1950 AHSME Problems/Problem 1

Problem

If $64$ is divided into three parts proportional to $2$ , $4$ , and $6$ , the smallest part is:

$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

If the three number are in proportion to $2:4:6$ , then they should also be in proportion to $1:2:3$ . This implies that the three numbers can be expressed as $x$ , $2x$ , and $3x$ . Add these values together to get: $x+2x+3x=6x=64$ Divide each side by 6 and get that $x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}$ which is answer choice $\boxed{C}$ .

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1950 AHSME Problems/Problem 1

Problem

Solution