Simple and Compound Interest · Study Notes · Edexcel IGCSE 4MA1 Higher

How money grows over time. Simple = linear. Compound = exponential. Master both for Paper 2H, plus depreciation as the 'compound decrease' twin.

At a glance

Simple interest: $I = \dfrac{Prt}{100}$ — linear, on principal only.
Compound interest: $A = P(1 + r/100)^{n}$ — interest earns interest.
Depreciation: $A = P(1 - r/100)^{n}$ — same shape, decreasing.
Multiplier method is the cleanest for compound problems.
Compound exceeds simple after year 1; gap grows over time.
Reverse problems use cube/ $n$ th root.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.13 — Calculate simple interest.
1.13 — Calculate compound interest.
1.13 — Calculate compound depreciation.
1.13 — Use the multiplier method consistently.

Simple interest

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Interest on the original principal only. Linear growth.

Definition. Simple interest is calculated on the ORIGINAL principal, regardless of accumulated interest. Same amount each year.

Formula:

$I = \dfrac{P \cdot r \cdot t}{100}$

where:

$P$ = principal (initial amount).
$r$ = annual interest rate as a NUMBER (5 for 5%).
$t$ = time in years.
$I$ = interest earned.

Total amount in account: $A = P + I$ .

Examples:

£ $1000$ at $5\%$ for $4$ years:

$I = \dfrac{1000 \times 5 \times 4}{100} = £200$ .
Total: £ $1200$ .

£ $3000$ at $4\%$ for $5$ years:

$I = \dfrac{3000 \times 4 \times 5}{100} = £600$ .
Total: £ $3600$ .

When is simple interest used?

Some short-term loans.
Some bonds.
Less common in modern banking — most savings accounts and mortgages are compound.

Worked qualitative. A bond pays $6\%$ simple interest per year on £ $1000$ . After how many years has the interest equalled the principal?

Set $I = P = 1000$ .
$1000 = 1000 \times 6 \times t / 100$ .
$1000 = 60t$ , so $t = 16.67$ years.
About $16$ years $8$ months.

Edexcel tip. Always check if 'simple' or 'compound' is specified. Simple is typically Foundation Tier; Higher Tier focuses on compound.

$I = Prt/100$ .
$r$ is the percentage NUMBER (5, not 0.05).
Same interest each year.
Used for some bonds and short-term loans.

Compound interest

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Interest added to principal, then earns its own interest. Exponential growth.

Definition. Compound interest is calculated on the principal PLUS all previously earned interest. Money grows exponentially.

Formula:

$A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$

where:

$A$ = final amount (principal + interest).
$P$ = principal.
$r$ = annual rate.
$n$ = number of years.

The multiplier: $1 + r/100$ tells you how much money grows each year.

Examples:

£ $5000$ at $4\%$ for $3$ years:

Multiplier: $1.04$ .
$A = 5000 \times 1.04^{3} = 5000 \times 1.124864 = £5624.32$ .

£ $10{,}000$ at $7\%$ for $5$ years:

$A = 10000 \times 1.07^{5} = 10000 \times 1.402552 = £14{,}025.52$ .

The 'rule of 72' (rough mental check):

Money roughly DOUBLES in $72/r$ years.
$5\%$ : doubles in ~ $14.4$ years.
$10\%$ : doubles in ~ $7.2$ years.

Compound vs simple — the gap.

£ $1000$ at $5\%$ for $10$ years:

Simple: $I = 1000 \times 5 \times 10 / 100 = £500$ . Total £ $1500$ .
Compound: $A = 1000 \times 1.05^{10} = £1628.89$ . Interest £ $628.89$ .
Compound exceeds simple by £ $128.89$ .

For longer periods or higher rates, the gap is dramatic.

Simple interest adds a fixed amount each year; compound interest curves upward as interest earns interest.

Worked qualitative. Why is compound so much better than simple over time?

Simple grows by $r\%$ of $P$ EACH year — constant.
Compound grows by $r\%$ of the current balance — which itself grows.
After $1$ year: same. After $2$ years: compound has earned interest on year-1 interest. Gap widens each year.
Famous: 'Compound interest is the eighth wonder of the world' — Albert Einstein (probably misattributed but memorable).

Edexcel tip. Always state the formula. $A = P(1 + r/100)^{n}$ . Show the substitution. Compute step by step. Round only at the end.

