Difference between revisions of "1986 AJHSME Problems/Problem 6"

Latest revision as of 20:10, 3 July 2013

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 $6$

$\boxed{\text{E}}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 ==See Also==
-[[1986 AJHSME Problems]]
+{{AJHSME box|year=1986|num-b=5|num-a=7}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1986 AJHSME Problems/Problem 6"

Latest revision as of 20:10, 3 July 2013

Problem

Solution

See Also