Difference between revisions of "1955 AHSME Problems/Problem 5"
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==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 5y varies inversely with the square of x, this particular one will be <math>5yx^2 = n</math>. | ||
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| + | We can plug in 16 for y and 1 for x, which makes n 80. The equation is now <math>5yx^2 = 80</math> | ||
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| + | When x = 8, we can solve for y using the equation <math>320y = 80</math>, which makes y <math>\textbf{(D)} \frac{1}{4}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 18:17, 2 August 2020
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 5y varies inversely with the square of x, this particular one will be 
.
We can plug in 16 for y and 1 for x, which makes n 80. The equation is now 
When x = 8, we can solve for y using the equation 
, which makes y 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
