1986 AJHSME Problems/Problem 6

Revision as of 09:26, 24 January 2009 by 5849206328x (talk | contribs) (Solution)

Problem

$\frac{2}{1-\frac{2}{3}}=$

$\text{(A)}\ -3 \qquad \text{(B)}\ -\frac{4}{3} \qquad \text{(C)}\ \frac{2}{3} \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 6$

Solution

Just simplify the bottom as $\frac{3}{3}-\frac{2}{3}=\frac{1}{3}$, getting us $\frac{2}{\frac{1}{3}}$, with which we multiply top and bottom by 3, we get $\frac{6}{1}$, or $6$

$\boxed{\text{E}}$

See Also

1986 AJHSME Problems