BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: A borrower takes out a loan to be repaid over 20 years. The first payment is 1102 payable at the end of the first month. Each subsequent monthly payment is five more than the previous month's payment.

Calculate the accumulated value of the payments at the end of 15 years using an annual effective interest rate of 6.5\%.

Options:
A
442,031
B
443,525
C
445,578
D
447,287
E
448,547

Correct answer: A

Solution: The effective monthly rate is \(1.065^{1 / 12}-1=0.0052617\). The accumulated value is
\(1097 s_{1800_{0.0052617}}+5(I S)_{1800_{0.0052617}}\)
\(=1097(298.733)+5 \frac{\left.\ddot{s}_{180}\right|_{0.0052617}-180}{0.0052617}\)
\(=327,710+5 \frac{300.3049-180}{0.0052617}=442,031\).
Question: A two-year loan of 100 is repaid with a payment of \(X\) at the end of the first year and \(2 X\) at the end of the second year. The annual effective interest rate charged by the lender is \(8 \%\) in the first year and \(i\) in the second year. The annual effective yield rate for the lender is \(10 \%\).

Calculate \(i\).

Options:
A
\(\quad 12.8 \%\)
B
\(\quad 12.9 \%\)
C
\(\quad 13.0 \%\)
D
\(\quad 13.1 \%\)
E
\(13.2 \%\)

Correct answer: E

Solution: Based on the effective yield rate, \(100=X / 1.1+2 X / 1.1^{2} \Rightarrow X=39.03\). After one year, the outstanding loan balance is \(100+8-39.03=68.97\). For the balance to be zero after two years, \(68.97(1+i)-2(39.03)=0 \Rightarrow i=78.06 / 68.97-1=0.1318=13.2 \%\).
Question: Determine which of the following conditions are necessary for an immunization strategy.
I. The present value of the cash inflow from the assets is equal to the present value of the cash outflow from the liabilities.
II. The price sensitivity to changes in interest rates is greater for assets than for liabilities.
III. The convexity of assets is less than the convexity of liabilities.

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 is true.
II is false, the price sensitivity of assets and liabilities must be equal.
III is false, the convexity of assets should be greater than the convexity of liabilities.
Question: Consider a 7-year loan to be repaid with equal payments made at the end of each year. The annual effective interest rate is \(10 \%\).

Calculate the Macaulay duration of the loan payments.

Options:
A
\(\quad 3.15\)
B
\(\quad 3.29\)
C
3.40
D
3.50
E
\(\quad 3.62\)

Correct answer: E

Solution: Let \(x\) be the annual payment amount. Macaulay duration is
\(\frac{\frac{1 x}{1.1}+\frac{2 x}{1.1^{2}}+\frac{3 x}{1.1^{3}}+\cdots+\frac{7 x}{1.1^{7}}}{\frac{x}{1.1}+\frac{x}{1.1^{2}}+\frac{x}{1.1^{3}}+\cdots+\frac{x}{1.1^{7}}}=\frac{17.6315}{4.8684}=3.62\)
Alternatively, the duration can be calculated as \((I a)_{7 \mid 0.1} / a_{7 \mid 0.1}\).
Question: A company has liabilities of 402.11 due at the end of each of the next three years.
The company will match the duration of its liabilities by investing a total of 1000 in one-year and three-year zero-coupon bonds. The annual effective yield of both bonds is \(10 \%\).

Calculate the amount the company will invest in one-year bonds.

Options:
A
366
B
402
C
442
D
500
E
532

Correct answer: E

Solution: PV of liabilities is \(402.11\left(1 / 1.1+1 / 1.1^{2}+1 / 1.1^{3}\right)=1000\). Duration of liabilities is 402.11 \(\left(1 / 1.1+2 / 1.1^{2}+3 / 1.1^{3}\right) / 1000=1.93653\). Let \(X\) be the investment in one-year bonds. To match duration, since zero-coupon bonds have duration \(=\) maturity, \(1.93653=[X+3(1000- X)] / 1000\). Then, \(2 X=3000-1936.53=1063.47\) and \(X=532\).
Question: A bond with a face value of 1000 and a redemption value of 1080 has an annual coupon rate of \(8 \%\) payable semiannually. The bond is bought to yield an annual nominal rate of \(10 \%\) convertible semiannually.

