Finance is where IB Mathematics Applications & Interpretation HL feels most like the real world: every loan, savings plan, pension and asset value is just money changing by a fixed factor each period. AI treats the topic less as algebra to grind by hand and more as a modelling exercise — set the problem up correctly, then let the calculator's finance (TVM) solver do the arithmetic. Because a GDC is always allowed, the marks live in choosing the right tool — a power formula, a geometric series, or the solver — getting the signs and periods right, and reading the answer back in context (dollars, months, whole payments).
This cheatsheet condenses the whole of SL 1.4 & 1.7 — compound interest and depreciation, the future and present value of annuities, loan repayments and amortisation, and every field of the finance solver ($N$, $I\%$, $PV$, $PMT$, $FV$, $P/Y$, $C/Y$) — onto one page, and flags the traps (sign errors, wrong compounding periods, rounding the term down) that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Compound interest: one idea, many products SL 1.4
Every product on this sheet rests on a single idea: money changes by a fixed multiplier each period. A one-off amount left to grow is a geometric model.
Multiplier:a rate of $r\%$ becomes the growth factor $1+\dfrac{r}{100}$, applied once per period
Future value:$FV=P\left(1+\dfrac{r}{100}\right)^{t}$ — principal $P$ after $t$ periods at $r\%$ per period
TrickTurn the rate into a multiplier first: $4\%$ per year means multiply by $1.04$ once each year, so $t$ years multiplies by $1.04^{t}$.
NoteA single amount uses this power formula. A stream of equal payments is an annuity — a geometric series whose value collapses to the present- and future-value formulas in §6–§9.
§2 — Compounding more than once a year SL 1.4
Interest is often added several times a year. Split the annual rate over $k$ periods and count $kt$ of them.
$k$ times a year:$FV=P\left(1+\dfrac{r}{100k}\right)^{kt}$ — $k$ compounding periods per year for $t$ years
Common $k$:annually $k=1$, quarterly $k=4$, monthly $k=12$, daily $k=365$
TrickDivide the rate by $k$ and multiply the years by $k$: $6\%$ compounded monthly uses $0.5\%$ per month over $12t$ months.
NoteMore frequent compounding earns a little more. The effective annual rate is $\left(1+\dfrac{r}{100k}\right)^{k}-1$ — so a nominal $6\%$ compounded monthly is really about $6.17\%$ a year.
§3 — Depreciation SL 1.4
An asset that loses value each year depreciates — the same power model, but the multiplier now drops below $1$.
Depreciation:$V=P\left(1-\dfrac{r}{100}\right)^{t}$ — value after $t$ years losing $r\%$ of value per year
TrickSubtract the rate from $1$: losing $15\%$ a year means multiplying by $0.85$ each year, so after $t$ years the value is $P\times0.85^{t}$.
TrapThe multiplier is $1-\frac{r}{100}$, e.g. $\times0.85$ for $15\%$. Writing $\times(-0.15)$ or $\times1.15$ is the classic slip — one makes the value negative, the other makes it grow.
§4 — Time to reach a target: take logs SL 1.4
Going forwards is one calculation; the sharper AI question runs backwards — how long until a sum reaches a target? The unknown $t$ is trapped in the exponent, so the only way down is to take logarithms of both sides.
$$P\left(1+\tfrac{r}{100}\right)^{t}=A\;\Longrightarrow\;t=\frac{\ln(A/P)}{\ln\!\left(1+\tfrac{r}{100}\right)}$$
TrapThis is a ratio of two separate logs, not $\ln\!\dfrac{A/P}{1+r/100}$. And because interest is credited only at each period end, round a fractional $t$ up to the next whole period.
NoteThis is a synthesis with Exponents & Logarithms (SL 1.5). If you prefer, skip the algebra and solve $P\left(1+\tfrac{r}{100}\right)^{t}=A$ on the GDC with the equation solver or by intersecting graphs.
§5 — The finance (TVM) solver SL 1.7
From here on the calculator's finance solver does the heavy lifting. It links five quantities: enter any four and solve for the fifth.
$N$:total number of payment periods — for monthly payments, $N=12\times\text{years}$
$I\%$:the nominal annual interest rate, entered as a whole number ($6$, not $0.06$)
$PV,\ PMT,\ FV$:present value (principal / lump sum today), the payment each period, and future value ($FV=0$ to clear a loan or empty a fund)
$P/Y,\ C/Y$:$P/Y=C/Y=12$ for monthly payments compounded monthly — the solver divides $I\%$ by this for you
TrapSign convention. Money you receive is positive; money you pay is negative. A loan has $PV>0$ (cash in) and $PMT<0$ (repayments out); a savings plan has $PV=0$ with deposits $PMT<0$. Mixing the signs is the single most common finance error.
