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Helpful Geometry formula sheet: http://gmatclub.com/forum/best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Last edited by Bunuel on 30 Jan 2015, 05:04, edited 3 times in total.

Consider 1)+2) from 2) we have \(6a=b\) plug it in 1) and solve for "a". Than plug it into 2). Now with known "a" and "b" you can find the value of \((a^(-2))(b^(-3))\) Remember that you don't need to solve it, you need to know that the equation could be solved. (C)

Putting the question in the right form can help you quickly arrive at the answer. What is \(\frac{1}{a^2b^3}?\)

1. \(\frac{1}{a^3b^2} = \frac{1}{36}\) Since a and b are real numbers so they can occur in many combinations to give 1/36. This statement alone is not sufficient.

2. \(\frac{a}{b} = \frac{1}{6}\) Again, a and b are real numbers and they can take many different values to give 1/6 (e.g. a = 1, b = 6 or a = 2, b = 12 etc). This statement alone is not sufficient.

You can easily get the value of \(\frac{1}{a^2b^3}\) by combining the two statements. \(\frac{1}{a^2b^3} = \frac{1}{a^3b^2} * \frac{a}{b} = \frac{1}{36} * \frac{1}{6}\) Hence they are sufficient together. Answer (C)
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Here is the way I thought: (a^-3)(b^-2)=(36^-1) (1/a^3)(1/b^2)=1/36 (a^3)(b^2)=36 (a^3)(b^2)=(3^2)(2^2) (a^3)(b^2)=(1^3)(6^2) a= 1 ; b = 6

How using fractions would make this reasoning wrong?

You are missing a point there: we are NOT told that \(a\) and \(b\) are integers, hence from \(a^3*b^2=36\) you cannot say for sure that \(a=1\) and \(b=6\). Because for ANY \(a\) there will exist some \(b\) which will satisfy \(a^3*b^2=36\) (and vise-versa). For example if \(a=2\) then \(b^3=9\) and \(b=\sqrt[3]{9}\).

Re: What is the value of a^(-2)*b^(-3)? [#permalink]

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13 Sep 2012, 06:53

Bunuel wrote:

What is the value of a^-2*b^-3?

Note that we are not told that \(a\) and \(b\) are integers.

\(a^{-2}*b^{-3}=\frac{1}{a^2b^3}=?\) So, basically we need to find the value of \(a^2b^3\).

(1) \(a^{-3}*b^{-2}=36^{-1}\) --> \(a^3b^2=36\). Not sufficient. (2) \(ab^{-1}=6\) --> \(\frac{b}{a}=\frac{1}{6}\). Not sufficient.

(1)+(2) Multiply (1) by (2): \(a^3b^2*\frac{b}{a}=a^2b^3=36*\frac{1}{6}\). Sufficient.

Answer: C.

Hi all, I would like to add to the explanation given by Bunuel. (1)as explained by bunuel \(a^3b^2=36\) - Insufficient ----> Why? Because we can be sure that a=1 but we b can be either +6 or -6 (2) b = 6a----Insufficient because this is just a ratio & nothing is mentioned about their values.

Hope it helps.
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Note that we are not told that \(a\) and \(b\) are integers.

\(a^{-2}*b^{-3}=\frac{1}{a^2b^3}=?\) So, basically we need to find the value of \(a^2b^3\).

(1) \(a^{-3}*b^{-2}=36^{-1}\) --> \(a^3b^2=36\). Not sufficient. (2) \(ab^{-1}=6\) --> \(\frac{b}{a}=\frac{1}{6}\). Not sufficient.

(1)+(2) Multiply (1) by (2): \(a^3b^2*\frac{b}{a}=a^2b^3=36*\frac{1}{6}\). Sufficient.

Answer: C.

Hi all, I would like to add to the explanation given by Bunuel. (1)as explained by bunuel \(a^3b^2=36\) - Insufficient ----> Why? Because we can be sure that a=1 but we b can be either +6 or -6 (2) b = 6a----Insufficient because this is just a ratio & nothing is mentioned about their values.

Hope it helps.

Let me correct you: we cannot be sure that \(a=1\) from (1). Since we are not told that \(a\) and \(b\) are integers, then \(a\) could, for example be 2 and \(b\) could be \(-\frac{3}{\sqrt{2}}\) or \(\frac{3}{\sqrt{2}}\).

Re: What is the value of a^(-2)*b^(-3)? [#permalink]

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10 Jul 2013, 07:01

1

This post received KUDOS

Bunuel wrote:

What is the value of a^-2*b^-3?

Note that we are not told that \(a\) and \(b\) are integers.

\(a^{-2}*b^{-3}=\frac{1}{a^2b^3}=?\) So, basically we need to find the value of \(a^2b^3\).

(1) \(a^{-3}*b^{-2}=36^{-1}\) --> \(a^3b^2=36\). Not sufficient. (2) \(ab^{-1}=6\) --> \(\frac{b}{a}=\frac{1}{6}\). Not sufficient.

(1)+(2) Multiply (1) by (2): \(a^3b^2*\frac{b}{a}=a^2b^3=36*\frac{1}{6}\). Sufficient.

Answer: C.

small typo here : statement 2 says : \(ab^{-1}=6^{-1}\) and not \(ab^{-1}=6\) so we have \(\frac{a}{b} = \frac{1}{6}\) so \(a^2b^3= a^3 *b^2 * \frac{b}{a} = 36 *6 = 216\) and not \(\frac{36}{6}\)

Note that we are not told that \(a\) and \(b\) are integers.

\(a^{-2}*b^{-3}=\frac{1}{a^2b^3}=?\) So, basically we need to find the value of \(a^2b^3\).

(1) \(a^{-3}*b^{-2}=36^{-1}\) --> \(a^3b^2=36\). Not sufficient. (2) \(ab^{-1}=6\) --> \(\frac{b}{a}=\frac{1}{6}\). Not sufficient.

(1)+(2) Multiply (1) by (2): \(a^3b^2*\frac{b}{a}=a^2b^3=36*\frac{1}{6}\). Sufficient.

Answer: C.

small typo here : statement 2 says : \(ab^{-1}=6^{-1}\) and not \(ab^{-1}=6\) so we have \(\frac{a}{b} = \frac{1}{6}\) so \(a^2b^3= a^3 *b^2 * \frac{b}{a} = 36 *6 = 216\) and not \(\frac{36}{6}\)

Hope it helps

Yes, exponent was missing in the second statement. Edited. Thank you.
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Re: What is the value of a^(-2)*b^(-3)? [#permalink]

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22 Jan 2015, 08:29

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