$A = P(1 + r/100)^{n}$ .
Multiplier each year.
Exceeds simple after year 1.
Doubles in roughly $72/r$ years.

Compound depreciation

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Same formula, with $1 - r/100$ as multiplier.

Definition. Compound depreciation: value decreases by a percentage each year. Same exponential structure as compound interest, but downward.

Formula:

$A = P \times \left(1 - \dfrac{r}{100}\right)^{n}$

Common contexts:

Cars losing value.
Equipment writing off.
Antiques (sometimes).
Buildings (sometimes).

Each year the value is multiplied by 0.85, so the curve falls steeply then flattens towards zero.

Example. Car £ $18000$ , $15\%$ depreciation/year, after $4$ years:

Multiplier: $1 - 0.15 = 0.85$ .
$A = 18000 \times 0.85^{4} = 18000 \times 0.522006 = £9396.10$ .

Difference from simple depreciation.

Simple depreciation (rare in real world): same amount each year. £ $18000$ at £ $2700$ /year for $4$ years would lose $4 \times 2700 = £10{,}800$ , leaving £ $7200$ . Different number!

Compound is more realistic — assets typically lose proportionally less as they age.

Worked qualitative. Why does compound depreciation feel slower than expected?

After year 1: car worth £ $15300$ .
After year 2: $15300 \times 0.85 = £13005$ .
The decrease is on a smaller amount each time.
Asymptotically approaches zero but never reaches it (mathematically).

Edexcel tip. Same setup as compound interest, just $1 - r/100$ instead of $1 + r/100$ . Identify whether the problem is growth or decline.

$A = P(1 - r/100)^{n}$ .
Multiplier slightly less than 1.
Common: cars, equipment.
Asymptotically approaches 0.

Reverse problems — find rate or time

▼

Use logs (or calculator) to back out $r$ or $n$ .

Reverse problem 1: Find $r$ given $A, P, n$ .

Set up: $A = P \times (1 + r/100)^{n}$ .

Solve:

$\dfrac{A}{P} = (1 + r/100)^{n}$ .
$\sqrt[n]{A/P} = 1 + r/100$ .
$r = 100 \times (\sqrt[n]{A/P} - 1)$ .

Example. £ $2000$ → £ $2315.25$ in $3$ years. Find rate.

$\dfrac{2315.25}{2000} = 1.157625$ .
Cube root: $1.157625^{1/3} = 1.05$ .
$r = 100 \times 0.05 = 5\%$ .

Reverse problem 2: Find $n$ given $A, P, r$ .

Set up: $A = P \times (1 + r/100)^{n}$ .

Solve using logs:

$A/P = (1 + r/100)^{n}$ .
$\log(A/P) = n \times \log(1 + r/100)$ .
$n = \log(A/P) / \log(1 + r/100)$ .

Example. £ $1000$ at $5\%$ until reaches £ $1500$ . Find $n$ .

$1500/1000 = 1.5 = 1.05^{n}$ .
$\log 1.5 = n \log 1.05$ .
$n = \log 1.5 / \log 1.05 \approx 0.176 / 0.0212 \approx 8.31$ years.
So takes about $9$ years.

Worked qualitative. Why might Edexcel give 'how many full years?' rather than exact $n$ ?

Logs give non-integer answers.
Real-world banks compound at year-end → only whole years matter.
'After how many years is value at least £ $X$ ?' rounds up.

Edexcel tip. For 'find $r$ ' use cube/ $n$ th root. For 'find $n$ ' use logs. Both are calculator skills on Paper 2H.

Find $r$ : take $n$ th root.
Find $n$ : use logs.
Round time UP to nearest year for 'reaches' problems.
Calculator paper only.

Quick recap

Simple: $I = Prt/100$ , linear.
Compound: $A = P(1 + r/100)^{n}$ , exponential.
Depreciation: $A = P(1 - r/100)^{n}$ .
Compound exceeds simple after year 1.
Reverse for $r$ : $n$ th root. Reverse for $n$ : logs.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Simple interest: $I = Prt/100$ .
Compound interest: $A = P(1 + r/100)^{n}$ .
Depreciation: $A = P(1 - r/100)^{n}$ .
Multiplier: $1 + r/100$ for growth, $1 - r/100$ for decline.
Rule of 72: doubles in $\sim 72/r$ years.

