pranab01 wrote:
Dick takes twice as long as Jane to run any given distance. Starting at the same moment, Dick and Jane run towards each other from opposite ends of the schoolyard, a total distance of x, at their respective constant rates until they meet
Quantity A |
Quantity B |
The fraction of the total distance x that is covered by Jane |
\({\frac{2}{3}}x\) |
GIVEN: Dick takes twice as long as Jane to run any given distance.In other words, Jane travels TWICE as fast as Dick does.
So, for every 1 meter that Dick travels, Jane travels 2 meters.
GIVEN: Starting at the same moment, Dick and Jane run towards each other from opposite ends of the schoolyard, a total distance of x, at their respective constant rates until they meetThis is a CLOSING GAP question.
So, for every 1 meter that Dick travels towards Jane, Jane travels 2 meters towards Dick.
In other words, every time the gap closes by 3 units of length, Jane travels 2 units and Dick travels 1 unit.
Since the initial gap is x, we know that the total COMBINED distances traveled by Dick and Jane is x.
This means Dick traveled 1/3 of x, and Jane traveled
2/3 of xWe get:
QUANTITY A: 2/3 of x
QUANTITY B: (2/3)x
Answer: C
Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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