1960 AHSME Problems/Problem 17

Problem 17

The formula $N=8 \times 10^{8} \times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. The lowest income, in dollars, of the wealthiest $800$ individuals is at least:

$\textbf{(A)}\ 10^4\qquad \textbf{(B)}\ 10^6\qquad \textbf{(C)}\ 10^8\qquad \textbf{(D)}\ 10^{12} \qquad \textbf{(E)}\ 10^{16}$

Solution

Plug $800$ for $N$ because $800$ is the number of people who has at least $x$ dollars. \[800 = 8 \times 10^{8} \times x^{-3/2}\] \[10^{-6} = x^{-3/2}\] In order to undo raising to the $-\frac{3}{2}$ power, raise both sides to the $-\frac{2}{3}$ power. \[(10^{-6})^{-2/3} = (x^{-3/2})^{-2/3}\] \[x = 10^4  \text{ dollars}\]

The answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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