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01 Oct 2019, 02:36
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B
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Difficulty:
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Question Stats:
81% (01:48) correct
19%
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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
M37-95
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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06 Dec 2021, 07:22
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Official Solution:
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
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
General Discussion
01 Oct 2019, 04:15
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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
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