How it’s examined

Interest appears every Paper 2H (calculator). Typical: compound interest for $n$ years (3-4 marks), depreciation (3 marks), reverse problem to find rate or years (4-5 marks). Examiner reports flag (1) using simple formula for compound problem, (2) wrong sign in depreciation multiplier, (3) rounding too early.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024, 4MA1/2H Jun 2023. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on simple and compound interest, written exactly the way a tutor would explain them at the board.

1Simple interest calculation
Foundation• simple interest
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Question
£ $3{,}000$ is invested at $4\%$ simple interest per year. How much interest is earned after $5$ years?
Step-by-step solution
1. Step 1
  Simple interest formula: $I = \dfrac{Prt}{100}$ .
2. Step 2
  Substitute: $I = \dfrac{3000 \times 4 \times 5}{100} = \dfrac{60{,}000}{100} = £600$ .
Answer
£ $600$
Examiner tip
Simple interest is LINEAR — same amount each year. The total amount in the account would be principal + interest = £ $3{,}600$ .
2Compound interest calculation
Higher• Adapted from 4MA1/2H Jan 2024 Q9• compound interest
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Question
£ $5{,}000$ is invested at $4\%$ compound interest per year. How much is in the account after $3$ years?
Step-by-step solution
1. Step 1
  Compound interest: $A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$ .
2. Step 2
  Substitute: $A = 5000 \times 1.04^{3}$ .
3. Step 3
  Compute: $1.04^{3} = 1.124864$ . So $A = 5000 \times 1.124864 = £5{,}624.32$ .
Answer
£ $5{,}624.32$
Examiner tip
Compound interest is EXPONENTIAL — interest earns interest. Always use the multiplier $1 + r/100$ , raised to the power of years.
3Compare simple vs compound interest
Higher• compare
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Question
£ $1{,}000$ is invested at $5\%$ per year. Calculate (a) the SIMPLE interest after $10$ years, (b) the value of the investment with COMPOUND interest after $10$ years. By how much does compound exceed simple?
Step-by-step solution
1. Step 1
  (a) Simple: $I = \dfrac{1000 \times 5 \times 10}{100} = £500$ . Total: £ $1{,}500$ .
2. Step 2
  (b) Compound: $A = 1000 \times 1.05^{10} = 1000 \times 1.62889 \approx £1{,}628.89$ . Interest earned: £ $628.89$ .
3. Step 3
  Compound exceeds simple by $1628.89 - 1500 = £128.89$ .
Answer
Simple: £ $500$ . Compound: £ $628.89$ interest. Difference: £ $128.89$ .
Examiner tip
Compound ALWAYS exceeds simple after the first year. The longer the investment, the bigger the gap (exponential vs linear).
4Compound depreciation
Higher• depreciation
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Question
A car worth £ $18{,}000$ depreciates by $15\%$ per year. What is its value after $4$ years?
Step-by-step solution
1. Step 1
  Depreciation: $A = P \times \left(1 - \dfrac{r}{100}\right)^{n}$ .
2. Step 2
  Substitute: $A = 18000 \times 0.85^{4}$ .
3. Step 3
  Compute: $0.85^{4} = 0.522006\ldots$ . So $A = 18000 \times 0.522 = £9{,}396$ (to nearest £).
Answer
£ $9{,}396$
Examiner tip
Depreciation uses $(1 - r/100)$ instead of $(1 + r/100)$ . Same formula structure.
5Find the interest rate
Higher• Adapted from 4MA1/2H Jun 2023 Q14• find rate
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Question
£ $2{,}000$ is invested for $3$ years at compound interest. The investment is worth £ $2{,}315.25$ at the end. Find the annual rate.
Step-by-step solution
1. Step 1
  Set up: $2315.25 = 2000 \times \left(1 + \dfrac{r}{100}\right)^{3}$ .
2. Step 2
  Divide: $\left(1 + \dfrac{r}{100}\right)^{3} = \dfrac{2315.25}{2000} = 1.157625$ .
3. Step 3
  Cube root: $1 + \dfrac{r}{100} = 1.157625^{1/3} = 1.05$ .
4. Step 4
  Solve: $r/100 = 0.05$ , so $r = 5\%$ .
Answer
$5\%$
Examiner tip
Reverse-engineering the rate uses the cube/ $n$ th root. On Paper 2H this is calculator-allowed.
6Reading the rate from a multiplier [1 mark]
Getting started• Paper 1H/2H style — 1 mark• depreciation, multiplier
▼
Question
The value of a painting is multiplied by $0.94$ at the end of every year.

