Posted: Thu Aug 16, 2007 1:20 pm Post subject: circle : Quant Question Archive [LOCKED]
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Posted: Thu Aug 16, 2007 1:20 pm Post subject: circle

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Manager
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Posted: Thu Aug 16, 2007 1:20 pm Post subject: circle [#permalink]

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New post 16 Aug 2007, 10:27
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Posted: Thu Aug 16, 2007 1:20 pm Post subject: circle

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Hi all, how do you crack such quections:

If A is the area and C is the circumference of a circle, which of the following is an expression for C in terms of A?
Please, explane your answer.
Source: GMAT courses


A - 2 sqrt P/A
B - 2 P/sqrt A
C - 2 sqrt A/P
D - 2P sqrtA
E - 2 sqrtPA


As A=Pr^2
n C= 2Pr
r= radius of circle

so form A..
r=sqrt(A/P)

put it into C
C= 2P*sqrt(A/P)
C = 2 sqrt(A*P)

Correct answer is E
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2 sqr AP [#permalink]

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New post 16 Aug 2007, 10:41
2 P sqr A/P
U can write it like 2* P*sqrtA/sqrtP

When u divide P to sqrtP, u ll get sqrtP....
so 2* sqrtA*sqrtP
=2 sqr AP
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 [#permalink]

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New post 16 Aug 2007, 10:51
i also get E) 2 sqrtPA for the answer

A = P*r^2
C = 2*P*r

we have to solve for r for area formula in order to get A:
A = P*r^2 -> r^2 = A / P -> r = sqrt A/P

now we have to plug back r into circumference to find the expression for C:

r = sqrt A/P
C = 2*P*r -> 2*P*(sqrt A/P)
2*P*(1/ sqrt P)(sqrt A) -> 2 sqrtP * sqrtA
2 sqrt PA
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New post 16 Aug 2007, 19:36
A = pi(r^2) --> r = sqrt(A/pi)
C = 2(pi)(r) --> C = 2(pi)(sqrt(A/pi) --> C = 2 sqrt(pi)sqrt(A) --> C = 2 sqrt(A)(pi)

Ans E
  [#permalink] 16 Aug 2007, 19:36
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Posted: Thu Aug 16, 2007 1:20 pm Post subject: circle

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