What is the easiest way to remember growth and decay multipliers?

For growth, add the rate to $1$: $1 + \frac{r}{100}$. For decay, subtract it from $1$: $1 - \frac{r}{100}$. For example, $7\%$ growth uses $1.07$, while $7\%$ decay uses $0.93$.

Why is repeated percentage change written using a power?

The power records how many times the multiplier is applied. A quantity growing by $4\%$ for five years is calculated using $P \times 1.04^5$.

How do I reverse compound growth or decay?

Divide the final amount by the complete multiplier. If $A$ is the final amount after $n$ periods, the original amount is $\frac{A}{m^n}$.

Should I round after each year in a growth and decay question?

No. Keep the full calculator value throughout and round only the final answer according to the question's instruction. Early rounding can change the final result.

Common Mistakes with Growth and Decay in GCSE Maths

A growth and decay question can look familiar enough to do quickly. That is often where the trouble begins. A student sees a percentage, remembers something about multipliers and enters numbers into a calculator before deciding what those numbers mean.

The most common mistakes with growth and decay in GCSE Maths come from using the wrong multiplier, treating compound change as simple change, counting the periods incorrectly or rounding too soon. The reliable method is to identify the starting amount, convert the percentage into a multiplier, raise that multiplier to the number of periods and round only the final answer.

Imagine a student called Aisha reviewing the marks she lost in a practice paper. Her calculator skills are fine. The real problem is that several small decisions were made too quickly. Here is how she can fix each one.

The growth and decay checklist

Before touching the calculator, write down these six things:

Identify the original amount.
Decide whether the quantity is growing or decaying.
Convert the percentage rate into a decimal multiplier.
Count the number of repeated periods carefully.
Use the multiplier as a power.
Keep the full calculator value and round at the end.

For an original amount $P$ , a decimal multiplier $m$ and $n$ periods, the general calculation is:

\times m^n

For growth at $r%r\%$ , the multiplier is:

\frac{r}{100}

For decay at $r%r\%$ , the multiplier is:

\frac{r}{100}

The Repeated Percentage Change lesson brings together the lesson, questions and supporting revision resources for this method.

Once this checklist is familiar, use the Maths Genie GCSE Maths lessons and GCSE Maths Question Bank to practise recognising the correct setup before calculating.

Mistake: adding the same percentage each time

Aisha sees an investment of $£240\pounds 240$ growing by $5%5\%$ each year for three years. She calculates $5%5\%$ of $£240\pounds 240$ , gets $£12\pounds 12$ and adds it three times:

\times 12 = 276

Why this happens

This treats the question as simple growth. However, repeated percentage change usually means the amount changes after every period. In the second year, the $5%5\%$ is calculated from the new balance, not the original $£240\pounds 240$ .

The correct method

A growth rate of $5%5\%$ gives the multiplier:

\frac{5}{100} = 1.05

There are three years, so raise the multiplier to the power of $3$ :

240 \times 1.05^3 = 277.83

The investment is worth $£277.83\pounds 277.83$ after three years. The power matters because it applies the percentage change to each new balance.

A student discovers that interest has grown into an enormous houseplant

Mistake: subtracting the rate instead of finding the multiplier

The next question says that a machine worth $£850\pounds 850$ depreciates by $12%12\%$ each year for four years. Aisha enters $850 \times 0.12^4$ .

Why this happens

The number $0.12$ represents the proportion being lost. It does not represent the proportion remaining. Decay questions require the amount left after the decrease.

The correct method

If $12%12\%$ is lost, then $88%88\%$ remains:

100%−12%=88%100\% - 12\% = 88\%

Convert $88%88\%$ into a decimal multiplier:

88100=0.88\frac{88}{100} = 0.88

Now apply this multiplier four times:

850 \times 0.88^4 = 509.741056

Rounded to the nearest penny, the machine is worth:

£509.74\pounds 509.74

For more targeted questions, the Compound Interest and Depreciation exam booklet provides structured practice on choosing and applying multipliers.

Mistake: using the number of dates instead of the number of changes

Suppose a population is recorded at the beginning of $2023$ and grows by $4%4\%$ annually. A question asks for its value at the beginning of $2026$ .

Aisha notices four year labels, $2023$ , $2024$ , $2025$ and $2026$ , so she uses a power of $4$ .