Write down the percentage decrease in its value each year. [1]
Step-by-step solution
1. Step 1
  B1 A multiplier of $0.94$ keeps $94\%$ of the value, so $100\% - 94\% = 6\%$ is lost each year.
  $(1 - 0.94) \times 100 = 6$
Answer
$6\%$
Examiner tip
Examiner tip: A multiplier below $1$ is a decrease; the decrease is the gap between the multiplier and $1$ .

Common mistake: Answering $94\%$ , which is the percentage of the value that remains, not the percentage lost.
7Finding the amount invested from simple interest [2 marks]
Getting started• Paper 1H/2H style — 2 marks• simple interest, reverse
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Question
Ravi invests some money at $4.5\%$ per year simple interest. After $3$ years he has earned £ $378$ in interest.

Work out how much money Ravi invested. [2]
Step-by-step solution
1. Step 1
  M1 Three years of simple interest is $3 \times 4.5\% = 13.5\%$ of the amount invested. So £ $378$ is $13.5\%$ of it.
  $378 \div (0.045 \times 3)$
2. Step 2
  A1 Evaluate.
  $378 \div 0.135 = 2800$
Answer
£ $2800$
Examiner tip
Examiner tip: Check it: $2800 \times 0.045 \times 3 = 378$ .

Common mistake: Working out $4.5\%$ of £ $378$ and adding it on. The £ $378$ is the interest, not the amount invested, so you must divide by $0.135$ .
8Overall change after three successive changes [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• repeated percentage change, multiplier
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Question
The price of a bicycle is increased by $15\%$ . In a sale, the new price is reduced by $20\%$ . After the sale, the price is increased by $10\%$ .

Work out the overall percentage change in the price of the bicycle. [3]
Step-by-step solution
1. Step 1
  M1 Write each change as a multiplier: $1.15$ , $0.8$ and $1.1$ .
2. Step 2
  M1 Multiply them to get one overall multiplier.
  $1.15 \times 0.8 \times 1.1 = 1.012$
3. Step 3
  A1 $1.012$ is greater than $1$ , so this is an increase.
  $(1.012 - 1) \times 100 = 1.2$
Answer
An increase of $1.2\%$
Examiner tip
Examiner tip: Say whether it is an increase or a decrease; a bare " $1.2$ " leaves the marker to guess.

Common mistake: Adding and subtracting the percentages: $15 - 20 + 10 = 5\%$ . Each change acts on a different price, so the multipliers must be multiplied.
9Compound interest on a loan with a repayment [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• compound interest, loan
▼
Question
Sam borrows £ $5000$ . Interest of $6\%$ per year is added to the amount owed at the end of each year.

At the end of the first year, after the interest is added, Sam pays back £ $1500$ .

Work out how much Sam owes at the end of the second year, after the interest is added. [4]
Step-by-step solution
1. Step 1
  M1 Add the first year of interest.
  $5000 \times 1.06 = 5300$
2. Step 2
  M1 Take off the repayment. Only this smaller amount earns interest in year $2$ .
  $5300 - 1500 = 3800$
3. Step 3
  M1 Add the second year of interest.
  $3800 \times 1.06$
4. Step 4
  A1 Evaluate.
  $= 4028$
Answer
£ $4028$
Examiner tip
Examiner tip: Work one year at a time whenever money goes in or out. A single power such as $1.06^{2}$ only works when nothing changes in between.

Common mistake: Writing $5000 \times 1.06^{2} - 1500 = 4118$ , which charges interest on the £ $1500$ after Sam has already paid it back.
10How long for an investment to double [5 marks]
Stretch• Paper 1H/2H style — 5 marks• compound interest, number of years
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Question
An investment earns $6.5\%$ per year compound interest.

