Author 
Message 
TAGS:

Hide Tags

Director
Joined: 29 Nov 2012
Posts: 775

A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
Updated on: 09 Jan 2013, 03:54
Question Stats:
66% (02:10) correct 34% (02:34) wrong based on 373 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Attachments
Question.png [ 51.23 KiB  Viewed 9938 times ]
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html
Originally posted by fozzzy on 08 Jan 2013, 21:16.
Last edited by fozzzy on 09 Jan 2013, 03:54, edited 2 times in total.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
16 Jun 2014, 22:03
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.
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 7777 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) Answer (C)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Senior Manager
Joined: 27 Jun 2012
Posts: 380
Concentration: Strategy, Finance

Re: A machine has 2 flat
[#permalink]
Show Tags
08 Jan 2013, 22:20
Attachment:
Circular Plates.jpg [ 118.78 KiB  Viewed 9898 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
_________________
Thanks, Prashant Ponde
Tough 700+ Level RCs: Passage1  Passage2  Passage3  Passage4  Passage5  Passage6  Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible  Official Guide 13 Questions Set Mapped: Click here Finance your Student loan through SoFi and get $100 referral bonus : Click here




Senior Manager
Joined: 27 Jun 2012
Posts: 380
Concentration: Strategy, Finance

Re: A machine has 2 flat
[#permalink]
Show Tags
08 Jan 2013, 21:18
Where is the screenshot!
_________________
Thanks, Prashant Ponde
Tough 700+ Level RCs: Passage1  Passage2  Passage3  Passage4  Passage5  Passage6  Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible  Official Guide 13 Questions Set Mapped: Click here Finance your Student loan through SoFi and get $100 referral bonus : Click here



Director
Joined: 29 Nov 2012
Posts: 775

Re: A machine has 2 flat
[#permalink]
Show Tags
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
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



Director
Joined: 29 Nov 2012
Posts: 775

Re: A machine has 2 flat
[#permalink]
Show Tags
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 answerGreat explanation! I didn't realize they were testing the concepts of polygon! kudos to you!
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



Board of Directors
Joined: 01 Sep 2010
Posts: 3304

Re: A machine has 2 flat
[#permalink]
Show Tags
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 answerGreat 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
_________________
COLLECTION OF QUESTIONS AND RESOURCES Quant: 1. ALL GMATPrep questions Quant/Verbal 2. Bunuel Signature Collection  The Next Generation 3. Bunuel Signature Collection ALLINONE WITH SOLUTIONS 4. Veritas Prep Blog PDF Version 5. MGMAT Study Hall Thursdays with Ron Quant Videos Verbal:1. Verbal question bank and directories by Carcass 2. MGMAT Study Hall Thursdays with Ron Verbal Videos 3. Critical Reasoning_Oldy but goldy question banks 4. Sentence Correction_Oldy but goldy question banks 5. Readingcomprehension_Oldy but goldy question banks



Intern
Joined: 08 Jan 2013
Posts: 4

Re: A machine has 2 flat
[#permalink]
Show Tags
09 Jan 2013, 16:30
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.



Manager
Joined: 04 Oct 2011
Posts: 193
Location: India
Concentration: Entrepreneurship, International Business
GPA: 3

Re: A machine has 2 flat
[#permalink]
Show Tags
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?
_________________
GMAT  Practice, Patience, Persistence Kudos if u like



Director
Joined: 29 Nov 2012
Posts: 775

Re: A machine has 2 flat
[#permalink]
Show Tags
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
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



Retired Moderator
Joined: 29 Oct 2013
Posts: 266
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

Re: A machine has 2 flat
[#permalink]
Show Tags
03 Jun 2014, 22:50
carcass wrote: 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 answerGreat 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!
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1829
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
04 Jun 2014, 23:52
Refer screenshot below: I just believe that it takes more time to read the problem than to solve it
Attachments
cir.jpg [ 40.46 KiB  Viewed 7917 times ]
_________________
Kindly press "+1 Kudos" to appreciate



Manager
Joined: 12 Feb 2012
Posts: 125

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
16 Jun 2014, 20:24
"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.
Posted from my mobile device



Intern
Joined: 23 Apr 2016
Posts: 22
Location: Finland
Concentration: General Management, International Business
GPA: 3.65

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
13 Nov 2016, 12:39
Can some one please explain to me what does this mean: "one plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. "
How can two holes be aligned in this scenario ... I am not able to figure this out.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
14 Nov 2016, 04:57
thapliya wrote: Can some one please explain to me what does this mean: "one plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. "
How can two holes be aligned in this scenario ... I am not able to figure this out. Look at the original diagram given. If you just put one plate perfectly on top of the other, in their current position, their top holes will be aligned.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Senior Manager
Joined: 06 Jun 2016
Posts: 259
Location: India
Concentration: Operations, Strategy
GMAT 1: 600 Q49 V23 GMAT 2: 680 Q49 V34
GPA: 3.9

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
14 Nov 2016, 05:32
thapliya wrote: Can some one please explain to me what does this mean: "one plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. "
How can two holes be aligned in this scenario ... I am not able to figure this out. one hole from each plate is coinciding that is how two holes are aligned in this scenario



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12660
Location: United States (CA)

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
29 Mar 2018, 12:56
Hi All, This question comes with some odd wording, but here is the "intent" of what it's asking: The two circles have "holes" drilled at different spots (4 on one circle; 5 on the other). The prompt tells us to line up 2 holes (1 from each circle), then asks what the LEAST number of degrees you'd have to spin either of the circles so that two DIFFERENT holes would line up. This question is based on the number of degrees in a circle: In the "Circle with 4 holes", the holes are at 0, 90, 180 and 270 In the "Circle with 5 holes", the holes are at 0, 72, 144, 216 and 288 If we line up the "0 holes", then we have to look for the LEAST difference between any OTHER pair of holes. You can find the least difference is two different sets: (90 and 72) and (270 and 288). Each has a difference of 18 degrees. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Senior Manager
Joined: 17 Mar 2014
Posts: 383

Re: A machine has two flat circular plates of the same diameter
[#permalink]
Show Tags
21 Jul 2018, 15:18
interesting question . Bunuel, karishma could you share more questions on this topic.




Re: A machine has two flat circular plates of the same diameter &nbs
[#permalink]
21 Jul 2018, 15:18






