Difference between revisions of "1955 AHSME Problems/Problem 5"
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== Problem == | == Problem == | ||
+ | <math>5y</math> varies inversely as the square of <math>x</math>. When <math>y=16, x=1</math>. When <math>x=8, y</math> equals: | ||
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+ | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024 </math> | ||
==Solution== | ==Solution== | ||
+ | An inverse variation can be expressed in the form <math>xy = n</math>, where <math>n</math> is any number (except perhaps zero). Since <math>5y</math> varies inversely with the square of <math>x</math>, this particular one will be <math>5yx^2 = n</math>. | ||
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+ | We can plug in <math>16</math> for <math>y</math> and <math>1</math> for <math>x</math>, which makes <math>n=80</math>. The equation is now <math>5yx^2 = 80</math>. | ||
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+ | When <math>x = 8</math>, we can solve for <math>y</math> using the equation <math>320y = 80</math>, which makes <math>y</math> <math>\boxed{\textbf{(D)} \frac{1}{4}}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 01:52, 14 October 2021
Problem
varies inversely as the square of
. When
. When
equals:
Solution
An inverse variation can be expressed in the form , where
is any number (except perhaps zero). Since
varies inversely with the square of
, this particular one will be
.
We can plug in for
and
for
, which makes
. The equation is now
.
When , we can solve for
using the equation
, which makes
.
See Also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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All AHSME Problems and Solutions |
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