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 $\boxed{\textbf{(A)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1960 AHSME Problems/Problem 17

Problem 17

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