At this yield rate, the present value of the redemption value is 601 on the purchase date.
Calculate the purchase price of the bond.

Options:
A
911
B
923
C
956
D
974
E
984

Correct answer: C

Solution: Let \(n\) be the term of the bond in half-years. We know that \(601=1080 v^{n}\) and thus \(v^{n}=601 / 1080\)
. Then \(a_{\bar{n} \mid 0.05}=\frac{1-v^{n}}{0.05}=\frac{1-601 / 1080}{0.05}=8.87037\). The purchase price of the bond is
\(40 a_{\bar{n} 0.05}+1080 v^{n}=40(8.87037)+610=956\).
Question: Bond X and Bond Y are \(n\)-year bonds with face amount of 10,000 and semiannual coupons, each yielding an annual nominal interest rate of \(7 \%\) convertible semiannually. Bond X has an annual coupon rate of \(6 \%\) and redemption value \(c\). Bond Y has an annual coupon rate of \(5 \%\) and redemption value \(c+50\). The price of Bond X exceeds the price of Bond Y by 969.52.

Calculate \(n\).

Options:
A
14
B
17
C
23
D
34
E
46

Correct answer: B

Solution: Let \(X\) be the price of Bond X . Then, for the two bonds:
\(X=10,000(0.03) a_{2 n \mid 0.035}+c(1.035)^{-2 n}\)
\(X-969.52=10,000(0.025) a_{2 n \mid 0.035}+(c+50)(1.035)^{-2 n}\).
Subtracting the second equation from the first gives
\(969.52=50 a_{2 n 0.035}-50(1.035)^{-2 n}\)
\(969.52=\frac{50}{0.035}\left[1-(1.035)^{-2 n}\right]-50(1.035)^{-2 n}\)
\(1.035^{-2 n}=459.05 / 1478.57=0.310469\)
\(n=-(0.5) \ln (0.310469) / \ln (1.035)=17\).
Question: A loan of 10,000 is repaid at an annual effective interest rate of \(8 \%\), with equal payments made at the end of each year for ten years. The lender immediately deposits each payment into an account earning an annual effective interest rate of \(10 \%\).

Calculate the total amount of interest earned by the lender during the term of the loan.

Options:
A
11,589
B
13,576
C
13,751
D
14,191
E
14,903

Correct answer: C

Solution: Payment equals: \(\left(10,000 / a_{100.08}\right)=1,490.29\)
Accumulated total equals: \(1,490.29 s_{100.10}=23,751.46\)
\(23,751.46-10,000=13,751.46\)
Question: A worker is starting an endowment fund for his charity. The fund will accumulate at a nominal annual interest rate of \(12 \%\) convertible monthly. Beginning today, the worker will deposit 500 monthly for ten years. No deposits or withdrawals will be made for the subsequent ten years. Exactly twenty years from today, monthly payments of \(X\) will be made to his charity and continue forever.

Calculate \(X\).

Options:
A
3270
B
3572
C
3758
D
3796
E
3834

Correct answer: D

Solution: Value of fund after 20 years: \(500 \ddot{s}_{\left.120\right|_{0.01}}(1.01)^{120}=383,404.42\)
\(383,404.42=\frac{X}{d}=\frac{X}{i}(1+i)=\frac{X}{0.01}(1.01)\)
\(X=3796.08\)
Question: Matthew deposits 100 into Fund \(X\), which earns a nominal annual rate of discount of \(12 \%\) convertible quarterly.

At the same time, Sarah deposits 200 into Fund Y, which earns a constant annual force of interest of 0.08 .

After \(n\) years, the balance of Fund X equals the balance of Fund Y .
Calculate \(n\).

Options:
A
14.5
B
16.6
C
17.6
D
18.1
E
18.6

Correct answer: B

Solution: \(100\left(1-\frac{0.12}{4}\right)^{-4 n}=200 e^{0.08 n}\)
\((-4 n) \ln 0.97=\ln (200 / 100)+(0.08 n)\)
\(0.0418 n=\ln 2\)
\(n=16.57\)
Question: A firm has a liability cash flow of 100 at the end of year two and a second liability cash flow of 200 at the end of year three.

