BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: Bruce deposits 100 into a bank account. His account is credited interest at an annual nominal rate of interest of \(4 \%\) convertible semiannually.

At the same time, Peter deposits 100 into a separate account. Peter's account is credited interest at an annual force of interest of \(\delta\).

After 7.25 years, the value of each account is the same.
Calculate \(\delta\).

Options:
A
0.0388
B
0.0392
C
0.0396
D
0.0404
E
0.0414

Correct answer: C

Solution: Given the same principal invested for the same period of time yields the same accumulated value, the two measures of interest \(i^{(2)}=0.04\) and \(\delta\) must be equivalent, which means:
\(\left(1+\frac{i^{(2)}}{2}\right)^{2}=e^{\delta}\) over a one-year period. Thus,
\(e^{\delta}=\left(1+\frac{i^{(2)}}{2}\right)^{2}=1.02^{2}=1.0404\)
\(\delta=\ln (1.0404)=0.0396\).
Question: A perpetuity costs 77.1 and makes end-of-year payments. The perpetuity pays 1 at the end of year 2,2 at the end of year \(3, \ldots, n\) at the end of year ( \(n+1\) ). After year ( \(n+1\) ), the payments remain constant at \(n\). The annual effective interest rate is \(10.5 \%\).

Calculate \(n\).

Options:
A
17
B
18
C
19
D
20
E
21

Correct answer: C

Solution: \[
\begin{aligned}
77.1 & =v(I a)_{\text {nl }}+\frac{n v^{n+1}}{i} \\
& =v\left[\frac{\ddot{a}_{\text {n }}-n v^{n}}{i}\right]+\frac{n v^{n+1}}{i} \\
& =\frac{a_{\text {n }}}{i}-\frac{n v^{n+1}}{i}+\frac{n v^{n+1}}{i} \\
& =\frac{a_{\text {n }}}{i}=\frac{1-v^{n}}{i^{2}}=\frac{1-v^{n}}{0.011025} \\
0.85003 & =1-v^{n} \\
1.105^{-n} & =0.14997 \\
n & =-\frac{\ln (0.14997)}{\ln (1.105)}=19
\end{aligned}
\]

To obtain the present value without remembering the formula for an increasing annuity, consider the payments as a perpetuity of 1 starting at time 2 , a perpetuity of 1 starting at time 3 , up to a perpetuity of 1 starting at time \(n+1\). The present value one period before the start of each perpetuity is \(1 / i\). The total present value is \((1 / i)\left(v+v^{2}+\cdots+v^{n}\right)=(1 / i) a_{\bar{n}}\).
Question: A 20-year loan of 1000 is repaid with payments at the end of each year.
Each of the first ten payments equals \(150 \%\) of the amount of interest due. Each of the last ten payments is \(X\).

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

Options:
A
32
B
57
C
70
D
97
E
117

Correct answer: D

Solution: For the first 10 years, each payment equals \(150 \%\) of interest due. The lender charges \(10 \%\), therefore \(5 \%\) of the principal outstanding will be used to reduce the principal.
At the end of 10 years, the amount outstanding is \(1000(1-0.05)^{10}=598.74\).
Thus, the equation of value for the last 10 years using a comparison date of the end of year 10 is \(598.74=X a_{10110 \%}=6.1446 X\)
\(X=97.44\).
Question: A perpetuity-immediate pays 100 per year. Immediately after the fifth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay \(X\) at the end of the first year. Each subsequent annual payment will be \(8 \%\) greater than the preceding payment.
The annual effective rate of interest is \(8 \%\).
Calculate \(X\).

Options:
A
54
B
64
C
74
D
84
E
94

Correct answer: A

Solution: The value of the perpetuity after the fifth payment is \(100 / 0.08=1250\). The equation to solve is:
\[
\begin{aligned}
1250 & =X\left(v+1.08 v^{2}+\cdots+1.08^{24} v^{25}\right) \\
& =X(v+v+\cdots+v)=X(25) / 1.08 \\
X & =50(1.08)=54
\end{aligned}
\]
Question: Ernie makes deposits of 100 at time 0 , and \(X\) at time 3 . The fund grows at a force of interest \(\delta_{t}=\frac{t^{2}}{100}, t>0\).
The amount of interest earned from time 3 to time 6 is also \(X\).
Calculate \(X\).

