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IB Mathematics SL 1.7 Amortization and annuities AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

Give answers in parts (a), (d)(i), (e) and (f) to the nearest dollar.
Peter invests \(37\,000\) AUD at \(6.4\%\) p.a. compounded quarterly.
(a) Find the value of the investment after \(2\) years. [3]
After \(m\) months, the fixed deposit exceeds \(50\,000\) AUD.
(b) Find the minimum value of \(m\in\mathbb{N}\). [4]
Peter plans to purchase an apartment costing \(200\,000\) AUD, pays \(25\%\) upfront and finances the rest.
(c) Write down the loan amount. [1]
The loan term is \(10\) years with equal end-of-month payments of \(1700\) AUD, compounded monthly.
(d) For this loan, find
(i) the total interest Peter pays;
(ii) the annual interest rate (nominal, compounded monthly). [5]
After \(5\) years of payments Peter settles the remainder with one final payment.
(e) Find the amount of this final payment. [3]
(f) Calculate how much Peter saves by making the single final payment after \(5\) years compared to continuing regular payments for the full term. [3]
▶️ Answer / Explanation (Detailed working)
Markscheme-style solution

Key ideas: Quarterly compounding: \(r_q=\frac{0.064}{4}=0.016\). Monthly annuity rate \(i=\frac{r}{12}\).

(a) Future value after \(2\) years with quarterly compounding:

\[ \text{FV}=37{,}000\left(1+\frac{0.064}{4}\right)^{2\times 4} =37{,}000\,(1.016)^{8}. \] \((1.016)^8=1.135402\ldots\) so \[ \text{FV}\approx 37{,}000\times 1.135402\ldots=\boxed{\$42{,}010}. \]
[3 marks]

(b) Minimum months \(m\) so that balance \(>50{,}000\) AUD. Compounded quarterly \(\Rightarrow\) number of quarters \(=\frac{m}{3}\):

\[ 37{,}000(1.016)^{m/3}>50{,}000 \;\Longrightarrow\; \frac{m}{3}>\frac{\ln(50{,}000/37{,}000)}{\ln(1.016)}=18.9692\ldots \]
Hence \(m>56.907\ldots\Rightarrow\boxed{m=57}\text{ (months, next multiple of 3).}\)
[4 marks]

(c) Loan amount: \[ 200{,}000\times(1-0.25)=\boxed{150{,}000\text{ AUD}}. \]

[1 mark]

(d)(i) Total paid over 10 years minus principal:

Number of payments \(N=12\times10=120\). Total paid \(=120\times1700=204{,}000\).
Interest \(=204{,}000-150{,}000=\boxed{\$54{,}000}\).

(d)(ii) Nominal annual rate \(r\) (compounded monthly) from \[ 150{,}000 =1700\left[\frac{1-(1+\tfrac{r}{12})^{-120}}{\tfrac{r}{12}}\right]. \] Solve with a GDC/solver: \[ r\approx 0.0646\ \Rightarrow\ \boxed{6.46\%\ \text{p.a.}}. \]

[5 marks]

(e) Final lump-sum after 60 payments equals outstanding balance \(B_{60}\):

With \(r=6.46\%\), monthly \(i=\tfrac{0.0646}{12}\): \[ B_{60} =150{,}000(1+i)^{60}-1700\left(\frac{(1+i)^{60}-1}{i}\right) \approx \boxed{\$86{,}973}. \]
[3 marks]

(f) Savings if settling at 5 years instead of full term:

Scheduled total over 10 years: \(204{,}000\).
Paid with early settlement: \(60\times1700+86{,}973=188{,}973\).
Savings \(=204{,}000-188{,}973=\boxed{\$15{,}027}\).
[3 marks]
Total Marks: 19
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