```Date: Jul 19, 2004 8:08 PM
Author: Domenico Rosa
Subject: Algebra 1 the vos Savant way

Marilyn vos Savant has written another convolutedsolution to the following problem (PARADE magazine, 18July 2004, Page 6). This provides another example ofthe importance of identifying suitable unknownquantities and defining variables, concepts that werereally stressed when Miss Phoebe Fitzpatrick taught meAlgebra 1 in 1962-63.The solution is much, much simpler if we let X be thenumber of minutes that the hour hand has moved whenthe two hands meet. As pointed out by Marilyn, theminutes hand will have moved 12X minutes. Since position of minute hand = position of hour hand, wehave the equation   12X = 15 + X,  which yields X = 15/11 minutes. Thus the two hands meet atapproximately 3:16.3636.Dom Rosa- -----------------------------------------------Starting at 3 o????clock, at what time will the minutehand catch up with the hour hand?  ????-Bill Beachey, Berne, Ind.At 16.3636 minutes after 3. That????s the short answer.Not very satisfying is it? So--as readers often askhow I solve problems--here????s the long answer.  Say that X is the distance that the hour handtravels in an hour and Y is the distance that theminute hand moves in the same time. The minute handmoves 12 times as fast, so Y equals 12X. Now say thatP is the fraction traveled by the hands in an hour.We????re looking for the point between 3 o????clock and 4o????clock at which PX and PY coincide. That would be thespot where 3X plus PX equals PY. As Y equals 12X, thismeans 3X plus PX equals P(12X).  We????re in the home stretch! Solving the equation 3X + PX = P(12X) gives us a fraction of 3/11. Whichmeans the minute hand will catch up with the hour handat 3:16.3636. Oh, well. I????ll bet this at least thiswill tamp down the tendency to ask how I solve problems.------- End of Forwarded Message
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