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555-605 Level|   Geometry|      
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Bunuel
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 28 Aug 2020, 03:11
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Genoa2000
Has anybody thought that this question was a "C-Trap" question?
It seemed to easy to be C IMO

BTW: Please Bunuel add tag GMATPrep
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Done. Thank you.
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Queenolin
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 31 Aug 2020, 18:20
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So please correct me if I understand wrong:

If a question asks about "values", meaning it just wants a real value, not a expression based on the given values "r1, r2, r3" ?

The "values" in the question means real numbers right?
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 27 Mar 2021, 12:42
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BrushMyQuant
Explaining the question:
Area of Circle 1 = \(A_1\) = \(Π (r_1)^2\)
Area of Circle 2 = \(A_1\) + \(A_2\) = \(Π (r_2)^2\)
Area of Circlet 3 = \(A_1\) + \(A_2\) + \(A_3\) = \(Π (r_3)^2\)

Solution
To get \(\frac{A_1}{A_2}\) and \(\frac{A_2}{A_3}\) we need to express all of these values in terms of one value (\(A_1\) or \(A_2\) or \(A_3\))
STAT1 \(A_2\) = \(A_3\) alone is not sufficient as I don't know relation of \(A_1\) with \(A_2\) to find \(\frac{A_1}{A_2}\)
STAT2 \(A_2 + A_3\) = 2 \(A_1\) alone is not sufficient as I don't know the individual relationship between \(A_1\),\(A_2\) and \(A_3\)

Combining both of them together we have
\(A_2\) = \(A_3\)
\(A_2\)+\(A_3\) = 2 \(A_1\)
=> 2\(A_3\) = 2\(A_1\) or \(A_1\) = \(A_3\) = \(A_2\)
So \(\frac{A_1}{A_2}\) and \(\frac{A_2}{A_3}\) both will be 1 each
So, both together are sufficient => Answer C

Hope it helps!
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Hmmm, my solution was identical except for this:

So \(\frac{A_1}{A_2}\) and \(\frac{A_2}{A_3}\) both will be 1 each

I said E.

Why? Because r3 > r2 > r1 so the areas can't possibly be in a 1:1:1 ratio.

What did I miss? BrentGMATPrepNow
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 27 Mar 2021, 13:27
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Hmmm, my solution was identical except for this:

So \(\frac{A_1}{A_2}\) and \(\frac{A_2}{A_3}\) both will be 1 each

I said E.

Why? Because r3 > r2 > r1 so the areas can't possibly be in a 1:1:1 ratio.

What did I miss? BrentGMATPrepNow[/quote]

I have a feeling that you misread the question.
A_2 isn't the area of the second circle;
A_2 = (the area of the second circle) - (the area of the first circle)
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 13 Oct 2022, 10:35
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We do not need to solve and find relationship among values of r1, r2, and r3 to get values of A1/A2 and A2/A3; however, below is complete solution.

Hope it helps.

As per question stem:

A3 = pi r3^2 - pi r2^2

A2 = pi r2^2 - pi r1^2

A1 = pi r1^2

Statement 1:

A2 = A3

pi r2^2 - pi r1^2 = pi r3^2 - pi r2^2

2(pi r2^2) = pi r3^2 + pi r1^2

Since, we do not know relationship between values of r1 with either r2 or r3, it is not possible to find value of A1/A2.

Statement 2:

We will have pi r3^2 - pi r2^2 + pi r2^2 - pi r1^2 = 2(pi r1^2)

or pi r3^2 - pi r1^2 = 2(pi r1^2)

or pi r3^2 = 3(pi r1^2)

Cancel pi and take square root both sides of equality we get

r3 = √3 * r1 ( radius cannot be negative)

Since, we do not know relationship between values of r2 with either r1 or r3, it is not possible to find value of A1/A2 and A2/A3.

Combining both statement:

We already got r3 = √3 * r1 from statement 2

From statement 1, we get 2(pi r2^2) = pi r3^2 + pi r1^2 (replace pi r3^2 with 3(pi r1^2))

2(pi r2^2) = 4 (pi r1^2)

Or pi r2^2 = 2 * pi r1^2

Cancel pi and take square root both sides of equality we get

r2 = √2 * r1

Now we have relationship among r1, r2, and r3

r2 = √2 * r1
r3 = √3 * r1

Select any values of r1 and we can get values of A1/A2 and A2/A3, for example r1 =1 then r2 = √2 and r3 = √3

Circle 1 area = pi
Circle 2 area = 2 pi
Circle 3 area = 3 pi

So A1 = 1 pi
A2 = Circle 2 area - Circle 1 area = 2 pi - pi = 1 pi
A3 = Circle 3 area - Circle 2 area = 3 pi - 2 pi = 1 pi

A1/A2 = 1/1 = 1
A3/A2 = 1/1 = 1

Both statement together are sufficient. Answer C.
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Circle 1, Circle 2 and Circle 3 have the same center, and have a radiu 16 Dec 2023, 00:56
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