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> | |
− | Solution | + | == 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) | ||
Line 21: | Line 22: | ||
Simplifying this we get, | Simplifying this we get, | ||
− | <math>\fbox{B) 160}</math> | + | <math>\fbox{\textbf{(B) } 160}</math> |
~lli, awanglnc | ~lli, awanglnc | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | {{AHSME 50p box|year=1959|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:32, 21 July 2024
Problem 20
It is given that varies directly as and inversely as the square of , and that when and . Then, when and , equals:
Solution
varies directly to (The inverse variation of y and the square of z)
We can write the expression
Now we plug in the values of when and .
This gives us
We can use this to find the value of when and
Simplifying this we get,
~lli, awanglnc
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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