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Alfred bikes from home to a restaurant at a constant rate of 10 miles

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Bunuel
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Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   01 Oct 2019, 02:36
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81% (01:49) correct 19% (02:11) wrong based on 179 sessions
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Alfred bikes from home to a restaurant at a constant rate of 10 miles per hour. Reaching the restaurant, he realizes that he forgot a wallet at home and returns back, by the same route, at a constant rate of 12 miles per hour. After picking the wallet in negligible time, he bikes to the restaurant again by the same route at a constant rate of 15 miles per hour. What is the average rate of Alfred for the entire, three-leg trip?

A. 11 miles per hour
B. 12 miles per hour
C. 12.5 miles per hour
D. 13 miles per hour
E. 14 miles per hour


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Bunuel
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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   06 Dec 2021, 07:22
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Alfred bikes from home to a restaurant at a constant rate of 10 miles per hour. Reaching the restaurant, he realizes that he forgot a wallet at home and returns back, by the same route, at a constant rate of 12 miles per hour. After picking the wallet in negligible time, he bikes to the restaurant again by the same route at a constant rate of 15 miles per hour. What is the average rate of Alfred for the entire, three-leg trip?

A. 11 miles per hour
B. 12 miles per hour
C. 12.5 miles per hour
D. 13 miles per hour
E. 14 miles per hour


Let the distance between the home and the restraint be \(d\) miles.

The average rate \(=\frac{total \ distance}{total \ time}=\frac{3d}{\frac{d}{10}+\frac{d}{12}+\frac{d}{15}}\)

\(d\) gets reduced and we get: \(= \frac{3}{\frac{1}{10}+\frac{1}{12}+\frac{1}{15}}=\frac{3}{\frac{15}{60}}=12\) miles.


Answer: B
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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   01 Oct 2019, 04:15
let the distance be D
time taken for the first-leg trip = \(\frac{D}{10}\)
time taken for the second-leg trip = \(\frac{D}{12}\)
time taken for the third-leg trip = \(\frac{D}{15}\)

--total distance travelled during 3 trip = 3D
--total time taken = \(\frac{D}{10}\)+\(\frac{D}{12}\)+\(\frac{D}{15}\) = \(\frac{15D}{60}\)

the average rate for three-leg trip =\(\frac{total-distance-travelled}{total-time-taken}\)

the average rate for three-leg trip = \(\frac{3D}{15D/60}\) = 12 miles per hour

B is the correct answer
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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   01 Oct 2019, 04:40
Bunuel wrote:
Alfred bikes from home to a restaurant at a constant rate of 10 miles per hour. Reaching the restaurant, he realizes that he forgot a wallet at home and returns back, by the same route, at a constant rate of 12 miles per hour. After picking the wallet in negligible time, he bikes to the restaurant again by the same route at a constant rate of 15 miles per hour. What is the average rate of Alfred for the entire, three-leg trip?

A. 11 miles per hour
B. 12 miles per hour
C. 12.5 miles per hour
D. 13 miles per hour
E. 14 miles per hour


Formula: When distance travelled is same, average speed is Harmonic mean of all the speeds
—> \(\frac{n}{S_{n}} = \frac{1}{S_{1}}+ \frac{1}{S_{2}} + . . . . + \frac{1}{S_{n}}\)
Let the average speed = x
—> \(\frac{3}{x}= \frac{1}{10}+ \frac{1}{12}+ \frac{1}{15}\)
—> \(\frac{3}{x}= \frac{(6 + 5 + 4)}{60}\)
—> \(\frac{3}{x}= \frac{15}{60}= \frac{1}{4}\)
—> \(x = 12\)

IMO Option B

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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   02 Oct 2019, 04:28
This can be solved by using the simple formula :-

3abc/(ab+bc+ca); when one travels at a speed a for 1/3rd distance, at speed b for 1/3rd distance and at speed c for the rest 1/3rd distance.

Here, the same distance is being covered 3 times hence the formula can be used.

(3*10*12*15)/(10X12 + 12X15 + 15X10)

=12 miles per hour.

Hence, B is the answer.
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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   07 Oct 2019, 20:12
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Bunuel wrote:
Alfred bikes from home to a restaurant at a constant rate of 10 miles per hour. Reaching the restaurant, he realizes that he forgot a wallet at home and returns back, by the same route, at a constant rate of 12 miles per hour. After picking the wallet in negligible time, he bikes to the restaurant again by the same route at a constant rate of 15 miles per hour. What is the average rate of Alfred for the entire, three-leg trip?

A. 11 miles per hour
B. 12 miles per hour
C. 12.5 miles per hour
D. 13 miles per hour
E. 14 miles per hour


We can use the formula: average = total distance/total time

average = 3d/(d/10 + d/12 + d/15)

Multiplying by 60/60, we have:

average = 180d/(6d + 5d + 4d)

average = 180d/15d = 12

Answer: B
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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink] New post   06 Dec 2021, 09:52
Bunuel wrote:
Alfred bikes from home to a restaurant at a constant rate of 10 miles per hour. Reaching the restaurant, he realizes that he forgot a wallet at home and returns back, by the same route, at a constant rate of 12 miles per hour. After picking the wallet in negligible time, he bikes to the restaurant again by the same route at a constant rate of 15 miles per hour. What is the average rate of Alfred for the entire, three-leg trip?

A. 11 miles per hour
B. 12 miles per hour
C. 12.5 miles per hour
D. 13 miles per hour
E. 14 miles per hour


M37-95

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Re: Alfred bikes from home to a restaurant at a constant rate of 10 miles [#permalink]
06 Dec 2021, 09:52
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