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If the perimeter of square region S and the perimeter of the

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If the perimeter of square region S and the perimeter of the  [#permalink]

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New post Updated on: 05 Aug 2012, 02:30
1
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A
B
C
D
E

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Question Stats:

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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

Originally posted by chintzzz on 19 Jun 2010, 22:23.
Last edited by Bunuel on 05 Aug 2012, 02:30, edited 1 time in total.
Edited the question.
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If the perimeter of square region S and the perimeter of the  [#permalink]

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New post 20 Jun 2010, 08:03
4
2
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.

Similar question: https://gmatclub.com/forum/if-the-perim ... 27004.html
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Re: pls clarify my answer  [#permalink]

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New post 19 Jun 2010, 23:44
perimeter of square = perimeter of circle

4 x = 2 pie r

x= side of square
r = radius of circle

solving above and putting pie = 22/7

7x = 11r

so x/r = 77/7

so (area of square)/(area of circle) = (x^2)/(pie r^2)

we already know x/r = 11/7 so (x^2)/(r^2) = (11^2)/(7^2)

so (x^2)/(pie r^2) = (11^2)/[(7^2) * (22/11)]

solving we will get 11/14 which is near to 3/4
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Re: Ratio  [#permalink]

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New post 10 Mar 2011, 10:51
1
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 the  [#permalink]

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New post 30 Jun 2015, 10:39
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  [#permalink]

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New post 30 Jun 2015, 10:45
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  [#permalink]

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New post 28 Sep 2015, 01:24
1
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  [#permalink]

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New post 28 Sep 2015, 03:07
2
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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Resources:
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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 the  [#permalink]

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