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GMAT Diagnostic Test Question 34 [#permalink]
 06 Jun 2009, 23:18
GMAT Diagnostic Test Question 34Field: geometry Difficulty: 600
A circle is inscribed in a half circle with a diameter of π. What is the ratio of the area of the half circle to the area not covered by the inscribed circle? A. 1: 1 B. 1: 2 C. 1: 4 D. 3: 4 E. 2: 1 REVISED VERSION OF THIS QUESTION IS HERE: gmat-diagnostic-test-question-79364.html#p1146441
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Re: GMAT Diagnostic Test Question 34 [#permalink]
10 Jun 2009, 20:25
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Explanation
Official Answer: EArea of the half circle = \frac{1}{2}*\pi r^2 = \frac{1}{2} * \pi (\frac{\pi}{2})^2 = \frac{\pi^3}{8}Area of the inscribed circle = \pi r^2 = \pi (\frac{\pi}{4})^2 = {\frac{\pi^3}{16}}Area of the half circle uncovered by the inscribed circle = {\frac{\pi^3}{8}} - {\frac{\pi^3}{16}} = {\frac{\pi^3}{16}}Ratio of the area of the half circle to the area not covered by the inscribed circle = ({\frac{\pi^3}{8}}) : ({\frac{\pi^3}{16}}) = 2:1.Therefore it is E.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
15 Jul 2009, 00:18
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Pardon me, just wanted to confirm the area of the smaller circle...
Shouldn't it be π^3 / 16?
This is because it's diameter is already π /2. So, the radius is π /4.
Hence, the area is π r^2 , which can be π (π/4)^2
What is your opinion?
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Re: GMAT Diagnostic Test Question 34 [#permalink]
15 Jul 2009, 00:28
cialit0506 wrote: Shouldn't it be π^3 / 16?
Thank you for posting it. There was an error in the explanation - it has been fixed and so has the answer choice. Please read through the explanation and see if it makes sense.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
15 Jul 2009, 02:41
The highest possible value in our difficulty rating is 750; the lowest is 600. bigoyal wrote: bb wrote: Difficulty: 600 Pardon me, if you find my question very basic. Can you please let me know the difficulty scale ? Surely my rating would be based on the difficulty level on the scale. _________________
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Re: GMAT Diagnostic Test Question 34 [#permalink]
16 Jul 2009, 01:35
dzyubam wrote: The highest possible value in our difficulty rating is 750; the lowest is 600. bigoyal wrote: bb wrote: Difficulty: 600 Pardon me, if you find my question very basic. Can you please let me know the difficulty scale ? Surely my rating would be based on the difficulty level on the scale. If 600 is lowest, I think this question is little difficult and time-consuming as compared to the easiest GMAT questions. I think, this questions should be of average difficulty level.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
16 Jul 2009, 01:48
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bigoyal wrote:
If 600 is lowest, I think this question is little difficult and time-consuming as compared to the easiest GMAT questions. I think, this questions should be of average difficulty level.
I would probably agree that it is a bit tough for a 600 level - will need to simplify.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
20 Aug 2009, 08:53
GMAT TIGER wrote: Explanation
Official Answer: EArea of the half circle = \frac{1}{2}*\pi r^2 = \frac{1}{2} * \pi (\frac{\pi}{2})^2 = \frac{\pi^3}{8}Area of the inscribed circle = \pi r^2 = \pi (\frac{\pi}{4})^2 = {\frac{\pi^3}{16}}Area of the half circle uncovered by the inscribed circle = {\frac{\pi^3}{8}} - {\frac{\pi^3}{16}} = {\frac{\pi^3}{16}}Ratio of the area of the half circle to the area not covered by the inscribed circle = ({\frac{\pi^3}{8}}) : ({\frac{\pi^3}{16}}) = 2:1.Therefore it is E. Hey GMAT TIGER, could you please explain how do you come from \frac{1}{2}*\pi r^2TO \frac{1}{2} * \pi (\frac{\pi}{2})^2Thx very much
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Re: GMAT Diagnostic Test Question 34 [#permalink]
23 Aug 2009, 12:02
Diameter of half circle = Pi => Radius of half circle = r = Pi/2 Area of half circle = (Pi * r^2) / 2 Plug in r and you get: Area of half circle = (Pi * (Pi/2)^2) / 2 Sorry, I don't know how to paste images on a post.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
25 Nov 2009, 11:00
I am new to the forum and just finished my diagnostic exam. I just want to make sure that my logic for this questions i right.
As soon as I finished reading the problem, I realized that the ratio has to be greater than 1 since the semicircle is obviously bigger than the area not covered by the inscribed circle. So the only right choice could have been E.
Am I missing something here?
