1953 AHSME Problems/Problem 17

Problem

A man has part of $ $4500$ invested at $4$ % and the rest at $6$ %. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:

$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of these}$

Solution

You are trying to find $\frac{2(0.06x)}{4500}$ , where $x$ is the principle for one investment. To find $x$ , solve $0.04(4500-x) = 0.06x$ . $X$ will come out to be $1800$ . Then, plug in x into the first equation, $\frac{2(0.06)(1800)}{4500}$ , to get $0.048$ . Finally, convert that to a percentage and you get $\boxed{\textbf{(B)}\ 4.8\%}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1953 AHSME Problems/Problem 17

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