NoteThe solver converts to the per-period rate $i=\dfrac{I\%}{100\times(C/Y)}$ automatically, so enter the whole percentage and set $P/Y=C/Y$ correctly — do not divide by $12$ yourself as well.
§6 — Savings plans: future value of an annuity SL 1.7
Pay a fixed amount into an account every period and each deposit compounds for the time it has left, so the fund outgrows the deposits alone.
Future value:$FV=D\cdot\dfrac{(1+i)^{n}-1}{i}$ — deposit $D$ each period, rate $i$ per period, $n$ deposits
On the GDC:$PV=0$, $PMT=-D$, enter $N$ and $I\%$, then solve for $FV$
Interest earned:$\text{interest}=FV-D\times n$ — the fund minus the plain total paid in
TrickThe gap between $FV$ and $D\times n$ is exactly the interest the plan earned — a very common part-(c).
TrapDeposits leave your pocket, so $PMT$ is negative and the resulting $FV$ comes back positive. Set $PV=0$ when the plan starts from nothing.
§7 — Present value & drawdowns SL 1.7
Turn it around: how much do you need today to fund a stream of equal withdrawals? Discount each future payment back to now.
Present value:$PV=W\cdot\dfrac{1-(1+i)^{-n}}{i}$ — lump sum needed now for $n$ withdrawals of $W$ at rate $i$
Drawdown:$PV=\text{fund}$, $FV=0$, enter $N$ and $I\%$, then solve for $PMT$
NoteA retirement drawdown is the mirror of a loan: a lump sum $PV$ is run down to $FV=0$ by equal withdrawals. Because interest keeps accruing on the shrinking balance, the sustainable withdrawal is more than simply $\text{fund}\div n$.
TrickIf the solver returns a negative $PMT$, that is only the cash-flow direction — report the size of the withdrawal, not the minus sign.
§8 — Loan repayments SL 1.7
A loan is an annuity seen from the lender's side: the amount borrowed is the present value, and equal repayments reduce it to zero.
Per month:$i=\dfrac{r}{1200}$ and $n=12\times\text{years}$ for a nominal $r\%$ compounded monthly
Repayment:$M=\dfrac{L\,i}{1-(1+i)^{-n}}$ — monthly payment on a loan $L$ over $n$ months
On the GDC:$PV=L$, $FV=0$, enter $N$ and $I\%$, then solve for $PMT$
NoteEach payment first covers that month's interest $i\times\text{balance}$; whatever is left chips away at the principal — which is why early payments are mostly interest and the balance falls slowly at first.
TrapThe repayment must exceed the first month's interest $i\times L$, or the balance never falls. Convert to the monthly rate $i=r/1200$ before substituting — don't feed an annual rate into a monthly formula.
§9 — Loan term, amortisation & total cost SL 1.7
The remaining loan questions all read off the same amortised model: how long to clear it, how much is still owed, and what it costs in total.
Months to repay:given a fixed payment, enter $PV$, $PMT=-\text{payment}$, $I\%$, $FV=0$ and solve for $N$; then round $N$ up
Balance owed:$B=L\cdot\dfrac{1-(1+i)^{-(N-m)}}{1-(1+i)^{-N}}$ — owed after $m$ of $N$ payments (the value of the payments not yet made)
Total cost:total paid $=PMT\times N$; total interest $=PMT\times N-PV$
TrapRound the number of payments up, never to nearest: at the floor value a positive balance still remains, so a whole extra (part-)payment is always needed to finish.
NoteLonger terms lower the monthly payment but raise the total interest — a core AI comparison. Keep the unrounded $PMT$ when computing total interest, or the answer can be several dollars out.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|
| "Value after $t$ years at $r\%$" | $P\left(1+\tfrac{r}{100}\right)^{t}$; to depreciate use $\left(1-\tfrac{r}{100}\right)^{t}$ | Multiplier below $1$ for depreciation |
| "Compounded monthly / quarterly" | $P\left(1+\tfrac{r}{100k}\right)^{kt}$ | Divide rate by $k$, multiply years by $k$ |
| "How long to reach $\$A$?" | Take logs: $t=\dfrac{\ln(A/P)}{\ln(1+r/100)}$ | Ratio of two logs; round $t$ up |
| "Savings plan — final value" | Solver: $PV=0$, $PMT=-D$, solve $FV$ | Deposits negative; interest $=FV-Dn$ |
| "Regular repayment / instalment" | Solver: $PV=L$, $FV=0$, solve $PMT$ | $PV>0$, $PMT<0$; set $P/Y=C/Y=12$ |
| "How many payments to clear it?" | Solver: $FV=0$, solve $N$ | Round $N$ up, not to nearest |
| "Balance still owed after $m$ payments" | Value of the remaining payments (or bal on the GDC) | Falls slowly early on — mostly interest |
| "Total interest / cost of the loan" | $PMT\times N-PV$ | Keep unrounded $PMT$; longer term $=$ more interest |
Worked Example — AI HL-Style Car Loan & Amortisation
Question (AI HL Paper 2 style — 6 marks)
Priya borrows $\$24\,000$ to buy a car. The loan is charged at a nominal annual interest rate of $6\%$ compounded monthly, and she repays it in equal monthly instalments over $5$ years.
