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A machine has two flat circular plates of the same diameter [#permalink]

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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.

Circular Plates.jpg [ 118.78 KiB | Viewed 5584 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 \(x = 18\)

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

The question might be a bit complicated to understand but the solution is very simple and straight forward:

Attachment:

Ques3.jpg [ 5.38 KiB | Viewed 3488 times ]

Let's look at how the holes are placed with respect to each other. The plate which has 4 holes has holes which are 90 degrees away from each other. The plate which has 5 holes has holes which are 360/5 = 72 degrees away from each other. Assume that the holes at the top are aligned.

This is how the rest of the holes are placed on the two plates: 5 hole plate - 72 degrees, 144 degrees, 216 degrees, 288 degrees 4 hole plate - 90 degrees, 180 degrees, 270 degrees

To align the first holes with each other, we can move the 5 hole plate clockwise 18 degrees. Or the second hole of 5 hole plate 36 degrees clockwise. The minimum movement required to align any two holes is 18 degrees (the difference between any two angles of the two plates)

Re: A machine has two flat circular plates of the same diameter [#permalink]

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16 Jun 2014, 20:24

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"Align a pair of circles..."

When I read this question, in my mind, I understood it as two circle from each plate have to be aligned. I spent 10 minutes tryin to figure out how the f@#$, you align two sets of holes on a plate with 4 90* seprated circle on top of 5 holed plate with 72* seperation. What they meant was a pair of holes from the TWO DIFFERENT PLATES.

Is this really math exam or an english exam. Honest to christ, I dont know.

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.

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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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
_________________

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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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
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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

There is a potential pitfall in this approach. We cannot rely on visual inspection that <BOQ is the least. In fact there are two different pairs of holes which are the least distance apart here and hence two least degree angles though both are 18 degrees. And hence your answer happens to be correct.

Here is a different approach which I believe is more water tight. Let plate A have 4 holes and B 5. So holes on plate will divide the circle in the following fractions: 1/4, 2/4, 3/4 and 4/4 Similarly holes on B will divide the circle in the following fractions: 1/5, 2/5, 3/5, 4/5 and 5/5 Lets multiply all values by the LCM of 4 & 5: Holes on A: 5/20, 10/20, 15/20, 20/20 and holes on B: 4/20, 8/20, 12/20, 16/20, 20/20 4/4(20/20) from and 5/5(20/20) are already coinciding. So the next closet pairs are : 1) 5/20 & 4/20 2) 15/20 & 16/20 Both these pairs are 1/20 fraction apart. On a circle this means 360*1/20 = 18 degrees apart Ans: C Hope it is clear!
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Re: A machine has two flat circular plates of the same diameter [#permalink]

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18 Aug 2015, 12:40

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