1971 AHSME Problems/Problem 10


Each of a group of $50$ girls is blonde or brunette and is blue eyed of brown eyed. If $14$ are blue-eyed blondes, $31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is

$\textbf{(A) }5\qquad \textbf{(B) }7\qquad \textbf{(C) }9\qquad \textbf{(D) }11\qquad  \textbf{(E) }13$


There are $31$ brunettes, so there are $50-31=19$ blondes. We know there are $14$ blue-eyed blondes, so there are $19-14=5$ brown-eyed blondes.

Next, we know that $18$ people are brown-eyed, so there are $18-5=13$ brown-eyed brunettes.

The answer is $\textbf{(E)} \quad 13.$

-edited by coolmath34