BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: A woman worked for 30 years before retiring. At the end of the first year of employment she deposited 5000 into an account for her retirement. At the end of each subsequent year of employment, she deposited \(3 \%\) more than the prior year. The woman made a total of 30 deposits.

She will withdraw 50,000 at the beginning of the first year of retirement and will make annual withdrawals at the beginning of each subsequent year for a total of 30 withdrawals. Each of these subsequent withdrawals will be \(3 \%\) more than the prior year. The final withdrawal depletes the account.

The account earns a constant annual effective interest rate.
Calculate the account balance after the final deposit and before the first withdrawal.

Options:
A
760,694
B
\(\quad 783,948\)
C
797,837
D
\(\quad 805,541\)
E
821,379

Correct answer: C

Solution: \(50,000\left[\frac{(1+i)^{30}-(1.03)^{30}}{(1+i)^{30}(i-0.03)}\right](1+i)=5,000\left[\frac{(1+i)^{30}-(1.03)^{30}}{i-0.03}\right]\)
\(50,000 /(1+i)^{29}=5,000\)
\((1+i)^{29}=10\)
\(i=10^{1 / 29}-1=0.082637\)
The accumulated amount is
\(50,000\left[\frac{(1.082637)^{30}-(1.03)^{30}}{(1.082637)^{30}(0.082637-0.03)}\right](1.082637)=797,836.82\)
Question: An insurance company purchases a perpetuity-due providing a geometric series of quarterly payments for a price of 100,000 based on an annual effective interest rate of \(i\). The first and second quarterly payments are 2000 and 2010, respectively.

Calculate \(i\).

Options:
A
\(\quad 10.0 \%\)
B
\(\quad 10.2 \%\)
C
\(\quad 10.4 \%\)
D
\(\quad 10.6 \%\)
E
\(10.8 \%\)

Correct answer: D

Solution: The first payment is 2,000 , and the second payment of 2,010 is 1.005 times the first payment.
Since we are given that the series of quarterly payments is geometric, the payments multiply by 1.005 every quarter.

Based on the quarterly interest rate, the equation of value is
\(100,000=2,000+2,000(1.005) v+2,000(1.005)^{2} v^{2}+2,000(1.005)^{3} v^{3}+\cdots=\frac{2,000}{1-1.005 v}\)
\(1-1.005 v=2,000 / 100,000 \Rightarrow v=0.98 / 1.005\).
The annual effective rate is \(v^{-4}-1=(0.98 / 1.005)^{-4}-1=0.10601=10.6 \%\).
Question: A bank agrees to lend 10,000 now and \(X\) three years from now in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual effective rate of \(6 \%\) for the first 5 years and at a force of interest \(\delta_{t}=\frac{1}{t+1}\) for \(t \geq 5\).

Calculate \(X\).

Options:
A
\(\quad 23,500\)
B
\(\quad 24,000\)
C
\(\quad 24,500\)
D
\(\quad 25,000\)
E
\(\quad 25,500\)

Correct answer: C

Solution: \(\left[10,000(1.06)^{5}+X(1.06)^{2}\right] e^{\int_{5}^{10} \frac{1}{t+1} d t}=75,000\)
\((13,382.26+1.1236 X) \frac{11}{6}=75,000\)
\(1.1236 X=27,526.83\)
\(X=24,498.78\)
Question: A company takes out a loan of \(15,000,000\) at an annual effective discount rate of \(5.5 \%\). You are given:
i) The loan is to be repaid with \(n\) annual payments of \(1,200,000\) plus a drop payment one year after the \(n\)th payment.
ii) The first payment is due three years after the loan is taken out.

Calculate the amount of the drop payment.