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Re: GMAT Diagnostic Test Question 34 [#permalink]
22 Dec 2009, 03:22
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Hey BB! I think this question is too easy because of the anwer choices. E is apparent. The level could be raised if you changed the answers (I mean add some ratios >1). But if you do this, the level would be higher than 600 cos. it's pretty time consuming... Agree? 
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Re: GMAT Diagnostic Test Question 34 [#permalink]
01 Oct 2010, 17:46
GMAT TIGER wrote: Explanation
Official Answer: EArea of the inscribed circle = \pi r^2 = \pi (\frac{\pi}{4})^2 = {\frac{\pi^3}{16}}Therefore it is E. Why can we assume the size of the inscribed circle?
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Re: GMAT Diagnostic Test Question 34 [#permalink]
04 Oct 2010, 03:39
The radius of this inscribed circle will be half of the radius of the half-circle it's inscribed into, hence \frac{\pi}{4}.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
05 Oct 2010, 10:01
Yea you have to divide the area of the half-circle because it's half a circle. That's how you get Pi/8
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Re: GMAT Diagnostic Test Question 34 [#permalink]
09 Dec 2010, 15:13
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I did this in much simpler way, hope it helps some of you. Since Pi is used as a diameter of an inscribed circle, and in order for us to calculate areas (where we need to use Pi again), i substituted Pi for 4. See the image below:  So now, we have a diameter of a smaller circle at 4, which means its radius is 2, and area 4Pi. Diameter of an inscribed circle will be the radius of a semi-circle. So, the area of a semi circle will pi 16Pi/2 = 8Pi. TO find the area not covered by the inscribed circle, we subtract the inscribed circle's area from the semi-circle's area: 8Pi - 4Pi = 4Pi. So, the ratio of the semi-circle to the non-covered region of the inscribed circle is 8Pi:4Pi or 2:1. Hope this helps to some of you guys.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
24 Feb 2011, 10:23
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Hey guys, Excuse me if I am misunderstanding something basic, but what does inscribe mean? Does it not mean that the smaller circle is just completely inside the half-circle (touching all sides possible) (Not necessarily in the centre). So, when I read the question, I can't figure where this smaller circle is exactly. As per my understanding, we can have infinite circles inscribed in the half-circle, touching all sides of the half-circle at various points, with of course, the biggest one in the centre. Please see attached image. So, why have we assumed that its centre is on the radius of the half circle?
Attachments
circle.jpg [ 22.42 KiB | Viewed 4598 times ]
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Re: GMAT Diagnostic Test Question 34 [#permalink]
08 May 2011, 05:19
The answer choices are poorly formulated, in my opinion. Because the half circle has to be bigger than the circle inscribed inside of it, the first number in the ratio must be larger than the second. Only option E is like this. The question is thus solved with no calculations.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
13 Feb 2012, 17:18
Interesting observation about the answer choices! I didn't even look at them and attacked the questions head on. Should have thought of that before, good strategy for next time.
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Re: GMAT Diagnostic Test Question 34 [#permalink]
22 Nov 2012, 06:24
What is the exact definition of inscribed?
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Re: GMAT Diagnostic Test Question 34 [#permalink]
22 Nov 2012, 06:51
bb wrote: GMAT Diagnostic Test Question 34Field: geometry Difficulty: 600
A circle is inscribed in a half circle with a diameter of π. What is the ratio of the area of the half circle to the area not covered by the inscribed circle? A. 1: 1 B. 1: 2 C. 1: 4 D. 3: 4 E. 2: 1 BELOW IS REVISED VERSION OF THIS QUESTION. A circle is inscribed right in the middle of a semicircle with a diameter of \pi as shown below. What is the ratio of the area of the semicircle to the area not covered by the inscribed circle?Attachment: 
Semicircle 2.png [ 14.79 KiB | Viewed 969 times ]
A. 4:1 B. 3:2 C. 3:1 D. 4:3 E. 2:1 Attachment: 
Semicircle.png [ 15.33 KiB | Viewed 970 times ]
Since the radius of the big circle ( \frac{\pi}{2}) is twice the radius of the inscribed circle ( \frac{\pi}{4}) then its area is 4 times greater then the area of the inscribed circle (because in the area formula the radius is squared. For example the area of a circle with radius of 2 is 4\pi, which is 4 times greater than the radius of a circle with the radius of 1, which is \pi). Thus the are of the semicircle is 4/2=2 times greater than the area of the inscribed circle: so, the area of the semicircle is 2 units, the area of the inscribed circle is 1 unit, and the area of the semicircle not covered by the inscribed circle is also 1 unit. Ratio: 2/1. Answer: E. Hope it's clear.
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Re: GMAT Diagnostic Test Question 34
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22 Nov 2012, 06:51
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