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 numbers 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 $\boxed{\textbf{(C)}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS