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Difference between revisions of "1955 AHSME Problems/Problem 5"

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==Solution==
 
==Solution==
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An inverse variation can be expressed in the form <math>xy = n</math>, where <math>n</math> is any number (except perhaps zero). Since 5y varies inversely with the square of x, this particular one will be <math>5yx^2 = n</math>.  
+
An inverse variation can be expressed in the form <math>xy = n</math>, where <math>n</math> is any number (except perhaps zero). Since <math>5y</math> varies inversely with the square of <math>x</math>, this particular one will be <math>5yx^2 = n</math>.  
  
−
We can plug in 16 for y and 1 for x, which makes n 80. The equation is now <math>5yx^2 = 80</math>
+
We can plug in <math>16</math> for <math>y</math> and <math>1</math> for <math>x</math>, which makes <math>n=80</math>. The equation is now <math>5yx^2 = 80</math>.
  
−
When x = 8, we can solve for y using the equation <math>320y = 80</math>, which makes y <math>\textbf{(D)} \frac{1}{4}</math>
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When <math>x = 8</math>, we can solve for <math>y</math> using the equation <math>320y = 80</math>, which makes <math>y</math> <math>\boxed{\textbf{(D)} \frac{1}{4}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 00:52, 14 October 2021

Problem

$5y$ varies inversely as the square of $x$. When $y=16, x=1$. When $x=8, y$ equals:

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024$

Solution

An inverse variation can be expressed in the form $xy = n$, where $n$ is any number (except perhaps zero). Since $5y$ varies inversely with the square of $x$, this particular one will be $5yx^2 = n$.

We can plug in $16$ for $y$ and $1$ for $x$, which makes $n=80$. The equation is now $5yx^2 = 80$.

When $x = 8$, we can solve for $y$ using the equation $320y = 80$, which makes $y$ $\boxed{\textbf{(D)} \frac{1}{4}}$.

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions


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