Math Test Problem

**11parcal** · 10-03-08, 11:20 AM

{[(1/3)^3][(1/2)^(-2)]}/{[(1/3)^(-3)][(1/4)^2)]}

This is a question I recently had on a math test, I looked at the answers and, after redoing the question several times concluded there was no correct answer the answers were:

a. 72
b. 1/36
c. 4/9
d. 8/9

Please tell me if I was correct in my assumption.

Having troubles with tex so I won't be using it

**Steve100** · 10-03-08, 11:30 AM

I got 16/27

**rpenner** · 10-03-08, 11:49 AM

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^3 \times \left(\frac{1}{2}\right)^{-2} } \over { \left(\frac{1}{3}\right)^{-3} \times \left(\frac{1}{4}\right)^2 }} & = & {{ 3^{-3} \times 2^2 } \over { 3^3 \times 4^{-2} }} \\ & = & {{ 4^2 \times 2^2 } \over { 3^3 \times 3^3 }} \\ & = & {{ 64 } \over { 729 }} \\ \end{eqnarray}$
which makes me wonder if the problem was copied correctly.

For example:

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{-4} \times \left(\frac{1}{2}\right)^{3} } \over { \left(\frac{1}{3}\right)^{-2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{4} \times 2^{-3} } \over { 3^{2} \times 4^{-3} }} \\ & = & {{ 3^4 \times 4^{3} } \over { 3^2 \times 2^3 }} \\ & = & 3^2 \times 2^3 \\ & = & 72 \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{-4} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{-3} }} & = & {{ 3^{-4} \times 2^{4} } \over { 3^{-2} \times 4^{3} }} \\ & = & {{ 3^2 \times 2^{4} } \over { 3^4 \times 4^3 }} \\ & = & {{ 1 } \over { 9 \times 4 }} \\ & = & {{ 1 } \over { 36 }} \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{4} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{-4} \times 2^{-4} } \over { 3^{-2} \times 4^{-3} }} \\ & = & {{ 4^3 \times 2^{-4} } \over { 3^4 \times 3^{-2} }} \\ & = & {{ 4 } \over { 9 }} \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{3} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{-4} \times 2^{-3} } \over { 3^{-2} \times 4^{-3} }} \\ & = & {{ 4^3 \times 2^{-3} } \over { 3^4 \times 3^{-2} }} \\ & = & {{ 8 } \over { 9 }} \\ \end{eqnarray}$

**11parcal** · 10-03-08, 12:43 PM

Originally Posted by rpenner

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^3 \times \left(\frac{1}{2}\right)^{-2} } \over { \left(\frac{1}{3}\right)^{-3} \times \left(\frac{1}{4}\right)^2 }} & = & {{ 3^{-3} \times 2^2 } \over { 3^3 \times 4^{-2} }} \\ & = & {{ 4^2 \times 2^2 } \over { 3^3 \times 3^3 }} \\ & = & {{ 64 } \over { 729 }} \\ \end{eqnarray}$
which makes me wonder if the problem was copied correctly.

For example:

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{-4} \times \left(\frac{1}{2}\right)^{3} } \over { \left(\frac{1}{3}\right)^{-2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{4} \times 2^{-3} } \over { 3^{2} \times 4^{-3} }} \\ & = & {{ 3^4 \times 4^{3} } \over { 3^2 \times 2^3 }} \\ & = & 3^2 \times 2^3 \\ & = & 72 \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{-4} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{-3} }} & = & {{ 3^{-4} \times 2^{4} } \over { 3^{-2} \times 4^{3} }} \\ & = & {{ 3^2 \times 2^{4} } \over { 3^4 \times 4^3 }} \\ & = & {{ 1 } \over { 9 \times 4 }} \\ & = & {{ 1 } \over { 36 }} \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{4} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{-4} \times 2^{-4} } \over { 3^{-2} \times 4^{-3} }} \\ & = & {{ 4^3 \times 2^{-4} } \over { 3^4 \times 3^{-2} }} \\ & = & {{ 4 } \over { 9 }} \\ \end{eqnarray}$

$\begin{eqnarray} {{ \left(\frac{1}{3}\right)^{4} \times \left(\frac{1}{2}\right)^{3} } \over { \left(\frac{1}{3}\right)^{2} \times \left(\frac{1}{4}\right)^{3} }} & = & {{ 3^{-4} \times 2^{-3} } \over { 3^{-2} \times 4^{-3} }} \\ & = & {{ 4^3 \times 2^{-3} } \over { 3^4 \times 3^{-2} }} \\ & = & {{ 8 } \over { 9 }} \\ \end{eqnarray}$

Thats what I got (The first one) so I'm not sure what the teacher is gonna do