Covers Lessons 1–5: simple and compound interest, effective annual rate, depreciation, and geometric sequences in finance.
Q1: B — $A = P(1+r)^n$ is compound interest.
Q2: C — $I = 12{,}000 \times 0.045 \times 3 = \$1{,}620$.
Q3: B — Monthly rate = $6\% / 12 = 0.5\%$.
Q4: C — $r_{\text{eff}} = (1 + 0.08/4)^4 - 1 = 8.24\%$.
Q5: A — Flat rate: $S = 30{,}000 - 6{,}000 \times 4 = \$6{,}000$.
Q6: D — Daily compounding always produces the highest EAR for a given nominal rate.
Q7: B — $n = \ln(2)/\ln(1.06) = 11.9$ years.
Q8: C — Common ratio is $(1+i) = 1.05$.
Q9 (3 marks): $P = 25{,}000 / (1.048)^5 = \$19{,}768.25$ [2]. They must deposit $\$19{,}768.25$ today [1].
Q10 (3 marks): (a) $S = 45{,}000(0.85)^5 = \$19{,}978.22$ [2]. (b) Total depreciation = $45{,}000 - 19{,}978.22 = \$25{,}021.78$ [1].
Q11 (4 marks): (a) Product A: $(1 + 0.058/2)^2 - 1 = 5.88\%$ [1]. Product B: $(1 + 0.057/12)^{12} - 1 = 5.85\%$ [1]. (b) Product A is better despite the lower nominal rate because its EAR is higher [1]. This demonstrates why consumers must compare EAR, not nominal rates [1].