The firm also has asset cash flows of \(X\) at the end of years one and five.
Using an annual effective interest rate of \(10 \%\), calculate the absolute value of the difference between the Macaulay durations of the asset and liability cash flows.

Options:
A
0.018
B
0.020
C
0.022
D
0.024
E
0.026

Correct answer: C

Solution: \(\bar{d}_{L}=\frac{2 v^{3} 3+1 v^{2} 2}{2 v^{3}+1 v^{2}}=\frac{6.16078}{2.32908}=2.6452\)
\(\bar{d}_{A}=\frac{X v^{5} 5+X v}{X v^{5}+X v}=\frac{4.0137}{1.5300}=2.6233\)
\(2.6452-2.6233=0.0219\)
Question: A company has liabilities of 1000 due at the end of each of the next three years. The company will exactly match the cash flows of assets and liabilities. The following zero-coupon bonds are available:

\begin{tabular}{|c|c|}
\hline Maturity & Yield \\
\hline 1-Year & \(8 \%\) \\
\hline 2-Year & \(9 \%\) \\
\hline 3-Year & \(10 \%\) \\
\hline
\end{tabular}

Calculate how much more will be invested in the 1 -year bond than in the 3 -year bond.

Options:
A
0
B
84
C
132
D
158
E
175

Correct answer: E

Solution: The amount invested in three-year bond is equal to the PV of the third year's payout, \(1000 /\left(1.1^{3}\right)=751.31\)
The amount invested in one-year bond is equal to the PV of the first year's payout, \(1000 / 1.08=925.93\)
\(925.93-751.31=174.61\)
Question: A perpetuity-immediate with annual payments is priced at \(X\) based on an annual effective interest rate of \(10 \%\). The amount of the first payment is 14,000 . Each payment, from the second through the twentieth, is \(4 \%\) larger than the previous payment. The \(21^{\text {st }}\) payment and each subsequent payment will be \(1 \%\) larger than the previous payment.

Calculate \(X\).

Options:
A
\(\quad 185,542\)
B
191,834
C
\(\quad 206,540\)
D
208,508
E
212,823

Correct answer: C

Solution: The PV of the first twenty payments is: \(14,000\left(\frac{1-\left(\frac{1.04}{1.10}\right)^{20}}{0.10-0.04}\right)=157,337.48\)
The PV of the remaining payments starting at time 21 is:
\(14,000(1.04)^{19}(1.01)\left(\frac{1}{0.10-0.01}\right)(1.10)^{-20}=49,202.44\)
Total equals 206,539.92.
Question: A four-year 1000 face amount bond, with an annual coupon rate of \(5 \%\) paid semiannually, has redemption value of \(C\). It is bought at a price to yield an annual nominal rate of \(6 \%\) convertible semiannually. If the term of the bond had been two years, the purchase price would have been \(7 \%\) less.

Calculate \(C\).

Options:
A
455
B
469
C
541
D
611
E
626

Correct answer: A

Solution: From the first bond: \(P=25 a_{\overline{8} \mid 0.03}+C v^{8}\)
From the second bond: \(0.93 P=25 a_{-40.03}+C v^{4}\)
Multiply the first equation by 0.93 and plug into the second equation:
\(0.93(25) a_{8 \mid 0.03}+0.93 C=25 a_{4 \mid 0.03}+C v^{4}\)
\(163.2078+0.73415 C=92.9275+0.88849 C\)
\(70.2804=0.15434 C\)
\(C=455.37\)
Question: Susan won the lottery today which will pay an annual perpetuity of \(X\), with the first payment occurring five years from today.

The perpetuity has a present value of 100,000 based on an annual effective interest rate of \(1 \%\) for the first ten years and \(5 \%\) for all years thereafter.

Calculate \(X\).