Options:
A
\(\quad 79,100\)
B
\(\quad 176,000\)
C
\(\quad 321,300\)
D
\(\quad 959,500\)
E
\(1,180,300\)

Correct answer: D

Solution: The effective annual interest rate is \(i=(1-d)^{-1}-1=(1-0.055)^{-1}-1=5.82 \%\)
The balance on the loan at time 2 is \(15,000,000(1.0582)^{2}=16,796,809\).
The number of payments is given by \(1,200,000 a_{\bar{n} \mid}=16,796,809\) which gives \(n=29.795 \Rightarrow 29\) payments of \(1,200,000\). The final equation of value is
\(1,200,000 a_{29}+X(1.0582)^{-30}=16,796,809\)
\(X=(16,796,809-16,621,012)(5.45799)=959,490\).
Question: Tim takes out an \(n\)-year loan with equal annual payments at the end of each year.
The interest portion of the payment at time ( \(n-1\) ) is equal to 0.5250 of the interest portion of the payment at time ( \(n-3\) ) and is also equal to 0.1427 of the interest portion of the first payment.

Calculate \(n\).

Options:
A
18
B
20
C
22
D
24
E
26

Correct answer: C

Solution: \(1-v^{2}=0.525\left(1-v^{4}\right) \Rightarrow 1=0.525\left(1+v^{2}\right) \Rightarrow v^{2}=0.90476 \Rightarrow v=0.95119\)
\(1-v^{2}=0.1427\left(1-v^{n}\right) \Rightarrow 1-\mathrm{v}^{n}=(1-0.90476) / 0.1427=0.667414 \Rightarrow v^{n}=0.332596\)
\(n=\ln (0.332596) / \ln (0.95119)=22\)
Question: A 5-year loan of 500,000 with an annual effective discount rate of \(8 \%\) is to be repaid by level end-of-year payments.

If the first four payments had been rounded up to the next multiple of 1,000 , the final payment would be \(X\).

Calculate \(X\).

Options:
A
103,500
B
111,700
C
115,200
D
125,200
E
127,500

Correct answer: D

Solution: The annual effective interest rate is \(0.08 /(1-0.08)=0.08696\). The level payments are \(500,000 / a_{\overline{5} \mid 0.08696}=500,000 / 3.9205=127,535\). This rounds up to 128,000 . The equation of value for \(X\) is
\(128,000 a_{40.08696}+X(1.08696)^{-5}=500,000\)
\(X=(500,000-417,466.36)(1.51729)=125,227\).
Question: A company plans to invest \(X\) at the beginning of each month in a zero-coupon bond in order to accumulate 100,000 at the end of six months. The price of each bond as a percentage of redemption value is given in the following chart:

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Maturity (months) & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline Price & \(99 \%\) & \(98 \%\) & \(97 \%\) & \(96 \%\) & \(95 \%\) & \(94 \%\) \\
\hline
\end{tabular}

Calculate \(X\) given that the bond prices will not change during the six-month period.

Options:
A
\(\quad 15,667\)
B
16,078
C
16,245
D
16,667
E
17,271

Correct answer: B

Solution: The accumulated value is the reciprocal of the price. The equation is
\(X[(1 / 0.94)+(1 / 0.95)+(1 / 0.96)+(1 / 0.97)+(1 / 0.98)+(1 / 0.99)]=100,000\).
\(X=16,078\)
Question: Jeff has 8000 and would like to purchase a 10,000 bond. In doing so, Jeff takes out a 10 year loan of 2000 from a bank and will make interest-only payments at the end of each month at a nominal rate of \(8.0 \%\) convertible monthly. He immediately pays 10,000 for a 10 -year bond with a par value of 10,000 and \(9.0 \%\) coupons paid monthly.

Calculate the annual effective yield rate that Jeff will realize on his 8000 over the 10 -year period.

Options:
A
\(\quad 9.30 \%\)
B
\(\quad 9.65 \%\)
C
\(10.00 \%\)
D
\(10.35 \%\)
E
\(10.70 \%\)

Correct answer: B

Solution: Jeff's monthly cash flows are coupons of \(10,000(0.09) / 12=75\) less loan payments of \(2000(0.08) / 12=13.33\) for a net income of 61.67 . At the end of the ten years (in addition to the 61.67) he receives 10,000 for the bond less a 2,000 loan repayment. The equation is \(8000=61.67 a_{120 l_{i}{ }^{(12)} / 12}+8000\left(1+i^{(12)} / 12\right)^{-120}\)
\(i^{(12)} / 12=0.00770875\)
\(i=1.00770875^{12}-1=0.0965=9.65 \%\).
Question: An investor owns a bond that is redeemable for 250 in 6 years from now. The investor has just received a coupon of \(c\) and each subsequent semiannual coupon will be \(2 \%\) larger than the preceding coupon. The present value of this bond immediately after the payment of the coupon is 582.53 assuming an annual effective yield rate of \(4 \%\).

