Solving Problems with Repeated Proportional Changes
Repeated proportional change happens when a value is increased or decreased by the same percentage more than once. Each change is applied to the current value, not the original one, so the effects compound.
The most efficient way to solve these problems is by using multipliers.
To apply a percentage increase, multiply by a number greater than \(1\).
To apply a percentage decrease, multiply by a number between \(0\) and \(1\).
$$
\text{multiplier for increase} = 1 + \frac{\text{percentage}}{100}
$$
$$
\text{multiplier for decrease} = 1 - \frac{\text{percentage}}{100}
$$
For repeated changes, multiply by the same multiplier each time.
For example, a value of \(200\) increases by 10% each year for 2 years:
$$
\text{multiplier} = 1.1
$$
$$
200 \times 1.1 \times 1.1 = 242
$$
So the final value is \(242\).
If a value decreases repeatedly, the method is the same.
For example, a population of \(5000\) decreases by 8% each year for 3 years:
$$
\text{multiplier} = 0.92
$$
$$
5000 \times 0.92 \times 0.92 \times 0.92 = 3893.44
$$
Repeated changes can also be written using powers.
$$
\text{final value} = \text{original value} \times \text{multiplier}^{\text{number of changes}}
$$
This shows clearly that repeated proportional change is not additive. For example, two 10% increases do not equal a 20% increase, because the second increase acts on a larger value.
Always read the question carefully to check how many times the change is applied and whether it is an increase or a decrease.