Options:
A
385
B
485
C
585
D
685
E
785

Correct answer: E

Solution: The accumulation function is \(a(t)=\exp \left[\int_{0}^{t}\left(s^{2} / 100\right) d s\right]=\exp \left(t^{3} / 300\right)\).
The accumulated value of 100 at time 3 is \(100 \exp \left(3^{3} / 300\right)=109.41743\).
The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3 . Thus
\((109.41743+X)[a(6) / a(3)-1]=X\)
\((109.41743+X)(2.0544332 / 1.0941743-1)=X\)
\((109.41743+X) 0.877613=X\)
\(96.026159=0.122387 X\)
\(X=784.61\).
Question: A 10-year loan of 2000 is to be repaid with payments at the end of each year. It can be repaid under the following two options:
(i) Equal annual payments at an annual effective interest rate of \(8.07 \%\).
(ii) Installments of 200 each year plus interest on the unpaid balance at an annual effective interest rate of \(i\).

The sum of the payments under option (i) equals the sum of the payments under option (ii).
Calculate \(i\).

Options:
A
\(\quad 8.75 \%\)
B
\(\quad 9.00 \%\)
C
\(\quad 9.25 \%\)
D
\(\quad 9.50 \%\)
E
\(\quad 9.75 \%\)

Correct answer: B

Solution: Option 1: \(2000=P a_{\overline{10} 0.0807}\)
\(P=299 \Rightarrow\) Total payments \(=2990\)
Option 2: Interest needs to be \(2990-2000=990\)
\(990=i[2000+1800+1600+\cdots+200]\)
\(=11,000 i\)
\(i=0.09=9.00 \%\)
Question: David can receive one of the following two payment streams:
(i) 100 at time 0,200 at time \(n\) years, and 300 at time \(2 n\) years
(ii) 600 at time 10 years

At an annual effective interest rate of \(i\), the present values of the two streams are equal.
Given \(v^{n}=0.76\), calculate \(i\).

Options:
A
\(\quad 3.5 \%\)
B
\(\quad 4.0 \%\)
C
\(\quad 4.5 \%\)
D
\(5.0 \%\)
E
\(5.5 \%\)

Correct answer: A

Solution: Equating present values:
\(100+200 v^{n}+300 v^{2 n}=600 v^{10}\)
\(100+200(0.76)+300(0.76)^{2}=600 v^{10}\)
\(425.28=600 v^{10}\)
\(0.7088=v^{10}\)
\(0.96617=v\)
\(1.03501=1+i\)
\(i=0.035=3.5 \%\).
Question: Payments are made to an account at a continuous rate of \((8 k+t k)\), where \(0 \leq t \leq 10\).
Interest is credited at a force of interest \(\delta_{t}=\frac{1}{8+t}\).
After time 10, the account is worth 20,000.
Calculate \(k\).

Options:
A
111
B
116
C
121
D
126
E
131

Correct answer: A

Solution: The accumulation function is:
\(a(t)=e^{\int_{0}^{t} \frac{1}{8+r} d r}=e^{\left.\ln (8+r)\right|_{0} ^{t}}=\frac{8+t}{8}\).
Using the equation of value at end of 10 years:
\(20,000=\int_{0}^{10}(8 k+t k) \frac{a(10)}{a(t)} d t=k \int_{0}^{10}(8+t) \frac{18 / 8}{(8+t) / 8} d t=k \int_{0}^{10} 18 d t\)
\(=180 k \Rightarrow k=\frac{20,000}{180}=111\).
Question: You have decided to invest in Bond X , an \(n\)-year bond with semi-annual coupons and the following characteristics:
(i) Par value is 1000 .
(ii) The ratio of the semi-annual coupon rate, \(r\), to the desired semi-annual yield rate, \(i\), is 1.03125 .
(iii) The present value of the redemption value is 381.50 .

Given \((1+i)^{-n}=0.5889\), calculate the price of bond X .

