Financial applications AI HL

IB Mathematics AI HL · Unit 1

Number and Algebra · Lesson 1.4

Financial Applications

Use geometric models to understand interest, depreciation, inflation, savings, and loans.

Unit 1 theme · God’s love

✦ Wisdom

Financial decisions affect our families and neighbours. Jesus taught us to care for people rather than place our hope in possessions. The love of God gives us a deeper security than money can provide, and it calls us to act with honesty and generosity.

Jesus invites you to come to him with your hopes, worries, and needs. You can trust him as your Savior and Lord, and ask him to guide the way you use what has been entrusted to you.

🎯 Learning objectives

By the end of this lesson, you should be able to:

distinguish simple interest from compound interest;
calculate future values when interest is compounded yearly, half-yearly, quarterly, or monthly;
model annual depreciation;
calculate the real value of an investment after accounting for inflation;
solve financial problems using a calculator or spreadsheet; and
interpret results and assess the assumptions of a financial model.

01 · Simple interest: a starting point

With simple interest, interest is calculated only on the original amount. Let P be the initial amount, r the annual interest rate written as a decimal, and t the number of years. The amount after t years is:

$\boxed{A=P(1+rt)}$

Worked example

€6000 earns simple interest at 4% per year for five years. Find its final value.

$A=6000[1+(0.04)(5)]$

$A=\boxed{€7200}$

The account earns €240 each year because 4% is always calculated from the original €6000.

02 · Compound interest

With compound interest, each period’s interest is added to the balance. Later interest is calculated on both the original deposit and earlier interest.

For annual compounding, let P be the initial amount, r the annual rate as a decimal, and t the number of years:

$\boxed{A=P(1+r)^t}$

Worked example

The same €6000 earns 4% compound interest annually for five years.

$A=6000(1.04)^5$

$A\approx\boxed{€7299.92}$

This exceeds the simple-interest result of €7200 by approximately €99.92.

Key distinction: Simple interest adds the same amount each year. Compound interest multiplies the balance by the same factor each year.

03 · Compounding more than once a year

If interest is compounded m times each year, divide the annual rate by m and multiply the number of years by m:

$\boxed{A=P\left(1+\frac{r}{m}\right)^{mt}}$

Compounding	Value of m
Yearly	1
Half-yearly	2
Quarterly	4
Monthly	12

Worked example

€5000 is invested at a nominal annual interest rate of 6%, compounded quarterly, for five years. Find the balance.

There are four interest periods per year. Each period earns 6% ÷ 4 = 1.5%.

$A=5000\left(1+\frac{0.06}{4}\right)^{4(5)}$

$A=5000(1.015)^{20}$

$A\approx\boxed{€6734.28}$

Check the timing: For five years of quarterly compounding, there are 20 periods, not five.

04 · Annual depreciation

Depreciation is a decrease in value over time. If an item loses d of its value each year, where d is written as a decimal, multiply by 1 − d each year.

$\boxed{V_t=P(1-d)^t}$

Worked example

A vehicle costs €24 000 and loses 12% of its value each year. Estimate its value after five years.

$V_5=24000(1-0.12)^5$

$V_5=24000(0.88)^5$

$V_5\approx\boxed{€12\,665.57}$

The value falls by 12% of its current value each year, so the euro amount lost changes from year to year.

05 · Inflation and real value

Nominal value is the amount of money shown in an account. Real value measures what that money can buy compared with today. If prices rise at a constant annual inflation rate i, divide a future nominal amount by (1 + i)^t to express it in today’s money.

$\boxed{\text{Real value in today's money} =\frac{\text{future nominal value}}{(1+i)^t}}$

Worked example

€8000 is invested at 5% compound interest annually for four years. Inflation averages 3% annually. Find the investment’s value after four years in today’s money.

$\text{Nominal value}=8000(1.05)^4=€9724.05$

$\text{Real value} =\frac{9724.05}{(1.03)^4} \approx\boxed{€8639.69}$

Although the account shows €9724.05, its purchasing power is equivalent to approximately €8639.69 today under this inflation model.

Important: Dividing by the inflation factor accounts for the way prices have changed. Simply subtracting 3% from 5% gives an approximation, not the exact real growth rate.

