Difference between revisions of "1986 AJHSME Problems/Problem 6"
5849206328x (talk | contribs) (New page: ==Problem== <math>\frac{2}{1-\frac{2}{3}}=</math> <math>\text{(A)}\ -3 \qquad \text{(B)}\ -\frac{4}{3} \qquad \text{(C)}\ \frac{2}{3} \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 6</math> ==...) |
Aquakitty11 (talk | contribs) |
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==Solution== | ==Solution== | ||
| − | {{ | + | 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> |
| + | |||
| + | <math>\boxed{\text{E}}</math> | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1986|num-b=5|num-a=7}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:10, 3 July 2013
Problem
Solution
Just simplify the bottom as 
, getting us 
, with which we multiply top and bottom by 3, we get 
, or 
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
