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

(Created page with "== Problem == Indicate in which one of the following equations <math> y</math> is neither directly nor inversely proportional to <math> x</math>: <math> \textbf{(A)}\ x \plus{} ...")
 
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Indicate in which one of the following equations <math> y</math> is neither directly nor inversely proportional to <math> x</math>:
 
Indicate in which one of the following equations <math> y</math> is neither directly nor inversely proportional to <math> x</math>:
  
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<math> \textbf{(A)}\ x \plus{} y \equal{} 0 \qquad\textbf{(B)}\ 3xy \equal{} 10 \qquad\textbf{(C)}\ x \equal{} 5y \qquad\textbf{(D)}\ 3x \plus{} y \equal{} 10</math>
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<math> \textbf{(A)}\ x + y = 0 \qquad\textbf{(B)}\ 3xy = 10 \qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x + y = 10</math>  
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<math> \textbf{(E)}\ \frac {x}{y} \equal{} \sqrt {3}</math>
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<math> \textbf{(E)}\ \frac {x}{y} = \sqrt {3}</math>
  
 
== Solution ==  
 
== Solution ==  
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{{solution}}
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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.
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<math> \textbf{(A)}\ x + y = 0\implies \frac{x}{y}=-1</math>
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<math>\textbf{(B)}\ 3xy = 10\implies xy=\frac{10}{3}</math>
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<math>\textbf{(C)}\ x = 5y\implies \frac{x}{y}=5</math>
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<math> \textbf{(E)}\ \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 + y = 10}</math>
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest 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.

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

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

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

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

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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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