1969 AMC 12 Problem 27

Attempt Problem 27 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in 22 hours, then the time, in hours, needed to traverse the nnth mile is:
2n−1\dfrac2{n-1}
n−12\dfrac{n-1}{2}
2n\dfrac2n
2n2n
2(n−1)2(n-1)
Answer: E
Concepts:distance rate and timerateratio and proportion
Small hint:
During the nnth mile, the number of miles already traveled is n−1n-1
Big hint:
Time for a fixed one-mile distance varies directly as n−1n-1
Solution:
For the nnth mile, vn=kn−1.v_n=\frac{k}{n-1}. Since the distance is one mile, the time is Tn=1vn=n−1k. T_n=\frac1{v_n}=\frac{n-1}{k}. The condition T2=2T_2=2 gives 1k=2.\frac{1}{k}=2. Therefore Tn=2(n−1).T_n=2(n-1).
Therefore, the correct answer is E.
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