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1955 AMC 12 Problem 33

Attempt Problem 33 of the 1955 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 1955 AMC 12 solutions, or check the answer key.

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

33.

Henry starts a trip when the hands of the clock are together between 88 a.m. and 99 a.m. He arrives at his destination between 22 p.m. and 33 p.m. when the hands of the clock are exactly 180∘180^\circ apart. The trip takes:

66 hr.

66 hr. 4371143\dfrac7{11} min.

55 hr. 1641116\dfrac4{11} min.

66 hr. 3030 min.

none of these

Answer: A
Concepts:clock handsangular speedelapsed time
Difficulty rating: 1880
Small Hint:

At the start, the minute hand must close a 240∘240^\circ gap at 5.5∘5.5^\circ per minute

Big Hint:

Compute the corresponding time after 2:002{:}00 when the hands are opposite and compare the two offsets

Solution:

At 8:00,8{:}00, the hour hand is 240∘240^\circ ahead. The minute hand gains at 5.5∘5.5^\circ per minute, so the hands coincide 2405.5=48011\frac{240}{5.5}=\frac{480}{11} minutes after 8.8. At 2:00,2{:}00, the hour hand is 60∘60^\circ ahead. For the hands to be 180∘180^\circ apart in the relevant direction, the minute hand must gain 240∘,240^\circ, again taking 48011\frac{480}{11} minutes. The start and finish therefore have the same minute offset within their hours, exactly six hours apart.

Thus, the correct answer is A.

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