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 Post subject: GMAT Logic
PostPosted: Sun May 23, 2010 12:23 pm 
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Joined: Tue Apr 13, 2010 8:48 am
Posts: 483
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 19?

A. a
B. b
C. c
D. d
E. e

(C) Think of the polygon as you would a wheel being rotated at its center. A wheel sweeps a 360º angle everytime it returns to its initial positon. Since this is a regular pentagon, we can incement our rotations by dividing 360/5 = 72. So everytime the center spins 72º, the polygon moves by one vertex clockwise.An alternative way to consider the problem is to realize that everytime n = 5, it returns to its original configuration. So, when n = 4, the position is occupied by the vertex that is one vertex away in the clockwise direction because it lags by one 72º turn. For more information on type of questions, order the online prep course.

Since the rotation is clockwise and the position b is 72 degrees short of returning to its original configuration, shouldn\'t its position be at a? (c would replace b at 21n)
 Post subject: Re: math: geometry question, pentagon.
PostPosted: Sun May 23, 2010 12:40 pm 
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Joined: Fri Apr 09, 2010 2:11 pm
Posts: 459
Imagine that we rotated the pentagon 20 × 72º clockwise. That makes it look the same:

If we rotate it (1 × 72) more degrees clockwise then it will look:

We see that in this case (n=21) point a stands at original position of point b.

If we rotate (1 × 72) less degrees clockwise than (20 × 72º) clockwise turn then in this case it will look:

For better understanding, please, take a look at this animation:
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