Rhonda runs at an average speed of 12 kilometers per hour, a

[BTGmoderatorAT]

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

OA: B
Is there a strategic approach to this question? Can any experts show?

[EconomistGMATTutor]

Hello ardz24.

Let's take a look at your question.

First, we have to convert the speed from kilometers per hour to kilometers per minute.

V_R----Runs----- 12 km/h ------- 12km/h * 1/60 (h/min) --------1/5 km/min= 0.2 k/min.

V_B---Bicycles ---- 30 km/h ------ 30km/h * 1/60 (h/min) ----- 1/2 km/min = 0.5 k/min.

Now, since the distance to work is the same, When she bicycles to work, her travel time is 2.25 minutes less than when she runs to work implies that $$0.2\cdot t_R=\ 0.5\cdot t_B\ \Leftrightarrow\ \ 0.2\cdot t_R=0.5\cdot\left(t_R-2.25\right)\ \Leftrightarrow\ 0.2t_R=0.5t_R-1.125$$ $$1.125=0.3t_R\ \Leftrightarrow\ t_R=3.75.$$ Now, we can find the distance using the following equation $$d=V_R\cdot t_R=0.2\cdot3.75\ =\ 0.75=\frac{3}{4}km.$$ This is why the correct answer is the option [spoiler]B=3/4[/spoiler].

I hope this answer may help you.

I'm available if you'd like a follow up.

Regards.

[Brent@GMATPrepNow]

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 work
It might be useful to start with a Word Equation
(Rhonda's running time in hours) = (Rhonda's cycling time in hours) + 2.25/60 hours
Aside: 2.25 minutes = 2.25/60 hours

travel time = distance/speed
We 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

Cheers,
Brent

[Scott@TargetTestPrep]

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

We are given that Rhonda runs at an average speed of 12 kilometers per hour and bicycles at an average speed of 30 kilometers per hour. We are also given that when she bicycles to work, her travel time is 2.25 minutes less than when she runs to work.

Since she runs and bicycles the same distance, we can let her distance from home to work = d.

Since time = distance/rate, her time running to work is d/12 and her time bicycling to work is d/30. We also need to convert 2.25 minutes to hours.

2.25 minutes = 2.25/60 = 9/240 = 3/80 hour

We can create the following equation and determine d:

d/12 = 3/80 + d/30

Multiplying the entire equation by 240, we have:

20d = 9 + 8d

12d = 9

d = 9/12 = 3/4

Answer: B