Module 9 of 47Lesson 4 of 8 · AI 1.4 — Financial applications: compound interest and depreciation

Course map · Topic 1: Number and Algebra

1.4 Financial applications: compound interest and depreciation

SL & HL10 worked questions

Money that grows (interest) or shrinks (depreciation) by the same percentage each period is a geometric sequence. In the exam you can use the formula or the Finance Solver — the Finance Solver is faster and safer.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ + 🖩 · Paper 1[3 marks]

Lesedi puts P 1 200 in a bank account that pays 5% per year, compounded annually. Find the amount in the account after 4 years. Give your answer to 2 decimal places.

Show the full step-by-step solution

Step 1🧠 ThinkWhat does “compounded annually” mean?

Once a year, the bank adds 5% to whatever is in the account. Adding 5% is the same as multiplying by

1.05

. So after 4 years we multiply by

1.05

four times.

Step 2✍️ By handWrite the formula and substitute

Here

PV = 1200

r = 5

k = 1

(once a year),

n = 4

FV = 1200\left(1+\dfrac{5}{100\times1}\right)^{1\times4}

FV = 1200\times1.05^{4}

FV = 1458.6075

FV = \text{P }1458.61

Step 3🖩 On the GDCType it straight into the calculator

1200×1.05›^💻 Shift+6, then →›4›enter💻 Enter

On a Calculator page, type the expression exactly as written. The ^ key makes the power.

1200 × 1.05⁴

✅ Check it Simple interest would give

1200 + 4\times60 = 1440

. Compound interest must give a bit more — and 1458.61 is a bit more. ✔

Final answer

\text{P }1458.61

Question 2●○○ Warm-up🖩 GDC · Paper 1[3 marks]

Thato invests P 8 000 at 6% per year compounded monthly. Use the Finance Solver to find the value of the investment after 5 years.

Show the full step-by-step solution

Step 1🧠 ThinkWork out N, the number of periods

Interest is added every month. 5 years × 12 months = 60 periods. So N = 60, and PpY = CpY = 12.

Step 2🖩 On the GDCOpen the Finance Solver

menu💻 click menu on keypad›8: Finance›1: Finance Solver…

menu → 8: Finance → 1: Finance Solver

Step 3🖩 On the GDCFill in every line, then solve for FV

60›tab💻 Tab›6›tab💻 Tab›-8000›tab💻 Tab›0›tab💻 Tab›tab💻 Tab›12›tab💻 Tab›12

PV is −8000 because Thato pays the money in. Pmt = 0 because he adds nothing extra. Now press ▲ until the cursor is on FV and press enter.

FV appears: 10790.80…

Final answer

\text{P }10\,790.80

⚠️ Common mistake Typing N = 5. N counts months here, not years.

Question 3●●○ Standard🖩 GDC · Paper 1[4 marks]

Boitumelo invests P 3 000 at 4.2% per year compounded quarterly. Find how many years it takes for the investment to first reach P 5 000.

Show the full step-by-step solution

Step 1🧠 ThinkThis time the unknown is N

We know the start (3000), the target (5000) and the rate. We leave N empty and let the calculator find it. Quarterly means 4 times a year, so PpY = CpY = 4.

Step 2🖩 On the GDCFill in the Finance Solver and solve for N

menu💻 click menu on keypad›8: Finance›1: Finance Solver…›tab💻 Tab›4.2›tab💻 Tab›-3000›tab💻 Tab›0›tab💻 Tab›5000›tab💻 Tab›4›tab💻 Tab›4

Move the cursor back up to N and press enter.

N = 48.905… quarters

Step 3✍️ By handTurn quarters into years — and round UP

Interest is only added at the end of each quarter. After 48 quarters the money is still under P 5 000, so we need the 49th quarter.

N = 48.905\ldots

quartersA1

\Rightarrow 49

quartersR1

49 \div 4 = 12.25

yearsA1

Final answer After 49 quarters, which is

12.25

years (12 years and 3 months).

⚠️ Common mistake Writing 12.2 years. The money only goes over P 5 000 when interest is added at the end of the 49th quarter.

Question 4●●○ Standard🖩 GDC · Paper 1[3 marks]

An amount of P 2 000 grows to P 2 600 in 5 years, with interest compounded monthly. Find the annual interest rate, correct to 2 decimal places.

