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Value of a^-2 * b^-3

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Source: — Data Sufficiency |
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by tohellandback » Thu Aug 06, 2009 6:21 pm
a^-2 * b^-3
=1/a^2 * 1/b^3

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

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

not sufficient coz we need b^3

2)ab^-1 = 6^-1
1/ab=1/6

1/a^b^2=1/36
not suffcient

combined
1/2

1/a=1
a=1
b=6

now u can find the value
answer C
The powers of two are bloody impolite!!
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by ShikenO/kau » Thu Aug 06, 2009 7:59 pm
We got to find the value of a^-2*b^-3

StatementI: a^-3*b^-2 = 36
Only 6 and 1 can fit into this equation.
But we dont know if a =1 or b=1...So this statement is insufficient.

StatementII:a^-1*b^-1 = 1/6
The following values can be combination of a and b to fit into this equation:
a.2,3
b.-2,-3
c.1,6
d.-1,-6
So this statement is insufficient.

Now combining both the statements:
We know a and b be must either 1 or 6.

Square the second statement
a^-2*b^-2 = 36^-1

Divide the first statement by second statement

(a^2*b^2)/(a^3*b^2)=36/36
a = 1
so b =6 and we will be able to find the answer for a^-2*b^-3
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by vikram_k51 » Fri Aug 07, 2009 11:44 am
What is the value of a^-2 * b^-3 ?
(1) a^-3 * b^-2 = 36^-1
(2) ab^-1 = 6^-1


Solving for a,b we get a=1 and b=6. We use both 1 and 2 for this soln and hence ans is C
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by psuv » Sat Aug 08, 2009 1:15 pm
What is the value of a^-2 * b^-3 ?
(1) a^-3 * b^-2 = 36^-1
(2) ab^-1 = 6^-1

Basically, we have to find value of 1/(a^2*b^3)

Statement 1: a^-3 * b^-2 = 36^-1 gives
1/(a^3*b^2) = 1/36 not sufficient, still need to know value of a/b to reach the answer

Statement 2: ab^-1 = 6^-1 gives
a/b = 1/6 Not sufficient
Combining statments 1 and 2- multiplying equations got in 1 and 2, we get the required value, so C is the answer.
Thanks,
Suv
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