Difference between revisions of "1964 AHSME Problems/Problem 5"

(Created page with "== Problem == If <math>y</math> varies directly as <math>x</math>, and if <math>y=8</math> when <math>x=4</math>, the value of <math>y</math> when <math>x=-8</math> is: <mat...")
 
 
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== Problem ==
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==Problem==
  
 
If <math>y</math> varies directly as <math>x</math>, and if <math>y=8</math> when <math>x=4</math>, the value of <math>y</math> when <math>x=-8</math> is:
 
If <math>y</math> varies directly as <math>x</math>, and if <math>y=8</math> when <math>x=4</math>, the value of <math>y</math> when <math>x=-8</math> is:
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== Solution ==
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==Solution==
  
 
If <math>y</math> varies directly as <math>x</math>, then <math>\frac{x}{y}</math> will be a constant for every pair of <math>(x, y)</math>.  Thus, we have:
 
If <math>y</math> varies directly as <math>x</math>, then <math>\frac{x}{y}</math> will be a constant for every pair of <math>(x, y)</math>.  Thus, we have:

Latest revision as of 10:57, 23 July 2019

Problem

If $y$ varies directly as $x$, and if $y=8$ when $x=4$, the value of $y$ when $x=-8$ is:

$\textbf{(A)}\ -16 \qquad \textbf{(B)}\ -4 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 4k, k= \pm1, \pm2, \dots \qquad \\ \textbf{(E)}\ 16k, k=\pm1,\pm2,\dots$


Solution

If $y$ varies directly as $x$, then $\frac{x}{y}$ will be a constant for every pair of $(x, y)$. Thus, we have:

$\frac{4}{8} = \frac{-8}{y}$

$y \cdot \frac{1}{2} = -8$

$y = -16$

$\boxed{\textbf{(A)}}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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