Why this happens

Dates are being counted instead of intervals. Growth occurs across the gaps between them.

The correct method

List the changes:

Beginning of $2023$ to beginning of $2024$ : one change.
Beginning of $2024$ to beginning of $2025$ : two changes.
Beginning of $2025$ to beginning of $2026$ : three changes.

The correct power is therefore $3$ . If the initial population is $2{,}500$ , the calculation is:

2500 \times 1.04^3 = 2812.16

If the context requires a whole number of people, interpret the rounding instruction carefully. The calculation and the interpretation are separate decisions.

Mistake: multiplying when the original amount is unknown

Aisha reaches a reverse growth problem: an account contains $£529.20\pounds 529.20$ after increasing by $8%8\%$ . She multiplies by $0.92$ because she wants to go backwards.

Why this happens

It feels natural to undo an $8%8\%$ increase with an $8%8\%$ decrease. But the two percentages would be based on different amounts. Reverse percentages require division by the original multiplier.

The correct method

An $8%8\%$ increase uses a multiplier of:

1.08

Let the original amount be $x$ . Then:

\times 1.08 = 529.20

Divide both sides by $1.08$ :

\frac{529.20}{1.08} = 490

The original amount was $£490\pounds 490$ . The same principle applies after several periods, but you divide by the complete repeated multiplier:

\frac{A}{m^n}

Mistake: assuming equal growth and decay cancel

A price of $£600\pounds 600$ rises by $10%10\%$ and then falls by $10%10\%$ . Aisha expects the answer to return to $£600\pounds 600$ .

Why this happens

The percentages look equal, but they act on different amounts. The decrease is applied after the price has grown.

The correct method

Use the growth multiplier followed by the decay multiplier:

600 \times 1.10 \times 0.90 = 594

The final price is $£594\pounds 594$ , not $£600\pounds 600$ . A quick multiplier check explains why:

1.10 \times 0.90 = 0.99

The combined effect leaves $99%99\%$ of the original amount, which is an overall decrease of $1%1\%$ .

Mistake: rounding during the calculation

Some growth and decay mistakes are not about the method at all. They happen because an intermediate answer is rounded and then used in the next stage.

Why this happens

Money is commonly written to two decimal places, so students may round after every year. That introduces a small error, which can become larger across several periods.

The correct method

Enter the complete calculation in one line:

\times m^n

Keep all the digits shown by the calculator. Round only the final answer, following the question's instruction. For money, this will often be to the nearest penny; for a population, it may be to the nearest whole number.

A quick summary of growth and decay mistakes

Mistake	Better decision
Adding the same percentage amount repeatedly	Use $\times m^n$ for compound change
Using the percentage lost as the decay multiplier	Subtract the rate from $1$
Counting dates rather than intervals	Count how many changes actually occur
Multiplying to find an original value	Divide by the multiplier
Assuming equal increases and decreases cancel	Multiply the two multipliers
Rounding after every period	Round only the final result

The pattern is simple: most lost marks begin before the calculator is used. Write the multiplier and number of periods explicitly, then calculate.

Exam-style practice

Try these without looking back at the examples.

Question one

A car is worth $£18,000\pounds 18{,}000$ and depreciates by $14%14\%$ each year. Find its value after three years, giving your answer to the nearest penny.

18000 \times 0.86^3 = 11448.792

The value is $£11,448.79\pounds 11{,}448.79$ .

Question two

A town has a population of $32{,}000$ . Its population grows by $2.5%2.5\%$ each year. Find the population after five years, giving your answer to the nearest whole number.

32000 \times 1.025^5 = 36205.0625\ldots

The population is approximately $36{,}205$ .

Question three

An investment is worth $£1,331\pounds 1{,}331$ after growing by $10%10\%$ annually for three years. Find its original value.

13311.103=1000\frac{1331}{1.10^3} = 1000

The original investment was $£1,000\pounds 1{,}000$ .

Build speed only after the setup is dependable. The Maths Genie Question Rush is useful when you are ready to make these decisions more quickly, while the GCSE Maths videos can help you revisit a method that still feels uncertain.