Options:
A
1019
B
1029
C
1050
D
1055
E
1072

Correct answer: D

Solution: Let C be the redemption value and \(v=1 /(1+i)\). Then
\[
\begin{aligned}
X & =1000 r a_{\overline{2 n i}}+C v^{2 n} \\
& =1000 r \frac{1-v^{2 n}}{i}+381.50 \\
& =1000(1.03125)\left(1-0.5889^{2}\right)+381.50 \\
& =1055.11
\end{aligned}
\]
Question: Project P requires an investment of 4000 today. The investment pays 2000 one year from today and 4000 two years from today.

Project Q requires an investment of \(X\) two years from today. The investment pays 2000 today and 4000 one year from today.

The net present values of the two projects are equal at an annual effective interest rate of \(10 \%\). Calculate \(X\).

Options:
A
5400
B
5420
C
5440
D
5460
E
5480

Correct answer: D

Solution: Equate net present values:
\(-4000+2000 v+4000 v^{2}=2000+4000 v-X v^{2}\)
\(\frac{4000+X}{1.21}=6000+\frac{2000}{1.1}\)
\(X=5460\).
Question: A perpetuity-immediate pays \(X\) per year. Brian receives the first \(n\) payments, Colleen receives the next \(n\) payments, and a charity receives the remaining payments. Brian's share of the present value of the original perpetuity is \(40 \%\), and the charity's share is \(K\).

Calculate \(K\).

Options:
A
\(24 \%\)
B
\(28 \%\)
C
\(32 \%\)
D
\(36 \%\)
E
\(40 \%\)

Correct answer: D

Solution: The present value of the perpetuity \(=X / i\). Let \(B\) be the present value of Brian's payments.
\(B=X a_{\bar{n}}=0.4 \frac{X}{i}\)
\(a_{\bar{n}}=\frac{0.4}{i} \Rightarrow 0.4=1-v^{n} \Rightarrow v^{n}=0.6\)
\(K=v^{2 n} \frac{X}{i}\)
\(K=0.36 \frac{X}{i}\),
Thus the charity's share is \(36 \%\) of the perpetuity's present value.
Question: At an annual effective interest rate of \(i, i>0 \%\), the present value of a perpetuity paying 10 at the end of each 3 -year period, with the first payment at the end of year 3, is 32 .

At the same annual effective rate of \(i\), the present value of a perpetuity paying 1 at the end of each 4 -month period, with first payment at the end of 4 months, is \(X\).

Calculate \(X\).

Options:
A
31.6
B
32.6
C
33.6
D
34.6
E
35.6

Correct answer: B

Solution: For the first perpetuity,
\(32=10\left(v^{3}+v^{6}+\cdots\right)=10 v^{3} /\left(1-v^{3}\right)\)
\(32-32 v^{3}=10 v^{3}\)
\(v^{3}=32 / 42\).
For the second perpetuity,
\(X=v^{1 / 3}+v^{2 / 3}+\cdots=v^{1 / 3} /\left(1-v^{1 / 3}\right)=(32 / 42)^{1 / 9} /\left[1-(32 / 42)^{1 / 9}\right]=32.599\).
Question: You are given the following information with respect to a bond:
(i) par value: 1000
(ii) term to maturity: 3 years
(iii) annual coupon rate: \(6 \%\) payable annually

You are also given that the one, two, and three year annual spot interest rates are \(7 \%, 8 \%\), and \(9 \%\) respectively.

Calculate the value of the bond.

Options:
A
906
B
926
C
930
D
950
E
1000

Correct answer: B

Solution: Using spot rates, the value of the bond is:
\(60 / 1.07+60 / 1.08^{2}+1060 / 1.09^{3}=926.03\).
Question: You are given the following information with respect to a bond:
(i) par value: 1000
(ii) term to maturity: 3 years
(iii) annual coupon rate: \(6 \%\) payable annually

You are also given that the one, two, and three year annual spot interest rates are \(7 \%, 8 \%\), and \(9 \%\) respectively.

The bond is sold at a price equal to its value.
Calculate the annual effective yield rate for the bond i .