Calculate \(c\).

Options:
A
32.04
B
32.68
C
40.22
D
48.48
E
49.45

Correct answer: A

Solution: The effective semi-annual yield rate is \(1.04=\left(1+\frac{i^{(2)}}{2}\right)^{2}=>\frac{i^{(2)}}{2}=1.9804 \%\). Then,
\(582.53=c(1.02) v+c(1.02 v)^{2}+\cdots+c(1.02 v)^{12}+250 v^{12}\)
\(=c \frac{1.02 v-(1.02 v)^{13}}{1-1.02 v}+250 v^{12}=12.015 c+197.579=>c=32.04\).
\(582.53=c \frac{1.02 v-(1.02 v)^{13}}{1-1.02 v}+250 v^{12}=12.015 c+197.579=>c=32.04\)
Question: Rhonda purchases a perpetuity providing a payment of 1 at the beginning of each year. The perpetuity's Macaulay duration is 30 years.

Calculate the modified duration of this perpetuity.

Options:
A
28.97
B
29.00
C
29.03
D
29.07
E
29.10

Correct answer: C

Solution: \(30=\operatorname{Mac} D=\frac{\sum_{n=0}^{\infty} n v^{n}}{\sum_{n=0}^{\infty} v^{n}}=\frac{I a_{\varpi}}{\ddot{a}_{\infty}}=\frac{1 /(d i)}{1 / d}=\frac{(1+i) / i^{2}}{(1+i) / i}=\frac{1}{i}\) and so \(i=1 / 30\).
\(M o d D=\frac{M a c D}{1+i}=\frac{30}{1+\frac{1}{30}}=29.032\).
Question: An insurance company wants to match liabilities of 25,000 payable in one year and 20,000 payable in two years with specific assets. The following assets are currently available:
i) One-year bond with an annual coupon of \(6.75 \%\) at par
ii) Two-year bond with annual coupons of \(4.50 \%\) at par
iii) Two-year zero-coupon bond yielding 5.00\% annual effective

Calculate the smallest amount the company needs to disburse today to purchase assets that will exactly match these liabilities.

Options:
A
41,220
B
41,390
C
41,560
D
41,660
E
41,750

Correct answer: C

Solution: The strategy is to use the two highest yielding assets: the one-year bond and the two-year zerocoupon bond. The cost of these bonds is \(25,000 / 1.0675+20,000 / 1.05^{2}=41,560\).
Question: Martha leaves an estate of 500,000 . Interest on this estate is paid to John for the first \(X\) years at the end of each year. Karen receives annual interest payments from the end of year \(X+1\) forever. At an annual effective interest rate of \(5 \%\), the present value of Karen's interest payments is 1.59 times the present value of John's.

Calculate \(X\).

Options:
A
6
B
7
C
8
D
9
E
10

Correct answer: E

Solution: Let \(P\) be the annual interest paid. The present value of John's payments is \(P a_{\bar{X} \mid 0.05}\). The present value of Karen's payments is \(P(1.05)^{-X} a_{-0.05}=P(1.05)^{-X} / 0.05\). Then, \(P(1.05)^{-X} / 0.05=1.59 P a_{\bar{X} \mid 0.05}\)
\(\frac{1.05^{-X}}{0.05}=1.59 \frac{1-1.05^{-X}}{0.05}\)
\(1.59=2.59(1.05)^{-X}\)
\(\ln 1.59=\ln 2.59-X \ln 1.05\)
\(X=10\).
Question: A company is considering investing in a particular project. The project requires an investment of \(X\) today. Additional investments are required at the beginning of each of the next five years, with each year's investment \(5 \%\) greater than the previous year's investment.

The investment is expected to produce an income of 100 per year at the end of each year forever, with the first payment expected at the end of the first year.