06 · Solving for time

Sometimes a question asks when an investment reaches a target or when an item falls below a stated value. Write the model, then solve with logarithms, a graph, a table, or your calculator. Check whether the context requires a whole number of completed periods.

Worked example

An investment grows by 4.5% annually. Find the first whole year by which its value has at least doubled.

$(1.045)^t\geq2$

$t\geq\frac{\log 2}{\log 1.045}\approx15.747$

After 15 complete years it has not doubled; after 16 complete years it has. The answer is 16 years.

07 · Advanced application: regular savings

Suppose a fixed amount is deposited at the end of each month and the account earns a constant monthly interest rate. Each deposit has a different amount of time to grow. Their future values form a geometric series.

Let D be the deposit per month, j the monthly interest rate, and n the number of deposits. Immediately after the nth deposit, the balance is:

$\boxed{A=D\frac{(1+j)^n-1}{j}}$

Worked example

A student deposits €200 at the end of every month for 24 months. The account pays 0.5% interest each month. Find the balance immediately after the final deposit.

$A=200\frac{(1.005)^{24}-1}{0.005}$

$A\approx\boxed{€5086.39}$

The student deposited €4800 in total. The remaining €286.39 is interest under this model.

08 · Advanced application: loan repayments(Not required)

For a loan with fixed payments at the end of each month, a financial calculator can find the monthly payment. If the initial loan is L, the monthly interest rate is j, and the number of monthly payments is n, the payment M is:

$\boxed{M=\frac{Lj}{1-(1+j)^{-n}}}$

After k payments, the remaining balance can be calculated using:

$\boxed{B_k=L(1+j)^k -M\frac{(1+j)^k-1}{j}}$

Worked example

A €200 000 loan is repaid over 20 years at a monthly rate of 0.4%. Estimate the fixed monthly payment.

$M=\frac{200000(0.004)} {1-(1.004)^{-240}}$

$M\approx\boxed{€1297.91\text{ per month}}$

The calculation assumes the rate remains fixed, payments occur on time, and there are no additional fees.

Financial calculator check: Record whether payments occur at the beginning or end of each period. That choice changes the answer.

✏️ IB-style Paper 1 practice

A graphic display calculator may be used. Show your model and give monetary answers to the nearest cent unless instructed otherwise.

€6400 is invested at 4.5% annual interest compounded quarterly for six years.
Find its value after six years. [3 marks]
A machine costs €28 000 and depreciates by 18% each year.

(a) Find its value after six years. [2 marks]

(b) Find the first whole year after which its value is less than €7000.
[4 marks]
€10 000 is invested at 5% annual compound interest for ten years.
Inflation is modelled at 3% per year.

(a) Find the nominal value after ten years. [2 marks]

(b) Find its value in today’s money. [3 marks]
A saver deposits €250 at the end of every month for 36 months.
The monthly interest rate is 0.5%. Find the balance immediately
after the 36th deposit. [4 marks]
An investment earns 4.5% compound interest annually.
Find the first whole year by which its value has doubled.
[4 marks]

Reveal Paper 1 solutions

$A=6400\left(1+\frac{0.045}{4}\right)^{4(6)} \approx\boxed{€8371.14}.$
(a)

$V_6=28000(0.82)^6\approx\boxed{€8512.19}.$

(b) Solving

$28000(0.82)^t<7000$

gives

$t>\frac{\log(0.25)}{\log(0.82)} \approx6.986.$

After six years the machine is worth about €8512.19;
after seven years it is worth about €6979.99.
The first whole year is therefore

$\boxed{7}.$
(a)

$10000(1.05)^{10}\approx\boxed{€16\,288.95}.$

(b)

$\frac{10000(1.05)^{10}}{(1.03)^{10}} \approx\boxed{€12\,120.51}$

in today’s money.
$A=250\frac{(1.005)^{36}-1}{0.005} \approx\boxed{€9834.03}.$
$(1.045)^t\geq2 \quad\Rightarrow\quad t\geq\frac{\log 2}{\log 1.045}\approx15.747.$

The first whole year is

$\boxed{16}.$

🌍 Applications and advanced challenges

Comparing investment offers. A family can invest €12 000
for eight years in either:

• Account A: 4.2% interest compounded annually;

• Account B: 4.1% nominal annual interest compounded monthly.