Options:
A
8.1\%
B
\(\quad 8.3 \%\)
C
\(8.5 \%\)
D
\(\quad 8.7 \%\)
E
\(8.9 \%\)

Correct answer: E

Solution: Using spot rates, the value of the bond is:
\(60 / 1.07+60 / 1.08^{2}+1060 / 1.09^{3}=926.03\). The annual effective rate is the solution to \(926.03=60 a_{\overrightarrow{3} i}+1000(1+i)^{-3}\). Using a calculator, the solution is \(8.9 \%\).
Question: The current price of an annual coupon bond is 100 . The yield to maturity is an annual effective rate of \(8 \%\). The derivative of the price of the bond with respect to the yield to maturity is -700 .

Using the bond's yield rate, calculate the Macaulay duration of the bond in years.

Options:
A
7.00
B
7.49
C
7.56
D
7.69
E
8.00

Correct answer: C

Solution: Duration is the negative derivative of the price multiplied by one plus the interest rate and divided by the price. Hence, the duration is \(-(-700)(1.08) / 100=7.56\).
Question: Bill buys a 10-year 1000 par value bond with semi-annual coupons paid at an annual rate of \(6 \%\). The price assumes an annual nominal yield of \(6 \%\), compounded semi-annually.

As Bill receives each coupon payment, he immediately puts the money into an account earning interest at an annual effective rate of \(i\).

At the end of 10 years, immediately after Bill receives the final coupon payment and the redemption value of the bond, Bill has earned an annual effective yield of \(7 \%\) on his investment in the bond.

Calculate \(i\).

Options:
A
\(\quad 9.50 \%\)
B
\(\quad 9.75 \%\)
C
\(10.00 \%\)
D
\(10.25 \%\)
E
\(10.50 \%\)

Correct answer: B

Solution: Because the yield rate equals the coupon rate, Bill paid 1000 for the bond. In return he receives 30 every six months, which accumulates to \(30 s_{\overline{20} j}\) where \(j\) is the semi-annual interest rate. The equation of value is \(1000(1.07)^{10}=30 s_{\overline{20 \mid j}}+1000 \Rightarrow s_{\overline{20 \mid j}}=32.238\). Using a calculator to solve for the interest rate produces \(j=0.0476\) and so \(i=1.0476^{2}-1=0.0975=9.75 \%\).
Question: A man turns 40 today and wishes to provide supplemental retirement income of 3000 at the beginning of each month starting on his 65th birthday. Starting today, he makes monthly contributions of \(X\) to a fund for 25 years. The fund earns an annual nominal interest rate of \(8 \%\) compounded monthly.

On his \(65^{\text {th }}\) birthday, each 1000 of the fund will provide 9.65 of income at the beginning of each month starting immediately and continuing as long as he survives.

Calculate \(X\).

Options:
A
\(\quad 324.70\)
B
\(\quad 326.90\)
C
328.10
D
\(\quad 355.50\)
E
\(\quad 450.70\)

Correct answer: A

Solution: To receive 3000 per month at age 65 the fund must accumulate to \(3,000(1,000 / 9.65)= 310,880.83\). The equation of value is \(310,880.83=X \bar{s}_{\left.\overline{300}\right|_{0.08 / 12}}=957.36657 X \Rightarrow 324.72\).
Question: Joe must pay liabilities of 2000 due one year from now and another 1000 due two years from now. He exactly matches his liabilities with the following two investments:

Mortgage I: A one year mortgage in which \(X\) is lent. It is repaid with a single payment at time one. The annual effective interest rate is \(6 \%\).

Mortgage II: A two-year mortgage in which \(Y\) is lent. It is repaid with two equal annual payments. The annual effective interest rate is \(7 \%\).

Calculate \(X+Y\).

Options:
A
2600
B
2682
C
2751
D
2825
E
3000

Correct answer: C

Solution: Because only Mortgage II provides a cash flow at time two, it must be considered first. The mortgage provides \(Y / a_{\overline{2} \mid 0.07}=0.553092 Y\) at times one and two. Therefore, \(0.553092 Y=1000\) for \(Y=1808.02\). Mortgage I must provide \(2000-1000=1000\) at time one and thus \(X= 1000 / 1.06=943.40\). The sum is 2751.42.
Question: Matt purchased a 20 -year par value bond with an annual nominal coupon rate of \(8 \%\) payable semiannually at a price of 1722.25 . The bond can be called at par value \(X\) on any coupon date starting at the end of year 15 after the coupon is paid. The lowest yield rate that Matt can possibly receive is a nominal annual interest rate of \(6 \%\) convertible semiannually.

