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Divide the sum of $ \frac{65}{12} $ and $ \frac{8}{3} $ by their difference.
Given:
$\frac{65}{12}$ and $\frac{8}{3}$
To do:
Here we have to divide the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference.
Solution:
Sum of $\frac{65}{12}$ and $\frac{8}{3}$:
$\frac{65}{12} \ +\ \frac{8}{3}$
$=\frac{65 + 8(4)}{12}$ (LCM of 12 and 3 is 12)
$=\frac{65 + 32}{12}$
$= \frac{97}{12}$
Difference of $\frac{65}{12}$ and $\frac{8}{3}$:
$\frac{65}{12} \ -\ \frac{8}{3}$
$=\ \frac{65 \ -\ 8( 4)}{12}$
$=\ \frac{65\ -\ 32}{12}$
$=\ \mathbf{\frac{33}{12}}$
Now, dividing the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference;
$=\ \frac{\frac{97}{12}}{\frac{33}{12}}$
$=\ \frac{97}{12} \ \times \ \frac{12}{33}$
$= \frac{97}{33}$
$=\ \mathbf{\frac{97}{33}}$
So, answer is $\frac{97}{33}$.
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