GMATPrepNow wrote:
Rhonda runs at an average speed of 12 kilometers per hour, and she bicycles at an average speed of 30 kilometers per hour. When she bicycles to work, her travel time is 2.25 minutes less than when she runs to work. The distance to work, in kilometers, is
A) 27/40
B) 3/4
C) 7/8
D) 11/12
E) 6/5
When Rhonda bicycles to work, her travel time is 2.25 minutes less than when she runs to workIt might be useful to start with a "Word Equation"
(
Rhonda's running time in hours) = (
Rhonda's cycling time in hours) + 2.25/60
Aside: 2.25 minutes = 2.25/60 hourstravel time = distance/speedWe know the running and cycling speeds, but we don't know the distance.
So, let d = distance to work
So, we get:
d/12 =
d/30 + 2.25/60
To eliminate the fractions, multiply both sides by 60 (the LCM of 12, 30 and 60)
We get: 5d = 2d + 2.25
Subtract 2d from both sides: 3d = 2.25
Solve: d = 2.25/3
Check answer choices . . . not there.
Looks like we need to rewrite 2.25/3 as an equivalent fraction.
If we take 2.25/3, and multiply top and bottom by 4 we get: 9/12, which is the same as 3/4
Answer: B
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