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Is the circumference of the largest circle above that can contain all

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Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 01 Aug 2017, 12:44
2
8
00:00
A
B
C
D
E

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  55% (hard)

Question Stats:

59% (02:06) correct 41% (02:00) wrong based on 128 sessions

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Is the circumference of the largest circle above that can contain all four circles in the figure above equivalent to \(16 + 16\sqrt{2}*\pi\)?

(1) The radius of one of the small circles is 8

(2) The area of all four circles is \(256\pi\)


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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 01 Aug 2017, 20:11
1
\(16 + 16\sqrt{2}*\pi\)?

Well (1) and (2) gives the same information. So answer must be D or E.

Working out, I got the diameter of larger circle as \(16 + 16\sqrt{2}\)

Circumference of the circle will be 2*pi*r or pi*d
= \(16\pi + 16\sqrt{2}*\pi\)

Which is not equal to one given in prompt.

Hence D.

Am I correct?
I will elaborate once I get to know this is correct :P
Bunuel: Your questions with suspense answers kill me :(
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 02 Aug 2017, 12:39
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Bunuel wrote:
Is the circumference of the largest circle above that can contain all four circles in the figure above equivalent to \(16 + 16\sqrt{2}*\pi\)?


I'm curious about this question, because it might be an interesting problem, but I can't even begin to guess what it means. It asks about a circle "above", and tells us that circle "can contain all four circles ... above", which would mean the circle the question is talking about can contain itself. That doesn't make any sense.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 11:44
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For a circle to encompass all 4 smaller circles, it has to be either tangent to those 4 or simply contain them. The question stem does not provide any info on this so I guess the answer is E?
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 12:43
IanStewart wrote:
Bunuel wrote:
Is the circumference of the largest circle above that can contain all four circles in the figure above equivalent to \(16 + 16\sqrt{2}*\pi\)?


I'm curious about this question, because it might be an interesting problem, but I can't even begin to guess what it means. It asks about a circle "above", and tells us that circle "can contain all four circles ... above", which would mean the circle the question is talking about can contain itself. That doesn't make any sense.


Same reaction here. Question is not clear. I dont see the "largest" circle"
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 14:28
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1
The answer is D.

If all the four centres of the four circles are joined , it will form a square, we will have a square of length 16 units, from this the diagonal would be 16 √2 , the diameter of the bigger circle would be 16+16 √2 (adding the addtional radii to formulate the diameter). (Please refer the attached figure). This would give us the circumference of the bigger circle as (16+16 √2) π.

Since both 1 and 2 give the same data, each statement alone is sufficient as we have the radii of each circle and by forming a square we can find the diameter of the larger circle, thereby finding out the circumference of the circle.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 14:39
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The answer should be B.

It doesn't say anywhere that all 4 circles have the same radius, so in statement I, we could have 1 circle at the provided radius and the other 3 at another. Statement II provides you with the area which you can calculate the required circumference.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 14:55
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nrxbra001 wrote:
The answer should be B.

It doesn't say anywhere that all 4 circles have the same radius, so in statement I, we could have 1 circle at the provided radius and the other 3 at another. Statement II provides you with the area which you can calculate the required circumference.



How would you calculate the circumference of the bigger circle with the sum of areas of smaller circles?
The radii of each circle might differ,in that case you would not be able to find the radii of the bigger circle.

If the circles are not of the same size, then the answer would be E.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 03 Aug 2017, 19:36
1
Statements I and II basically give you the same information:

The area of all four circles is 256π: 4π\(r^{2}\)= 256π -> \(r^{2}\) = 64 -> r = 8

The distance X from the center of the imaginary bigger circle to the center of any smaller circle can be calculated using the Pythagorean theorem:
\(16^{2}\) = \(x^{2}\) + \(x^{2}\) -> 2\(x^{2}\) = 256 -> \(x^{2}\) = 128

The circumference of the larger circle therefore is:
2π*(x + r) -> 2π (8\(\sqrt{2}\) + 8) -> 16\(\sqrt{2}\)π + 16

Therefore, the correct answer is D.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 04 Aug 2017, 04:07
We need to know whether circles are identical or not? Incomplete question
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 04 Aug 2017, 06:50
Bunuel wrote:
Is the circumference of the largest circle above that can contain all four circles in the figure above


I suspect sharmili has correctly guessed the intentions of the question designer, and if so, the question should be asking about the smallest circle that is not depicted above (the 'largest circle' that could contain all the other circles is infinitely big, so asking for the largest circle doesn't make sense), and there seems to be a π missing in the number at the end of the question. We do need information about whether the small circles are identical to solve anything here.
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Re: Is the circumference of the largest circle above that can contain all  [#permalink]

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New post 14 Sep 2019, 21:37
Bunuel wrote:
Image
Is the circumference of the largest circle above that can contain all four circles in the figure above equivalent to \(16 + 16\sqrt{2}*\pi\)?

(1) The radius of one of the small circles is 8

(2) The area of all four circles is \(256\pi\)


Attachment:
gmat-tip-circles-1.png


Hi Bunuel
The circles are nowhere mentioned as identical ones. How do I know that it isn't a usual 700 question trap.
That's why I choose option 'E'.
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Re: Is the circumference of the largest circle above that can contain all   [#permalink] 14 Sep 2019, 21:37
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