Exam FM · practice
Exam FM — Loans (Amortization & Sinking Funds) Practice Flashcards
Thirty original SOA/CAS Exam FM-style multiple-choice problems on loan repayment — amortization payments, prospective and retrospective outstanding balances, the interest-principal split, total interest, drop and balloon payments, refinancing, sinking funds, and amortization-versus-sinking-fund comparisons — each with a fully worked solution.
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- Calculation: paymentA loan of $\$25{,}000$ is to be repaid with $15$ equal annual payments, the first one year after the loan is made. The annual effective interest rate is $7\%$. Calculate the level annual payment. (A) $\$2{,}565$ (B) $\$2{,}745$ (C) $\$2{,}937$ (D) $\$3{,}113$ (E) $\$3{,}295$**Answer: (B).** The payment satisfies $L = P\,a_{\overline{15|}}$ at $i=0.07$. First $a_{\overline{15|}} = \dfrac{1-1.07^{-15}}{0.07}$. Since $1.07^{-15}\approx 0.362446$, $a_{\overline{15|}}\approx \dfrac{0.637554}{0.07}\approx 9.107914$. Then $P = \dfrac{25000}{9.107914}\approx \$2{,}744.87$. The answer is **(B) $\$2{,}745$**. Distractor (A) divides by $\ddot a_{\overline{15|}} = (1.07)\,a_{\overline{15|}}$ (treating the payments as an annuity-due); (C) uses $n=13$; (E) uses $n=12$.
- Calculation: paymentA loan is repaid with level annual payments of $\$4{,}000$ for $20$ years at an annual effective rate of $6\%$, the first payment due in one year. Calculate the amount of the loan. (A) $\$43{,}290$ (B) $\$45{,}879$ (C) $\$48{,}632$ (D) $\$80{,}000$ (E) $\$147{,}142$**Answer: (B).** The loan is the present value of the payments: $L = 4000\,a_{\overline{20|}}$ at $i=0.06$. $a_{\overline{20|}} = \dfrac{1-1.06^{-20}}{0.06}$. With $1.06^{-20}\approx 0.311805$, $a_{\overline{20|}}\approx \dfrac{0.688195}{0.06}\approx 11.469921$. $L = 4000 \times 11.469921 \approx \$45{,}879.68$. The answer is **(B) $\$45{,}879$**. Distractor (D) is the undiscounted total $20\times4000$; (E) is the accumulated value $4000\,s_{\overline{20|}}$; (C) uses the annuity-due factor.
- Calculation: balanceA $\$60{,}000$ loan is amortized with $12$ level annual payments at an annual effective rate of $8\%$. Calculate the outstanding balance immediately after the $5$th payment. (A) $\$35{,}000$ (B) $\$38{,}642$ (C) $\$41{,}452$ (D) $\$44{,}305$ (E) $\$46{,}913$**Answer: (C).** Payment: $P = \dfrac{60000}{a_{\overline{12|}}}$ at $8\%$. $a_{\overline{12|}} = \dfrac{1-1.08^{-12}}{0.08}\approx 7.536078$, so $P\approx \$7{,}961.70$. Prospective balance after $5$ payments uses the $7$ remaining payments: $B_{5} = P\,a_{\overline{7|}}$, with $a_{\overline{7|}} = \dfrac{1-1.08^{-7}}{0.08}\approx 5.206370$. $B_{5} = 7961.70 \times 5.206370 \approx \$41{,}451.6$. The answer is **(C) $\$41{,}452$**. Distractor (A) is the straight-line balance $60000\times7/12$ (ignoring interest); the higher options (D)–(E) correspond to using too few elapsed payments (a balance taken before the $5$th payment).
- Interest & principal splitA loan of $\$100{,}000$ is repaid with $25$ level annual payments at an annual effective rate of $5\%$. Calculate the interest portion of the first payment. (A) $\$2{,}095$ (B) $\$4{,}095$ (C) $\$5{,}000$ (D) $\$7{,}095$ (E) $\$7{,}333$**Answer: (C).** The interest in any payment is charged on the balance at the start of that period. For the first payment that balance is the full loan $L$: $I_{1} = i\,L = 0.05 \times 100{,}000 = \$5{,}000$. The answer is **(C) $\$5{,}000$**. The payment itself is $P = 100000/a_{\overline{25|}} = 100000/14.093945 \approx \$7{,}095.25$ (distractor (D)); the principal portion of payment 1 is $P - I_1 \approx \$2{,}095.25$ (distractor (A)). The interest portion is simply $iL$.
