February 26, 2001
Problem of the Month

The Hands Of A Clock

The hands of a clock are together at 12 noon.

At what time in the afternoon will they next be together again?
(Be exact and explain)

Solution to the Problem:

The circle on the face of the clock is divided into 60 equal parts.

The minute hand travels around the circle at a speed of 60 spaces per hour.

The hour hand follows it at a speed of 5 spaces per hour.

So the minute hand gets ahead at a speed of 55 spaces per hour.

The two hands will be together again when the minute hand is a full lap, or 60 spaces, ahead.

To get that far ahead will take 60/55 hours, which is 1 1/11 hours, or 1 hour 5 5/11 minutes.

So the hands will be together at 5 5/11 minutes past 1 o'clock.

Another solution is to use a RATE-TIME-DISTANCE table.

Let y = the spaces traveled by the hour hand.
Then y + 60 = the spaces traveled by the minute hand.
Let x = hours traveled by both the hour and the minute hands.

	Rate	Time	Distance
Minute Hand	60 sp/hr	x hr	y + 60 sp
Hour Hand	5 sp/hr	x hr	y sp

This yields 2 equations:
60 x = y + 60
5 x = y

Solving simultaneously, x = 60/55 hours or
approximately 1:05:27 (hr:min:sec).

Correctly solved by:

1. Rick Jones	Kennett Square, Pennsylvania
2. Sasha Joseph	Hebron, Maine
3. Ryan Charest	Winchester, Virginia
4. David Powell	Winchester, Virginia
5. Walt Arrison	Philadelphia, Pennsylvania
6. Jim Kennedy	Limerick, Ireland
7. Travis Pullins	----------
8. James Alarie	University of Michigan -- Flint
9. Albert Van Steendam	Wichelen, Belgium
10. Margaret Glendening	San Rafael, California
11. David Dixon	Bennettsville, South Carolina
12. Keith Mealy	Cincinnati, Ohio
13. Bill Hall	Wellington, Florida
14. Kasturi Rajendran	----------

Send any comments or questions to: David Pleacher