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 $\boxed{\textbf{(B)}\ 4.8\%}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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