Options:
A
4100
B
4224
C
4357
D
4401
E
5696

Correct answer: B

Solution: \(100,000=X a_{60.01} v_{0.01}^{4}+\frac{X}{0.05} v_{0.01}^{10}\)
\(100,000=X(5.79548)(0.96098)+X(20)(0.90529)\)
\(100,000=X(23.6751)\)
\(X=4223.85\)
Question: A loan of \(X\) is repaid with level payments of \(R\) payable at the end of each year for \(n\) years.
You are given:
i) The interest paid in year 1 is 797.50.
ii) The principal repaid in year \(n-4\) is 865 .
iii) The principal outstanding at the end of year \(n-1\) is 1144.50.

Determine \(X\).

Options:
A
\(\quad 9,500\)
B
10,000
C
10,500
D
11,000
E
11,500

Correct answer: D

Solution: \(O B_{n-1}=R a_{1}=1144.5=R v\)
\(P_{n-4}=R v^{n-(n-4)+1}=R v^{5}=865\)
\(\frac{R v^{5}}{R v}=v^{4}=\frac{865}{1144.5}=0.75579\)
\(v^{4}=0.75579, \quad v=0.93240, \quad i=0.07251\)
\(I_{1}=X(0.07251)=797.50\)
\(X=10,999.02\)
Question: You have the option of purchasing one of the following two annuities-immediate:
i) The first annuity makes annual payments of 1000 for 20 years.
ii) The second annuity is a perpetuity that also has annual payments. The payment in each of the first 10 years is 600 . Beginning in year 11, the payments increase to 1200 , and remain at 1200 forever.

At an annual effective interest rate of \(i>0\), both annuities have a present value of \(X\).
Calculate \(X\).

Options:
A
8700
B
8750
C
8800
D
8850
E
8900

Correct answer: B

Solution: \(1000 \cdot a_{\overline{201 i}}=\frac{600}{i}+\frac{600 v^{10}}{i}\)
\(5\left(\frac{1-v^{20}}{i}\right)=\frac{3}{i}\left(1+v^{10}\right)\)
\(5-5 v^{20}=3+3 v^{10}\)
\(0=5 v^{20}+3 v^{10}-2\)
Let \(x=v^{10}=\frac{-3 \pm \sqrt{9+4(5)(2)}}{2(5)}=\frac{-3 \pm 7}{10}=0.4 \Rightarrow i=9.59582 \%\)
\(X=\frac{600}{0.0959582}(1+0.4)=8753.8\)
Question: Erin borrows \(X\) at an annual effective rate of \(5 \%\). If the principal and all accumulated interest is paid in one lump sum at the end of 20 years, 1000 more in interest would be paid than if the loan was repaid with level payments at the end of each of the first ten years.

Calculate \(X\).

Options:
A
389
B
540
C
704
D
736
E
954

Correct answer: D

Solution: If paid in one lump sum, the total interest paid is \(X\left(1.05^{20}-1\right)=1.65330 X\).
With level payments for 10 years, the total interest paid is
\(10\left(\frac{X}{a_{\overline{100} 0.05}}\right)-X=0.29505 X\).
Then,
\(1.65330 X=1000+0.29505 X\)
\(X=736.24\).
Question: A ten-year bond paying annual coupons of \(X\) has a face amount of 1000 , a price of 450 , and an annual effective yield rate of \(10 \%\).

A second ten-year bond has the same face amount and annual effective yield rate as the first bond. This second bond pays semi-annual coupons of \(\frac{X}{2}\). The price of the second bond is \(P\).

Calculate \(P\).

Options:
A
439
B
442
C
452
D
457
E
461

Correct answer: C

Solution: \(450=X a_{-100_{0.10}}+1000 v^{10}\)
Using BA II Plus calculator:
\(X=10.50\)
\(1.10=(1+j)^{2}, j=0.048881\)
\(P=5.25 a_{\left.\overline{20}\right|_{0.048881}}+1000 v^{20}\)
\(P=451.64\)
Question: An insurance company has liabilities due at the end of each of the next two years. The liability due at the end of the second year is twice that for the first year.