Calculate \(X\).

Options:
A
1400
B
1420
C
1440
D
1460
E
1480

Correct answer: C

Solution: Given the coupon rate is greater than the yield rate, the bond sells at a premium. Thus, the minimum yield rate for this callable bond is calculated based on a call at the earliest possible date because that is most disadvantageous to the bond holder (earliest time at which a loss occurs). Thus, \(X\), the par value, which equals the redemption value because the bond is a par value bond, must satisfy
Price \(=1722.25=0.04 X a_{\overline{30 \mid 0.03}}+X v_{0.03}^{30}=1.196 X \Rightarrow X=1440\).
Question: The annual force of interest credited to a savings account is defined by
\[
\delta_{t}=\frac{\frac{t^{2}}{100}}{3+\frac{t^{3}}{150}}
\]
with \(t\) in years. Austin deposits 500 into this account at time 0 .
Calculate the time in years it will take for the fund to be worth 2000.

Options:
A
6.7
B
8.8
C
14.2
D
16.5
E
18.9

Correct answer: E

Solution: \(2000=500 \exp \left(\int_{0}^{t} \frac{\frac{r^{2}}{100}}{3+\frac{r^{3}}{150}} d r\right)\)
\(4=\exp \left(0.5 \int_{0}^{t} \frac{\frac{r^{2}}{50}}{3+\frac{r^{3}}{150}} d r\right)=\exp \left[\left.0.5 \ln \left(3+\frac{r^{3}}{150}\right)\right|_{0} ^{t}\right]\)
\(4=\exp \left[0.5 \ln \left(1+\frac{t^{3}}{450}\right)\right]=\left(1+\frac{t^{3}}{450}\right)^{\frac{1}{2}}\)
\(16=\left(1+\frac{t^{3}}{450}\right)\)
\(t=18.8988\)
Question: Turner buys a new car and finances it with a loan of 22,000 . He will make \(n\) monthly payments of 450.30 starting in one month. He will make one larger payment in \(n+1\) months to pay off the loan. Payments are calculated using an annual nominal interest rate of \(8.4 \%\), convertible monthly. Immediately after the 18th payment he refinances the loan to pay off the remaining balance with 24 monthly payments starting one month later. This refinanced loan uses an annual nominal interest rate of \(4.8 \%\), convertible monthly.

Calculate the amount of the new monthly payment.

Options:
A
668
B
693
C
702
D
715
E
742

Correct answer: D

Solution: \(O B_{18}=22,000(1.007)^{18}-450.30 s_{18 \mid 0.007}=16,337.10\)
\(16,337.10=P a_{240.004}\)
\(P=715.27\)
Question: The prices of zero-coupon bonds are:

\begin{tabular}{|l|l|}
\hline Maturity & Price \\
\hline 1 & 0.95420 \\
\hline 2 & 0.90703 \\
\hline 3 & 0.85892 \\
\hline
\end{tabular}

Calculate the one-year forward rate, deferred two years.

Options:
A
0.048
B
0.050
C
0.052
D
0.054
E
0.056

Correct answer: E

Solution: \(\left(1+s_{3}\right)^{3}=\left(1+s_{2}\right)^{2}\left(1+{ }_{1} f_{2}\right)\)
\(0.85892=\frac{1}{\left(1+\mathrm{s}_{3}\right)^{3}}, s_{3}=0.052\)
\(0.90703=\frac{1}{\left(1+s_{2}\right)^{2}}, s_{2}=0.050\)
\(1.052^{3}=1.050^{2}\left(1+{ }_{1} f_{2}\right)\)
\({ }_{1} f_{2}=0.056\)
Question: Sam buys an eight-year, 5000 par bond with an annual coupon rate of \(5 \%\), paid annually. The bond sells for 5000 . Let \(d_{1}\) be the Macaulay duration just before the first coupon is paid. Let \(d_{2}\) be the Macaulay duration just after the first coupon is paid.

