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Difference between revisions of "1986 AJHSME Problems/Problem 6"
m (→‎Solution)
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==Solution==
 
==Solution==
  
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A rather simple problem, if you have a basic understanding of fractions.
+
Just simplify the bottom as <math>\frac{3}{3}-\frac{2}{3}=\frac{1}{3}</math>, getting us <math>\frac{2}{\frac{1}{3}}</math>, with which we multiply top and bottom by 3, we get <math>\frac{6}{1}</math>, or <math>6</math>
  
−
Just simplify the bottom, getting us <math>\frac{2}{\frac{1}{3}}</math>, with which we multiply both sides by 3, getting us <math>\frac{6}{1}</math>, or <math>6</math>
+
<math>\boxed{\text{E}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 09:26, 24 January 2009

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

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