At an annual effective interest rate of \(10.25 \%\), the project has a net present value of zero.
Calculate \(X\).

Options:
A
183
B
192
C
205
D
215
E
225

Correct answer: A

Solution: The present value of the income is \(100 a_{-\infty .1025}=100 / 0.1025=975.61\). The present value of the investment is
\(X\left[1+1.05 / 1.1025+(1.05 / 1.1025)^{2}+(1.05 / 1.1025)^{3}+(1.05 / 1.1025)^{4}+(1.05 / 1.1025)^{5}\right]\)
\(=X\left[1+1.05^{-1}+1.05^{-2}+1.05^{-3}+1.05^{-4}+1.05^{-5}\right]=X \frac{1-1.05^{-6}}{1-1.05^{-1}}=5.3295 X\).
Then \(975.61=5.3295 X\) for \(X=183.06\).
Question: A borrower took out a loan of 100,000 and promised to repay it with a payment at the end of each year for 30 years.

The amount of each of the first ten payments equals the amount of interest due. The amount of each of the next ten payments equals \(150 \%\) of the amount of interest due. The amount of each of the last ten payments is \(X\).

The lender charges interest at an annual effective rate of \(10 \%\).
Calculate \(X\).

Options:
A
\(\quad 3,204\)
B
\(\quad 5,675\)
C
7,073
D
\(\quad 9,744\)
E
11,746

Correct answer: D

Solution: The loan balance after 10 years is still 100,000 . For the next 10 payments, the interest paid is \(10 \%\) of the outstanding balance and therefore the principal repaid is \(5 \%\) of the outstanding balance. After 10 years the oustanding balance is \(100,000(0.95)^{10}=59,874\). Then, \(X=59,874 / a_{10 \mid 0.1}=59,874 / 6.14457=9,744\).
Question: An entrepreneur takes out a business loan for 60,000 with a nominal annual interest rate compounded monthly. The loan is scheduled to be paid off with level monthly payments, plus a final drop payment. All payments will be made at the end of the month.

The principal portion of the payment is 1,400 for the first month and 1,414 for the second month.
Calculate the drop payment.

Options:
A
780
B
788
C
1183
D
1676
E
1692

Correct answer: E

Solution: Let \(j\) be the monthly rate and \(X\) be the level monthly payment. The principal repaid in the first payment is \(1400=X-60,000 j\). The principal repaid in the second payment is \(1414=X- (60,000-1400) j\). Substituting \(X=1400+60,000 j\) from the first equation gives \(1414=1400+ 60,000 j-58,600 j\) or \(14=1400 j\) and thus \(j=0.01\) and \(X=2000\). Let \(n\) be the number of payments. Then \(60,000=2000 a_{n 0.01}\) and the calculator (or algebra) gives \(n=35.8455\). The equation for the drop payment, \(P\), is \(60,000=2000 a_{-\left.35\right|_{0.01}}+P v^{36}=58,817.16+0.698925 P\) for \(P =1692\).
Question: A student borrows money to pay for university tuition. He borrows 1000 at the end of each month for four years. No payments are made to repay the loan until the end of five years.

The loan accumulates interest at a \(6 \%\) nominal interest rate convertible monthly for the first two years and at an \(8 \%\) nominal interest rate convertible monthly for the following two years.

Calculate the loan balance at the end of four years immediately following the receipt of the final 1000.

Options:
A
\(\quad 54,098\)
B
55,224
C
55,762
D
56,134
E
56,350

Correct answer: C

Solution: The accumulated value is \(1000\left(s_{\left.24\right|_{0.06 / 12}}(1+0.08 / 12)^{24}+s_{\left.24\right|_{0.08 / 12}}\right)=1000(25.4320(1.1729)+25.9332)=55,762\).
Question: A loan is amortized with level monthly payments at an annual effective interest rate of \(10 \%\). The amount of principal repaid in the 6th month is 500.

Calculate the principal repaid in the 30th month.

