Question

18 POINTS! At the AoPS office, mice vary inversely with cats, that is, mice=k/cats, for some value of k. When there are 3r-19 cats, there are 2r+1 mice, and when there are 6r-27 mice, there are r-5 cats. Find k. Please help, and do not answer with an "I don't know" or "Sorry, this is too hard." Thank you all!

Answer 1

Answer:

k=30

Step-by-step explanation:

The number of mice vary inversely with cats. This is written as:

$Mice \propto \dfrac{1}{Cat} \\Mice = \dfrac{k}{Cat}$

When there are 3r-19 cats, there are 2r+1 mice

$2r+1 = \dfrac{k}{3r-19}\\ Cross multiply \\k=(3r-19)(2r+1)$

When there are 6r-27 mice, there are r-5 cats.

$6r-27 = \dfrac{k}{r-5}\\ Cross multiply \\k=(6r-27)(r-5)$

Taking the values of k above, we have:

$k=(3r-19)(2r+1) =(6r-27)(r-5)\\(3r-19)(2r+1) =(6r-27)(r-5)\\6r^2+3r-38r-19=6r^2-30r-27r+135\\ Collect like terms\\6r^2-6r^2+3r+30r-38r+27r-19-135=0\\22r-154=0\\22r=154\\ Divide both sides by 22\\r=7$

Therefore:

$k=(3r-19)(2r+1) \\=(3*7-19)(2*7+1)\\=(21-19)(14+1)\\=2*15\\k=30$