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The operation above rotates the regular pentagon clockwise by (72n) de

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The operation above rotates the regular pentagon clockwise by (72n) de  [#permalink]

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New post 08 Jul 2015, 03:53
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

61% (01:30) correct 39% (02:07) wrong based on 92 sessions

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Re: The operation above rotates the regular pentagon clockwise by (72n) de  [#permalink]

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New post 08 Jul 2015, 04:20
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Bunuel wrote:
Image
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4?

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

Kudos for a correct solution.

Attachment:
pentagon.gif


The central Angle will be divided in to 5 equal parts when all vertices are joined with the centre of the Pentagon

Therefore Each angle = 360/5 = 72 Degree

i.e. When the Pentagon Rotates by 72 degrees then A replaced the Point B
i.e. When the Pentagon Rotates by 72*2 degrees then E replaced the Point B
i.e. When the Pentagon Rotates by 72*3 degrees then D replaced the Point B
i.e. When the Pentagon Rotates by 72*4 degrees then C replaced the Point B

Answer: Option C
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Re: The operation above rotates the regular pentagon clockwise by (72n) de  [#permalink]

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New post 13 Jul 2015, 02:21
Bunuel wrote:
Image
The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4?

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

Kudos for a correct solution.

Attachment:
pentagon.gif


800score Official Solution:

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 increment our rotations by dividing 360/5 = 72.

So everytime the center spins 72º clockwise, the polygon moves by one vertex. So, when n = 4 the pentagon rotates by 4 vertices and the initial b position is replaced by c or answer choice (C).

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.
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Re: The operation above rotates the regular pentagon clockwise by (72n) de  [#permalink]

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New post 14 Jul 2015, 05:24
Since 360/5 = 72, and the equation rotates the corners by 72 degrees, then for 4*72 each number will be moved 4 corners ahead.

So, b will go to a and c will go to b.
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Re: The operation above rotates the regular pentagon clockwise by (72n) de  [#permalink]

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New post 23 Aug 2018, 00:21
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Re: The operation above rotates the regular pentagon clockwise by (72n) de   [#permalink] 23 Aug 2018, 00:21
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