Turn corrected mistakes into secure marks

Aisha's errors were not fixed by pressing calculator buttons faster. They were fixed by pausing long enough to identify the multiplier, the periods and whether the question was moving forwards or backwards. Knowing that method is the first step; following it consistently under exam pressure takes focused practice.

Start the matching Maths Genie lesson on repeated percentage change, then apply each correction in the GCSE Maths Question Bank.

The growth and decay checklist

Before touching the calculator, write down these six things:

Identify the original amount.
Decide whether the quantity is growing or decaying.
Convert the percentage rate into a decimal multiplier.
Count the number of repeated periods carefully.
Use the multiplier as a power.
Keep the full calculator value and round at the end.

For an original amount $P$ , a decimal multiplier $m$ and $n$ periods, the general calculation is:

\times m^n

For growth at $r%r\%$ , the multiplier is:

\frac{r}{100}

For decay at $r%r\%$ , the multiplier is:

\frac{r}{100}

The Repeated Percentage Change lesson brings together the lesson, questions and supporting revision resources for this method.

Once this checklist is familiar, use the Maths Genie GCSE Maths lessons and GCSE Maths Question Bank to practise recognising the correct setup before calculating.

Mistake: adding the same percentage each time

Aisha sees an investment of $£240\pounds 240$ growing by $5%5\%$ each year for three years. She calculates $5%5\%$ of $£240\pounds 240$ , gets $£12\pounds 12$ and adds it three times:

\times 12 = 276

Why this happens

The correct method

A growth rate of $5%5\%$ gives the multiplier:

\frac{5}{100} = 1.05

There are three years, so raise the multiplier to the power of $3$ :

240 \times 1.05^3 = 277.83

The investment is worth $£277.83\pounds 277.83$ after three years. The power matters because it applies the percentage change to each new balance.

A student discovers that interest has grown into an enormous houseplant

Mistake: subtracting the rate instead of finding the multiplier

The next question says that a machine worth $£850\pounds 850$ depreciates by $12%12\%$ each year for four years. Aisha enters $850 \times 0.12^4$ .

Why this happens

The number $0.12$ represents the proportion being lost. It does not represent the proportion remaining. Decay questions require the amount left after the decrease.

The correct method

If $12%12\%$ is lost, then $88%88\%$ remains:

100%−12%=88%100\% - 12\% = 88\%

Convert $88%88\%$ into a decimal multiplier:

88100=0.88\frac{88}{100} = 0.88

Now apply this multiplier four times:

850 \times 0.88^4 = 509.741056

Rounded to the nearest penny, the machine is worth:

£509.74\pounds 509.74

For more targeted questions, the Compound Interest and Depreciation exam booklet provides structured practice on choosing and applying multipliers.

Mistake: using the number of dates instead of the number of changes

Suppose a population is recorded at the beginning of $2023$ and grows by $4%4\%$ annually. A question asks for its value at the beginning of $2026$ .

Aisha notices four year labels, $2023$ , $2024$ , $2025$ and $2026$ , so she uses a power of $4$ .

Why this happens

Dates are being counted instead of intervals. Growth occurs across the gaps between them.

The correct method

List the changes:

Beginning of $2023$ to beginning of $2024$ : one change.
Beginning of $2024$ to beginning of $2025$ : two changes.
Beginning of $2025$ to beginning of $2026$ : three changes.

The correct power is therefore $3$ . If the initial population is $2{,}500$ , the calculation is:

2500 \times 1.04^3 = 2812.16

If the context requires a whole number of people, interpret the rounding instruction carefully. The calculation and the interpretation are separate decisions.

Mistake: multiplying when the original amount is unknown

Aisha reaches a reverse growth problem: an account contains $£529.20\pounds 529.20$ after increasing by $8%8\%$ . She multiplies by $0.92$ because she wants to go backwards.

Why this happens

It feels natural to undo an $8%8\%$ increase with an $8%8\%$ decrease. But the two percentages would be based on different amounts. Reverse percentages require division by the original multiplier.

The correct method

An $8%8\%$ increase uses a multiplier of:

1.08

Let the original amount be $x$ . Then:

\times 1.08 = 529.20

Divide both sides by $1.08$ :

\frac{529.20}{1.08} = 490

The original amount was $£490\pounds 490$ . The same principle applies after several periods, but you divide by the complete repeated multiplier:

\frac{A}{m^n}

Mistake: assuming equal growth and decay cancel

A price of $£600\pounds 600$ rises by $10%10\%$ and then falls by $10%10\%$ . Aisha expects the answer to return to $£600\pounds 600$ .