(a) Explain why the number of years it takes for the investment to double does not depend on the amount invested. [1]

(b) Find the least number of whole years it takes for the investment to at least double in value. [4]
Step-by-step solution
1. Step 1
  B1 (a) Doubling means the amount is multiplied by $2$ . Since $P \times 1.065^{n} = 2P$ gives $1.065^{n} = 2$ , the amount $P$ cancels.
2. Step 2
  M1 (b) Find the least $n$ for which $1.065^{n} \geqslant 2$ . Try a sensible value.
  $1.065^{10} = 1.877\ldots$
3. Step 3
  M1 Still less than $2$ ; try the next year.
  $1.065^{11} = 1.999\,15\ldots$
4. Step 4
  A1 Just below $2$ after $11$ years (write enough decimal places to show it). After $12$ years it is above.
  $1.065^{12} = 2.129\ldots$
5. Step 5
  A1 So the investment first doubles after $12$ years.
Answer
(a) $P \times 1.065^{n} = 2P$ so $P$ cancels (b) $12$ years
Examiner tip
Examiner tip: The year- $11$ value is $1.999\,15\ldots$ ; if you round it to $2.00$ you will wrongly conclude $11$ . Write at least $4$ decimal places near the target.

Common mistake: Using simple interest, $100 \div 6.5 = 15.4$ , so $16$ years. Compound growth doubles much sooner than that.

Question 1
Paper 1H/2H style — 1 mark1 mark
Anna invests € $12\,000$ at $2.5\%$ per year simple interest.

Write down the interest she earns each year. [1]
Model answer
$12\,000 \times 0.025 = 300$

€ $300$
Why this scores
Mark scheme:
- B1 for $300$ cao
Why this scores full marks: Simple interest is the same every year, so one year's interest is simply $2.5\%$ of the amount invested. The candidate gives the interest, not the new balance of € $12\,300$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
Leo invests € $5000$ at a fixed rate of simple interest. After $4$ years the account holds € $5720$

Work out the rate of interest per year. [2]
Model answer
Total interest $= 5720 - 5000 = 720$

Interest per year $= 720 \div 4 = 180$

$\frac{180}{5000} \times 100 = 3.6$

$3.6\%$
Why this scores
Mark scheme:
- M1 for $\frac{5720 - 5000}{4 \times 5000}$ oe, or $180 \div 5000$
- A1 for $3.6$ cao
Why this scores full marks: The candidate finds the interest, splits it evenly over the $4$ years (because it is simple interest) and expresses one year's interest as a percentage of the original amount.

Where marks go: Taking the fourth root of $\frac{5720}{5000}$ gives $3.42\%$ , the compound rate, and scores nothing here.
Question 3
Paper 1H/2H style — 2 marks2 marks
The value of a house increases by $3.8\%$ per year for $2$ years.

Calculate the total percentage increase in the value of the house over the $2$ years. Give your answer correct to $3$ significant figures. [2]
Model answer
$1.038^{2} = 1.077\,444$

$(1.077\,444 - 1) \times 100 = 7.7444$

$7.74\%$
Why this scores
Mark scheme:
- M1 for $1.038^{2}$ oe
- A1 for $7.74$ (awrt $7.74$ )
Why this scores full marks: No house price is given, so the candidate works with the multiplier alone and converts it to a percentage increase, rounding only at the end.

Where marks go: $2 \times 3.8 = 7.6\%$ ignores the interest-on-interest; giving $107.7\%$ answers "what percentage of the original" rather than the increase.
Question 4
Paper 1H/2H style — 3 marks3 marks
A bike loses $12\%$ of its value each year. After $2$ years it is worth £ $348.48$

Work out the value of the bike when it was new. [3]
Model answer
$0.88^{2} = 0.7744$

$348.48 \div 0.7744 = 450$

£ $450$
Why this scores
Mark scheme:
- M1 for $0.88^{2}$ ( $= 0.7744$ ) or $0.88$ used as a multiplier
- M1 for $348.48 \div 0.7744$ oe, e.g. $348.48 \div 0.88 \div 0.88$
- A1 for $450$ cao
Why this scores full marks: The candidate reverses the depreciation by dividing by the multiplier, twice, and can check the answer: $450 \times 0.88^{2} = 348.48$ .

