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 <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 00:52, 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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
