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Fraction

by ketkoag » Sat Mar 28, 2009 3:03 am
What is the value of (a^-2)(b^-3) ?
(1) (a^-3)(b-^2) = 36^-1
(2) ab^-1 = 6^-1

OA: C
Please explain why ans A is incorrect..
I marked A as the answer coz if we take statement first then 1/a^3b^2 = 1/36^1 that means this is possible only when a = 1 and b = 6.
So we can solve the question asked.

Please let me know if i am wrong.
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by DanaJ » Sat Mar 28, 2009 4:42 am
The reason why your way of solving this problem is incorrect is because it isn't stated anywhere in the problem that a and b have to be integers and this complicates things. For instance, a could be third degree root of 2 and b could be sqrt(18).

Now, let's see what you've got:

You're looking for (a^-2)*(b^-3) or [(a^2)(b^3)]^-1, which means you're actually looking for (a^2)*(b^3).

1. is equivalent to (a^3)*(b^2) = 36. This isn't enough to find (a^2)*(b^3).

2. is equivalent to ab = 6. Again, not enough.

But put the two together and you've found what you're looking for: divide eqn 1 to eqn 2 to get that (a^2)*b = 36/6 = 6. Since ab = 6, this means that a can only be 1, with b = 6. Because you have the individual values of a and b, the rest isn't difficult to determine.
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by cramya » Sat Mar 28, 2009 9:20 am
To find:

1/a^2 * 1/b^3


Stmt I

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

Prime factorize 36 = 2^2 * 3^2

We can see that a has to be 1 since no cube appears in the prime factorization.

However b can be -6 or 6 which changes what we need to find.

INSUFF

Stmt II

a/b = 1/6

a=6 b=36

a=6*6 b=36*6

This changes the value of what we need to find.

INSUFF

Together:

a=1

a/b = 1/6 implies 6a=b therefore b=6

We can find one single determinable value for 1/a^2 * 1/b^3


SUFF

Choose C

Regards,
CR
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