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

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What is the value of (a^-2)(b^-3)? 1) (a^-3)(b^-2)=36^-1 2) [#permalink]

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12 Mar 2006, 09:59
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What is the value of (a^-2)(b^-3)?

1) (a^-3)(b^-2)=36^-1
2) a(b^-1)=6^-1
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12 Mar 2006, 10:22
I think its A. The only combination satisfying statement 1 seems to be when a=1 and b=6. is this the right answer?
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12 Mar 2006, 10:30
Gordon wrote:
I think its A. The only combination satisfying statement 1 seems to be when a=1 and b=6. is this the right answer?

But a and b are not necessarily integers.
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12 Mar 2006, 10:33
vivek can you explain why you think it's C?
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12 Mar 2006, 10:38
buckkitty wrote:
What is the value of (a^-2)(b^-3)?

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

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

That's why it's C.
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12 Mar 2006, 19:04
(a^-2)(b^-3) = 1/a^2(b^3)

(1) Not sufficient.

(2) Only know the ratio a/b. Insufficient.

Using both, we can work out 1/a^2(b^3).

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

Sufficient.

Ans C(a^-2)(b^-3) = 1/a^2(b^3)
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12 Mar 2006, 22:06
It C

When we multiply both statements we can get the value of

(a^-2)(b^-3)
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13 Mar 2006, 13:48
C. different approach.

1. a = 2^(2/3) and b = 3 OR a = 2 and b = 3^(2/3)...INSUFF.
2. b = 6a....INSUFF.

1+2 ...we can find a and b...SUFF.

buckkitty wrote:
What is the value of (a^-2)(b^-3)?

1) (a^-3)(b^-2)=36^-1
2) a(b^-1)=6^-1
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13 Mar 2006, 13:58
yes! and thanks for pointing this out, I had managed to overlook this. This seems much simpler to me....I am much less of a math guru than all of you
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13 Mar 2006, 19:47
My approach was

given 1/a^3 x 1/b^2 = 1/36 => (1)
a/b = 1/6 => (2) these are not sufficient by themselves

combining

(1) can be re-written as

1/a^2 x 1/b^2 = a/36

or

1/a^2 x 1/b^3 = a/b * 1/36

Substituting value for a/b from (2)
we have
1/a^2 x 1/b^3 = 1/6 x 1/36

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16 Mar 2006, 00:47
laxieqv wrote:
buckkitty wrote:
What is the value of (a^-2)(b^-3)?

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

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

That's why it's C.

Can someone explain why the above is true? I still don't understand why.... Pls help )
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16 Mar 2006, 06:46
Ans: C

(i) Only implies a^3 * b ^ 2 = 36 ; not sufficient. Eliminate A and D
(ii) Only implies a/b = 6 ; not suff Eliminate B
(i) and (ii)
Substitue a/b = 6 in (i), we get b^5=216*6, hence b = 6 ONLY => a = 1
16 Mar 2006, 06:46
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