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

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==Solution==
 
==Solution==
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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>, we also have <math>y \neq 0</math>, so <math>y=8</math>. Hence <math>x=16\cdot 8 = \textbf{(E)}\; 128</math>.
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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>, we also have <math>y \neq 0</math>, so <math>y=8</math>. Hence <math>x=16\cdot 8 = \boxed{\textbf{(E)}\ 128}</math>.
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1983|before=First Question|num-a=2}}
 
{{AHSME box|year=1983|before=First Question|num-a=2}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:38, 19 February 2019

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 $x \neq 0$, we also have $y \neq 0$, so $y=8$. Hence $x=16\cdot 8 = \boxed{\textbf{(E)}\ 128}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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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