Why this happens

The percentages look equal, but they act on different amounts. The decrease is applied after the price has grown.

The correct method

Use the growth multiplier followed by the decay multiplier:

600 \times 1.10 \times 0.90 = 594

The final price is $£594\pounds 594$ , not $£600\pounds 600$ . A quick multiplier check explains why:

1.10 \times 0.90 = 0.99

The combined effect leaves $99%99\%$ of the original amount, which is an overall decrease of $1%1\%$ .

Mistake: rounding during the calculation

Some growth and decay mistakes are not about the method at all. They happen because an intermediate answer is rounded and then used in the next stage.

Why this happens

Money is commonly written to two decimal places, so students may round after every year. That introduces a small error, which can become larger across several periods.

The correct method

Enter the complete calculation in one line:

\times m^n

A quick summary of growth and decay mistakes

Mistake	Better decision
Adding the same percentage amount repeatedly	Use $\times m^n$ for compound change
Using the percentage lost as the decay multiplier	Subtract the rate from $1$
Counting dates rather than intervals	Count how many changes actually occur
Multiplying to find an original value	Divide by the multiplier
Assuming equal increases and decreases cancel	Multiply the two multipliers
Rounding after every period	Round only the final result

The pattern is simple: most lost marks begin before the calculator is used. Write the multiplier and number of periods explicitly, then calculate.

Exam-style practice

Try these without looking back at the examples.

Question one

A car is worth $£18,000\pounds 18{,}000$ and depreciates by $14%14\%$ each year. Find its value after three years, giving your answer to the nearest penny.

18000 \times 0.86^3 = 11448.792

The value is $£11,448.79\pounds 11{,}448.79$ .

Question two

A town has a population of $32{,}000$ . Its population grows by $2.5%2.5\%$ each year. Find the population after five years, giving your answer to the nearest whole number.

32000 \times 1.025^5 = 36205.0625\ldots

The population is approximately $36{,}205$ .

Question three

An investment is worth $£1,331\pounds 1{,}331$ after growing by $10%10\%$ annually for three years. Find its original value.

13311.103=1000\frac{1331}{1.10^3} = 1000

The original investment was $£1,000\pounds 1{,}000$ .

Turn corrected mistakes into secure marks

Start the matching Maths Genie lesson on repeated percentage change, then apply each correction in the GCSE Maths Question Bank.

The growth and decay checklist

Mistake: adding the same percentage each time

Why this happens

The correct method

Mistake: subtracting the rate instead of finding the multiplier

Why this happens

The correct method

Mistake: using the number of dates instead of the number of changes

Why this happens

The correct method

Mistake: multiplying when the original amount is unknown

Why this happens

The correct method

Mistake: assuming equal growth and decay cancel

Why this happens

The correct method

Mistake: rounding during the calculation

Why this happens

The correct method

A quick summary of growth and decay mistakes

Exam-style practice

Question one

Question two

Question three

Turn corrected mistakes into secure marks

Frequently Asked Questions

What is the easiest way to remember growth and decay multipliers?

What is the easiest way to remember growth and decay multipliers?

Why is repeated percentage change written using a power?

Why is repeated percentage change written using a power?

How do I reverse compound growth or decay?

How do I reverse compound growth or decay?

Should I round after each year in a growth and decay question?

Should I round after each year in a growth and decay question?

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The growth and decay checklist

Mistake: adding the same percentage each time

Why this happens

The correct method

Mistake: subtracting the rate instead of finding the multiplier

Why this happens

The correct method

Mistake: using the number of dates instead of the number of changes

Why this happens

The correct method

Mistake: multiplying when the original amount is unknown

Why this happens

The correct method

Mistake: assuming equal growth and decay cancel

Why this happens

The correct method

Mistake: rounding during the calculation

Why this happens

The correct method

A quick summary of growth and decay mistakes

Exam-style practice

Question one

Question two

Question three

Turn corrected mistakes into secure marks

Ready to breeze through your GCSEs?

Ready to breeze through your GCSEs?