BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: The amount that must be invested at \(i \%(0 \%<i \%<10 \%)\) to accumulate to \(Y\) at the end of three years at an annual rate of:
i) simple interest of \(i \%\) is \(Q\)
ii) compound interest of \(i \%\) is \(R\)
iii) simple discount of \(i \%\) is \(S\)
iv) compound discount of \(i \%\) is \(T\)

Determine which of the following is true.

Options:
A
\(\quad Q<R<S<T\)
B
\(\quad R<Q<S<T\)
C
\(S<T<R<Q\)
D
\(\quad T<S<Q<R\)
E
\(\quad T<S<R<Q\)

Correct answer: C

Solution: Given that the problem states that the inequality is true for all interest rates from \(0 \%\) to \(10 \%\) and all values of \(Y\), it is sufficient to determine it for one set of values. Select \(i=7 \%\) and \(Y=121\).
Then,
\(Q=121 /(1+3(0.07))=100\)
\(R=121 /(1.07)^{3}=98 / 77\)
\(S=121(1-0.07(3))=95.59\)
\(T=121(0.93)^{3}=97.33\)
Hence,
S \(<\) T \(<\) R \(<\) Q
Question: An investor's retirement account pays an annual nominal interest rate of \(4.2 \%\), convertible monthly.

On January 1 of year \(y\), the investor's account balance was \(X\). The investor then deposited 100 at the end of every quarter. On May 1 of year ( \(y+10\) ), the account balance was \(1.9 X\).

Determine which of the following is an equation of value that can be used to solve for \(X\).

Options:
A
\(\quad \frac{1.9 X}{(1.0105)^{\frac{124}{3}}}+\sum_{k=1}^{42} \frac{100}{(1.0105)^{k-1}}=X\)
B
\(\quad X+\sum_{k=1}^{42} \frac{100}{(1.0035)^{3(k-1)}}=\frac{1.9 X}{(1.0035)^{124}}\)
C
\(\quad X+\sum_{k=1}^{41} \frac{100}{(1.0035)^{3 k}}=\frac{1.9 X}{(1.0035)^{124}}\)
D
\(\quad X+\sum_{k=1}^{41} \frac{100}{(1.0105)^{k-1}}=\frac{1.9 X}{(1.0105)^{\frac{124}{3}}}\)
E
\(\quad X+\sum_{k=1}^{42} \frac{100}{(1.0105)^{k-1}}=\frac{1.9 X}{(1.0105)^{\frac{124}{3}}}\)

Correct answer: C

Solution: The monthly interest rate is \(0.042 / 12=0.0035\). The quarterly interest rate is \(1.0035^{3}-1=0.0105\) . The investor makes 41 quarterly deposits and the ending date is 124 months from the start. Using January 1 of year \(y\) as the comparison date produces the following equation:
\(X+\sum_{k=1}^{41} \frac{100}{1.0105^{k}}=\frac{1.9 X}{1.0035^{124}}\)
Substituting \(1.0105=1.0035^{3}\) gives answer (C).
Question: The first payment of a five-year annuity is due in five years in the amount of 1000. The subsequent four annual payments increase by 500 each year. The annual effective interest rate is \(i\).

Determine which of the following formulas gives the present value of the annuity.

Options:
A
\(\quad v^{6}\left[500 a_{5 i}+500(I a)_{5 i}\right]\)
B
\(\quad v^{6}\left[500 \ddot{a}_{\left.5\right|_{i}}+500(I \ddot{a})_{\left.5\right|_{i}}\right]\)
C
\(\quad v^{5}\left[500 a_{\overline{5}_{i}}+500(I \ddot{a})_{\overline{5}_{i}}\right]\)
D
\(\quad v^{5}\left[500 \ddot{a}_{5 i}+500(I \ddot{a})_{5 i}\right]\)
E
\(\quad v^{5}\left[1000 \ddot{a}_{\left.5\right|_{i}}+500(I \ddot{a})_{\left.5\right|_{i}}\right]\)

Correct answer: D

Solution: Using time 5 as the first reference point, then bringing that value back to time 0 :
\(v^{5}\left[500 \ddot{a}_{5 i}+500(I \ddot{a})_{5 i}\right]\)
This combines a five-year level annuity-due of 500 plus an increasing annuity-due starting with 500 and increasing by 500 .
Question: You are given the following information about a company's liabilities:
- Present value: 9697
- Macaulay duration: 15.24
- Macaulay convexity: 242.47

The company decides to create an investment portfolio by making investments into two of the following three zero-coupon bonds: 5 -year, 15 -year, and 20 -year. The company would like its position to be Redington immunized against small changes in yield rate.

The annual effective yield rate for each of the bonds is \(7.5 \%\).
Determine which of the following portfolios the company should create.

Options:
A
Invest 3077 for the 5-year bond and 6620 for the 20-year bond.
B
Invest 6620 for the 5 -year bond and 3077 for the 20-year bond.
C
Invest 465 for the 15-year bond and 9232 for the 20-year bond.
D
Invest 4156 for the 15-year bond and 5541 for the 20-year bond.
E
Invest 9232 for the 15-year bond and 465 for the 20-year bond.

Correct answer: A

Solution: Let \(x, y\), and \(z\) represent the amounts invested in the 5-year, 15-year, and 20-year zero-coupon bonds, respectively. Note that in this problem, one of these three variables is 0 .
The present value, Macaulay duration, and Macaulay convexity of the assets are, respectively, \(x+y+z, \frac{5 x+15 y+20 z}{x+y+z}, \frac{5^{2} x+15^{2} y+20^{2} z}{x+y+z}\).
We are given that the present value, Macaulay duration, and Macaulay convexity of the liabilities are, respectively, 9697, 15.24, and 242.47.
Since present values and Macaulay durations need to match for the assets and liabilities, we have the two equations
\(x+y+z=9697, \quad \frac{5 x+15 y+20 z}{x+y+z}=15.24\)
Note that 5 and 15 are both less than the desired Macaulay duration 15.24, so z cannot be zero. So try either the 5 -year and 20 -year bonds (i.e. \(y=0\) ), or the 15 -year and 20 -year bonds (i.e. \(x= 0)\).
In the former case, substituting \(y=0\) and solving for \(x\) and \(z\) yields
\(x=\frac{(20-15.24) 9697}{20-5}=3077.18\) and \(z=\frac{(15.24-5) 9697}{20-5}=6619.82\).
We need to check if the Macaulay convexity of the assets exceeds that of the liabilities.
The Macaulay convexity of the assets is \(\frac{5^{2}(3077.18)+20^{2}(6619.82)}{9697}=281.00\), which exceeds
the Macaulay convexity of the liabilities, 242.47 . The company should invest 3077 for the 5 -year bond and 6620 for the 20 -year bond.
Note that setting \(x=0\) produces \(y=9231.54\) and \(z=465.46\) and the convexity is 233.40, which is less than that of the liabilities.