1965 AMC 12 Problem 28

Attempt Problem 28 of the 1965 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 1965 AMC 12 solutions, or check the answer key.

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

28.

An escalator (moving staircase) of nn uniform steps visible at all times descends at constant speed. Two boys, AA and Z,Z, walk down the escalator steadily as it moves, AA negotiating twice as many escalator steps per minute as Z.Z. AA reaches the bottom after taking 2727 steps while ZZ reaches the bottom after taking 1818 steps. Then nn is:
6363
5454
4545
3636
3030
Answer: B
Concepts:relative speedsystem of equations
Small hint:
Let ZZ’s stepping rate be vv and the escalator rate be ee
Big hint:
Equate 27+27e2v27+\frac{27e}{2v} and 18+18ev18+\frac{18e}{v}
Solution:
Let ZZ take vv steps per minute, so AA takes 2v,2v, and let the escalator contribute ee steps per minute. Their travel times are 272v\frac{27}{2v} and 18v.\frac{18}{v}. Thus n=27+27e2v=18+18ev. n=27+\frac{27e}{2v} =18+\frac{18e}{v}. This gives ev=2,\frac{e}{v}=2, and then n=18+18(2)=54.n=18+18(2)=54.
Therefore, the correct answer is B.
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