The company uses a combination of the following two bonds to match the liabilities. The second bond has annual coupons.

\begin{tabular}{|c|c|c|}
\hline Term to maturity (in years) & Annual coupon rate & Annual effective yield \\
\hline 1 & \(0 \%\) & \(6 \%\) \\
\hline 2 & \(5 \%\) & \(8 \%\) \\
\hline
\end{tabular}

Assume both bonds are redeemed at par and the company invests 9465 in the two-year bond.
Calculate the amount the company should invest in the one-year bond to construct an exactly matched portfolio.

Options:
A
4245
B
4398
C
4481
D
4717
E
4953

Correct answer: C

Solution: Let \(Y\) indicate the nominal value of the two-year bond, then:
\(9465=\frac{0.05 Y}{1.08}+\frac{1.05 Y}{1.08^{2}}\), so \(Y=10,000\).
Thus, the amount of liability at the end of the second year is 10,500 .
Hence, the liability at the end of the first year is:
\(\frac{10,500}{2}=5250\).
So, the amount invested in the one-year bond is:
\(\frac{5250-10,000(0.05)}{1.06}=4481\).
Question: A perpetuity-immediate pays 20 for 10 years, decreases by 1 per year for 19 years, and then pays 1 per year thereafter. At an annual effective interest rate of \(6 \%\), the present value is equal to \(X\).

Calculate \(X\).

Options:
A
208
B
213
C
218
D
223
E
228

Correct answer: D

Solution: \(20 a_{\overline{10} \mid 0.06}+v^{10}(D a)_{\overline{19} \mid 0.06}+v^{29}\left(\frac{1}{0.06}\right)\)
\(=147.20+v^{10} 130.70+v^{29} 16.67\)
\(=147.20+72.98+3.08=223.26\)
Question: At the beginning of each year, a payment of 5000 is invested in a fund. The payments earn an annual effective interest rate of \(8 \%\). At the end of each year, the interest is reinvested in the fund at an annual effective interest rate of \(5 \%\). The amount in the fund at the end of ten years, immediately prior to the \(11^{\text {th }}\) annual payment, is \(X\).

Calculate \(X\) to the nearest 100 .

Options:
A
\(\quad 67,600\)
B
70,300
C
\(\quad 75,700\)
D
78,200
E
80,700

Correct answer: C

Solution: \(X=5000\left[10+0.08(I s)_{\overline{10 \mid 0.05}}\right]\)
\(X=5000\left[10+0.08\left(\frac{\ddot{S}_{\overline{10} 0.05}-10}{0.05}\right)\right]\)
\(X=75,654.30\)
Question: A loan with an annual nominal interest rate of \(9 \%\), convertible monthly, is repaid with 60 monthly payments. The first payment is 1000 and each succeeding payment is \(2 \%\) less than the previous payment.

Calculate the outstanding loan balance immediately after the \(40^{\text {th }}\) payment is made.

Options:
A
6889
B
7289
C
7344
D
7407
E
7862

Correct answer: A

Solution: The outstanding loan balance at any point in time is equal to the present value (at that same point in time) of the remaining installment payments.
\(B_{40}=1000(0.98)^{40}\left[\frac{1-\left(\frac{0.98}{1.0075}\right)^{20}}{0.0075+0.02}\right]=6889.11\)
Question: A 15-year loan with an annual effective interest rate of \(6 \%\) has payments of 400 at the end of each year. At the time of the fifth payment, the borrower pays an extra 1000 and then repays the balance over five years with a revised annual payment of \(X\).

Calculate \(X\).

Options:
A
\(\quad 264.13\)
B
435.38
C
461.51
D
556.46
E
\(\quad 698.90\)

Correct answer: C

Solution: \(O B_{5}=400 a_{\overline{10} 0.06}-1000=1944.03\)
\(1944.03=X a_{5} 0.06\)
\(X=461.51\)
Question: A loan of 1500 is to be repaid with payments made at the end of each year for 20 years. There are two repayment options:
i) equal payments at an annual effective interest rate of \(5 \%\)
ii) nonlevel payments of interest on the unpaid balance at an annual effective interest rate of \(i\), plus 75 of principal

The sum of the payments is the same under the two options.
Calculate \(i\).

