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What is the value of a^-2*b^-3 ? (1) (a^-3)*(b^-2) = 36^-1

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What is the value of a^-2*b^-3 ? (1) (a^-3)*(b^-2) = 36^-1 [#permalink] New post 12 Oct 2005, 12:13
What is the value of a^-2*b^-3 ?

(1) (a^-3)*(b^-2) = 36^-1
(2) ab^-1 = 6^-1

Last edited by trickygmat on 14 Oct 2005, 06:50, edited 1 time in total.
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 [#permalink] New post 12 Oct 2005, 12:24
I think it's C (both). What's OA
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 [#permalink] New post 12 Oct 2005, 12:25
Could you please repost the question? Should we interpret:
a^-3b^-2 = 36^-1
as (a^-3) * (b^-2) ? or (a ^ (-3 (b^-2) ) ) ?

I guess it should be former.
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[#permalink] New post 13 Oct 2005, 05:50
C

1. insuff. (a^-3)(b^-2)= 1/36, from this we can gather (a^3)(b^2)=36. Still we can't gatehr any info because the stem has (a^-2)(b^-3). 2 variables, one equation...insuff.

2. insuff. again 2 variables, one equation....insuff

together, 2 equations, 2 variables....suff.
C
OA?
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 [#permalink] New post 13 Oct 2005, 06:04
If we read statement 1 as a raised to the power of -3b, raised to the power of -2, then wouldn`t it be sufficient?

A?

Last edited by GMATT73 on 13 Oct 2005, 06:41, edited 1 time in total.
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 [#permalink] New post 13 Oct 2005, 06:39
a^-3b^-2 = 36^-1 = (a^-3) * (b^-2) = 1/36
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Re: d/s - value [#permalink] New post 13 Oct 2005, 14:09
trickygmat wrote:
What is the value of a^-2*b^-3 ?

(1) a^-3b^-2 = 36^-1
(2) ab^-1 = 6^-1


It's A

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

a^(-3b*-2)=6^-2

a^6b=6^-2

so a=6

and 6b=-2 or b= -1/3

So sufficient

Statement II

ab^-1 = 6^-1

1/ab=1/6
or ab=6 two variables so insufficient

Hence insufficient

Please provide OA
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 [#permalink] New post 13 Oct 2005, 14:28
Answer should be C
(considering a^-3b^-2 = 36^-1 as (a^-3) * (b^-2) from trickygmat's post)

Multiply stmt 1 * stmt 2 we get
(a^-2)*(b^-3) = (36^-1)(6^-1) = (216^-1)

ALI, I think you made a mistake here:
a^-3b^-2=36^-1
a^(-3b*-2)=6^-2

a^-3b^-2=36^-1 should be (a^-3) * (b^-2) = 36^-1
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 [#permalink] New post 14 Oct 2005, 07:35
sudhagar, could you pls show the working... how did you arrive at 1/(a^2*b^3) = 1/216???
  [#permalink] 14 Oct 2005, 07:35
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