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

1) or 2) alone ins

Consider 1)+2)
from 2) we have \(6a=b\)
plug it in 1) and solve for "a". Than plug it into 2).
Now with known "a" and "b" you can find the value of \((a^(-2))(b^(-3))\)
Remember that you don't need to solve it, you need to know that the equation could be solved.
(C)

What is \(a^{-2}b^{-3}=\frac{1}{a^2b^3}=\frac{1}{(ab)^2b}\)

1) a^-3b^-2 = 36^-1
\(a^{-3}b^{-2} = 36^{-1}\)
\(\frac{1}{a^3b^2} = \frac{1}{36}\)
\(\frac{1}{a^3b^2} = \frac{1}{36}\)
\(\frac{1}{(ab)^2a} = \frac{1}{36}\)
\(\frac{1}{(ab)^2} = \frac{a}{36}\)
\(\frac{1}{(ab)^2b}=\frac{a}{36b}\)--------------------1

\(a^3b^2=36\)
Possible values of a and b;
\(a=1; b=6\)
\(a=2; b=\sqrt{\frac{36}{8}}\)
\(a=0.1; b=\sqrt{\frac{36}{0.001}}\)

Not Sufficient.

2) ab^-1 = 6^-1
\(ab^{-1} = 6^{-1}\)
\(\frac{a}{b} = \frac{1}{6}\)------------------------2

a=1; b=6
a=2; b=12
a=3; b=18
a=3.2; b=19.2
Not Sufficient.

Combining both and using 1 and 2:

\(\frac{1}{(ab)^2b}=\frac{a}{36b}=\frac{1}{36*6}=\frac{1}{216}\)
Sufficient.

Ans: "C"

a. gives values (1,6) and (1,-6)
b gives values (1,6) and (-1,-6)

a+b gives (1,6) hence C

Putting the question in the right form can help you quickly arrive at the answer.
What is \(\frac{1}{a^2b^3}?\)

1. \(\frac{1}{a^3b^2} = \frac{1}{36}\)
Since a and b are real numbers so they can occur in many combinations to give 1/36. This statement alone is not sufficient.

2. \(\frac{a}{b} = \frac{1}{6}\)
Again, a and b are real numbers and they can take many different values to give 1/6 (e.g. a = 1, b = 6 or a = 2, b = 12 etc). This statement alone is not sufficient.

You can easily get the value of \(\frac{1}{a^2b^3}\) by combining the two statements.
\(\frac{1}{a^2b^3} = \frac{1}{a^3b^2} * \frac{a}{b} = \frac{1}{36} * \frac{1}{6}\)
Hence they are sufficient together. Answer (C)

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 a point there: we are NOT told that \(a\) and \(b\) are integers, hence from \(a^3*b^2=36\) you cannot say for sure that \(a=1\) and \(b=6\). Because for ANY \(a\) there will exist some \(b\) which will satisfy \(a^3*b^2=36\) (and vise-versa). For example if \(a=2\) then \(b^3=9\) and \(b=\sqrt[3]{9}\).

Similar question to practice: if-3-a-4-b-c-what-is-the-value-of-b-1-5-a-25-2-c-106047.html

Hope it helps.

Answer is C .

From 1: 1/a^3 *1/b^2= 1/36 Not sufficient
From 2 : 1/ab =1/6 means a, b can be :2,3 ; 3,2 ;6,1;1,6 . not sufficient

Combining both statements only a=1 and b=6 fulfills the statement 1 so answer is C

Try and fail but never fail to try.

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 \(\frac{3}{\sqrt{2}}\).

Hope it's clear.

P.S. This post might also help: what-is-the-value-of-a-2-b-104673.html#p1047996

Bunuel,

Even if we were told a&b are postive integers does knowing a^3*b^2=constant ever gaurentee knowing the value of a^2*a^3??

Posted from my mobile device

Well, if we were told that a and b are positive integers, then from a^3b^2=36 it follows that a=1 and b=6.

Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?

It would make it much clearer if you edited the original question to indicate for statement two "a * b^(-1)". I got confused and thought the whole term ab was taken to the negative first power.

In this case it would be (ab)^(-1), not ab^(-1). Still edited.

\(a*b^{(-1)}=6^{(-1)}\);

\(a*\frac{1}{b}=\frac{1}{6}\);

\(\frac{a}{b}=\frac{1}{6}\);

\(\frac{b}{a}=6\).

Hope it's clear.

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