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

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
minutes 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
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Starting at 3 o????clock, at what time will the minute
hand 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 ask
how I solve problems--here????s 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.
We????re looking for the point between 3 o????clock and 4
o????clock 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).
We????re 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. I????ll bet this at least this
will tamp down the tendency to ask how I solve problems.

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