- Calculation: splitA $\$30{,}000$ loan is amortized with $10$ level annual payments at an annual effective rate of $9\%$. Calculate the principal repaid in the $4$th payment. (A) $\$2{,}346$ (B) $\$2{,}557$ (C) $\$2{,}787$ (D) $\$3{,}037$ (E) $\$4{,}675$**Answer: (B).** Payment: $P = \dfrac{30000}{a_{\overline{10|}}}$ at $9\%$. $a_{\overline{10|}} = \dfrac{1-1.09^{-10}}{0.09}\approx 6.417658$, so $P\approx \$4{,}674.60$. For a level-payment loan the principal in payment $t$ has the closed form $PR_{t} = P\,v^{\,n-t+1}$. Here $n=10$, $t=4$, so the exponent is $n-t+1 = 7$: $PR_{4} = 4674.60\,(1.09)^{-7}$. $1.09^{-7}\approx 0.547034$, so $PR_{4}\approx 4674.60 \times 0.547034 \approx \$2{,}557.17$. The answer is **(B) $\$2{,}557$**. Distractor (A) uses $v^{8}$ (principal of payment 3); (C) uses $v^{6}$ (payment 5); (E) is the payment $P$ itself.
- Interest & principal splitA loan is repaid with $20$ level annual payments at an annual effective rate of $6\%$. The principal repaid in the $8$th payment is $\$1{,}200$. Calculate the principal repaid in the $14$th payment. (A) $\$1{,}200$ (B) $\$1{,}431$ (C) $\$1{,}517$ (D) $\$1{,}702$ (E) $\$1{,}804$**Answer: (D).** For a level-payment loan the principal portions form a geometric sequence with ratio $(1+i)$: $PR_{t+k} = PR_{t}\,(1+i)^{k}$. From payment $8$ to payment $14$ is $k=6$ periods: $PR_{14} = PR_{8}\,(1.06)^{6} = 1200 \times 1.418519 \approx \$1{,}702.2$. The answer is **(D) $\$1{,}702$**. The lower distractors come from using too small a gap (e.g. $(1.06)^{4}$ or $(1.06)^{5}$ instead of $(1.06)^{6}$); (E) uses $(1.06)^{7}$, an off-by-one in the other direction.
- Total interestA $\$50{,}000$ loan is amortized with level annual payments over $18$ years at an annual effective rate of $7\%$. Calculate the total interest paid over the life of the loan. (A) $\$24{,}750$ (B) $\$31{,}500$ (C) $\$39{,}471$ (D) $\$50{,}000$ (E) $\$63{,}000$**Answer: (C).** Total interest equals total payments minus the loan principal. Payment: $P = \dfrac{50000}{a_{\overline{18|}}}$ at $7\%$. $a_{\overline{18|}} = \dfrac{1-1.07^{-18}}{0.07}\approx 10.059087$, so $P\approx \$4{,}970.63$. Total payments $= 18 \times 4970.63 = \$89{,}471.3$. Total interest $= 89471.3 - 50000 \approx \$39{,}471$. The answer is **(C) $\$39{,}471$**. Do **not** compute interest as $i\,L\times n = 0.07\times50000\times18 = \$63{,}000$ (distractor (E)) — that assumes the full balance stays outstanding the whole term, which is the sinking-fund pattern, not amortization.
- Calculation: splitA $\$40{,}000$ loan is repaid with $12$ level annual payments at an annual effective rate of $8\%$. Calculate the total principal repaid in payments $5$ through $9$ inclusive. (A) $\$13{,}205$ (B) $\$14{,}995$ (C) $\$16{,}823$ (D) $\$18{,}480$ (E) $\$20{,}120$**Answer: (C).** The total principal repaid between two payments equals the drop in outstanding balance: $\sum_{k=5}^{9} PR_{k} = B_{4} - B_{9}$. Payment: $P = \dfrac{40000}{a_{\overline{12|}}} = \dfrac{40000}{7.536078} \approx \$5{,}307.80$. Prospective balances: $B_{4} = P\,a_{\overline{8|}} = 5307.80 \times 5.746639 \approx \$30{,}502.0$. $B_{9} = P\,a_{\overline{3|}} = 5307.80 \times 2.577097 \approx \$13{,}678.7$. Principal in payments 5–9 $= 30502.0 - 13678.7 \approx \$16{,}823.3$. The answer is **(C) $\$16{,}823$**. The key is that summed principal equals the *drop in balance* $B_{4}-B_{9}$; the lower distractors drop a payment from the range (e.g. counting only payments 6–9 instead of 5–9).