Money grows over time - and the formula is one you already know how to solve.
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.
The percent gate — use it before the formula
Read $3\%$Divide by $100$Write $r=0.03$Then use $I=Prt$
On paper, write the conversion even when the decimal feels obvious. It keeps the percent sign out of the multiplication.
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?
Mia deposits €1 200 at $2\%$ simple interest per year for $3$ years. Before calculating, write $2\%=0.02$ on paper. (a) Calculate the interest earned. (b) Calculate the total amount.
$I = 1200 \times 0.02 \times 3$
calculate I → A = P + I →
Problem 2
Mia's grandmother lends her €500 at $1.5\%$ simple interest per year for $2$ years. Before calculating, write $1.5\%=0.015$ on paper. (a) Calculate the interest owed. (b) Calculate the total repayment.
$I = 500 \times 0.015 \times 2$
calculate I → total →
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.)
An account earns €240 interest in $4$ years at $3\%$ per year. Find the principal $P$.
write equation → solve for P →
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 →
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
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 →
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 →
Problem 9 Open
Account A from Problem 8 uses €1 200 at $4\%$. Imagine it uses compound interest instead. (a) Calculate the compound total after $5$ years using $1200\times1.04^5$. (b) Compare it with the €1 440 simple-interest total. (c) Explain why the gap grows over time.
multiply by 1.04 five times → compare to simple → explain exponential growth →
Quick review · read simple interest as a rate
1. Name axes2. Find Δy3. Find Δx4. Divide + units
Problem 10 · Balance graph
A simple-interest account has a balance of €560 at year $2$ and €680 at year $6$. Let $x=$ time in years and $y=$ account balance in euros. (a) Calculate $\Delta y$. (b) Calculate $\Delta x$. (c) Calculate the slope with units. (d) Complete: The account balance ______ by ______ euros per year. (e) Answer: Should this ordinary euro-per-year rate be converted to a percentage? YES / NO, and why?
$300 = 1500 \times 0.04 \times t = 60t$ → $t = 300/60 = 5$ years
Problem 6
$I = 2180 - 2000 = 180$ · $180 = 2000 \times r \times 3 = 6000r$ → $r = 180/6000 = 0.03 = 3\%$ per year
Problem 7a
She needs $3500-2800=€700$ more.
Problem 7b
$700=2800\times0.025\times t=70t$ → $t=10$ years.
Problem 8a
A: $1200+1200(0.04)(5)=€1\,440$. B: $1000+1000(0.05)(5)=€1\,250$. Account A is still larger after $5$ years.
Problem 8b
$1200+48t=1000+50t$ → $200=2t$ → $t=100$ years.
Problem 9a
Compound total: $1200\times1.04^5\approx€1\,460$.
Problem 9b
Simple total is €1,440, so compound interest gives about €20 more after 5 years.
Problem 9c
Compound interest earns interest on earlier interest, so its yearly additions grow instead of staying equal.
Problem 10a
$\Delta y=680-560=€120$.
Problem 10b
$\Delta x=6-2=4$ years.
Problem 10c
$m=\frac{120}{4}=€30$/year.
Problem 10d
The account balance increases by €30 per year.
Problem 10e
NO. The axes use different units, euros and years, so the rate remains €30/year.
Coming up next → Lesson 24: 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 24, 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.