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Rhonda runs at an average speed of 12 kilometers

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Rhonda runs at an average speed of 12 kilometers

by Brent@GMATPrepNow » Mon Feb 13, 2017 10:39 am
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

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by DavidG@VeritasPrep » Mon Feb 13, 2017 1:21 pm
Brent@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

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Rhonda's running speed = 12km/hour or 12km/60minutes or 1km/5minutes.
Biking speed = 30km/hour = 30km/60minutes = 1km/2minutes

If it takes her t minutes to run to work, it would take her t - 2.25 minutes, or t - 9/4 to bike.

Running she covers (1/5)t kilometers. Biking she covers (1/2)(t-9/4) kilometers. So (1/5)t = (1/2)(t-9/4)
(1/5)t = (1/2)t - 9/8
(1/5)t - (1/2)t = - 9/8
(-3/10)t = -9/8
t = 90/24 = 15/4

If she covers (1/5)t running, she'll cover (1/5)* (15/4) = 15/20 = 3/4 km. The answer is B
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by Brent@GMATPrepNow » Mon Feb 13, 2017 2:24 pm
Brent@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
Another approach:

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
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
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by GMATGuruNY » Tue Feb 14, 2017 3:36 am
Brent@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
We can PLUG IN THE ANSWERS, which represent the distance.
2.25 minutes = 9/4 minutes = (9/4)/60 hours = 3/80 hours.
When the correct answer choice is plugged in, the difference between the two times will be 3/80 hours.

B: 3/4
At a speed of 12 kph, the time to bike 3/4 kilometers = d/r = (3/4)/12 = 1/16 hours.
At a speed of 30 kph, the time to walk 3/4 kilometers = d/r = (3/4)/30 = 1/40 hours.
Time difference = 1/16 - 1/40 = 5/80 - 2/80 = 3/80 hours.
Success!

The correct answer is B.
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by Scott@TargetTestPrep » Wed Feb 15, 2017 3:08 pm
Brent@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
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

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by Matt@VeritasPrep » Wed Feb 15, 2017 6:03 pm
Let's use D = RT.

Distance, running:
D = distance
R = 12 km/hr = 1/5 km/min
T = m

Distance, biking:
D = distance
R = 30 km/hr = 1/2 km/min
T = m - 2.25

Since the two distances are the same, D = D and RT = RT, giving us

(1/5)m = (1/2)(m - 2.25)

(1/5)m = (1/2)m - 1.125

1.125 = (3/10)m

m = 3.75

From there, D = (1/5)m = (1/5) * 3.75 = 0.75, or B.
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by Matt@VeritasPrep » Wed Feb 15, 2017 6:07 pm
Another approach that's quicker to solve but harder to conceive in the first place:

Suppose Rhonda is cloned, and the two Rhondas head to work, one running, one biking. (What could be less glamorous than cloning yourself only to head to work! But hey ...) As before, we'll put their rates in km/min, so the runner = 1/5 km/min and the biker = 1/2 km/min.

Let's say it takes m minutes for biking Rhonda to get to work. After m minutes, running Rhonda has traveled (1/5)m and biking Rhonda has traveled (1/2)m.

That means the entire distance = (1/2)m. It will take running Rhonda 2.25 more minutes to get to work, and her remaining distance = (1/2)m - (1/5)m. So we've got

2.25 = (1/2)m - (1/5)m

3.75 = m

and from there the solution is the same, (1/5) * 3.75.
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