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 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 | ---------- |