1950 AMC 12 Problem 1

Attempt Problem 1 of the 1950 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1950 AMC 12 solutions, or check the answer key.

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64

is divided into three parts proportional to

2,

4,

and

6,

the smallest part is:

5\dfrac{1}{3}

11

10\dfrac{2}{3}

5

None of these answers

Add the three numbers in the ratio to find the total number of equal shares

The smallest part is

\dfrac{2}{2+4+6}

64

The ratio contains

2+4+6=12

equal shares. The smallest part is therefore

64\left(\frac{2}{12}\right)=\frac{32}{3}=10\frac{2}{3}.

Thus, the correct answer is C.

Problem 1 in other years