Difference between revisions of "1959 AHSME Problems/Problem 20"

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== Problem 20 ==
 
It is given that <math>x</math> varies directly as <math>y</math> and inversely as the square of <math>z</math>, and that <math>x=10</math> when <math>y=4</math> and <math>z=14</math>. Then, when <math>y=16</math> and <math>z=7</math>, <math>x</math> equals:
 
It is given that <math>x</math> varies directly as <math>y</math> and inversely as the square of <math>z</math>, and that <math>x=10</math> when <math>y=4</math> and <math>z=14</math>. Then, when <math>y=16</math> and <math>z=7</math>, <math>x</math> equals:
  
 
  <math>\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120</math>
 
  <math>\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120</math>
  
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== Solution ==
  
 
<math>x</math> varies directly to <math>\frac{y}{z^2}</math> (The inverse variation of y and the square of z)
 
<math>x</math> varies directly to <math>\frac{y}{z^2}</math> (The inverse variation of y and the square of z)
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~lli, awanglnc
 
~lli, awanglnc
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== See also ==
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{{AHSME 50p box|year=1959|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 12:41, 9 April 2024

Problem 20

It is given that $x$ varies directly as $y$ and inversely as the square of $z$, and that $x=10$ when $y=4$ and $z=14$. Then, when $y=16$ and $z=7$, $x$ equals:

$\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120$

Solution

$x$ varies directly to $\frac{y}{z^2}$ (The inverse variation of y and the square of z)

We can write the expression

$x = \frac{ky}{z^2}$

Now we plug in the values of $x=10$ when $y=4$ and $z=14$.

This gives us $k=490$

We can use this to find the value of $x$ when $y=4$ and $z=14$

$x=\frac{490\cdot4}{14^2}$

Simplifying this we get,

$\fbox{B) 160}$

~lli, awanglnc

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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