Options:
A
\(\quad 5.26 \%\)
B
\(\quad 5.51 \%\)
C
\(5.76 \%\)
D
\(\quad 6.01 \%\)
E
\(\quad 6.26 \%\)

Correct answer: C

Solution: Option 1
\(1500=P a_{\overline{20} 0_{0.05}}\)
\(P=120.3639\)
Total Payment \(=120.3639(20)=2407.28\)

Option 2
Total Payment \(=75 * 20+1500 i+(1500-75) i+\)
\[
\begin{aligned}
& (1500-2 * 75) \cdot i+\cdots+(1500-19 * 75) i \\
= & 1500+1500 i(20)-75 i \cdot \sum_{k=0}^{19} k \\
= & 1500+30,000 i-75 i \cdot\left(\frac{19(20)}{2}\right) \\
= & 1500+15,750 i
\end{aligned}
\]
\(2407.28=1500+15,750 i\)
\(i=0.0576\)
Question: The table gives information about the assets and liabilities of three companies.

\begin{tabular}{|c|l|c|c|l|c|c|}
\hline & \multicolumn{3}{|c|}{ Assets } & \multicolumn{3}{c|}{ Liabilities } \\
\hline Company & \begin{tabular}{l} 
Present \\
Value
\end{tabular} & \begin{tabular}{l} 
Modified \\
Duration
\end{tabular} & Convexity & \begin{tabular}{l} 
Present \\
Value
\end{tabular} & \begin{tabular}{l} 
Modified \\
Duration
\end{tabular} & Convexity \\
\hline U & 200,000 & 10.2 & 150 & 200,000 & 9.3 & 148 \\
\hline V & 300,000 & 8.6 & 85 & 300,000 & 8.6 & 80 \\
\hline W & 400,000 & 15.8 & 300 & 400,000 & 15.8 & 305 \\
\hline
\end{tabular}

Determine which of the following statements are true.
(I) Company U's position is immunized against small interest rate changes.
(II) Company V's position is immunized against small interest rate changes.
(III) Company W's position is immunized against small interest rate changes.

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: B

Solution: Besides present values of assets and liabilities matching, 1) their modified durations must also match, and 2) the convexity of the assets must exceed the convexity of the liabilities, in order for the company's position to be immunized against small changes in interest rate. Only company V satisfies all these conditions.
Question: On the first of the month, Chuck took out a business loan for 50,000 , with payments at the end of each month based on an annual nominal interest rate compounded monthly. Each monthly payment is equal to 800 , except for a final drop payment.

Immediately after the first payment the balance owed was 49,800.
Calculate the number of payments that Chuck needed to pay off the loan.

Options:
A
115
B
117
C
119
D
121
E
123

Correct answer: B

Solution: Let \(j\) be the monthly interest rate.
\(50,000(1+j)-800=49,800\)
\(j=0.012\)
\(50,000=800 a_{\bar{n} 0.012}\)
\(n=116.22\)

Drop payment at payment number 117.
Question: A scholarship foundation is committed to make a payment of 20,000 exactly two years from today. It plans to use two zero-coupon bonds to fully immunize the liability based on a flat curve and an annual effective yield rate of \(5.5 \%\). One of the bonds will have a one-year maturity and the other will have a three-year maturity.

Calculate the face value of the one-year zero-coupon bond.

Options:
A
\(\quad 9,479\)
B
\(\quad 9,779\)
C
10,079
D
10,379
E
10,679

Correct answer: A

Solution: Let \(h(i)=P V_{A}(i)-P V_{L}(i)\).
Full immunization of a single liability requires both equations:
\(h(i)=0, h^{\prime}(i)=0\)
\(A_{1} v+A_{3} v^{3}-20,000 v^{2}=0\)
\(A_{1} v^{2}+3 A_{3} v^{4}-40,000 v^{3}=0\)
\(v=\frac{1}{1.055}\).
Solve these two equations in two unknowns to get \(A_{1}=9478.67\)
Question: A loan of 20,000 is to be repaid with payments at the end of each year for ten years. The first payment is \(X\). Each subsequent payment is \(6 \%\) greater than the preceding payment.