Show the full step-by-step solution

Step 1🧠 ThinkUnknown = I(%)

Everything else is known: N = 5 × 12 = 60, PV = −2000, FV = 2600, PpY = CpY = 12.

Step 2🖩 On the GDCSolve for I(%)

menu💻 click menu on keypad›8: Finance›1: Finance Solver…›60›tab💻 Tab›tab💻 Tab›-2000›tab💻 Tab›0›tab💻 Tab›2600›tab💻 Tab›12›tab💻 Tab›12

Put the cursor on the I(%) line and press enter.

I(%) = 5.2587…

Final answer

5.26\%

per year

⚠️ Common mistake Getting an error because both PV and FV were typed as positive — the solver cannot find a rate where you get money back without paying any in.

Question 5●○○ Warm-up✍️ + 🖩 · Paper 1[3 marks]

A car is bought for P 180 000. It loses 15% of its value every year. Find its value after 6 years, to the nearest pula.

Show the full step-by-step solution

Step 1🧠 ThinkLosing 15% means keeping 85%

If the car loses 15% each year, it keeps

100\% - 15\% = 85\%

. Keeping 85% means multiplying by

0.85

. Do this 6 times.

Step 2✍️ By handSubstitute into the depreciation formula

V = 180\,000\times(1-0.15)^{6}

V = 180\,000\times0.85^{6} = 67\,886.91\ldots

V \approx \text{P }67\,887

Step 3🖩 On the GDCOr use the Finance Solver with a negative rate

Depreciation is just “negative interest”: type I(%) = −15, with PpY = CpY = 1 (once a year).

I(%) = −15 gives the same answer

Final answer

\text{P }67\,887

Question 6●●○ Standard🖩 GDC · Paper 1[3 marks]

For the car in Question 5 (bought for P 180 000, losing 15% per year), find the number of complete years after which its value first falls below half of what was paid.

Show the full step-by-step solution

Step 1🧠 ThinkHalf of the price

Half of 180 000 is 90 000. We want N when FV = 90 000 and the rate is −15%.

Step 2🖩 On the GDCSolve for N

menu💻 click menu on keypad›8: Finance›1: Finance Solver…›tab💻 Tab›-15›tab💻 Tab›-180000›tab💻 Tab›0›tab💻 Tab›90000›tab💻 Tab›1›tab💻 Tab›1

Cursor on N, then enter.

N = 4.265… years

Step 3✍️ By handRound up and check

N = 4.265\ldots

After 4 years:

180\,000\times0.85^{4} = 93\,961.13 \gt 90\,000

After 5 years:

180\,000\times0.85^{5} = 79\,866.96 \lt 90\,000

Final answer After 5 complete years.

Question 7●●○ Standard🖩 GDC · Paper 2[5 marks]

Kabo has P 3 000 to invest for 10 years. Bank A offers 5.5% per year compounded annually. Bank B offers 5.4% per year compounded monthly.

(a) Find the value of the investment with each bank.[4]
(b) State which bank Kabo should choose, and how much more he would have.[1]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCBank A: N = 10, PpY = CpY = 1

Bank A: FV = 5124.43

Step 2🖩 On the GDCBank B: N = 120, PpY = CpY = 12

Just change N, I(%), PpY and CpY — the rest stays. Then solve FV again.

Bank B: FV = 5141.79

Part (b)

Step 1✍️ By handCompare

5141.79 - 5124.43 = 17.36

Bank B is better by P 17.36A1

💡 Remember A lower rate can still win if interest is added more often. That is why you must always calculate — never just compare the percentages.

Final answer (a) Bank A:

\text{P }5124.43

; Bank B:

\text{P }5141.79

. (b) Bank B, by

\text{P }17.35

–

17.36

Question 8●●● Exam-level🖩 GDC · Paper 2[5 marks]

Neo invests P 10 000 at 6% per year compounded annually for 8 years. The rate of inflation is 2.5% per year.

(a) Find the value of the investment after 8 years.[2]
(b) By using a real interest rate, find the real value of the investment after 8 years.[3]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCNormal Finance Solver calculation

FV = 15 938.48

Part (b)

Step 1🧠 ThinkWhat is “real value”?

Prices go up 2.5% every year, so each pula buys a little less. The real growth of the money is roughly

6\% - 2.5\% = 3.5\%

per year. We use this real rate instead of 6%.

