E1.13, E1.17 - Reverse percentages and repeated change | Cambridge IGCSE Mathematics

A percentage change uses the value at the start of that change. You will recover an original amount, then model repeated growth and decay, including population change and depreciation. These Extended methods connect reverse percentages in E1.13 with exponential change in E1.17.

Recovering an original value

A shop reduces a price by $20\%$ , then increases the reduced price by $20\%$ . Does the price return to where it started? The second percentage acts on a different amount. Keep that distinction in mind when you try to undo a change.

You already know that a $20\%$ decrease multiplies by $0.8$ , while a $20\%$ increase multiplies by $1.2$ . Multiplication and division undo each other. To recover the original value, divide the final value by the multiplier that produced it. This requires a nonzero multiplier. A $100\%$ reduction leaves zero whatever the original amount was, so that final zero cannot uniquely reveal the original.

Reverse percentage

A reverse percentage calculation finds the original amount when you know the amount after a stated percentage change. For example, a price after a $20\%$ discount represents $80\%$ of its original price, not $100\%$ .

Write down what percentage of the original you have before choosing a calculation. For a profit percentage, check its base: a selling price after a $20\%$ profit on cost is $120\%$ of the cost price.

Worked example 1: undoing a discount

Question. A bag costs $\$ 96 $after a$ 20%$ discount. Find its original price.

Plan. Recognise $\$ 96 $as$ 80% $of the original, then divide by$ 0.8$.

\begin{aligned} 100-20&=80 && \text{percentage remaining}\\ 96\div0.8&=960\div8 && \text{scale both numbers by ten}\\ &=120 && \text{recover the whole amount} \end{aligned}

Answer. The original price was $\$ 120$.

Check. $20\%$ of $120$ is $24$ , and $120-24=96$ .

Keep the final price and percentage the same, but change a decrease to an increase. The final price now represents more than the whole original amount.

Worked example 2: undoing a profit increase

Question. A trader sells a bag for $\$ 96 $, making a$ 20%$ profit on its cost price. Find the cost price.

Plan. Recognise the selling price as $120\%$ of cost, then divide by $1.2$ .

\begin{aligned} 100+20&=120 && \text{percentage of cost received}\\ 96\div1.2&=960\div12 && \text{scale both numbers by ten}\\ &=80 && \text{recover the cost} \end{aligned}

Answer. The cost price was $\$ 80$.

Check. $20\%$ of $80$ is $16$ ; $80+16=96$ . The original is below the increased price, as it should be.

Check the direction of the original change before trying to reverse it.

Watch out

$96\times1.2$ does not undo a $20\%$ discount. It adds $20\%$ of $96$ , whereas the discount removed $20\%$ of the original. Undo $\times0.8$ with $\div0.8$ .

Use the final percentage of the original as your starting point in this check.

Repeated growth

Repeated percentage growth applies the same multiplier to the new amount each time. A population model that grows by $10\%$ each year adds more people in its second year because its starting population is larger.

For example, $2000$ grows first to $2200$ . The next $10\%$ is $220$ , rather than $200$ . This is the same compounding idea used for compound interest, applied to other quantities.

Repeated percentage growth

For an initial amount $A$ , an increase of $r\%$ each period gives

\text{amount after }n\text{ periods}=A\left(1+\frac{r}{100}\right)^n

Here $n$ is the number of completed periods. The rate and period must stay the same. The power means repeated multiplication; when $n=0$ , the amount is $A$ .

Not in the List of formulas: learn the multiplier method.

The formula describes a model. A forecast population is not guaranteed to follow a fixed percentage every year. If a whole-number population is requested, round the final forecast, keeping the intermediate values unrounded.

Worked example 3: two years of growth

Question. A population is $2000$ . A model predicts growth of $10\%$ each year. Find the population after two years.

Plan. Multiply by $1.1$ once for each year.

\begin{aligned} 2000\times1.1&=2200 && \text{after one year}\\ 2200\times1.1&=2420 && \text{after two years}\\ 2000\times1.1^2&=2420 && \text{write repeated multiplication compactly} \end{aligned}

Answer. The model predicts $2420$ people.

