Direct Variation Practice Worksheet with equations and word problems.

It looks like you've uploaded a worksheet titled "Direct Variation Practice" with problems related to direct variation, where one variable is directly proportional to another. Let's go through each problem step by step and solve them.

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🔷 What is Direct Variation?

When we say y varies directly as x, it means:

$$
y = kx
$$

Where:
- $ y $ and $ x $ are variables,
- $ k $ is the constant of proportionality.

To solve these problems, we:
1. Use given values to find $ k $.
2. Then use $ k $ to find the unknown value.

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Let’s solve each problem:

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Problem 1:

> Suppose $ y $ varies directly as $ x $. If $ y = 3 $ when $ x = 15 $, find $ y $ when $ x = 4 $.

Step 1: Find $ k $

$$
y = kx \Rightarrow 3 = k(15) \Rightarrow k = \frac{3}{15} = \frac{1}{5}
$$

Step 2: Use $ k = \frac{1}{5} $ to find $ y $ when $ x = 4 $

$$
y = \frac{1}{5} \cdot 4 = \frac{4}{5}
$$

✔ Answer: $ y = \frac{4}{5} $

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Problem 2:

> Suppose $ y $ varies directly as $ x $. If $ y = 34 $ when $ x = 10 $, find $ y $ when $ x = 17 $.

$$
34 = k(10) \Rightarrow k = \frac{34}{10} = 3.4
$$

Now find $ y $ when $ x = 17 $:

$$
y = 3.4 \cdot 17 = 57.8
$$

✔ Answer: $ y = 57.8 $

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Problem 3:

> Suppose $ y $ varies directly as $ x $. If $ x = 13 $ when $ y = 12 $, find $ y $ when $ x = 25 $.

Wait — this says "if $ x = 13 $ when $ y = 12 $" → same thing: $ y = 12 $, $ x = 13 $

So:
$$
12 = k(13) \Rightarrow k = \frac{12}{13}
$$

Now find $ y $ when $ x = 25 $:

$$
y = \frac{12}{13} \cdot 25 = \frac{300}{13} \approx 23.08
$$

✔ Answer: $ y = \frac{300}{13} $ or approximately $ 23.08 $

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Problem 4:

> Suppose $ y $ varies directly as $ x $. If $ x = 24 $ when $ y = 8 $, find $ y $ when $ x = 39 $.

$$
8 = k(24) \Rightarrow k = \frac{8}{24} = \frac{1}{3}
$$

Now $ y = \frac{1}{3} \cdot 39 = 13 $

✔ Answer: $ y = 13 $

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Problem 5:

> Suppose $ y $ varies directly as $ x $. If $ x = 27 $ when $ y = 2 $, find $ y $ when $ x = 2 $.

$$
2 = k(27) \Rightarrow k = \frac{2}{27}
$$

Now $ y = \frac{2}{27} \cdot 2 = \frac{4}{27} $

✔ Answer: $ y = \frac{4}{27} $

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Problem 6:

> Suppose $ y $ varies directly as $ x $. If $ y = 38 $ when $ x = 4 $, find $ y $ when $ x = 6 $.

$$
38 = k(4) \Rightarrow k = \frac{38}{4} = 9.5
$$

Now $ y = 9.5 \cdot 6 = 57 $

✔ Answer: $ y = 57 $

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Problem 7:

> Suppose $ y $ varies directly as $ x $. If $ y = 150 $ when $ x = 5 $, find $ y $ when $ x = 5 $.

Wait — both $ x $ and $ y $ are 5 in the second case?

No — let’s read carefully:

- Given: $ y = 150 $ when $ x = 5 $
- Find $ y $ when $ x = 5 $

That’s the same x-value, so $ y $ should be the same.

But let’s check:

$$
150 = k(5) \Rightarrow k = 30
$$

Then $ y = 30 \cdot 5 = 150 $

✔ Answer: $ y = 150 $

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Problem 8:

> Suppose $ y $ varies directly as $ x $. If $ y = 4 $ when $ x = 32 $, find $ y $ when $ x = 175 $.

$$
4 = k(32) \Rightarrow k = \frac{4}{32} = \frac{1}{8}
$$

Now $ y = \frac{1}{8} \cdot 175 = \frac{175}{8} = 21.875 $

✔ Answer: $ y = 21.875 $ or $ \frac{175}{8} $

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Problem 9 (Word Problem):

> The number of pencils sold varies directly as the cost. If 5 pencils cost $2.45, find the cost of 8 pencils.

Let:
- $ c $ = cost
- $ p $ = number of pencils

Since cost varies directly with number of pencils:

$$
c = kp
$$

Given: $ c = 2.45 $, $ p = 5 $

$$
2.45 = k(5) \Rightarrow k = \frac{2.45}{5} = 0.49
$$

So cost per pencil is $ 0.49.

Now find cost for $ p = 8 $:

$$
c = 0.49 \cdot 8 = 3.92
$$

✔ Answer: $ \$3.92 $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ y = \frac{4}{5} $ |
| 2 | $ y = 57.8 $ |
| 3 | $ y = \frac{300}{13} \approx 23.08 $ |
| 4 | $ y = 13 $ |
| 5 | $ y = \frac{4}{27} $ |
| 6 | $ y = 57 $ |
| 7 | $ y = 150 $ |
| 8 | $ y = 21.875 $ or $ \frac{175}{8} $ |
| 9 | $ \$3.92 $ |

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