Algebra II Worksheet 10B featuring combined variation problems.

Let's solve each problem from Algebra II - Worksheet 10B: Combined Variation step by step. We'll use the principles of joint variation, direct variation, and inverse variation.

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Key Concepts:

- Direct variation: $ y \propto x $ → $ y = kx $
- Inverse variation: $ y \propto \frac{1}{x} $ → $ y = \frac{k}{x} $
- Joint variation: $ y \propto xz $ → $ y = kxz $
- Combined variation: Mix of direct and inverse, e.g., $ y \propto \frac{x}{z} $ → $ y = \frac{kx}{z} $

We always find the constant of proportionality (k) using given values, then use it to find unknowns.

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Problem 1

> $ y $ varies jointly as $ x $ and $ z $. If $ y = 5 $ when $ x = 3 $ and $ z = 4 $, find $ y $ when $ x = 6 $ and $ z = 8 $.

Step 1: Joint variation:
$ y = kxz $

Plug in known values:
$$
5 = k(3)(4) = 12k \Rightarrow k = \frac{5}{12}
$$

Now find $ y $ when $ x = 6 $, $ z = 8 $:
$$
y = \frac{5}{12}(6)(8) = \frac{5}{12} \cdot 48 = 20
$$

✔ Answer: 20

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Problem 2

> $ y $ varies jointly as $ x $ and $ z $. If $ y = 12 $ when $ x = 4 $ and $ z = 3 $, find $ y $ when $ x = 9 $ and $ z = 8 $.

$ y = kxz $

$$
12 = k(4)(3) = 12k \Rightarrow k = 1
$$

Now:
$$
y = (1)(9)(8) = 72
$$

✔ Answer: 72

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Problem 3

> $ y $ varies directly as $ x $ and inversely as $ z $. If $ y = 5 $ when $ x = 3 $ and $ z = 4 $, find $ y $ when $ x = 6 $ and $ z = 8 $.

So: $ y = \frac{kx}{z} $

$$
5 = \frac{k(3)}{4} \Rightarrow 5 \cdot 4 = 3k \Rightarrow 20 = 3k \Rightarrow k = \frac{20}{3}
$$

Now:
$$
y = \frac{(20/3)(6)}{8} = \frac{120/3}{8} = \frac{40}{8} = 5
$$

✔ Answer: 5

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Problem 4

> $ y $ varies directly as $ x^2 $ and inversely as $ z $. If $ y = 12 $ when $ x = 2 $ and $ z = 7 $, find $ y $ when $ x = 3 $ and $ z = 9 $.

So: $ y = \frac{kx^2}{z} $

$$
12 = \frac{k(2)^2}{7} = \frac{4k}{7} \Rightarrow 12 \cdot 7 = 4k \Rightarrow 84 = 4k \Rightarrow k = 21
$$

Now:
$$
y = \frac{21(3)^2}{9} = \frac{21 \cdot 9}{9} = 21
$$

✔ Answer: 21

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Problem 5

> $ A $ varies jointly as $ b $ and $ h $. If $ A = 16 $ when $ b = 2 $ and $ h = 8 $, find $ A $ when $ b = 8 $ and $ h = 16 $.

$ A = kbh $

$$
16 = k(2)(8) = 16k \Rightarrow k = 1
$$

Now:
$$
A = (1)(8)(16) = 128
$$

✔ Answer: 128

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Problem 6

> $ y $ varies jointly as $ x $ and $ \sqrt{z} $. If $ y = 6 $ when $ x = 3 $ and $ z = 9 $, find $ y $ when $ x = 4 $ and $ z = 36 $.

So: $ y = kx\sqrt{z} $

First, $ \sqrt{9} = 3 $, so:
$$
6 = k(3)(3) = 9k \Rightarrow k = \frac{6}{9} = \frac{2}{3}
$$

Now $ \sqrt{36} = 6 $, so:
$$
y = \frac{2}{3}(4)(6) = \frac{2}{3} \cdot 24 = 16
$$

✔ Answer: 16

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Problem 7

> $ y $ varies jointly as $ \sqrt[3]{x} $ and $ z^2 $. If $ y = 3 $ when $ x = 8 $ and $ z = 4 $, find $ y $ when $ x = 27 $ and $ z = 6 $.