Options:
A
500
B
555
C
605
D
705
E
805

Correct answer: C

Solution: Each month the principal paid increases by \(1.1^{1 / 12}\). Thus, the amount of principal paid increases to \(500\left(1.1^{1 / 12}\right)^{30-6}=500(1.1)^{2}=605\).
Question: Bond X is a 20 -year bond with annual coupons and the following characteristics:
i) \(\quad\) Par value is 1000 .
ii) The annual coupon rate is \(10 \%\).
iii) Bond X is callable at par on any of the last five coupon dates.

Calculate the maximum purchase price for Bond X that will guarantee an annual effective yield rate of at least \(5 \%\).

Options:
A
1519
B
1542
C
1570
D
1596
E
1623

Correct answer: B

Solution: Because the yield is less than the coupon rate, the bond sells at a premium and the worst case for the buyer is an early call. Hence the price should be calculated based on the bond being called at time 16. The price is \(100 a_{\overline{16 \mid 0.05}}+1000(1.05)^{-16}=100(10.0378)+458.11=1542\). (When working with callable bonds, the maximum a buyer will pay is the smallest price over the various call dates. Paying more may not earn the desired yield.)
Question: You are given the following information about a 20 -year bond with face amount 7500 :
i) The bond has an annual coupon rate of \(7.4 \%\) paid semiannually.
ii) The purchase price results in an annual nominal yield rate to the investor of 5.3\% convertible semiannually.
iii) The amount for amortization of premium in the fourth coupon payment is 28.31.

Calculate the redemption value of the bond.

Options:
A
7660
B
7733
C
7795
D
7879
E
7953

Correct answer: A

Solution: The book value after the third coupon is \(7500(0.037) a_{\overline{370} 0.0265}+C(1.0265)^{-37}=6493.05+0.379943 C\) and after the fourth coupon it is \(7500(0.037) a_{\overline{360.0265}}+C(1.0265)^{-36}=6387.61+0.390012 C\). Then,
\(6493.05+0.379943 C-(6387.61+0.390012 C)=28.31\)
\(105.44-0.010069 C=28.31\)
\(C=7660.15\).
Question: You are given the following information about a 20 -year bond with face amount 1000 :
i) The bond has an annual coupon rate of \(r\) payable semiannually and is redeemable at par.
ii) The nominal annual yield rate convertible semiannually is \(7.2 \%\).
iii) The amount for accumulation of discount in the seventh coupon payment is 4.36.

Calculate \(r\).

Options:
A
\(2.1 \%\)
B
\(\quad 4.0 \%\)
C
\(4.3 \%\)
D
\(6.0 \%\)
E
\(6.9 \%\)

Correct answer: C

Solution: The book value after the sixth coupon is \(1000(r / 2) a_{\overline{340.036}}+1000(1.036)^{-34}=9716.01 r+300.45\). After the seventh coupon it is \(1000(r / 2) a_{\overline{330.036}}+1000(1.036)^{-33}=9565.79 r+311.26\). Then, \(4.36=9565.79 r+311.26-(9716.01 r+300.45)=10.81-150.22 r r=(10.81-4.36) / 150.22=0.0429\).
Question: Bond A and Bond B are both annual coupon, five-year, 10,000 par value bonds bought to yield an annual effective rate of \(4 \%\).
i) Bond A has an annual coupon rate of \(r \%\), a redemption value that is \(10 \%\) below par, and a price of \(P\).
ii) Bond B has an annual coupon rate of \((r+1) \%\), a redemption value that is \(10 \%\) above par, and a price of \(1.2 P\).

Calculate \(r\) \%.

Options:
A
\(\quad 5.85 \%\)
B
\(\quad 6.85 \%\)
C
\(7.85 \%\)
D
\(8.85 \%\)
E
\(\quad 9.85 \%\)

Correct answer: B

Solution: The two equations are:
\(P=(10,000 r) a_{5 \mid 0.04}+9,000(1.04)^{-5}=44,518.22 r+7,397.34\)
\(1.2 P=[10,000(r+0.01)] a_{510.04}+11,000(1.04)^{-5}=44,518.22 r+9,486.38\).
Subtracting the first equation from the second gives \(0.2 P=2089.04\) for \(P=10,445\).20. Inserting this in the first equation gives \(r=(10,445.20-7,397.34) / 44,518.22=0.0685\).
Question: You are given the following information about an \(n\)-year bond, where \(n>10\) :
i) The bond pays \(8 \%\) semiannual coupons and has face amount 1000 .
ii) The bond is redeemable at par.
iii) The bond is callable at par 5 years after issue or 10 years after issue.
iv) \(\quad P\) is the price to guarantee a yield of \(6.8 \%\) convertible semiannually and \(Q\) is the price to guarantee a yield of \(8.8 \%\) convertible semiannually.
v) \(\quad|\quad| P-Q \mid=123.36\).

