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1994 AJHSME Problem 13
1994 AJHSME Problem 13
Source:
July 9, 2011
Problem Statement
The number halfway between
1
6
\dfrac{1}{6}
6
1
and
1
4
\dfrac{1}{4}
4
1
is
(A)
1
10
(B)
1
5
(C)
5
24
(D)
7
24
(E)
5
12
\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}
(A)
10
1
(B)
5
1
(C)
24
5
(D)
24
7
(E)
12
5
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