We are given a variation problem involving
direct and
inverse variation.
---
Problem Statement:
> $ y $ varies
directly as $ x $ and
inversely as the square of $ z $.
> When $ x = 32 $, $ y = 6 $, and $ z = 4 $, find $ x $ when $ y = 10 $ and $ z = 3 $.
---
Step 1: Understand the Variation
- "Varies directly as $ x $" → $ y \propto x $
- "Varies inversely as the square of $ z $" → $ y \propto \frac{1}{z^2} $
Combining both:
$$
y \propto \frac{x}{z^2}
$$
So we can write:
$$
y = k \cdot \frac{x}{z^2}
$$
where $ k $ is the constant of proportionality.
---
Step 2: Use Given Values to Find $ k $
Given:
$ x = 32 $, $ y = 6 $, $ z = 4 $
Plug into the equation:
$$
6 = k \cdot \frac{32}{4^2} = k \cdot \frac{32}{16} = k \cdot 2
$$
Solve for $ k $:
$$
k = \frac{6}{2} = 3
$$
So the equation becomes:
$$
y = 3 \cdot \frac{x}{z^2}
$$
---
Step 3: Find $ x $ when $ y = 10 $, $ z = 3 $
Use the equation:
$$
10 = 3 \cdot \frac{x}{3^2} = 3 \cdot \frac{x}{9}
$$
Simplify:
$$
10 = \frac{3x}{9} = \frac{x}{3}
$$
Multiply both sides by 3:
$$
x = 30
$$
---
✔ Final Answer:
$$
\boxed{x = 30}
$$
---
Summary:
- We used the combined variation formula: $ y = k \cdot \frac{x}{z^2} $
- Found $ k = 3 $ using initial values
- Then solved for $ x $ with new $ y $ and $ z $
- Final answer: $ \boxed{30} $
Parent Tip: Review the logic above to help your child master the concept of direct inverse and joint variation worksheet answers.
