1959 AHSME Problems/Problem 20

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:

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

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1959 AHSME Problems/Problem 20

Problem 20

Solution

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