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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 monirjewel on 10 Jul 2013, 20:08, edited 2 times in total.

Re: 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 one point: we are not told that a and b are integers, hence from a^3*b^2=36 you can not say for sure that a=1 and b=6. Because for ANY a there will exits 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: 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. _________________

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Re: 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: 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: 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

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