Math worksheet on direct and inverse variation with ten problems and a grading scale.

Let's solve each problem step by step, explaining the reasoning and calculations involved.

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Problem 1:

Suppose F varies directly as A and N. Find the constant of proportionality when A = 182, F = 365, and N = 80.

- Direct variation with two variables:
If $ F \propto A \cdot N $, then:
$$
F = k \cdot A \cdot N
$$
where $ k $ is the constant of proportionality.

- Plug in the values:
$$
365 = k \cdot 182 \cdot 80
$$
$$
365 = k \cdot 14560
$$
$$
k = \frac{365}{14560}
$$

- Simplify:
$$
k = \frac{365 \div 5}{14560 \div 5} = \frac{73}{2912}
$$

✔ Answer: $ k = \frac{73}{2912} $ or approximately $ 0.02506 $

---

Problem 2:

If J varies inversely as K and J = -148 when K = 28, find J when K = 3.

- Inverse variation:
$$
J = \frac{k}{K}
$$

- Use given values to find $ k $:
$$
-148 = \frac{k}{28} \Rightarrow k = -148 \cdot 28 = -4144
$$

- Now find J when $ K = 3 $:
$$
J = \frac{-4144}{3} \approx -1381.33
$$

✔ Answer: $ J = -\frac{4144}{3} $ or approximately $ -1381.\overline{3} $

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Problem 3:

If E varies directly as D, and E = 16 when D = 36, then what is E when D = 38?

- Direct variation:
$$
E = kD
$$

- Find $ k $:
$$
16 = k \cdot 36 \Rightarrow k = \frac{16}{36} = \frac{4}{9}
$$

- Now find E when $ D = 38 $:
$$
E = \frac{4}{9} \cdot 38 = \frac{152}{9} \approx 16.89
$$

✔ Answer: $ E = \frac{152}{9} $ or approximately $ 16.89 $

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Problem 4:

If Q = 10 when W = 4, and Q varies directly as W, find Q when W = 14.

- Direct variation:
$$
Q = kW
$$

- Find $ k $:
$$
10 = k \cdot 4 \Rightarrow k = \frac{10}{4} = 2.5
$$

- Now find Q when $ W = 14 $:
$$
Q = 2.5 \cdot 14 = 35
$$

✔ Answer: $ Q = 35 $

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Problem 5:

Variable A varies directly with P and Q. If A = 42 when P = 8 and Q = 9, find A when P = 44 and Q = 7.

- Joint variation:
$$
A = k \cdot P \cdot Q
$$

- Use known values:
$$
42 = k \cdot 8 \cdot 9 = k \cdot 72 \Rightarrow k = \frac{42}{72} = \frac{7}{12}
$$

- Now find A when $ P = 44 $, $ Q = 7 $:
$$
A = \frac{7}{12} \cdot 44 \cdot 7 = \frac{7}{12} \cdot 308 = \frac{2156}{12} = 179.666...
$$

- Simplify:
$$
\frac{2156}{12} = \frac{539}{3} \approx 179.67
$$

✔ Answer: $ A = \frac{539}{3} $ or approximately $ 179.67 $

---

Problem 6:

B varies inversely with S. If B is 62 when S = 11, find B when S = 34.

- Inverse variation:
$$
B = \frac{k}{S}
$$

- Find $ k $:
$$
62 = \frac{k}{11} \Rightarrow k = 62 \cdot 11 = 682
$$

- Now find B when $ S = 34 $:
$$
B = \frac{682}{34} = 20.0588...
$$

- Simplify:
$$
\frac{682}{34} = \frac{341}{17} \approx 20.06
$$

✔ Answer: $ B = \frac{341}{17} $ or approximately $ 20.06 $

---

Problem 7:

Variable L varies directly and jointly with A and Q. If L = 402 when A = 18 and Q = 14, find L when A = 10 and Q = 32.

- Joint variation:
$$
L = k \cdot A \cdot Q
$$

- Find $ k $:
$$
402 = k \cdot 18 \cdot 14 = k \cdot 252 \Rightarrow k = \frac{402}{252} = \frac{67}{42}
$$

- Now find L when $ A = 10 $, $ Q = 32 $:
$$
L = \frac{67}{42} \cdot 10 \cdot 32 = \frac{67}{42} \cdot 320 = \frac{21440}{42} = \frac{10720}{21} \approx 510.48
$$

✔ Answer: $ L = \frac{10720}{21} $ or approximately $ 510.48 $

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Problem 8:

If E varies directly as D, and E = 94 and D = 8, then find the constant of proportionality.

- Direct variation:
$$
E = kD
$$

- Solve for $ k $:
$$
94 = k \cdot 8 \Rightarrow k = \frac{94}{8} = 11.75
$$

✔ Answer: $ k = 11.75 $ or $ \frac{47}{4} $

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Problem 9:

The number of hours, h, it takes for a block of ice to melt varies inversely as the temperature, t. If it takes 4 hours for a square inch of ice to melt at 32°, find the constant of proportionality.

- Inverse variation:
$$
h = \frac{k}{t}
$$

- Given: $ h = 4 $, $ t = 32 $
$$
4 = \frac{k}{32} \Rightarrow k = 4 \cdot 32 = 128
$$

✔ Answer: $ k = 128 $

---

Problem 10:

The weight of a body varies inversely as the square of its distance from the center of the earth. If the radius of the earth is 9200 miles, how much would a 260-pound man weigh 1800 miles above the surface of the earth?

- Let:
- $ W $ = weight
- $ d $ = distance from center of Earth
- $ W \propto \frac{1}{d^2} \Rightarrow W = \frac{k}{d^2} $

- At the surface: $ d = 9200 $ miles, $ W = 260 $ lb
$$
260 = \frac{k}{(9200)^2} \Rightarrow k = 260 \cdot (9200)^2
$$

- Compute $ k $:
$$
(9200)^2 = 84,640,000
$$
$$
k = 260 \cdot 84,640,000 = 21,996,800,000
$$

- Now, 1800 miles above surface → $ d = 9200 + 1800 = 11,000 $ miles
$$
W = \frac{21,996,800,000}{(11,000)^2} = \frac{21,996,800,000}{121,000,000}
$$

- Divide:
$$
W = \frac{21,996,800,000}{121,000,000} = 181.808... \approx 181.8 \text{ pounds}
$$

✔ Answer: The man would weigh approximately 181.8 pounds

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ k = \frac{73}{2912} $ |
| 2 | $ J = -\frac{4144}{3} $ |
| 3 | $ E = \frac{152}{9} $ |
| 4 | $ Q = 35 $ |
| 5 | $ A = \frac{539}{3} $ |
| 6 | $ B = \frac{341}{17} $ |
| 7 | $ L = \frac{10720}{21} $ |
| 8 | $ k = \frac{47}{4} $ |
| 9 | $ k = 128 $ |
| 10 | $ \approx 181.8 $ pounds |

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