Proportion and inverse proportion

It is given that $x$ is directly proportional to $y^2$ and $y$ is inversely proportional to $z$. If $x=20$ and $y=2$ when $z=5$

(A) the value of $y$ when $z=20$
(B) the value of $x$ when $y=3$
(C) an equation relating $x$ and $z$
(D) the value of $z$ when $x = \frac 5 4$, given that $z>0$

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So you are given the two relations to start:

$$x \propto y^{2} \implies x=cy^{2} \text{ for some }c\in\mathbb{R} \\ y \propto \frac{1}{z} \implies y=\frac{k}{z} \text{ for some } c\in\mathbb{R}$$

You are then given that when $z=5$, $y=2$ and $x=20$. Therefore, starting with our $y$ relation to find the value of $k$, we get:

$$2=\frac{k}{5} \implies k = 10$$

Therefore we have found our constant of proportionality for our $y$ variable. Next we have our $x$ relation to find the constant of proportionality for:

$$20=c(2)^{2}=4c \implies c=\frac{20}{4}=5$$

Now you have found these you should be able to work through the rest of the problems applying similar logic. Leave a comment if you get stuck!

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Thanks alot what I was really confuse about was whether the x variable and y variable is the same or not – user82844 Jun 17 '13 at 17:55
@user82844 No problem, glad to help! If this answered your question could you click on the tick next to the answer to mark the question as answered, so it no longer shows up in the unanswered section? Thanks! – Shaktal Jun 17 '13 at 18:13

The first statement is telling you that

$$x \ = \ A \cdot y^2 \ \ \text{and} \ \ y \ = \ \frac{B}{z} \ ,$$

with $A$ and $B$ being "proportionality constants" that you will need to determine by using the information given in the beginning of the second sentence. Once you know those values, you will be able to work out the answers to parts (A) through (D).

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if $x\propto y\implies x=ky$ (k is a constant ) $$\dfrac {x_1}{x_2}=\dfrac {{y_1}^2}{{y_2}^2}$$

partB

$$\dfrac {20}{x_2}=\dfrac {2^2}{3^2}\implies x_2=\dfrac{20\times 9}{4}\implies x_2=45$$

part A

$$\dfrac{y_1}{y_2}=\dfrac{z_2}{z_1}$$ $$\dfrac{2}{y_2}=\dfrac{20}{5}\implies y_2=0.5$$

part c

$$x\propto y^2\implies x=ky^2\implies y=\sqrt {\dfrac xk}$$ and $$y\propto \dfrac 1z\implies y=\dfrac cz\implies \sqrt{\dfrac xk}=\dfrac cz\implies \sqrt x\propto\dfrac 1z\implies x=\dfrac {1}{\sqrt z}$$

I think you can now solve part D on you own

part D

$$\dfrac{x_1}{x_2}=\sqrt{\dfrac {z_2}{z_1}}$$ $$\dfrac{20}{\frac 54}=\sqrt{\dfrac {z_2}{5}}$$ $$\dfrac{20^2}{({\frac 54})^2}={\dfrac {z_2}{5}}$$ $$z_2=\dfrac{400\times 5\times 16}{25}$$ $$z_2=1280$$

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