The loan payments are based on an annual effective interest rate of \(10 \%\).
Calculate the loan balance immediately after the eighth payment.

Options:
A
6685
B
6937
C
7353
D
7794
E
8088

Correct answer: C

Solution: \(20,000=X\left[\frac{1-\left(\frac{1.06}{1.10}\right)^{10}}{0.10-0.06}\right]\)
\(X=2584.39\)
\(L B_{8}=X\left(\frac{1.06^{8}}{1.10}+\frac{1.06^{9}}{1.10^{2}}\right)=7353\)
Question: Today's deposit of 9550 earns an annual effective interest rate of \(i\) for five years. At the end of the fifth year, the entire accumulated balance is reinvested into a 20 -year annuity-due. The annuity-due has level annual payments of 756.97 at an annual effective interest rate of \(4 \%\).

Calculate \(i\).

Options:
A
\(\quad 1.5 \%\)
B
\(2.3 \%\)
C
\(\quad 7.7 \%\)
D
\(\quad 12.0 \%\)
E
\(19.7 \%\)

Correct answer: B

Solution: \(9550(1+i)^{5}=756.97 \ddot{a}_{200.04}\)
\(9550(1+i)^{5}=10,698.97\)
\((1+i)^{5}=1.12031\)
\(i=0.02298\)
Question: An actuary invests 1000 at the end of each year for 30 years. The investments will earn interest at a \(4 \%\) annual effective interest rate and, at the end of each year, the interest will be reinvested at a \(3 \%\) annual effective interest rate.

Calculate the accumulated value of the investment at the end of the 30-year period.

Options:
A
\(\quad 51,625\)
B
53,434
C
55,260
D
58,437
E
\(\quad 58,938\)

Correct answer: B

Solution: \(\left[40(I s)_{290.03}+30(1000)\right]=\left[40 \frac{\ddot{s}_{290.03}-29}{0.03}+30(1000)\right]=53,433.89\)
Question: A purchase of 5020 is paid off with a loan at an annual effective interest rate of \(6.25 \%\).
Payments are made at the end of each year until the loan is paid off. Each payment is 360 except for a final balloon payment, which is less than 720 .

Calculate the amount of the balloon payment.

Options:
A
611
B
630
C
649
D
667
E
685

Correct answer: C

Solution: \(5020=360 a_{\bar{n} 0.0625}\)
\(n=33.85\)
Use \(n=33\)
\(5020=360 a_{\overline{33} \mid 0.0625}+X v^{33}\)
\(5020=4980.95+X v^{33}\)
\(39.05=X v^{33}\)
\(X=288.75\)
\(B=360+288.75=648.75\)
Question: A 15-year bond with semiannual coupons is purchased for 2895.28. The bond is redeemable for 1000. The first coupon payment is equal to \(R\) and each subsequent coupon is \(1 \%\) larger than the previous coupon payment. The annual nominal yield rate on this bond is \(6.2 \%\) convertible semiannually.

Calculate \(R\).

Options:
A
110
B
114
C
122
D
128
E
132

Correct answer: B

Solution: \(2895.28=R v+R(1.01) v^{2}+R(1.01)^{2} v^{3}+\ldots+R(1.01)^{29} v^{30}+1000 v^{30}\)
\(\Rightarrow R\left[\frac{1-\left(\frac{1.01}{1.031}\right)^{30}}{0.031-0.01}\right]+1000 v^{30}=2895.28\)
\(=>R=113.75\)
Question: An annuity writer sells an annual perpetuity-due, with a first payment of 3000 . The payments increase by \(7 \%\) annually.

The purchase price, \(P\), of the perpetuity is based on a \(12 \%\) annual force of interest.
Calculate \(P\).

Options:
A
41,096
B
52,177
C
\(\quad 58,829\)
D
\(\quad 60,000\)
E
\(\quad 67,200\)

Correct answer: C

Solution: \(i=e^{0.12}-1\)
\(i=0.12749685\)
\(3000\left[\frac{1}{0.12749685-0.07}\right](1.12749685)\)
\(=58,829.14\)