Check. The increases are $200$ and $220$ . Adding them to $2000$ gives $2420$ .

Change only the number of years. A third year needs a third multiplication, not another $10\%$ of the initial population.

Worked example 4: one extra period

Question. Use the same model to find the population after three years.

Plan. Take the two-year population and apply the same multiplier once more.

\begin{aligned} 2420\times1.1&=2420+242 && \text{add ten per cent of the new amount}\\ &=2662 && \text{after three years}\\ 2000\times1.1^3&=2662 && \text{three growth periods} \end{aligned}

Answer. The model predicts $2662$ people.

Check. The third increase is $242$ , larger than $220$ . Also, $2662\div1.1=2420$ , reversing the last year.

The increasing base also changes the overall percentage gain.

Watch out

$10\%+10\%+10\%=30\%$ does not describe three years of compound growth. Here the overall increase is $662\div2000\times100=33.1\%$ , because each increase has a new base.

Keep the rate and number of periods, but try a different initial population.

Repeated decay and depreciation

Repeated percentage decay removes the same fraction of the current amount each period. Depreciation is a fall in an asset's value. A machine losing $10\%$ of its value each year loses a smaller amount of money each successive year.

A decrease of $r\%$ leaves the fraction $1-\frac r{100}$ . For ordinary decay with $0<r<100$ , this multiplier lies strictly between $0$ and $1$ . Starting with a positive amount gives a smaller positive amount after every period. The unrounded model never reaches exactly zero in a finite number of periods. A $100\%$ decrease is the different boundary case that leaves zero immediately.

Worked example 5: two years of depreciation

Question. A machine is worth $\$ 800 $. It depreciates by$ 10%$ of its current value each year. Find its value after two years.

Plan. Retain $90\%$ twice, using the multiplier $0.9$ .

\begin{aligned} 800\times0.9&=720 && \text{value after one year}\\ 720\times0.9&=648 && \text{value after two years}\\ 800\times0.9^2&=648 && \text{two decay periods} \end{aligned}

Answer. The value is $\$ 648$.

Check. The losses are $80$ and $72$ . Thus $800-80-72=648$ ; the second loss is smaller.

Now extend the model by one year. Only the number of periods changes.

Worked example 6: one extra year of decay

Question. Find the value of the same machine after three years.

Plan. Multiply the two-year value by $0.9$ once more.

\begin{aligned} 648\times0.9&=648-64.8 && \text{lose ten per cent of the current value}\\ &=583.2 && \text{value after three years}\\ 800\times0.9^3&=583.2 && \text{three decay periods} \end{aligned}

Answer. The value is $\$ 583.20$.

Check. $583.2\div0.9=648$ , reversing the last year. The value is positive and below $648$ .

For changing rates, use one multiplier per change instead of one repeated power. This resolves the opening puzzle: a $20\%$ decrease followed by a $20\%$ increase gives $0.8\times1.2=0.96$ . The price finishes at $96\%$ of its original value, so it is $4\%$ lower.

Watch out

Writing $800(1-0.3)=560$ for three annual $10\%$ decreases treats every loss as $10\%$ of $800$ . The correct model is $800\times0.9^3$ , because each loss uses the current value.

A larger decay rate still follows the same multiplier reasoning.

Quick Recap

To recover an original amount, divide by its percentage-change multiplier.
Repeated growth multiplies by $1+\frac r{100}$ each period.
Repeated decay multiplies by $1-\frac r{100}$ each period; the change in amount usually gets smaller.
Use the number of completed periods as the power, and round only the final answer when required.

A $20\%$ reduction followed by a $20\%$ increase leaves $96\%$ of the starting price, so the two changes do not cancel.

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E1.13, E1.17 - Reverse percentages and repeated change

Recovering an original value

Repeated growth

Repeated decay and depreciation

Quick Recap

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