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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
Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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
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
Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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
Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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 = \(\frac{Average Distance}{{Average Speed}}\)
Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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 B
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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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
Re: Rhonda runs at an average speed of 12 kilometers per hour, [#permalink]
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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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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