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Direct Proportion

If \(y\) is directly proportional to \(x\), then when \(x\) is multiplied by a number, \(y\) is multiplied by the same number.

\[ y \propto x \] \[ y = kx \]

Example

The number of miles covered is directly proportional to the time spent walking.

If 5 miles are covered in 2 days, how many miles will be covered in 7 days?

\[ y \propto x \] \[ y = kx \] \[ 5 = k \cdot 2 \] \[ k = \frac{5}{2} = 2.5 \] \[ y = 2.5x \] \[ y = 2.5 \times 7 = 17.5 \]

17.5 miles are covered in 7 days.

Example

The cost of bricks is directly proportional.

Fred buys 10 bricks for £5.

George buys 5 bricks for £2.50.

How much does Harry pay for 18 bricks?

\[ \text{Cost} \propto \text{Number of bricks} \] \[ C = k n \] \[ 5 = k \cdot 10 \] \[ k = \frac{5}{10} = 0.5 \] \[ C = 0.5n \] \[ C = 0.5 \times 18 = 9 \]

Harry pays £9 for 18 bricks.

Indirect Proportion

If \(y\) is indirectly proportional to \(x\), then when \(x\) is multiplied by a number, \(y\) is divided by that number — and vice versa.

As one number goes up, the other goes down.

\[ y \propto \frac{1}{x} \] \[ y = \frac{k}{x} \]

Example

The number of days taken to dig a ditch is inversely proportional to the number of men digging.

If it takes 4 days for 10 men to dig a trench, how long will it take 8 men?

\[ y = \frac{k}{x} \] \[ 4 = \frac{k}{10} \] \[ k = 40 \] \[ y = \frac{40}{x} \] \[ y = \frac{40}{8} = 5 \]

It takes 8 men 5 days.

Squared Proportion

If \(y\) is directly proportional to \(x^2\),

then when \(x\) is multiplied by a number, \(y\) is multiplied by the square of that number.

\[ y \propto x^2 \] \[ y = kx^2 \]

Example

The distance moved by an accelerating car is directly proportional to the square of the time of acceleration.

After 2 seconds, the car has travelled 12 m. What is the distance travelled after 9 seconds?

\[ y = kx^2 \] \[ 12 = k \cdot 2^2 \] \[ 12 = 4k \] \[ k = 3 \] \[ y = 3x^2 \] \[ y = 3 \times 9^2 \] \[ y = 3 \times 81 = 243 \]

The car has travelled 243 m.

Joint Variation

Example

\(y\) varies as \(x^2\) and inversely as \(z\).

Given \(y = 12\) when \(x = 6\) and \(z = 9\), find \(y\) when \(x = 10\) and \(z = 15\).

\[ y \propto \frac{x^2}{z} \] \[ y = k \frac{x^2}{z} \] \[ 12 = k \cdot \frac{6^2}{9} \] \[ 12 = k \cdot \frac{36}{9} \] \[ 12 = 4k \] \[ k = 3 \] \[ y = 3 \cdot \frac{x^2}{z} \] \[ y = 3 \cdot \frac{10^2}{15} \] \[ y = 3 \cdot \frac{100}{15} \] \[ y = 3 \cdot \frac{20}{3} = 20 \]

So \(y = 20\).

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