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If the perimeter of square region S and the perimeter of circular regi
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Updated on: 06 Nov 2018, 02:49
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If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to: A. \(\frac {3}{2}\) B. \(\frac {4}{3}\) C. \(\frac {3}{4}\) D. \(\frac {2}{3}\) E. \(\frac {1}{2}\)
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Originally posted by Dumpling on 22 Feb 2006, 12:30.
Last edited by Bunuel on 06 Nov 2018, 02:49, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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If the perimeter of square region S and the perimeter of circular regi
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Re: Ratio
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10 Mar 2011, 09:51
If Perimeter of square = Perimeter of Circle, then: 4a = 2\(\pi\)r, where a = side of square and r is radius of the circle a/r = \(\pi\)/2
Area of S/Area of C = \(a^2\)/ \(\pi\) \(r^2\) = \((\pi/2)^2\) * 1/\(\pi\) = \(\pi\)/4 = 3.14/4 = \(\approx\) 3/4 = C



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Re: If the perimeter of square region S and the perimeter of circular regi
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09 Mar 2014, 02:07
Please refer diagram below:
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Re: If the perimeter of square region S and the perimeter of circular regi
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09 Apr 2015, 22:12
Hi All, This question is perfect for TESTing VALUES. We're told that the PERIMETER of a square is equal to the PERIMETER (meaning the 'circumference') of a circle. We're asked to figure out the approximate ratio of the area of the square to the area of the circle. Let's TEST VALUES. Since we're dealing with a circle, let's work "pi" into our math right from the beginning.... Perimeter = 4pi For the Square: Perimeter = 4pi Side length = pi Area = (pi)^2 For the Circle: Perimeter = 4pi Radius = 2 Area = 4pi Area of Square/Area of Circle = (pi)^2/4pi = pi/4 = about 3.14/4 = about 3/4 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If the perimeter of square region S and the perimeter of the
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30 Jun 2015, 09:39
Bunuel wrote: chintzzz wrote: I got an answer of B but the official answer is different. please explain
If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to A.3/2 B.4/3 C.3/4 D.2/3 E.1/2 \(P_{square}=4x=2\pi{r}= P_{circle}\) > \(x=\frac{\pi{r}}{2}\).\(\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Answer: C. Can you explain this portion: Pcircle > x=πr2.



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Re: If the perimeter of square region S and the perimeter of the
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30 Jun 2015, 09:45
xLUCAJx wrote: Bunuel wrote: chintzzz wrote: I got an answer of B but the official answer is different. please explain
If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to A.3/2 B.4/3 C.3/4 D.2/3 E.1/2 \(P_{square}=4x=2\pi{r}= P_{circle}\) > \(x=\frac{\pi{r}}{2}\).\(\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Answer: C. Can you explain this portion: Pcircle > x=πr2. Equating the perimeter of square (4x, where x is the length of the side) and the perimeter of the circle (circumference = \(2\pi{r}\)), we get \(4x=2\pi{r}\), which gives \(x=\frac{\pi{r}}{2}\). Hope it's clear.
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If the perimeter of square region S and the perimeter of the
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28 Sep 2015, 00:24
Bunuel wrote: chintzzz wrote: I got an answer of B but the official answer is different. please explain
If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to A.3/2 B.4/3 C.3/4 D.2/3 E.1/2 \(P_{square}=4x=2\pi{r}=P_{circle}\) > \(x=\frac{\pi{r}}{2}\). \(\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Answer: C. Apologies for what is probably a very simple question, but could someone please explain this step here? \(\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}\) Thank you
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Re: If the perimeter of square region S and the perimeter of the
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28 Sep 2015, 02:07
DropBear wrote: Bunuel wrote: chintzzz wrote: I got an answer of B but the official answer is different. please explain
If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to A.3/2 B.4/3 C.3/4 D.2/3 E.1/2 \(P_{square}=4x=2\pi{r}=P_{circle}\) > \(x=\frac{\pi{r}}{2}\). \(\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Answer: C. Apologies for what is probably a very simple question, but could someone please explain this step here? \(\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}\) Thank you \(x=\frac{\pi{r}}{2}\). \(\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{(\frac{\pi{r}}{2})^2}{\pi{r^2}}=\frac{\frac{\pi^2 r^2}{4}}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Hope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If the perimeter of square region S and the perimeter of circular regi
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07 May 2016, 06:04
I reached till (22 * 7)/ 4...Is there a quick way to estimate/ calculate the final answer? Bunuel wrote: Stiv wrote: If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to:
A. \(\frac {3}{2}\)
B. \(\frac {4}{3}\)
C. \(\frac {3}{4}\)
D. \(\frac {2}{3}\)
E. \(\frac {1}{2}\) Given: \(P_{square}=4x=2\pi{r}=P_{circle}\) > \(x=\frac{\pi{r}}{2}\). \(\frac{A_{square}}{A_{sdquare}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})\). Answer: C.



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Re: If the perimeter of square region S and the perimeter of circular regi
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07 May 2016, 10:03
Hi ameyaprabhu, The questions that you'll face on the Official GMAT are almost all based on patterns of some type (even if you don't immediately realize that a pattern is there). In the Quant section, TESTing VALUES can often be used to define a pattern, so you might try using that approach in these types of situations. My explanation (2 posts above your post) shows how to approach this prompt in that way. GMAT assassins aren't born, they're made, Rich
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