Calculate \(\frac{d_{1}}{d_{2}}\).

Options:
A
\(\quad 0.91\)
B
0.93
C
0.95
D
0.97
E
1.00

Correct answer: C

Solution: Let \(d_{0}\) be the Macaulay duration at time 0 .
\(d_{0}=\ddot{a}_{8 \mid 0.05}=6.7864\)
\(d_{1}=d_{0}-1=5.7864\)
\(d_{2}=\ddot{a}_{\left.7\right|_{0.05}}=6.0757\)
\(\frac{d_{1}}{d_{2}}=\frac{5.7864}{6.0757}=0.9524\)
This solution employs the fact that when a coupon bond sells at par the duration equals the present value of an annuity-due. For the duration just before the first coupon the cash flows are the same as for the original bond, but all occur one year sooner. Hence the duration is one year less.

Alternatively, note that the numerators for \(d_{1}\) and \(d_{2}\) are identical. That is because they differ only with respect to the coupon at time 1 (which is time 0 for this calculation) and so the payment does not add anything. The denominator for \(d_{2}\) is the present value of the same bond, but with 7 years, which is 5000 . The denominator for \(d_{1}\) has the extra coupon of 250 and so is 5250. The desired ratio is then \(5000 / 5250=0.9524\).
Question: Bill and Joe each put 10 into separate accounts at time \(t=0\), where \(t\) is measured in years. Bill's account earns interest at a constant annual effective interest rate of \(K / 25, K>0\).
Joe's account earns interest at a force of interest, \(\delta_{t}=\frac{1}{K+0.25 t}\).
At the end of four years, the amount in each account is \(X\).
Calculate \(X\).

Options:
A
20.7
B
21.7
C
22.7
D
23.7
E
24.7

Correct answer: A

Solution: Equating the accumulated values after 4 years provides an equation in \(K\).
\(10\left(1+\frac{K}{25}\right)^{4}=10 \exp \left(\int_{0}^{4} \frac{1}{K+0.25 t} d t\right)\)
\(4 \ln (1+0.04 K)=\int_{0}^{4} \frac{1}{K+0.25 t} d t=\left.4 \ln (\mathrm{~K}+0.25 \mathrm{t})\right|_{0} ^{4}=4 \ln (K+1)-4 \ln (K)=4 \ln \frac{K+1}{K}\)
\(1+0.04 K=\frac{K+1}{K}\)
\(0.04 K^{2}=1\)
\(K=5\).
Therefore, \(X=10(1+5 / 25)^{4}=20.74\).
Question: John made a deposit of 1000 into a fund at the beginning of each year for 20 years.
At the end of 20 years, he began making semiannual withdrawals of 3000 at the beginning of each six months, with a smaller final withdrawal to exhaust the fund. The fund earned an annual effective interest rate of \(8.16 \%\).

Calculate the amount of the final withdrawal.

Options:
A
561
B
1226
C
1430
D
1488
E
2240

Correct answer: C

Solution: The accumulated value is \(1000 \ddot{s}_{200.0816}=50,382.16\). This must provide a semi-annual annuitydue of 3000 . Let \(n\) be the number of payments. Then solve \(3000 \ddot{a}_{\bar{n} 0.04}=50,382.16\) for \(n=26.47\). Therefore, there will be 26 full payments plus one final, smaller, payment. The equation is \(50,382.16=3000 \ddot{a}_{260.04}+X(1.04)^{-26}\) with solution \(X=1430\). Note that the while the final payment is the 27 th payment, because this is an annuity-due, it takes place 26 periods after the annuity begins.
Question: The present value of a perpetuity paying 1 every two years with first payment due immediately is 7.21 at an annual effective rate of \(i\).

Another perpetuity paying \(R\) every three years with the first payment due at the beginning of year two has the same present value at an annual effective rate of \(i+0.01\).

Calculate \(R\).

