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Difference between revisions of "1983 AHSME Problems/Problem 1"

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From <math>\frac{x}{4} = 4y</math>, we get <math>x=16y</math>. Plugging in the other equation, <math>\frac{16y}{2} = y^2</math>, so <math>y^2-8y=0</math>. Factoring, we get <math>y(y-8)=0</math>, so the solutions are <math>0</math> and <math>8</math>. Since <math>x \neq 0</math>, <math>x=8</math>. <math>y=16\cdot 8 = \textbf{(E)}\; 128</math>.

Revision as of 14:16, 28 June 2015

Problem

If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$, then $x$ equals

$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$

Solution

From $\frac{x}{4} = 4y$, we get $x=16y$. Plugging in the other equation, $\frac{16y}{2} = y^2$, so $y^2-8y=0$. Factoring, we get $y(y-8)=0$, so the solutions are $0$ and $8$. Since $x \neq 0$, $x=8$. $y=16\cdot 8 = \textbf{(E)}\; 128$.

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