Covers Lessons 1–4: simple and compound interest, effective annual rate, depreciation, and geometric sequences in finance.
8 multiple-choice questions, graded for your checkpoint score with immediate feedback. Then complete 3 written responses; these are saved for you to self-review against model answers and are not included in the MC score.
Suggested time: 30 minutes. Show formula substitutions in written responses. Keep full precision through calculations and round currency once, at the end, to the nearest cent unless a prompt says otherwise. State whether payments occur at period start or end.
Write your working, then open the matching model answer to self-check.
A saver wants AUD 25,000 in 5 years. The account earns 4.8% p.a. compounded annually. Find the deposit needed today. Give a formula and round the final currency value to cents.
A machine costs AUD 45,000 and depreciates by 15% of its current value each year. Find its value after 5 years and the total decrease in value.
Compare Product A at 5.8% p.a. compounded half-yearly and Product B at 5.7% p.a. compounded monthly. Calculate each EAR to 2 decimal places and identify the larger.
Question: A saver invests AUD 12,000 at 4.5% p.a. simple interest for 3 years. How much interest is earned? Key: A. Simple interest is I=Prt=12,000×0.045×3=AUD 1,620. The final balance would be AUD 13,620.
Question: A principal of AUD 10,000 is invested at 5% p.a., compounded annually for 3 years. What is its value, to the nearest cent? Key: B. Use A=P(1+r)^n: 10,000(1.05)^3=AUD 11,576.25.
Question: A nominal rate of 6% p.a. is compounded monthly. What is the monthly periodic rate? Key: C. Divide the nominal annual rate by 12: 6%/12=0.5% per month.
Question: A nominal rate of 8% p.a. is compounded quarterly. What is the effective annual rate, to 2 decimal places? Key: D. EAR=(1+0.08/4)^4−1=0.08243216…, or 8.24%.
Question: An asset worth AUD 20,000 depreciates by 15% of its current value each year. What is its value after 4 years? Key: B. Reducing-balance depreciation gives 20,000(0.85)^4=AUD 10,440.50.
Question: A balance follows the recurrence Aₙ₊₁=1.05Aₙ, with A₀=AUD 2,000. What are the first three balances A₁,A₂,A₃? Key: C. Multiply by 1.05 each year: 2,100; 2,205; 2,315.25. This is a geometric sequence with common ratio 1.05.
Question: At 6% p.a. compounding annually, approximately how long does it take an investment to double? Key: D. Solve 2=(1.06)^n: n=ln(2)/ln(1.06)≈11.90 years. The balance first exceeds double after 12 whole years.
Question: Product A pays 5.8% p.a. compounded half-yearly; Product B pays 5.7% p.a. compounded monthly. Which has the higher EAR? Key: A. A: (1+0.058/2)^2−1=5.8841%. B: (1+0.057/12)^12−1=5.8513%. Product A has the larger effective annual rate.
$PV=\frac{25{,}000}{(1.048)^5}=\$19{,}775.78$. Discounting gives the deposit needed today.
$V_5=45{,}000(0.85)^5=\$19{,}966.74$. The decrease is $45{,}000-19{,}966.74=\$25{,}033.26$. Round only the final currency values.
Product A: $(1+0.058/2)^2-1=0.058841=5.88\%$. Product B: $(1+0.057/12)^{12}-1\approx0.058513=5.85\%$. Product A has the slightly higher EAR.
Tick when you have finished all questions and reviewed your answers.