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1989 AJHSME Problem 12
1989 AJHSME Problem 12
Source:
June 25, 2011
Problem Statement
1
−
1
3
1
−
1
2
=
\frac{1-\frac{1}{3}}{1-\frac{1}{2}} =
1
−
2
1
1
−
3
1
=
(A)
1
3
(B)
2
3
(C)
3
4
(D)
3
2
(E)
4
3
\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{3}{2} \qquad \text{(E)}\ \frac{4}{3}
(A)
3
1
(B)
3
2
(C)
4
3
(D)
2
3
(E)
3
4
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