Direct and Inverse Variation Worksheet | PDF - Free Printable

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 $

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Problem 2:

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

- Inverse variation: $ J \propto \frac{1}{K} $ → $ 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.

- $ Q \propto W $ → $ 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.

- $ A \propto P \cdot Q $ → $ A = kPQ $

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

- Now find $ A $ when $ P = 44 $, $ Q = 7 $:
$$
A = \frac{7}{12} \cdot 44 \cdot 7 = \frac{7}{12} \cdot 308 = \frac{2156}{12} = 179.\overline{6}
$$

Or simplified:
$$
\frac{2156}{12} = \frac{539}{3} \approx 179.67
$$

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

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Problem 6:

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

- $ B \propto \frac{1}{S} $ → $ B = \frac{k}{S} $

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

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

Let’s simplify:
$$
\frac{682}{34} = \frac{341}{17} \approx 20.06
$$

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

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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 \propto A \cdot Q $ → $ L = kAQ $

- 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 $ \approx 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.

- $ E = kD $
- $ 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.

- $ h \propto \frac{1}{t} $ → $ h = \frac{k}{t} $

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

✔ Answer: $ k = 128 $

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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 $ be weight, $ d $ be distance from center of Earth.
- $ W \propto \frac{1}{d^2} $ → $ W = \frac{k}{d^2} $

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

- Compute $ d $ at 1800 miles above:
$$
d = 9200 + 1800 = 11000 \text{ miles}
$$

- New weight:
$$
W = \frac{k}{(11000)^2} = \frac{260 \cdot (9200)^2}{(11000)^2}
$$

- Simplify:
$$
W = 260 \cdot \left(\frac{9200}{11000}\right)^2 = 260 \cdot \left(\frac{92}{110}\right)^2 = 260 \cdot \left(\frac{46}{55}\right)^2
$$

- Compute $ \left(\frac{46}{55}\right)^2 = \frac{2116}{3025} $

- $ W = 260 \cdot \frac{2116}{3025} $

- Calculate:
$$
\frac{260 \cdot 2116}{3025} = \frac{549,160}{3025} \approx 181.5 \text{ pounds}
$$

✔ Answer: Approximately $ 181.5 $ pounds

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ k = \frac{73}{2912} $ |
| 2 | $ J = -\frac{4144}{3} \approx -1381.33 $ |
| 3 | $ E = \frac{152}{9} \approx 16.89 $ |
| 4 | $ Q = 35 $ |
| 5 | $ A = \frac{539}{3} \approx 179.67 $ |
| 6 | $ B = \frac{341}{17} \approx 20.06 $ |
| 7 | $ L = \frac{10720}{21} \approx 510.48 $ |
| 8 | $ k = \frac{47}{4} = 11.75 $ |
| 9 | $ k = 128 $ |
| 10 | $ \approx 181.5 $ pounds |

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