Compound interest - Answer key

Fluency

$I = 4000 \times 0.05 \times 3 = 600$ dollars. Total: $4600$ .
$A = 4000 \times 1.05^3 = 4000 \times 1.157625 = 4630.50$ dollars.
Compound gives $4630.50 - 4600 = 30.50$ dollars more.
$A = 10000 \times 1.03^6 = 10000 \times 1.19405 \approx 11940.52$ dollars.
$A = 1800 \times 0.75^2 = 1800 \times 0.5625 = 1012.50$ dollars.
$A = 5000\!\left(1 + \dfrac{0.08}{4}\right)^{8} = 5000 \times 1.02^8 = 5000 \times 1.17166 \approx 5858.30$ dollars.
$A = 3000 \times 1.1^4 = 3000 \times 1.4641 = 4392.30$ dollars.
$A = 7000 \times 1.06^5 = 7000 \times 1.33823 \approx 9367.58$ . Interest $= 9367.58 - 7000 = 2367.58$ dollars.

Reasoning

$A = 12000\!\left(1 + \dfrac{0.045}{12}\right)^{36} = 12000 \times 1.00375^{36} \approx 12000 \times 1.14396 \approx 13727.55$ dollars.
$28000 \times 0.85^t < 10000$ . $0.85^t < 0.3571$ . By trial: $0.85^6 \approx 0.3771$ , $0.85^7 \approx 0.3206$ . First below $10000 after 7 whole years.
Option 1: $8000 \times 1.06^5 = 8000 \times 1.33823 \approx 10705.80$ . Option 2: $8000 \times 1.004833^{60} \approx 8000 \times 1.33556 \approx 10684.48$ . Option 1 ( $6\%$ annually) gives about $21 more.
$15000 = P \times 1.07^{10}$ . $1.07^{10} \approx 1.96715$ . $P = \dfrac{15000}{1.96715} \approx 7625.85$ dollars.
$4500 = 9000 \times (1-r)^4$ . $(1-r)^4 = 0.5$ . $1 - r = 0.5^{0.25} \approx 0.8409$ . $r \approx 0.159$ , so approximately $15.9\%$ per year.
$A = 20000\!\left(1 + \dfrac{0.032}{4}\right)^{32} = 20000 \times 1.008^{32} \approx 20000 \times 1.29098 \approx 25819.63$ . Interest $= 25819.63 - 20000 = 5819.63$ dollars.

Reasoning

Simple interest after $T$ years: $A_S = P(1 + rT)$ , which is linear in $T$ . Compound interest: $A_C = P(1+r)^T$ , which is exponential. For small $T$ , $(1+r)^T \approx 1 + rT$ (nearly equal). As $T$ increases, the exponential term grows faster than the linear term because each year’s interest is applied to an ever-larger base. The gap is $P[(1+r)^T - (1 + rT)]$ , which increases with $T$ .
Effective annual rate $= (1.015)^{12} - 1 \approx 1.19562 - 1 = 0.19562$ , or about $19.6\%$ per year.
Bank A: $20000 \times 1.05^{10} = 20000 \times 1.62889 \approx 32577.89$ . Bank B: $20000\!\left(1 + \dfrac{0.049}{365}\right)^{3650} \approx 20000 \times e^{0.49} \approx 20000 \times 1.63232 \approx 32646.33$ . Bank B gives slightly more (about $68 extra). Mia should choose Bank B.
Need $50000 \times 0.8^n < 5000$ . $0.8^n < 0.1$ . By trial: $0.8^{10} \approx 0.10737$ , $0.8^{11} \approx 0.08590$ . Replace after 11 whole years.

Reasoning

Simple interest for $2$ years: $I_S = P \times r \times 2 = 2Pr$ . Compound interest for $2$ years: $A_C = P(1+r)^2 = P(1 + 2r + r^2)$ . So $I_C = A_C - P = 2Pr + Pr^2$ . Difference: $I_C - I_S = (2Pr + Pr^2) - 2Pr = Pr^2$ .
$15000 \times 1.08^t > 25000$ . $1.08^t > \tfrac{5}{3} \approx 1.6667$ . By trial: $1.08^6 \approx 1.5869$ , $1.08^7 \approx 1.7138$ . Debt exceeds $25000 after 7 whole years. Rule of 70: doubling time $\approx \tfrac{70}{8} = 8.75$ years; the debt reaches $\tfrac{5}{3}$ of the original (not double) so it happens sooner than the doubling time, which is consistent.
Need $10000 \times 1.06^t > 15000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.5$ . $1.02913^t > 1.5$ . By trial: $1.02913^{14} \approx 1.494$ , $1.02913^{15} \approx 1.538$ . After 15 whole years the first investment first exceeds the second.

Year 10 core - answers