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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