# Thread: Math Test Problem

1. ## Math Test Problem

{[(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

2. I got 16/27

3. $\begin{eqnarray}
{{ \left(\frac{1}{3}\right)^3 \times \left(\frac{1}{2}\right)^{-2} } \over { \left(\frac{1}{3}\right)^{-3} \times \left(\frac{1}{4}\right)^2 }}
& = &
{{ 3^{-3} \times 2^2 } \over { 3^3 \times 4^{-2} }} \\
& = &
{{ 4^2 \times 2^2 } \over { 3^3 \times 3^3 }} \\
& = &
{{ 64 } \over { 729 }} \\
\end{eqnarray}$

which makes me wonder if the problem was copied correctly.

For example:

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

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

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

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

4. 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

5. Originally Posted by 11parcal
Thats what I got (The first one) so I'm not sure what the teacher is gonna do
Print out this entire thread and show it to your teacher.

6. Don't print out my post, I posted that when your equation was different.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•