In order to solve for $\displaystyle z$, first set up the variation equation for $\displaystyle x,$ $\displaystyle y,$ and $\displaystyle z$:

$\displaystyle z = \frac{(K)(2y)}{3x}$

where $\displaystyle K$ is the constant of variation. The $\displaystyle x$ term varies indirectly with $\displaystyle z$ and is therefore in the denominator.

Next, we solve for $\displaystyle K$ based on the initial values of the variables:

$\displaystyle 2 = \frac{(K)(2\times 4)}{(3\times 8)}$

$\displaystyle 2 = \frac{8K}{24}$

$\displaystyle 48 = 8K$

$\displaystyle 6=K$

Now that we have the value of $\displaystyle K$, we can solve for $\displaystyle z$ in the second scenario:

$\displaystyle z=\frac{(6)(2y)}{3x}$

$\displaystyle z=\frac{12y}{3x}$

$\displaystyle z=\frac{4y}{x}$

$\displaystyle z=\frac{4(6)}{3}$

$\displaystyle z=\frac{24}{3}$

$\displaystyle z=8$

The problem follows the formula 

$\displaystyle P = \frac{k}{n}$ 

where P is the number of slices you get, n is the number of people, and k is the constant of variation.

Setting P=3 and n = 16 yields k=48.

$\displaystyle 48=P(n)$

Now we substitute 12 in for n and solve for P

$\displaystyle 48=P(12)$

$\displaystyle \frac{48}{12}=P$

$\displaystyle 4=P$

Therefore with 12 people, you get 4 slices.

The problem follows the formula 

 $\displaystyle R = \frac{k}{n}$

where R is the number of raffle tickets you get, n is the number of people, and k is the constant of variation.

Setting R=20 and n = 150 yields k=3000.

$\displaystyle 3000=R(n)$

Plugging in 100 for n and solving for R you get:

$\displaystyle 3000=R(100)$

$\displaystyle \frac{3000}{100}=R$

$\displaystyle 30=R$

The answer R is 30 tickets.

The problem follows the formula 

 $\displaystyle B = \frac{k}{n}$

where B is the budget per committee, n is the number of committees, and k is the constant of variation.

Setting B=500 and n = 4 yields k=2000.

Now using the following equation we can plug in our n of 2 and solve for B.

$\displaystyle 2000=B(n)$

$\displaystyle 2000=B(2)$

$\displaystyle \frac{2000}{2}=B$

$\displaystyle 1000=B$

The answer of B is $1000.

The problem follows the formula 

 $\displaystyle H = \frac{k}{n}$

where H is the number of hours to complete the job, n is the number of people hired, and k is the constant of variation.

Setting H=6 and n = 8 yields k=48.

Therefore using the following equation we can plug 16 in for n and solve for H.

$\displaystyle 48=H(16)$

$\displaystyle \frac{48}{16}=H$

$\displaystyle 3=H$

Therefore H is 3 hours.

$\displaystyle x$ varies inversely with $\displaystyle y$, so the variation equation can be written as:

$\displaystyle x = k\cdot\frac{1}{y}$

$\displaystyle k$ can be solved for, using the first scenario:

$\displaystyle 15 = k\cdot\frac{1}{2}$

$\displaystyle k = 15\cdot2 = 30$

Using this value for $\displaystyle k$ = 30 and $\displaystyle y$ = 90, we can solve for $\displaystyle x$:

$\displaystyle x = 30\cdot\frac{1}{90} = \frac{30}{90} = \frac{3}{9} = \frac{1}{3}$

From the first sentence, we can write the equation of variation as:

$\displaystyle \small x = k \cdot \frac{A}{\sqrt{B}}$

We can then check each of the possible answer choices by substituting the values into the variation equation with the values given for $\displaystyle \small x$ and $\displaystyle \small k$.

$\displaystyle \small 3 = 2 \cdot \frac{3}{\sqrt{4}} = 2 \cdot \frac{3}{2} = 3$ 

Therefore the equation is true if $\displaystyle \small A = 3$ and $\displaystyle \small B = 4$

$\displaystyle \small 3 = 2 \cdot \frac{7.5}{\sqrt{25}} = 2 \cdot \frac{7.5}{5} = 2 \cdot 1.5 = 3$

Therefore the equation is true if $\displaystyle \small A = 7.5$ and $\displaystyle \small B = 25$

$\displaystyle \small 3 = 2 \cdot \frac{9}{\sqrt{36}} = 2 \cdot \frac{9}{6} = 2 \cdot 1.5 = 3$

Therefore the equation is true if $\displaystyle \small A = 9$ and $\displaystyle \small B = 36$

The correct answer choice is then "All of these answers are correct" 

From the relationship of $\displaystyle \small x$, $\displaystyle \small y$, and $\displaystyle \small z$; the equation of variation can be written as:

$\displaystyle \small x = k \cdot y \cdot z^{2}$

Using the values given in the first scenario, we can solve for k:

$\displaystyle \small 3 = k \cdot \frac{1}{3} \cdot 3^{2} = k \cdot \frac{1}{3} \cdot 9 = k \cdot 3$

$\displaystyle \small k = 1$

Using the value of k and the values of y and z, we can solve for x:

$\displaystyle \small x = 1 \cdot 12 \cdot 2^{2} = 12 \cdot 4 = 48$

First, we can create an equation of variation from the the relationships given:

$\displaystyle y = k \cdot \frac{1}{x \cdot \sqrt{z}}$

Next, we substitute the values given in the first scenario to solve for $\displaystyle \small k$:

$\displaystyle \frac{2}{9} = k \cdot \frac{1}{4 \cdot \sqrt{81}} = k \cdot \frac{1}{4 \cdot 9} = k \cdot \frac{1}{36}$

$\displaystyle \small \rightarrow k = 36 \cdot \frac{2}{9} = \frac{36}{9} \cdot 2 = 4 \cdot 2 = 8$

Using the value for $\displaystyle k$, we can now use the second values for $\displaystyle x$ and $\displaystyle z$ to solve for $\displaystyle y$:

$\displaystyle \small Y = 8 \cdot \frac{1}{8 \cdot \sqrt{4}} = 8 \cdot \frac{1}{8 \cdot 2} = 8 \cdot \frac{1}{16} = \frac{1}{2}$

From the relationship given, we can set up the variation equation

$\displaystyle \small x = k \cdot \sqrt{y} \cdot z^{2}$

Using the first relationship, we can then solve for $\displaystyle \small k$

$\displaystyle \small 5 = k \cdot \sqrt{100} \cdot 12^{2}$

$\displaystyle \small 5 = k \cdot 10 \cdot 144$

$\displaystyle \small k = \frac{5}{10 \cdot 144} = \frac{1}{2\cdot144} = \frac{1}{288}$

Now using the values from the second relationship, we can solve for x

$\displaystyle \small x = \frac{1}{288}\cdot\sqrt4\cdot6^2 = \frac{1}{288}\cdot2\cdot36 = \frac{72}{288} = \frac{1}{4}$