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A machine has two flat circular plates of the same diameter

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A machine has two flat circular plates of the same diameter [#permalink] New post 08 Jan 2013, 21:16
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A machine has two flat circular plates of the same diameter both plates have holes of one inch diameter that are equally placed and are the same from the edges as shown above. one plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. If one plate remains stationary what is the least number of degrees that the other plate must be rotated so that a different pair of holes is perfectly aligned.

A. 6
B. 12
C. 18
D. 24
E. 36
[Reveal] Spoiler: OA

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Last edited by fozzzy on 09 Jan 2013, 03:54, edited 2 times in total.
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Re: A machine has 2 flat [#permalink] New post 08 Jan 2013, 21:18
Where is the screenshot!
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Re: A machine has 2 flat [#permalink] New post 08 Jan 2013, 21:19
PraPon wrote:
Where is the screenshot!


sorry it was a bit blurry tried changing it doesn't look any better :(
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Re: A machine has 2 flat [#permalink] New post 08 Jan 2013, 22:20
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Attachment:
Circular Plates.jpg
Circular Plates.jpg [ 118.78 KiB | Viewed 953 times ]

As the holes are equidistant on both the plates, they form pentagon and square on their respective plate (as shown in the figure)

Lets consider two circular plates as below:
Plate1: Consider holes A & B are on Pentagon in plate1
Plate2: Consider holes P & Q are on square in plate2

Its given that two plates are kept over other and two of their holes are coinciding. Lets assume holes A & P are coinciding.
From the diagram you can see, holes B & Q can be aligned with minimum rotation. i.e. we need to find out <BOQ = x =?

Central angle formed by two vertices of pentagon i.e. <AOB = y = \frac{360}{5} = 72
Central angle formed by two vertices of square i.e. <POQ = \frac{360}{4} = 90

<AOB + <BOQ = <AOQ
i.e. y + x = 90
i.e. 72 + x = 90
Resolve to x = 18

Hence choice (C) is the answer
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Re: A machine has 2 flat [#permalink] New post 09 Jan 2013, 02:29
PraPon wrote:
Attachment:
Circular Plates.jpg

As the holes are equidistant on both the plates, they form pentagon and square on their respective plate (as shown in the figure)

Lets consider two circular plates as below:
Plate1: Consider holes A & B are on Pentagon in plate1
Plate2: Consider holes P & Q are on square in plate2

Its given that two plates are kept over other and two of their holes are coinciding. Lets assume holes A & P are coinciding.
From the diagram you can see, holes B & Q can be aligned with minimum rotation. i.e. we need to find out <BOQ = x =?

Central angle formed by two vertices of pentagon i.e. <AOB = y = \frac{360}{5} = 72
Central angle formed by two vertices of square i.e. <POQ = \frac{360}{4} = 90

<AOB + <BOQ = <AOQ
i.e. y + x = 90
i.e. 72 + x = 90
Resolve to x = 18

Hence choice (C) is the answer


Great explanation! I didn't realize they were testing the concepts of polygon! kudos to you!
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Re: A machine has 2 flat [#permalink] New post 09 Jan 2013, 09:12
fozzzy wrote:
PraPon wrote:
Attachment:
Circular Plates.jpg

As the holes are equidistant on both the plates, they form pentagon and square on their respective plate (as shown in the figure)

Lets consider two circular plates as below:
Plate1: Consider holes A & B are on Pentagon in plate1
Plate2: Consider holes P & Q are on square in plate2

Its given that two plates are kept over other and two of their holes are coinciding. Lets assume holes A & P are coinciding.
From the diagram you can see, holes B & Q can be aligned with minimum rotation. i.e. we need to find out <BOQ = x =?

Central angle formed by two vertices of pentagon i.e. <AOB = y = \frac{360}{5} = 72
Central angle formed by two vertices of square i.e. <POQ = \frac{360}{4} = 90

<AOB + <BOQ = <AOQ
i.e. y + x = 90
i.e. 72 + x = 90
Resolve to x = 18

Hence choice (C) is the answer


Great explanation! I didn't realize they were testing the concepts of polygon! kudos to you!


see the new Gmatclub math book. is well explained inside
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Re: A machine has 2 flat [#permalink] New post 09 Jan 2013, 16:30
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The four equidistant circular rings on the first circular plate divides it into four equal parts; so each part is 1/4
The five equidistant rings on the second cirular plate divides it into five equal parts; each part is 1/5


If one of the circular plate is overlaid on the other with a pair of circular holes aligned, the holes will be 1/4 minus 1/5 = 1/20 apart from each other. If we have to align a different pair of circles, we have to move the circular plate on the top 1/20 * 360 degrees.
1/20*360 = 18 degrees.

Further explanation: I interpreted this as a form of a circle and its sector problem. Area covered by a sector is given by A = angle/360 * area of the circle
If we take area of the circular plate as 1, the area covered by each sector will be 1/20. We are asked to find out the angle subtended by the sector between the holes.

1/20 = angle/360 *1

angle = 360/20


So C is the answer.

Hope this approach helps.
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Re: A machine has 2 flat [#permalink] New post 09 Jan 2013, 19:19
fozzzy wrote:
A machine has two flat circular plates of the same diameter both plates have holes of one inch diameter that are equally placed and are the same from the edges as shown above. one plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. If one plate remains stationary what is the least number of degrees that the other plate must be rotated so that a different pair of holes is perfectly aligned.

6
12
18
24
36


First plate (4 holes) = 360/4 = 90
Second Plate (5 holes) = 360/5 = 72

So least degree to get a hole aligned with other = 90 - 72 =18

Guys Please validate this!
Will this method be useful for similar type problems ?
Or did i go wrong anywhere?
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Re: A machine has 2 flat [#permalink] New post 09 Jan 2013, 21:57
chvrmohan4 wrote:
The four equidistant circular rings on the first circular plate divides it into four equal parts; so each part is 1/4
The five equidistant rings on the second cirular plate divides it into five equal parts; each part is 1/5


If one of the circular plate is overlaid on the other with a pair of circular holes aligned, the holes will be 1/4 minus 1/5 = 1/20 apart from each other. If we have to align a different pair of circles, we have to move the circular plate on the top 1/20 * 360 degrees.
1/20*360 = 18 degrees.

Further explanation: I interpreted this as a form of a circle and its sector problem. Area covered by a sector is given by A = angle/360 * area of the circle
If we take area of the circular plate as 1, the area covered by each sector will be 1/20. We are asked to find out the angle subtended by the sector between the holes.

1/20 = angle/360 *1

angle = 360/20


So C is the answer.

Hope this approach helps.


Really nice solution! very well thought! kudos to you
Re: A machine has 2 flat   [#permalink] 09 Jan 2013, 21:57
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