1955 AMC 12 Problem 33

Question

Henry starts a trip when the hands of the clock are together between

8

a.m. and

9

a.m. He arrives at his destination between

2

p.m. and

3

p.m. when the hands of the clock are exactly

180^\circ

apart. The trip takes:

6

hr.

6

hr.

43\dfrac7{11}

min.

5

hr.

16\dfrac4{11}

min.

6

hr.

30

min.

none of these

At the start, the minute hand must close a

240^\circ

gap at

5.5^\circ

per minute

Compute the corresponding time after

2{:}00

when the hands are opposite and compare the two offsets

At

8{:}00,

the hour hand is

240^\circ

ahead. The minute hand gains at

5.5^\circ

per minute, so the hands coincide

\frac{240}{5.5}=\frac{480}{11}

minutes after

8.

At

2{:}00,

the hour hand is

60^\circ

ahead. For the hands to be

180^\circ

apart in the relevant direction, the minute hand must gain

240^\circ,

again taking

\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.

Problem 33 in other years