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Difference between revisions of "1951 AHSME Problems/Problem 17"

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Notice that for any directly or inversely proportional values, it can be expressed as <math>\frac{x}{y}=k</math> or <math>xy=k</math>. Now we try to convert each into its standard form counterpart.
 
Notice that for any directly or inversely proportional values, it can be expressed as <math>\frac{x}{y}=k</math> or <math>xy=k</math>. Now we try to convert each into its standard form counterpart.
  
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<math> \textbf{(A)}\ x \plus{} y \equal{} 0\implies \frac{x}{y}=-1</math>
+
<math> \textbf{(A)}\ x + y = 0\implies \frac{x}{y}=-1</math>
  
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<math>\textbf{(B)}\ 3xy \equal{} 10\implies xy=\frac{10}{3}</math>
+
<math>\textbf{(B)}\ 3xy = 10\implies xy=\frac{10}{3}</math>
  
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<math>\textbf{(C)}\ x \equal{} 5y\implies \frac{x}{y}=5</math>
+
<math>\textbf{(C)}\ x = 5y\implies \frac{x}{y}=5</math>
  
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<math> \textbf{(E)}\ \frac {x}{y} \equal{} \sqrt {3}\implies \frac {x}{y} \equal{} \sqrt {3}</math>
+
<math> \textbf{(D)}\ \frac {x}{y} = \sqrt {3}\implies \frac {x}{y} = \sqrt {3}</math>
  
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As we can see, the only equation without a "standard" form is <math>\textbf{(D)}</math>, so our answer is <math>\boxed{\textbf{(D)}\ 3x \plus{} y \equal{} 10}</math>
+
As we can see, the only equation without a "standard" form is <math>\textbf{(D)}</math>, so our answer is <math>\boxed{\textbf{(D)}\ 3x + y = 10}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 19:18, 30 April 2015

Problem

Indicate in which one of the following equations $y$ is neither directly nor inversely proportional to $x$:

$\textbf{(A)}\ x + y = 0 \qquad\textbf{(B)}\ 3xy = 10 \qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x + y = 10$ $\textbf{(E)}\ \frac {x}{y} = \sqrt {3}$

Solution

Notice that for any directly or inversely proportional values, it can be expressed as $\frac{x}{y}=k$ or $xy=k$. Now we try to convert each into its standard form counterpart.

$\textbf{(A)}\ x + y = 0\implies \frac{x}{y}=-1$

$\textbf{(B)}\ 3xy = 10\implies xy=\frac{10}{3}$

$\textbf{(C)}\ x = 5y\implies \frac{x}{y}=5$

$\textbf{(D)}\ \frac {x}{y} = \sqrt {3}\implies \frac {x}{y} = \sqrt {3}$

As we can see, the only equation without a "standard" form is $\textbf{(D)}$, so our answer is $\boxed{\textbf{(D)}\ 3x + y = 10}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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