Marilyn vos Savant has written another convoluted solution to the following problem (PARADE magazine, 18 July 2004, Page 6). This provides another example of the importance of identifying suitable unknown quantities and defining variables, concepts that were really stressed when Miss Phoebe Fitzpatrick taught me Algebra 1 in 1962-63.
The solution is much, much simpler if we let X be the number of minutes that the hour hand has moved when the two hands meet. As pointed out by Marilyn, the minute hand will have moved 12X minutes. Since position of minute hand = position of hour hand, we have the equation 12X = 15 + X, which yields X = 15/11 minutes. Thus the two hands meet at approximately 3:16.3636.
Dom Rosa - -----------------------------------------------
Starting at 3 oÃÂclock, at what time will the minute hand catch up with the hour hand? ÃÂ-Bill Beachey, Berne, Ind.
At 16.3636 minutes after 3. ThatÃÂs the short answer. Not very satisfying is it? So--as readers often ask how I solve problems--hereÃÂs the long answer. Say that X is the distance that the hour hand travels in an hour and Y is the distance that the minute hand moves in the same time. The minute hand moves 12 times as fast, so Y equals 12X. Now say that P is the fraction traveled by the hands in an hour. WeÃÂre looking for the point between 3 oÃÂclock and 4 oÃÂclock at which PX and PY coincide. That would be the spot where 3X plus PX equals PY. As Y equals 12X, this means 3X plus PX equals P(12X). WeÃÂre in the home stretch! Solving the equation 3X + PX = P(12X) gives us a fraction of 3/11. Which means the minute hand will catch up with the hour hand at 3:16.3636. Oh, well. IÃÂll bet this at least will tamp down the tendency to ask how I solve problems.