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

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==Solution==
 
==Solution==
 
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>.
 
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>.
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==See Also==
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{{AHSME box|year=1983|before=First Question|num-a=26}}
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{{MAA Notice}}

Revision as of 06:37, 18 May 2016

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 $x \neq 0$, $x=8$. $y=16\cdot 8 = \textbf{(E)}\; 128$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 26
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
All AHSME Problems and Solutions


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