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Helpful Geometry formula sheet: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, 06:04, 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)

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

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03 May 2011, 17:48

Expert's post

dc123 wrote:

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

1) a^-3b^-2 = 36^-1

2) ab^-1 = 6^-1

the Ans says C but doesnt only B work?

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) _________________

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

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22 Feb 2012, 02:49

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Expert's post

wizard wrote:

I am a bit confused on (1)

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, 07:53

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

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13 Sep 2012, 07:59

Expert's post

fameatop wrote:

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.

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, 08:01

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 \(\frac{36}{6}\)

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

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10 Jul 2013, 08:04

Expert's post

stne wrote:

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 \(\frac{36}{6}\)

Hope it helps

Yes, exponent was missing in the second statement. Edited. Thank you. _________________

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

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22 Jan 2015, 09:29

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