Where marks go: Multiplying by $1.12^{2}$ gives £ $437.13$ . Adding $12\%$ back on is not the reverse of taking $12\%$ off, the same error examiners report on reverse percentages.
Question 5
Paper 1H/2H style — 3 marks3 marks
The price of a coat is increased by $x\%$ . The new price is then decreased by $10\%$ . Overall, the price has increased by $3.5\%$

Find the value of $x$ . [3]
Model answer
$\left(1 + \frac{x}{100}\right) \times 0.9 = 1.035$

$1 + \frac{x}{100} = 1.035 \div 0.9 = 1.15$

$x = 15$
Why this scores
Mark scheme:
- M1 for $\left(1 + \frac{x}{100}\right) \times 0.9 = 1.035$ oe, or $1.035$ and $0.9$ both seen as multipliers
- M1 for $1.035 \div 0.9$ ( $= 1.15$ )
- A1 for $15$ cao
Why this scores full marks: The candidate writes the overall change as a product of multipliers, then divides to undo the $10\%$ decrease.

Where marks go: $3.5 + 10 = 13.5$ adds percentages that act on different amounts; $1.035 \times 1.1$ (multiplying where dividing is needed) gives $13.85$ .
Question 6
Paper 1H/2H style — 4 marks4 marks
Mia has £ $10\,000$ to save for $3$ years.

Account A pays $3\%$ per year compound interest.

Account B pays $5\%$ compound interest in the first year and $2\%$ per year compound interest in each of the next $2$ years.

Calculate which account gives more money at the end of the $3$ years, and by how much. [4]
Model answer
Account A:

$10\,000 \times 1.03^{3} = 10\,927.27$

Account B:

$10\,000 \times 1.05 \times 1.02^{2} = 10\,924.20$

$10\,927.27 - 10\,924.20 = 3.07$

Account A gives £ $3.07$ more.
Why this scores
Mark scheme:
- M1 for $10\,000 \times 1.03^{3}$ oe
- M1 for $10\,000 \times 1.05 \times 1.02^{2}$ oe
- A1 for $10\,927.27$ and $10\,924.2(0)$
- A1 for Account A by £ $3.07$
Why this scores full marks: Both final amounts are calculated in full, then compared, and the conclusion names the account and the difference.

Where marks go: Adding the rates, $5 + 2 + 2 = 9\%$ against $3 \times 3 = 9\%$ , and saying the accounts are the same. With figures this close, only the full calculation separates them.
Question 7
Paper 1H/2H style — 4 marks4 marks
Carl invests some money at a fixed rate of compound interest. He makes no other payments into or out of the account.

At the end of year $2$ the account holds £ $6724$ . At the end of year $3$ it holds £ $6892.10$

(a) Work out the rate of interest. [2]

(b) Work out the amount Carl invested. [2]
Model answer
(a)

$6892.10 \div 6724 = 1.025$

$2.5\%$

(b)

$6724 \div 1.025^{2} = 6400$

£ $6400$
Why this scores
Mark scheme:
- M1 for $6892.10 \div 6724$ oe
- A1 for $2.5$ cao
- M1 for $6724 \div 1.025^{2}$ ft their rate, or $6892.10 \div 1.025^{3}$
- A1 for $6400$ cao
Why this scores full marks: One year of growth links the two balances, so their ratio is the multiplier. The candidate then divides by it twice to go back two years.

Where marks go: Giving the difference, £ $168.10$ , as the answer to (a); it must be divided by $6724$ to become a rate. In (b), multiplying by $0.975^{2}$ instead of dividing by $1.025^{2}$ gives £ $6392.00$ .
Question 8
Paper 1H/2H style — 5 marks5 marks
Gabi and Hugo each invest € $4000$ .

Gabi's account pays $6\%$ per year simple interest.

Hugo's account pays $5\%$ per year compound interest.

Find the first year at the end of which Hugo's account holds more money than Gabi's. [5]
Model answer
After $n$ years: Gabi $= 4000 + 240n$ , Hugo $= 4000 \times 1.05^{n}$

$n = 8$ : Gabi $= 5920$ , Hugo $= 5909.82\ldots$ (Gabi more)

$n = 9$ : Gabi $= 6160$ , Hugo $= 6205.31\ldots$ (Hugo more)

Year $9$
Why this scores
Mark scheme:
- M1 for $4000 \times 0.06 = 240$ per year, or $4000 + 240n$ oe
- M1 for $4000 \times 1.05^{n}$ for any $n \geqslant 2$
- M1 for both amounts at $n = 8$ ( $5920$ and $5909.8\ldots$ )
- M1 for both amounts at $n = 9$ ( $6160$ and $6205.3\ldots$ )
- A1 for $9$ dep on all previous M marks
Why this scores full marks: The candidate writes a rule for each account, then tests the years either side of the change-over, so the answer is proved rather than asserted.

