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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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

*kudos for all correct solutions


\(Average speed (run) =\) \(12 \frac{km}{hr}\)

\(Average speed (bicycle) =\) \(30 \frac{km}{hr}\)
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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ziyuenlau wrote:
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

*kudos for all correct solutions


Average speed (run) = \(12 \frac{km}{hr}\)

Average speed (bicycle) = \(30 \frac{km}{hr}\)

\(Distance = Speed*Time\)

Average Time = \(\frac{Average Distance}{{Average Speed}}\)

\(Time (Bicycle) = Time (Run) - 2.25 min\)

\([\frac{(distance)}{{30km/hr}}]*(\frac{60min}{1hr})\)\(=[\frac{(distance)}{{12km/hr}}]*(\frac{60min}{hr}) - 2.25min\)

\(2*distance = 5*distance - 2.25\)

\(3*distance = 2.25\)

\(distance = \frac{3}{4}\)
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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Let x minutes takes to reach office while running

So (x -2.25) minutes by bicycle to reach office

Distance by run = Distance by bicycle

x*12 = 30*(x - 2.25)

Solve it. x = 15/4 minutes i.e 1/16 hours

Distance = 12 *x or 30*x - 2.25) = 12 * 1/16 = 3/4
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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

*kudos for all correct solutions


let t=running time
12/30=(t-3/80)/t
t=1/16 hr
1/16 hr*12 kph=3/4 k distance
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
GMATPrepNow wrote:
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 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


Hi GMATPrepNow

Why do you divide 2.25 by 60 ? what is the point? And why by 60 ? where from did you get 60?

thank you :-)
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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

*kudos for all correct solutions


Let d be the distance.

\(\frac{d}{12} - \frac{d}{30} = \frac{2.25}{60}\)

\(d = 0.75\)
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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dave13 wrote:
GMATPrepNow wrote:
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 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


Hi GMATPrepNow

Why do you divide 2.25 by 60 ? what is the point? And why by 60 ? where from did you get 60?

thank you :-)


Time in hours = (time in minutes)/60

For example, 30 minutes = 30/60 hours = 1/2 hour
And 45 minutes = 45/60 hours = 3/4 hour
And 10 minutes = 10/60 hours = 1/6 hour

So, 2.25 minutes = 2.25/60 hours
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
KARTAB wrote:
Let x minutes takes to reach office while running

So (x -2.25) minutes by bicycle to reach office

Distance by run = Distance by bicycle

x*12 = 30*(x - 2.25)

Solve it. x = 15/4 minutes i.e 1/16 hours

Distance = 12 *x or 30*x - 2.25) = 12 * 1/16 = 3/4



You are the only one who has tried solving it via equating distance and not the time. However, how do you get to 15/4? Shouldn't you be dividing 2.25 minutes by 60 to convert into hours or shouldn't you be converting speed per hour into speed per minute?
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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

*kudos for all correct solutions


Let the distance be X

X/12 - X/30 = 2.25/60.

Solving for X, we get X = 0.75

B.
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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
Given a constant Distance, speed is directly proportional to Time traveled.

If speed increases by (n / d)

Then Time traveled will DECREASE by n / (d + n)


Her speed increases by: (30 - 12) / 12 = 3/2

This Increase in Speed by 3/2 = n/d

Corresponds to a DECREASE in Time traveled by: 3 / (2 + 3) = 3/5


This decrease in time traveled of 2.25 min (Base Time T) ———> corresponds to a (3/5) Decrease in the Base Time T

9/4 min = (3/5) * T

T = 15/4 minutes ———> this is the time it takes her to travel the distance at Original Speed of 12 kmph


D = S * T ——-remember to convert the Units of minutes into hours


(12 kmph) * (15/4) * (1/60) =

Now it becomes easier to cancel NUM and DEN values

3/4 of a kilometer

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Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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