Mia is saving for her own horse. Her grandmother offers a loan. A Spanish bank pays interest on savings. Whether you're borrowing or saving, simple interest follows a single formula — and it's really just proportionality and equation-solving in disguise.
Simple interest formula:
$$I = P \times r \times t$$
$I$ = interest earned (€) ·
$P$ = principal (starting amount, €) ·
$r$ = interest rate per year (as a decimal: $3\% = 0.03$) ·
$t$ = time in years
Total amount: $A = P + I = P(1 + rt)$
To find any unknown: substitute the known values and solve the equation — just like Lessons 01–05.
Worked Examples
Worked Example 1 — Find interest and total
Mia deposits €800 at $3\%$ per year for $4$ years. Find the interest and the total amount.
$$I = P \times r \times t = 800 \times 0.03 \times 4$$
$$I = 800 \times 0.12 = 96$$
$$A = 800 + 96 = €896$$
Worked Example 2 — Find the principal
An account earns $€150$ interest in $2$ years at $5\%$ per year. Find the principal.
$$I = P \times r \times t$$
$$150 = P \times 0.05 \times 2 \quad | \text{ simplify right side}$$
$$150 = 0.10 \times P \quad | \div 0.10$$
$$P = 1\,500$$
Worked Example 3 — Find the time
Mia has €600. She needs €750. The account pays $2.5\%$ per year. How many years?
$$I = 750 - 600 = 150$$
$$150 = 600 \times 0.025 \times t$$
$$150 = 15t \quad | \div 15$$
$$t = 10 \text{ years}$$
Warm-Up — Find interest and total (formula given)
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Problem 1
Mia deposits €1 200 in a German savings account at $2\%$ per year for $3$ years. Find the interest earned and the total amount.
$I = 1200 \times 0.02 \times 3$
calculate I → A = P + I →
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Problem 2
Mia's grandmother lends her €500 at a friendly rate of $1.5\%$ per year. Mia pays it back after $2$ years. How much interest does she owe? What is the total repayment?
$I = 500 \times 0.015 \times 2$
calculate I → total →
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Problem 3
A Spanish bank pays $3.5\%$ simple interest per year. Mia invests €900 for $18$ months. How much interest does she earn? (Careful: 18 months $=$ ? years.)
$t = 18/12 = ?$ years · $I = 900 \times 0.035 \times t$
convert time → calculate I →
Core Problems — find the unknown variable
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Problem 4
An account earns €240 interest in $4$ years at $3\%$ per year. Find the principal $P$.
write equation → solve for P →
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Problem 5
Mia's account has €1 500 principal at $4\%$ per year. She wants to earn €300 in interest. How many years will this take? Write and solve an equation.
I = P r t → substitute → solve for t →
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Problem 6
An account turned €2 000 into €2 180 in 3 years. What was the annual interest rate? Write and solve an equation for $r$.
find I first → write I = Prt → solve for r → convert to % →
From here — write the equation yourself, then solve
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Problem 7
Mia is saving to buy a horse that costs €3 500. She already has €2 800 in a savings account at $2.5\%$ per year.
(a) How much more does she need?
(b) How many years will she need to save (using simple interest on the €2 800) to reach €3 500? Round to one decimal place.
find I needed → write equation → solve for t →
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Problem 8 Challenge
Two savings accounts offer different deals:
Account A: €1 200 at $4\%$ per year simple interest.
Account B: €1 000 at $5\%$ per year simple interest.
(a) Find the total in each account after 5 years.
(b) After how many years does Account B overtake Account A? (Set up an equation where $A_B > A_A$ and solve for $t$.)
total after 5 years each → set equal → solve for crossover t →
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Problem 9 Open
Simple interest is linear — the amount grows by the same euros every year (constant rate of change). Compound interest grows faster because each year's interest is added to the principal before calculating the next year's interest.
Using Account A from Problem 8 as a model: if instead of simple interest the account uses compound interest at $4\%$, the amount after each year is $A \times 1.04$. Calculate the compound total after $5$ years (multiply by $1.04$ five times). How does it compare to the simple interest total? Explain why the gap grows over time.
multiply by 1.04 five times → compare to simple → explain exponential growth →
Show answers
Problem 1
$I = 1200 \times 0.02 \times 3 = €72$ · $A = 1200 + 72 = €1\,272$
Problem 2
$I = 500 \times 0.015 \times 2 = €15$ · Total repayment $= €515$
Problem 3
$t = 1.5$ years · $I = 900 \times 0.035 \times 1.5 = €47.25$
Problem 4
$240 = P \times 0.03 \times 4 = 0.12P$ → $P = 240/0.12 = €2\,000$
Problem 5
$300 = 1500 \times 0.04 \times t = 60t$ → $t = 300/60 = 5$ years
Problem 6
$I = 2180 - 2000 = 180$ · $180 = 2000 \times r \times 3 = 6000r$ → $r = 180/6000 = 0.03 = 3\%$ per year
Problem 7
(a) Need $3500 - 2800 = €700$ more. (b) $700 = 2800 \times 0.025 \times t = 70t$ → $t = 10$ years.
Problem 8
(a) A: $1200 + 1200(0.04)(5) = 1200 + 240 = €1\,440$ · B: $1000 + 1000(0.05)(5) = 1000 + 250 = €1\,250$ · After 5 years, A is still larger. (b) Set $1200 + 48t = 1000 + 50t$ → $200 = 2t$ → $t = 100$ years. It takes a very long time — the higher rate doesn't overcome the starting gap!
Problem 9
Compound: $1200 \times 1.04^5 = 1200 \times 1.2167 \approx €1\,460$ · Simple: $€1\,440$ · Difference: $≈ €20$ after 5 years. The gap grows because compound interest earns interest on the interest — an exponential process (like geometric sequences, Lesson 8–9). Over longer periods the gap becomes enormous.
Coming up next → Lesson 22: Graphing Lines ($y = mx + b$)
Simple interest is linear: $A = P + Prt = P(1 + rt)$. That has the same shape as $y = mx + b$. In Lesson 22, you'll graph lines using slope and y-intercept, connect them to savings and business models, and write equations from contexts you've already mastered.