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06 Dec 2021, 07:21
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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
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
06 Dec 2021, 07:21
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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
