Difference between revisions of "Proportion/Intermediate"
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==Problem== | ==Problem== | ||
| − | + | <math>x</math> is directly proportional to the sum of the squares of <math>y</math> and <math>z</math> and inversely proportional to <math>y</math> and the square of <math>z</math>. If <math>x = 8</math> when <math>y = 1/2</math> and <math>z = \sqrt {3}/2</math>, find <math>y</math> when <math>x = 1</math> and <math>z = 6</math>. (Thanks to Bicameral of the AoPS forum for this one) | |
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==Solution== | ==Solution== | ||
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| + | First, let's write these out algebraically: the first is <math>x=k(y^2+z^2)</math>, the second is <math>xy=k_1</math>, and the third is <math>xz^2=k_2</math>. Plugging the values into the first equation gives <math>8=k((\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2)=k(\frac{1}{4}+\frac{3}{4})=k</math>, so <math>k=8</math>. The second gives <math>8*\frac{1}{2}=4=k_1</math>, and the third gives <math>8*(\frac{\sqrt{3}}{2})^2=8*3/4=6=k_2</math>. We can use any of these three to solve the problem. Using the second gives us <math>1*y=y=4</math>, so the answer is <math>y=4</math>. | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 13:24, 26 February 2013
Problem
is directly proportional to the sum of the squares of 
and 
and inversely proportional to and the square of . If 
when 
and 
, find when 
and 
. (Thanks to Bicameral of the AoPS forum for this one)
Solution
First, let's write these out algebraically: the first is 
, the second is 
, and the third is 
. Plugging the values into the first equation gives 
, so 
. The second gives 
, and the third gives 
. We can use any of these three to solve the problem. Using the second gives us 
, so the answer is 
.
