Comparing Loan Offers with Compound Interest
A family needs to borrow $8,000 for home repairs. Bank A offers 6.0% annual interest compounded monthly, repaid as a single lump sum after 3 years. Bank B offers 5.6% annual interest compounded continuously for the same 3-year term but charges a $150 arrangement fee added to the principal at the start. A relative offers to lend the $8,000 at a flat 5.5% simple interest per year with no fees. The family wants the cheapest total repayment but is unsure how to compare compounding methods fairly.
1. Calculate the total repayment for each of the three options, showing the formula used for each.
Read the worked answer
Bank A (monthly compounding): A = P(1 + r/n)^(nt) = 8000(1 + 0.06/12)^(12·3) = 8000(1.005)^36.
(1.005)^36 ≈ 1.19668, so A ≈ $9,573.44.
Bank B (continuous, fee added): principal = 8000 + 150 = 8150. A = Pe^(rt) = 8150·e^(0.056·3) = 8150·e^0.168.
e^0.168 ≈ 1.18293, so A ≈ 8150 × 1.18293 ≈ $9,640.88.
Relative (simple interest): A = P(1 + rt) = 8000(1 + 0.055·3) = 8000(1.165) = $9,320.00.
Ordering: relative ($9,320) < Bank A ($9,573.44) < Bank B ($9,640.88).
2. The family could repay Bank A early after 18 months. Compute the payoff then and assess whether early repayment changes the ranking.
Read the worked answer
Bank A after 18 months: A = 8000(1.005)^18.
(1.005)^18 ≈ 1.09393, so payoff ≈ $8,751.44.
Bank B after 18 months: A = 8150·e^(0.056·1.5) = 8150·e^0.084 ≈ 8150 × 1.08762 ≈ $8,864.10.
Relative after 18 months: 8000(1 + 0.055·1.5) = 8000(1.0825) = $8,660.00.
Ranking unchanged: relative < Bank A < Bank B.
Note: early repayment of Bank A avoids the remaining 18 months of interest, saving about $9,573.44 - $8,751.44 = $822 versus full term. The relative's simple interest accrues linearly, so its advantage shrinks slightly over shorter terms but it remains cheapest here because its rate is lowest and no fees apply.