Step 2🖩 On the GDCSame solver, but I(%) = 3.5

Real value = 13 168.09

Final answer (a)

\text{P }15\,938.48

(b) real value

\approx \text{P }13\,168.09

(in today's money).

Question 9●●● Exam-level🖩 GDC · Paper 2[7 marks]

Kagiso deposits P 15 000 into a savings account paying 3.8% per year compounded monthly.

(a) Find the value of the account after 7 years, to the nearest pula.[2]
(b) Find the total interest earned in these 7 years.[1]
(c) Find the number of months until the account first doubles in value. Give your answer in years and months.[4]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCN = 7 × 12 = 84

FV = 19 562.80

Part (b)

Step 1✍️ By handInterest = what you end with − what you put in

19\,563 - 15\,000 = \text{P }4\,563

Part (c)

Step 1🧠 ThinkDoubles means FV = 30 000

Leave N empty, put FV = 30 000, and solve for N.

Step 2🖩 On the GDCSolve for N

N = 219.23… months

Step 3✍️ By handRound up and convert

N = 219.23\ldots \Rightarrow 220

monthsA1

220 = 18\times12 + 4

18 years and 4 monthsA1

Final answer (a)

\text{P }19\,563

(b)

\text{P }4\,563

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[6 marks]

A laptop is bought for P 14 000. Its value decreases by 22% each year.

(a) Find the value of the laptop after 3 years.[2]
(b) Find the percentage of its original value that has been lost after 3 years.[2]
(c) Find after how many complete years the laptop is first worth less than P 2 000.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handKeep 78% each year

V = 14\,000\times0.78^{3}

V = \text{P }6\,643.73

Part (b)

Step 1✍️ By handLoss as a fraction of the original

\dfrac{14\,000 - 6\,643.73}{14\,000}\times100\%

= 52.5\%

(3 s.f.)A1

💡 Not 66%! Three losses of 22% do not add up to 66%, because each 22% is taken from a smaller value.

Part (c)

Step 1🖩 On the GDCFinance Solver with I(%) = −22

N = 7.83… years

Step 2✍️ By handRound up

N = 7.83\ldots \Rightarrow 8

complete yearsA1

Final answer (a)

\text{P }6\,643.73

(b)

52.5\%

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Compound interest: interest is added to the money, and next time you earn interest on the interest too.
The formula $FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}$ : $PV$ = starting money, $r$ = yearly rate %, $k$ = number of times interest is added per year, $n$ = number of years.
k values: yearly $k=1$ , half-yearly $k=2$ , quarterly $k=4$ , monthly $k=12$ .
Depreciation (value going down by r% each year): value $= PV\left(1-\dfrac{r}{100}\right)^{n}$ . On the Finance Solver just type a negative interest rate.
Finance Solver signs: money you pay out is negative, money you receive is positive. So PV is negative, FV is positive.
Inflation: to find the real value of an investment, use the real interest rate ≈ interest rate − inflation rate.
Money answers: give to 2 decimal places (or the nearest whole unit if the question says so). Time answers: round up to a whole period when interest is only added at the end of a period.

Compound interest

FV = PV\times\left(1+\dfrac{r}{100k}\right)^{kn}

📘 In the formula booklet

Depreciation (by r% per year)

V = V_0\times\left(1-\dfrac{r}{100}\right)^{n}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Open the Finance Solver (Calculator page)

menu💻 click menu on keypad›8: Finance›1: Finance Solver…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 8: Finance.
Choose 1: Finance Solver….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

menu → 8: Finance → 1: Finance Solver

Worked example: Question 2 — Open the Finance Solver

Move between boxes

tab💻 Tab

Press tab to move to the next box.

Solve for the empty unknown: put the cursor on that line, then

enter💻 Enter

Press enter.

⚠️ Where students lose marks

Typing PV as positive — then FV comes out negative. Not wrong maths, but it confuses many students. Negative in, positive out.
Forgetting that N is the number of periods, not years: 5 years compounded monthly → N = 60.
Leaving PpY = 1 when the interest is monthly. For this subtopic PpY and CpY are both equal to k.
Rounding time down. If 48.9 quarters are needed, the money only reaches the target after the 49th quarter.

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL AI 1.4 — Financial applications: compound interest and depreciation

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

1.4 Financial applications: compound interest and depreciation

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks