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

(Created page with "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=1...")
 
(Fixing latex for our own solution)
Line 21: Line 21:
 
Simplifying this we get,
 
Simplifying this we get,
  
−
<math>\box B) 160</math>
+
<math>\fbox{B) 160}</math>
 +
 
 +
~lli, awanglnc

Revision as of 19:31, 8 July 2023

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 1:

$x$ 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

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