Calculate \(n\).

Options:
A
11
B
15
C
19
D
22
E
26

Correct answer: C

Solution: When the yield is \(6.8 \%<8 \%\), the bond is sold at a premium and hence an early call is most disadvantageous. Therefore, \(P=40 a_{\overline{10} 0.034}+1000(1.034)^{-10}=1050.15\). When the yield is \(8.8 \%> 8 \%\), the bond is sold at discount. Hence, \(Q<1000<P\). and thus \(Q=1050.15-123.36=926.79\). Also, because the bond is sold at a discount, the latest call is the most disadvantageous. Thus, \(926.79=40 a_{2 n l 0.044}+1000(1.044)^{-2 n}=\frac{40}{0.044}+(1.044)^{-2 n}\left(1000-\frac{40}{.044}\right)=909.09+90.90(1.044)^{-2 n}\)
\(17.70=90.90(1.044)^{-2 n}\)
\(2 n=-\ln (17.70 / 90.90) / \ln (1.044)=38\)
\(n=19\).
Question: A company must pay liabilities of 1000 at the end of year 1 and \(X\) at the end of year 2 . The only investments available are:
i) One-year zero-coupon bonds with an annual effective yield of \(5 \%\)
ii) Two-year bonds with a par value of 1000 and \(10 \%\) annual coupons, with an annual effective yield of \(6 \%\)

The company constructed a portfolio that creates an exact cash flow matching strategy for these liabilities. The total purchase price of this portfolio is 1783.76.

Calculate the amount invested in the one-year zero-coupon bonds.

Options:
A
784
B
831
C
871
D
915
E
935

Correct answer: C

Solution: Let \(A\) be the redemption value of the zero-coupon bonds purchased and \(B\) the number of twoyear bonds purchased. The total present value is:
\(1783.76=A / 1.05+B\left(100 / 1.06+1100 / 1.06^{2}\right)=0.95238 A+1073.3357 B\).
To exactly match the cash flow at time one, \(A+100 B=1000\). Substituting \(B=10-0.01 A\) in the first equation gives \(1783.76=0.95238 A+10733.357-10.733357 A\) for \(A=8949.597 / 9.780977 =915\). The amount invested is then \(915 / 1.05=871\).
Question: A 20-year bond priced to have an annual effective yield of \(10 \%\) has a Macaulay duration of 11 . Immediately after the bond is priced, the market yield rate increases by \(0.25 \%\). The bond's approximate percentage price change, using a first-order modified approximation, is \(X\).

Calculate \(X\).

Options:
A
\(-2.22 \%\)
B
\(-2.47 \%\)
C
\(-2.50 \%\)
D
\(-2.62 \%\)
E
\(-2.75 \%\)

Correct answer: C

Solution: The modified duration is \(11 / 1.10=10\). Then, \(P(0.1025) \approx P(0.10)[1-(0.1025-0.10) 10]=0.975 P(0.10)\). Therefore, the approximate percentage price change is \(100(0.975-1)=-2.50 \%\).
Question: A company owes 1000 one year from now and 1000 two years from now.
Which of the following demonstrates a strategy to use exact cash-flow matching between assets and liabilities?
I. The company purchases a one-year zero-coupon bond and a two-year zero-coupon bond, each with a face amount of 1000 .
II. The company deposits 1859.41 into an account that currently earns an annual effective interest rate of \(5 \%\) that is subject to change in one year.
III. The company purchases an asset that has the same duration as the liabilities and a larger convexity.

Options:
A
I only
B
II only
C
III only
D
I, II, and III
E
The correct answer is not given by (A), (B), (C) or (D).

