IB Maths AI HL Financial Applications Paper 2 & 3 TVM solver ~7 min read

Amortisation

Amortisation is paying off a loan with fixed regular repayments while interest keeps building on the balance still owed. These questions are solved with your GDC’s finance (TVM) solver — the skill is entering every value, with the right signs, into the right field.

📘 What you need to know

What is amortisation?

Amortisation is the process of paying off a loan with fixed, regular repayments — for example repaying a mortgage at a set amount each month. Even though every repayment is the same size, the lender adds interest to the balance still owed, so each payment first covers that interest and only then reduces the loan.

Over time the balance falls, and once it reaches zero the loan is fully paid off. Because interest is charged along the way, the total amount repaid is always more than the sum borrowed.

Using the GDC’s finance solver

Amortisation questions are answered with the GDC’s finance / TVM (time value of money) solver. Enter every value the question gives, leave the one you want blank, and let the GDC solve it.

The TVM solver — one field per known value NI%PV PMTFVP/YC/Y ? 640000 −6000 1212 unknown — leave blank the loan — positive repayment — negative FV = 0: loan clearedP/Y = C/Y = 12 for monthly repayments compounded monthly
A sample finance solver set-up for a $40 000 loan at 6% repaid at $600 a month. The blank field is what the GDC solves for.

The fields mean: N is the number of repayment periods, I% the nominal annual rate, PV the loan amount, PMT the repayment per period, FV the final balance (0 when paid off), and P/Y, C/Y the payments and compounding periods per year.

Total repayments and interest

Once the GDC gives N and the repayment, the total amount repaid is simply the number of repayments times the repayment. The total interest is how much that total exceeds the original loan.

Total repaid and total interest total repaid = N × (repayment) total interest = total repaid − amount borrowed the total repaid is always greater than the loan — the gap is the cost of borrowing

🧭 Recipe — solving an amortisation problem

  1. Identify the loan — the amount borrowed, the rate, the repayment and the timeframe.
  2. Open the finance / TVM solver and enter every known value.
  3. Apply the signs: PV is positive (money received), PMT is negative (money repaid).
  4. Set FV = 0 (loan fully paid) and P/Y = C/Y = periods per year.
  5. Leave the unknown blank and solve; write out every input for method marks.

Worked examples

WE 1

Finding the repayment time

Tom takes out a loan of $40 000 at a nominal annual interest rate of 6%, compounded monthly. He repays $600 at the end of each month. Find how long, in years and months, it takes to repay the loan.

enter into the TVM solver, solving for N I% = 6, PV = 40000, PMT = −600, FV = 0, P/Y = C/Y = 12 GDC gives N = 81.30 months convert: 81.30 ÷ 12 = 6.775 years 0.775 × 12 ≈ 9 months 6 years and 9 months the loan amount is positive, the repayment negative — money flowing opposite ways.
WE 2

Finding the monthly repayment

A loan of $25 000 at a nominal annual rate of 7.2%, compounded monthly, is to be repaid over 5 years. Find the monthly repayment.

5 years monthly ⇒ N = 60; solve for PMT N = 60, I% = 7.2, PV = 25000, FV = 0, P/Y = C/Y = 12 GDC gives PMT = −497.39 PMT is negative — it is money repaid monthly repayment = $497.39 N = 5 × 12 = 60 — always count the periods, not the years.
WE 3

Finding the size of the loan

Hannah can afford to repay $450 per month for 6 years on a loan at a nominal annual rate of 5.4%, compounded monthly. Find the largest loan she can take out, to the nearest dollar.

6 years monthly ⇒ N = 72; solve for PV N = 72, I% = 5.4, PMT = −450, FV = 0, P/Y = C/Y = 12 GDC gives PV = 27622.39 largest loan ≈ $27622 PV comes out positive — it is money received by the borrower.
WE 4

Total repaid and total interest

A car loan of $15 000 at a nominal annual rate of 8.4%, compounded monthly, is repaid by monthly instalments of $480. (a) Find the number of months to repay it. (b) Find the total amount repaid. (c) Find the total interest paid.

(a) solve for N I% = 8.4, PV = 15000, PMT = −480, FV = 0, P/Y = C/Y = 12 N ≈ 35.4 months (b) total repaid = N × repayment ≈ 35.4 × 480 ≈ $16987 (c) interest = total repaid − loan (a) ≈ 35.4 months · (b) ≈ $16987 · (c) ≈ $1987 the $1987 gap between repaid and borrowed is the cost of the loan.
WE 5

Comparing two repayment plans

Liam borrows $20 000 at a nominal annual rate of 6.6%, compounded monthly. Plan A repays $400 per month, Plan B repays $500 per month. How much less total interest does Liam pay under Plan B?

Plan A: solve for N, then interest N ≈ 58.6; total ≈ 23452; interest ≈ $3452 Plan B: solve for N, then interest N ≈ 45.3; total ≈ 22649; interest ≈ $2649 difference in interest Plan B saves ≈ $803 larger repayments clear the loan sooner, so less interest builds up.
WE 6

Full question: a mortgage

A mortgage of $150 000 is taken at a nominal annual rate of 4.8%, compounded monthly, and repaid over 20 years. (a) Find the monthly repayment. (b) Find the total amount repaid. (c) Find the total interest paid.

(a) 20 years monthly ⇒ N = 240; solve for PMT N = 240, I% = 4.8, PV = 150000, FV = 0, P/Y = C/Y = 12 PMT = −973.44 ⇒ repayment $973.44 (b) total = 240 × 973.44 ≈ $233625 (c) interest = 233625 − 150000 (a) $973.44 · (b) $233625 · (c) $83625 over 20 years the interest alone is more than half the sum borrowed.

💡 Top tips

âš  Common mistakes

Next up: Annuities — the mirror image of a loan. Instead of repaying money you borrowed, you receive regular payments from a sum invested. The same TVM solver is used, but the signs are reversed.

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