(a) Find the final value of each account. [4 marks]

(b) Identify the better offer under these assumptions and find
the difference in final value. [2 marks]
Buying a vehicle. A car costs €32 000. Its value falls
by 15% each year.

(a) Find its predicted value after four years.
[2 marks]

(b) Find the first whole year after which its predicted value
is below €10 000. [4 marks]

(c) Explain why an actual resale price may differ from the model.
[1 mark]
Saving for a goal. A student deposits €150 at the end
of each month into an account paying 0.3% interest monthly.

(a) Find the balance immediately after 48 deposits.
[4 marks]

(b) Find the interest earned and explain why it is less than
48 months of interest on the full amount deposited.
[3 marks]
Loan comparison. A €120 000 loan has a fixed monthly
interest rate of 0.5% and is repaid in 180 equal monthly payments,
made at the end of each month.

(a) Find the monthly payment. [4 marks]

(b) Calculate the total of all 180 payments using your
unrounded payment. [2 marks]

(c) Hence estimate the total interest paid, assuming the rate
remains fixed and there are no fees. [2 marks]
Inflation and purchasing power. A salary increases
from €40 000 by 2% each year for six years. Prices rise by 3%
each year over the same period.

(a) Find the nominal salary after six years.
[2 marks]

(b) Express this salary in today’s money.
[3 marks]

(c) State whether the modelled purchasing power has increased
or decreased. [1 mark]

Reveal application solutions

Account A:

$A_A=12000(1.042)^8\approx €16\,673.74.$

Account B:

$A_B=12000\left(1+\frac{0.041}{12}\right)^{96} \approx €16\,651.37.$

Account A gives approximately

$\boxed{€22.37}$

more.
(a)

$V_4=32000(0.85)^4 \approx\boxed{€16\,704.20}.$

(b) Solve

$32000(0.85)^t<10000.$

This gives

$t>\frac{\log(10000/32000)}{\log(0.85)} \approx7.16,$

so the first whole year is

$\boxed{8}.$

(c) Condition, mileage, demand, and market prices can affect
the actual resale value.
(a)

$A=150\frac{(1.003)^{48}-1}{0.003} \approx\boxed{€7731.76}.$

(b) Total deposits are

$48(150)=€7200,$

so interest is approximately

$\boxed{€531.76}.$

Later deposits have been in the account for fewer months.
(a) Using the unrounded value in the calculation,

$M=\frac{120000(0.005)}{1-(1.005)^{-180}} \approx\boxed{€1012.63\text{ per month}}.$

(b)

$180M\approx\boxed{€182\,273.07}.$

(c) Interest is approximately

$182273.07-120000=\boxed{€62\,273.07}.$

Note: A real lender normally rounds each monthly
payment to cents and adjusts the final payment, so its exact
schedule may differ slightly.
(a)

$N=40000(1.02)^6 \approx\boxed{€45\,046.50}.$

(b)

$R=\frac{40000(1.02)^6}{(1.03)^6} \approx\boxed{€42\,492.28}$

in today’s money.

(c) This is above the original €40 000, so the modelled
purchasing power has increased.

✓ Lesson recap

Simple interest is based on the original amount:
$A=P(1+rt).$
Compound interest multiplies the changing balance:
$A=P\left(1+\frac{r}{m}\right)^{mt}.$
Annual depreciation uses the factor
$1-d.$
To express future money in today’s purchasing power, divide
by the inflation factor.
For loans and regular savings, check the interest period and
whether payments occur at the beginning or end of that period.

Financial Applications

✦ Wisdom

🎯 Learning objectives

01 · Simple interest: a starting point

02 · Compound interest

03 · Compounding more than once a year

04 · Annual depreciation

05 · Inflation and real value

06 · Solving for time

07 · Advanced application: regular savings

08 · Advanced application: loan repayments(Not required)

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

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alphy school publishers

IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Financial applications AI HL

✦ Wisdom

🎯 Learning objectives

01 · Simple interest: a starting point

02 · Compound interest

03 · Compounding more than once a year

04 · Annual depreciation

05 · Inflation and real value

06 · Solving for time

07 · Advanced application: regular savings

08 · Advanced application: loan repayments(Not required)

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

Related

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alphy school publishers

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