Correct answer: A

Solution: I provides cash flows that exactly matches the liabilities. II only has \(\mathrm{PV}(\mathrm{A})=\mathrm{PV}(\mathrm{B})\), which is not sufficient for exact matching. III describes Redington immunization, not exact matching.
Question: Two 15-year par value bonds, \(X\) and \(Y\), each pay an annual coupon of 200 at the end of the year. The face amount of Bond X is one-half the face amount of Bond Y .

At an annual effective yield of \(i\), the price of Bond X is 2695.39 and the price of Bond Y is 3490.78.

Calculate the coupon rate for Bond X.

Options:
A
\(\quad 6.3 \%\)
B
\(\quad 7.4 \%\)
C
\(\quad 8.8 \%\)
D
\(\quad 10.0 \%\)
E
\(11.4 \%\)

Correct answer: D

Solution: Let \(F\) be the face amount of Bond X. Then, \(2695.39=200 a_{15 \mid}+F v^{15}\) and \(3490.78=200 a_{15 \mid}+2 F v^{15}\).
Subtract the first equation from the second to obtain \(795.39=F v^{15}\).
Then for bond \(X, 2695.39=200 a_{15 \mid}+795.39 \Rightarrow a_{15 \mid}=(2695.39-795.39) / 200=9.5\). This implies \(i=0.0634\). Then \(9.5=\left(1-v^{15}\right) / 0.0634 \Rightarrow v^{15}=1-0.0634(9.5)=0.3977\) and \(F=795.39 / 0.3977=2000\). The coupon rate is \(200 / 2000=10.0 \%\).
Question: An annuity provides level payments of 1000 every six months for a fixed period.
Using an annual effective interest rate of \(i\), the future value of this annuity at the time of the last payment is \(19,549.25\) and the present value of this annuity at the time of the first payment is 7,968.89.

Calculate \(i\).

Options:
A
\(\quad 7.4 \%\)
B
\(\quad 8.5 \%\)
C
\(\quad 15.4 \%\)
D
\(\quad 17.0 \%\)
E
\(17.7 \%\)

Correct answer: E

Solution: Let \(n\) be the number of payments and let \(j\) be the interest rate per half-year. Because the given values are \(n-1\) half-years apart, \(7,968.89(1+j)^{n-1}=19,549.25\). Also,
\(7,968.89=1,000 \ddot{a}_{\bar{n}}=1,000\left(a_{\overline{n-1}}+1\right)=1,000\left(\frac{1-v^{n-1}}{j}+1\right)=1,000\left(\frac{1-7,968.89 / 19,549.25}{j}+1\right)\).
Then,
\(j=\frac{1-7,968.89 / 19,549.25}{\frac{7,968.89}{1,000}-1}=0.085\)
for \(i=(1.085)^{2}-1=0.1772=17.7 \%\).
Question: Let A and B be bonds with semiannual coupons as described in the table below:

\begin{tabular}{c|ccccc} 
Bond & Price & \begin{tabular}{c} 
Annual \\
coupon rate
\end{tabular} & Par & \begin{tabular}{c} 
Years to \\
redemption
\end{tabular} & \begin{tabular}{c} 
Annual nominal yield rate \\
convertible semiannually
\end{tabular} \\
\hline A & \(X\) & \(8 \%\) & 1000 & 5 & \(6 \%\) \\
B & \(X\) & \(y\) & 1000 & 5 & \(7 \%\)
\end{tabular}

Calculate \(y\).

Options:
A
\(\quad 8.45 \%\)
B
\(\quad 8.65 \%\)
C
\(8.85 \%\)
D
\(\quad 9.05 \%\)
E
\(\quad 9.25 \%\)

Correct answer: D

Solution: Let \(C\) be the amount of the semiannual coupon for bond B .
\(X=40 a_{10 \mid 0.03}+1000(1.03)^{-10}=1085.30\)
\(X=1085.30=C a_{100.035}+1000(1.035)^{-10}=8.3166 C+708.9188\)
\(C=(1085.30-708.9188) / 8.3166=45.2566\)
\(y=\frac{45.2566 \times 2}{1000}=0.0905=9.05 \%\).