1. **Problem statement:** Ray took a 60-month fixed installment loan of 12000 with monthly payments of 232. He wants to repay the loan in full after 11 payments instead of the 12th. We need to find the APR (annual percentage rate) of the loan. 2. **Formula used:** The loan is an installment loan with fixed monthly payments. The formula for the monthly payment $P$ on a loan amount $L$ with monthly interest rate $i$ over $n$ months is: $$P = L \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ 3. **Known values:** - $L = 12000$ - $P = 232$ - $n = 60$ 4. **Goal:** Find monthly interest rate $i$, then APR = $i \times 12 \times 100$%. 5. **Rearranging the formula to solve for $i$ is complex, so we use iterative or numerical methods.** 6. **Approximate $i$ by trial or financial calculator:** Try $i=0.01$ (1% monthly): $$232 \stackrel{?}{=} 12000 \times \frac{0.01(1.01)^{60}}{(1.01)^{60} - 1}$$ Calculate denominator: $$(1.01)^{60} = 1.8194$$ Calculate numerator: $$0.01 \times 1.8194 = 0.018194$$ Calculate fraction: $$\frac{0.018194}{1.8194 - 1} = \frac{0.018194}{0.8194} = 0.02221$$ Calculate payment: $$12000 \times 0.02221 = 266.52$$ Too high, so $i$ is less than 1%. Try $i=0.008$ (0.8% monthly): $$(1.008)^{60} = 1.6010$$ $$0.008 \times 1.6010 = 0.012808$$ $$\frac{0.012808}{1.6010 - 1} = \frac{0.012808}{0.6010} = 0.02131$$ $$12000 \times 0.02131 = 255.72$$ Still too high. Try $i=0.006$ (0.6% monthly): $$(1.006)^{60} = 1.4258$$ $$0.006 \times 1.4258 = 0.008555$$ $$\frac{0.008555}{1.4258 - 1} = \frac{0.008555}{0.4258} = 0.02008$$ $$12000 \times 0.02008 = 240.96$$ Still too high. Try $i=0.005$ (0.5% monthly): $$(1.005)^{60} = 1.3489$$ $$0.005 \times 1.3489 = 0.006744$$ $$\frac{0.006744}{1.3489 - 1} = \frac{0.006744}{0.3489} = 0.01932$$ $$12000 \times 0.01932 = 231.84$$ Close to 232, so $i \approx 0.005$ monthly. 7. **Calculate APR:** $$\text{APR} = i \times 12 \times 100 = 0.005 \times 12 \times 100 = 6\%$$ **Final answer:** APR = 6%