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

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

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New post 13 Feb 2017, 11:39
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Question Stats:

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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

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

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New post 13 Feb 2017, 12:14
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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 distance be x then

x/12 = x/30+2.25/60
solving
x=3/4 km

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

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New post 13 Feb 2017, 15:27
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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


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

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

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New post 13 Feb 2017, 19:06
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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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Rhonda runs at an average speed of 12 kilometers per hour,  [#permalink]

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New post 13 Feb 2017, 19:11
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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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New post 24 Feb 2017, 09:55
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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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New post 24 Feb 2017, 11:56
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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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New post 12 Dec 2017, 10:58
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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]

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New post 03 Feb 2018, 06:13
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]

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New post 03 Feb 2018, 09:53
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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New post 05 Feb 2018, 08:49
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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, &nbs [#permalink] 05 Feb 2018, 08:49
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