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Topic: Algebra 1 the vos Savant way
Replies: 0

 Domenico Rosa Posts: 455 Registered: 12/4/04
Algebra 1 the vos Savant way
Posted: Jul 20, 2004 10:11 AM

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 oÃÂclock, at what time will the minute hand catch up with
the hour hand?
ÃÂ-Bill Beachey, Berne, Ind.

At 16.3636 minutes after 3. ThatÃÂs the short answer. Not very
satisfying is it? So--as readers often ask how I solve
problems--hereÃÂ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. WeÃÂre looking for
the point between 3 oÃÂclock and 4 oÃÂ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).
WeÃÂ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. IÃÂll bet this at least will
tamp down the tendency to ask how I solve problems.

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