Options:
A
1.23
B
1.56
C
1.60
D
1.74
E
\(\quad 1.94\)

Correct answer: D

Solution: For the first perpetuity,
\(\frac{1}{(1+i)^{2}-1}+1=7.21\)
\(\frac{1}{6.21}=(1+i)^{2}-1\)
\(i=0.0775\).
For the second perpetuity,
\(R\left[\frac{1}{(1.0775+0.01)^{3}-1}+1\right](1.0875)^{-1}=7.21\)
\(1.286139 R=7.21(1.0875)(0.286139)\)
\(R=1.74\).
Question: A borrower takes out a 15-year loan for 65,000, with level end-of-month payments. The annual nominal interest rate of the loan is \(8 \%\), convertible monthly.

Immediately after the 12th payment is made, the remaining loan balance is reamortized. The maturity date of the loan remains unchanged, but the annual nominal interest rate of the loan is changed to \(6 \%\), convertible monthly.

Calculate the new end-of-month payment.

Options:
A
528
B
534
C
540
D
546
E
552

Correct answer: E

Solution: The monthly payment on the original loan is \(\frac{65,000}{a_{\overline{180 \mid 8 / 12 \%}}}=621.17\). After 12 payments the outstanding balance is \(621.17 a_{\overline{168 \mid} 8 / 12 \%}=62,661.40\). The revised payment is \(\frac{62,661.40}{a_{\overline{168 \mid} 6 / 12 \%}}=552.19\).
Question: An investor purchases a 10-year callable bond with face amount of 1000 for price \(P\). The bond has an annual nominal coupon rate of \(10 \%\) paid semi-annually.

The bond may be called at par by the issuer on every other coupon payment date, beginning with the second coupon payment date.

The investor earns at least an annual nominal yield of \(12 \%\) compounded semi-annually regardless of when the bond is redeemed.

Calculate the largest possible value of \(P\).

Options:
A
885
B
892
C
926
D
965
E
982

Correct answer: A

Solution: For a bond bought at discount, the minimum price will occur at the latest possible redemption date. \(P=50 a_{\overline{20 \mid 0.06}}+1000(1.06)^{-20}=885.30\). (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: Let \(S\) be the accumulated value of 1000 invested for two years at a nominal annual rate of discount \(d\) convertible semiannually, which is equivalent to an annual effective interest rate of \(i\).

Let \(T\) be the accumulated value of 1000 invested for one year at a nominal annual rate of discount \(d\) convertible quarterly.
\(S / T=(39 / 38)^{4}\).
Calculate \(i\).

Options:
A
\(\quad 10.0 \%\)
B
\(10.3 \%\)
C
\(10.8 \%\)
D
\(\quad 10.9 \%\)
E
\(11.1 \%\)

Correct answer: C

Solution: \(\frac{(1-d / 2)^{-4}}{(1-d / 4)^{-4}}=\left(\frac{39}{38}\right)^{4} \Rightarrow \frac{1-d / 2}{1-d / 4}=\frac{38}{39} \Rightarrow 39-39(d / 2)=38-38(d / 4)\)
\(d(39 / 2-38 / 4)=39-38\)
\(d=1 /(19.5-9.5)=0.1\)
\(1+i=(1-d / 2)^{-2}=.95^{-2}=1.108 \Rightarrow i=10.8 \%\).
Question: An investor owns a bond that is redeemable for 300 in seven years. The investor has just received a coupon of 22.50 and each subsequent semiannual coupon will be \(X\) more than the preceding coupon. The present value of this bond immediately after the payment of the coupon is 1050.50 assuming an annual nominal yield rate of \(6 \%\) convertible semiannually.

Calculate \(X\).

Options:
A
7.54
B
10.04
C
22.37
D
34.49
E
43.98

Correct answer: A

Solution: \(1050.50=22.50 a_{\left.14\right|_{0.03}}+X(I a)_{\left.14\right|_{0.03}}+300(1.03)^{-14}\)
\(=22.50 a_{14 \mid 0.03}+X\left(\frac{\ddot{a}_{14 \mid 0.03}-14(1.03)^{-14}}{0.03}\right)+300(1.03)^{-14}\)
\(=22.50(11.2961)+X(79.3102)+198.3353\)
\(X=7.54\)