 possible ways to assign rates for them.

Of these 16 possibilities, the only way John travels further than Amanda is if John = 6 mph and Amanda = 5 mph; John = 6 mph and Amanda = 4 mph; or if John = 5 mph and Amanda = 4 mph. Since there are 3 favorable outcomes in a total of 16 outcomes, the probability is 3/16.

Answer: A

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