So: $ y = k \cdot \sqrt[3]{x} \cdot z^2 $

Note: $ \sqrt[3]{8} = 2 $, $ z^2 = 16 $

$$
3 = k(2)(16) = 32k \Rightarrow k = \frac{3}{32}
$$

Now: $ \sqrt[3]{27} = 3 $, $ z^2 = 36 $

$$
y = \frac{3}{32} \cdot 3 \cdot 36 = \frac{3}{32} \cdot 108 = \frac{324}{32} = 10.125
$$

Or as fraction: $ \frac{324}{32} = \frac{81}{8} $

✔ Answer: $ \frac{81}{8} $ or 10.125

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Problem 8

> $ y $ varies directly as $ \sqrt{x} $ and inversely as $ z $. If $ y = 10 $ when $ x = 9 $ and $ z = 12 $, find $ y $ when $ x = 16 $ and $ z = 10 $.

So: $ y = \frac{k\sqrt{x}}{z} $

$ \sqrt{9} = 3 $

$$
10 = \frac{k(3)}{12} \Rightarrow 10 \cdot 12 = 3k \Rightarrow 120 = 3k \Rightarrow k = 40
$$

Now: $ \sqrt{16} = 4 $, $ z = 10 $

$$
y = \frac{40 \cdot 4}{10} = \frac{160}{10} = 16
$$

✔ Answer: 16

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Problem 9

> $ x $ varies jointly as $ y^3 $ and $ \sqrt{z} $. If $ x = 7 $ when $ y = 2 $ and $ z = 4 $, find $ x $ when $ y = 3 $ and $ z = 9 $.

So: $ x = k y^3 \sqrt{z} $

First: $ y^3 = 8 $, $ \sqrt{4} = 2 $

$$
7 = k(8)(2) = 16k \Rightarrow k = \frac{7}{16}
$$

Now: $ y^3 = 27 $, $ \sqrt{9} = 3 $

$$
x = \frac{7}{16} \cdot 27 \cdot 3 = \frac{7}{16} \cdot 81 = \frac{567}{16} = 35.4375
$$

✔ Answer: $ \frac{567}{16} $ or 35.4375

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Problem 10

> $ x $ varies directly as $ y^3 $ and inversely as $ \sqrt{z} $. If $ x = 7 $ when $ y = 2 $ and $ z = 4 $, find $ x $ when $ y = 3 $ and $ z = 9 $.

So: $ x = \frac{k y^3}{\sqrt{z}} $

$ y^3 = 8 $, $ \sqrt{4} = 2 $

$$
7 = \frac{k(8)}{2} = 4k \Rightarrow k = \frac{7}{4}
$$

Now: $ y^3 = 27 $, $ \sqrt{9} = 3 $

$$
x = \frac{(7/4)(27)}{3} = \frac{189/4}{3} = \frac{189}{12} = 15.75
$$

✔ Answer: $ \frac{63}{4} $ or 15.75

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Problem 11

> The number of girls varied directly as the number of boys and inversely as the number of teachers. When there were 50 girls, there were 20 teachers and 10 boys. How many boys were there when there were 10 girls and 100 teachers?

Let:
- $ G $ = number of girls
- $ B $ = number of boys
- $ T $ = number of teachers

Given: $ G \propto \frac{B}{T} $ → $ G = \frac{kB}{T} $

Use first case: $ G = 50 $, $ B = 10 $, $ T = 20 $

$$
50 = \frac{k(10)}{20} = \frac{k}{2} \Rightarrow k = 100
$$

Now: $ G = 10 $, $ T = 100 $, find $ B $

$$
10 = \frac{100 \cdot B}{100} = B
$$

✔ Answer: 10 boys

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Problem 12

> Strawberries varied jointly as plums and tomatoes. If 500 strawberries went with 4 plums and 25 tomatoes, how many plums would go with 40 strawberries and 2 tomatoes?

Let:
- $ S $ = strawberries
- $ P $ = plums
- $ T $ = tomatoes

$ S \propto P \cdot T $ → $ S = kPT $

Given: $ S = 500 $, $ P = 4 $, $ T = 25 $

$$
500 = k(4)(25) = 100k \Rightarrow k = 5
$$

Now: $ S = 40 $, $ T = 2 $, find $ P $

$$
40 = 5 \cdot P \cdot 2 = 10P \Rightarrow P = 4
$$

✔ Answer: 4 plums

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 20 |
| 2 | 72 |
| 3 | 5 |
| 4 | 21 |
| 5 | 128 |
| 6 | 16 |
| 7 | $ \frac{81}{8} $ or 10.125 |
| 8 | 16 |
| 9 | $ \frac{567}{16} $ or 35.4375 |
| 10 | $ \frac{63}{4} $ or 15.75 |
| 11 | 10 boys |
| 12 | 4 plums |

Let me know if you'd like these formatted as a PDF or need explanations for any specific step!