Where marks go: Stopping at a year where Hugo is ahead without showing he was behind the year before. Using $1.05 \times 8$ instead of $1.05^{8}$ for eight years of compound interest.
Question 9
Paper 1H/2H style — 5 marks5 marks
Kofi has £ $12\,000$ . He invests £ $x$ in Account A, which pays $4\%$ per year simple interest, for $3$ years. He invests the rest in Account B, which pays $5\%$ per year compound interest, for $2$ years.

He earns £ $1300$ interest altogether.

Work out how much Kofi invests in each account. [5]
Model answer
Interest from A $= x \times 0.04 \times 3 = 0.12x$

Interest from B $= (12\,000 - x)(1.05^{2} - 1) = 0.1025(12\,000 - x)$

$0.12x + 0.1025(12\,000 - x) = 1300$

$0.12x + 1230 - 0.1025x = 1300$

$0.0175x = 70 \quad\Rightarrow\quad x = 4000$

Account A: £ $4000$ , Account B: £ $8000$
Why this scores
Mark scheme:
- M1 for $0.12x$ oe for the interest from Account A
- M1 for $(12\,000 - x) \times 0.1025$ oe for the interest from Account B
- M1 for a correct equation, e.g. $0.12x + 0.1025(12\,000 - x) = 1300$
- M1 for correctly rearranging to $0.0175x = 70$ oe
- A1 for £ $4000$ and £ $8000$
Why this scores full marks: Each account gets the right kind of interest, the compound interest is written as the multiplier minus $1$ , and both amounts are stated at the end. Check: $480 + 820 = 1300$ .

Where marks go: Writing Account B's interest as $1.05^{2}(12\,000 - x)$ , which is the final amount, not the interest. Stopping at $x = 4000$ without giving the amount in Account B.
Question 10
Paper 1H/2H style — 5 marks5 marks
A car was worth £ $18\,000$ . In the first year its value fell by $r\%$ . In the second year its value fell by $2r\%$ . At the end of the second year the car was worth £ $12\,960$

Find the value of $r$ . Show your working clearly. [5]
Model answer
$18\,000 \times \frac{100 - r}{100} \times \frac{100 - 2r}{100} = 12\,960$

$(100 - r)(100 - 2r) = 12\,960 \times \frac{10\,000}{18\,000} = 7200$

$10\,000 - 300r + 2r^{2} = 7200$

$r^{2} - 150r + 1400 = 0$

$(r - 10)(r - 140) = 0$

$r = 10$ or $r = 140$ , but a fall of $140\%$ (or $280\%$ ) is impossible, so $r = 10$
Why this scores
Mark scheme:
- M1 for $\left(1 - \frac{r}{100}\right)\left(1 - \frac{2r}{100}\right)$ oe as a two-year multiplier
- M1 for setting it equal to $\frac{12\,960}{18\,000}$ ( $= 0.72$ ) oe
- M1 for a correct three-term quadratic, e.g. $r^{2} - 150r + 1400 = 0$ oe
- M1 for a correct method to solve their quadratic
- A1 for $10$ only, with $140$ rejected
Why this scores full marks: The candidate multiplies the two multipliers, expands to a quadratic, solves it with working, and rejects the root that makes no sense in context.

Where marks go: Adding the falls to get $3r\%$ , so $3r = 28$ ; and leaving both roots, which loses the final mark. A bare answer of $r = 10$ with no working scores 0.

4MA1 Simple and Compound Interest

At a glance

What you’ll learn

Simple interest

Compound interest

Compound depreciation

Reverse problems — find rate or time

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Simple interest calculation

2Compound interest calculation

3Compare simple vs compound interest

4Compound depreciation

5Find the interest rate

6Reading the rate from a multiplier [1 mark]

7Finding the amount invested from simple interest [2 marks]

8Overall change after three successive changes [3 marks]

9Compound interest on a loan with a repayment [4 marks]

10How long for an investment to double [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10