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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, 05:04, edited 3 times in total.

Re: What is the value of a^(-2)*b^(-3)? [#permalink]
22 Feb 2012, 01:49

1

This post received KUDOS

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]
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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

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Re: What is the value of a^(-2)*b^(-3)? [#permalink]
13 Sep 2012, 06:59

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

Re: What is the value of a^(-2)*b^(-3)? [#permalink]
10 Jul 2013, 07:04

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]
22 Jan 2015, 08:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: What is the value of a^(-2)*b^(-3)? [#permalink]
29 Jan 2015, 19:47

It would make it much clearer if you edited the original question to indicate for statement two "a * b^(-1)". I got confused and thought the whole term ab was taken to the negative first power.

Re: What is the value of a^(-2)*b^(-3)? [#permalink]
30 Jan 2015, 05:04

Expert's post

Derkus wrote:

It would make it much clearer if you edited the original question to indicate for statement two "a * b^(-1)". I got confused and thought the whole term ab was taken to the negative first power.

In this case it would be (ab)^(-1), not ab^(-1). Still edited. _________________

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