Difference between revisions of "1960 AHSME Problems/Problem 17"

(Solution to Problem 17)
 
m (See Also)
Line 21: Line 21:
  
 
==See Also==
 
==See Also==
{{AHSME box|year=1960|num-b=16|num-a=18}}
+
{{AHSME 40p box|year=1960|num-b=16|num-a=18}}

Revision as of 20:14, 10 May 2018

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions
Invalid username
Login to AoPS