(a) Find her monthly repayment.
(b) Find the total interest she pays over the life of the loan.
(c) Find the amount still owing immediately after her 24th payment.
Solution
- (a) Monthly rate $i=\dfrac{6}{1200}=0.005$ and $N=12\times5=60$ months. In the finance solver enter $PV=24\,000$, $I\%=6$, $N=60$, $FV=0$, $P/Y=C/Y=12$ and solve: $PMT=-463.99$. She repays $\$463.99$ per month. (M1)(A1) (the minus sign just means cash flowing out.)
- (b) Keep the full-precision payment, $PMT=463.9872\ldots$: total paid $=463.9872\ldots\times60=\$27\,839.23$, so interest $=27\,839.23-24\,000=\$3\,839.23$. (M1)(A1) (pre-rounding to $\$463.99$ first gives $\$3\,839.40$ — about $17$ cents too high.)
- (c) The balance is the value of the $36$ payments still to come: $B=24\,000\cdot\dfrac{1-1.005^{-36}}{1-1.005^{-60}}=\$15\,251.73$ (or read bal(24) on the GDC). (M1)(A1)
Examiner's note: hold the unrounded $PMT$ from the solver until the very end — rounding to the nearest cent before $\times N$ can throw the total interest out by cents or even dollars. Watch the signs ($PV$ positive, $PMT$ negative), state the units (dollars, months), and note the balance in (c) is barely a third paid off after $40\%$ of the term, because early payments are mostly interest.
Common Student Questions
When do I use present value versus future value of an annuity?
Ask which way the money moves in time. If you are paying money in now and want to know what it grows to later — a savings plan — you want future value ($FV=D\cdot\frac{(1+i)^n-1}{i}$). If you want the lump sum needed today to support future payments out — a loan or a pension drawdown — you want present value ($PV=W\cdot\frac{1-(1+i)^{-n}}{i}$). In the solver it is the same tool either way; you just choose whether $PV$ or $FV$ is the unknown.
Why is the payment (or deposit) negative in the finance solver?
The solver uses a sign convention: money you receive is positive and money you pay out is negative, so that every problem balances to zero over its life. For a loan you receive the principal ($PV>0$) and pay instalments back ($PMT<0$); for a savings plan your deposits leave your pocket ($PMT<0$) and the fund you collect at the end is positive ($FV>0$). If your answer comes out with the "wrong" sign, it usually just means you set that cash flow up as the opposite direction — report its size.
Do P/Y and C/Y always have to be 12?
No — they must match how often the payment is made and how often interest is compounded. Use $12$ for monthly, $4$ for quarterly, $2$ for half-yearly and $1$ for annual. They are usually equal, and then the solver turns your nominal $I\%$ into the per-period rate for you. Set them before entering the other values, and don't also divide $I\%$ by the frequency by hand or you will apply the division twice.
When I solve for the number of payments, why round up instead of to the nearest whole number?
Because at the floor value the loan is not quite cleared — a small balance still remains — so one more payment is unavoidable, even if it is a smaller final one. Rounding $N$ to the nearest whole number would leave money owing whenever the true value ends in anything below $.5$, which is wrong in context. The rule is simple: a fractional number of payments always rounds up to the next whole payment.
Does compounding more often always earn more, and what is the effective annual rate?
More frequent compounding does earn a little more for the same nominal rate, because interest starts earning interest sooner — but the gain shrinks quickly and levels off (the limit is continuous compounding). The fair way to compare two schemes is the effective annual rate, $\left(1+\frac{r}{100k}\right)^{k}-1$: a nominal $6\%$ compounded monthly works out at about $6.17\%$ a year, so it beats a flat $6\%$ compounded annually but only by a small margin.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
- Notes PDF — every compound-interest, annuity and amortisation method in full, with the exact finance-solver keystrokes and sign conventions.
- Tutorial booklet — 25+ AI HL-style questions from basic compound interest to full loan-amortisation and pension-drawdown applications.
- Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every finance method.
- Predicted-paper questions — the exact loan, annuity and depreciation question types most likely in the next session.
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