Solve the actuarial science questions in the image.

Steps to Solve

1. Solve Question 1: Find $\delta$ when an investment doubles in 8.45 years

Since the investment doubles, we have $2 = e^{\delta t}$, where $t = 8.45$. Then $\ln(2) = \delta t$, so $\delta = \frac{\ln(2)}{t}$

2. Calculation for Question 1

Substituting $t=8.45$, we get $\delta = \frac{\ln(2)}{8.45} \approx 0.08203$.

3. Solve Question 2: Determine accumulated value at time 4 of $350 invested at time 0

We are given $A(0) = 100$, $A(1) = 108$, $A(2) = 119$. We want to find $350 \cdot \frac{A(4)}{A(0)}$. We need to estimate $A(4)$.

Let's analyze the growth: From $t=0$ to $t=1$, the growth is $108/100 = 1.08$. From $t=1$ to $t=2$, the growth is $119/108 \approx 1.1019$.

We can approximate the growth from $t=2$ to $t=3$ as $1.12$ and from $t=3$ to $t=4$ as $1.14$. Then $A(3) \approx 119 \cdot 1.12 = 133.28$ and $A(4) \approx 133.28 \cdot 1.14 = 151.94$. So $A(4) \approx 152$.

Therefore, the accumulated value at time 4 of $350 invested at time 0 is approximately $350 \cdot \frac{152}{100} = 350 \cdot 1.52 = 532$.

Since we do not have enough amount function information, we will assume simple interest for each year (i.e. calculate $i_1$ and $i_2$ and then derive $i_3$ and $i_4$) for the single year, not compounded. Then $100(1+i_1) = 108 \implies i_1 = 0.08$ so $i_1 = 8%$ Also $108(1+i_2) = 119 \implies i_2 = 11/108 \approx 0.10185$ so $i_2 \approx 10.19%$ We can estimate $i_3 \approx 12%$ and $i_4 \approx 14%$.

Then $A(3) = 119(1+0.12) = 119(1.12) = 133.28$ and $A(4) = 133.28(1+0.14) = 133.28(1.14) = 151.9392 \approx 151.94$

The accumulated value at time 4 of $350 invested at time 0 is $350 \cdot A(4)/A(0) = 350 \cdot 151.94/100 = 350 \cdot 1.5194 = 531.79$

4. Solve Question 3: Balance after 5 years with 0.4% interest compounded quarterly

The nominal annual interest rate is 0.4%, so the quarterly interest rate is $\frac{0.4%}{4} = 0.1% = 0.001$. The number of quarters in 5 years is $5 \cdot 4 = 20$. The balance is $1000(1 + 0.001)^{20} = 1000(1.001)^{20} \approx 1000(1.02018) \approx 1020.18$.

5. Solve Question 4: Find original investment if it accumulates to $500 at time 4 and $1000 at time 10

Let the original investment be $X$. Then $X \cdot a(4) = 500$ and $X \cdot a(10) = 1000$. Thus, $\frac{a(10)}{a(4)} = \frac{1000}{500} = 2$.

We have $a(10) = a(4) \cdot 2$. Since $a(10)$ is the accumulated value at time 10 and $a(4)$ is the accumulated value at time 4, it took 6 years to double. Assume exponential growth. Then $a(t) = e^{\delta t}$. So $e^{10\delta} = 2 e^{4\delta}$. $e^{6\delta} = 2$, so $6\delta = \ln 2$ and $\delta = \frac{\ln 2}{6}$. Then $a(4) = e^{4 \cdot \frac{\ln 2}{6}} = e^{\frac{2}{3} \ln 2} = e^{\ln 2^{2/3}} = 2^{2/3} \approx 1.5874$. Since $X \cdot a(4) = 500$, we have $X = \frac{500}{a(4)} = \frac{500}{2^{2/3}} = \frac{500}{1.5874} \approx 314.98 \approx 315$.

6. Solve Question 5: Find discount rate r for cash payment

The store receives 95% of the purchase price one-half month later for credit card purchases. The effective annual rate is 12%, so the monthly effective rate is $i$ where $(1+i)^{12} = 1.12$, or $i = (1.12)^{1/12} - 1 \approx 0.0094888$. The half-month effective rate $j$ is such that $(1+j)^2 = 1+i = 1.0094888$, so $j = \sqrt{1.0094888} - 1 \approx 0.004733$. Let the purchase price be 1. Paying with credit card yields $0.95$ after one-half month. The present value of this is $0.95/(1+j) = 0.95/1.004733 \approx 0.94552$. The cash payment yields $1-r$ immediately. So $1-r = 0.94552$. Then $r = 1 - 0.94552 = 0.05448$, so $r \approx 5.45%$.

7. Solve Question 6: Find the accumulated value of $500 invested for 2T years

The accumulated value of $1200$ invested at time 0 for $T$ years is $1320$. Using simple interest, $1200(1+rT) = 1320$, so $1+rT = \frac{1320}{1200} = 1.1$. Then $rT = 0.1$. Since $r = \frac{0.1}{T}$. We want the accumulated value of $500$ invested for $2T$ years. This is $500(1 + r(2T)) = 500(1 + 2rT) = 500(1 + 2 \cdot 0.1) = 500(1.2) = 600$.

8. Solve Question 7: Find X given interest of $108 and discount of $100

Interest on $X$ for one year is $108$, so $X \cdot i = 108$, where $i$ is the interest rate. The equivalent discount on $X$ is $100$, so $X \cdot d = 100$, where $d$ is the discount rate. These are equivalent, so $i = \frac{d}{1-d}$, so $i = \frac{100/X}{1 - 100/X} = \frac{100}{X-100}$. Since $X \cdot i = 108$, we have $X \cdot \frac{100}{X-100} = 108$, so $100X = 108(X-100) = 108X - 10800$, which gives $8X = 10800$, so $X = \frac{10800}{8} = 1350$.

9. Solve Question 8: Find k given $\delta = kt$ and $A(5) = 250$ for an initial deposit of 100

The accumulated value is $A(t) = A(0) e^{\int_0^t \delta(s) ds}$. Here, $A(0) = 100$ and $A(5) = 250$. So $A(5) = 100 e^{\int_0^5 ks ds} = 100 e^{k s^2/2 \vert_0^5} = 100 e^{25k/2} = 250$. Then $e^{25k/2} = \frac{250}{100} = 2.5$. Taking the natural logarithm, $\frac{25k}{2} = \ln(2.5)$, so $k = \frac{2 \ln(2.5)}{25} \approx \frac{2 \cdot 0.91629}{25} \approx 0.0733$.

10. Solve Question 9: Find annual effective rate of interest

Nominal discount rate of 4.8% payable monthly means $d^{(12)} = 0.048$. Thus $d = d^{(12)}/12 = 0.048/12 = 0.004$ is the monthly discount rate. $1-d = 1/ (1+i)$, thus $1-0.004 = 1/(1+i)$, and $i = 0.004016$. The monthly interest rate is 0.004016 The effective annual rate of interest is $(1+i)^{12} - 1 = (1.004016)^{12} - 1 \approx 0.04907$. Thus the effective interest rate is 4.91%.

11. Solve Question 10: Find effective interest rate during the 4th year

$A(3) = 6000$, $A(4) = 6198.6$, and $A(5) = 6762.6$. We want to find the effective interest rate during the 4th year, which is $i_4 = \frac{A(4) - A(3)}{A(3)} = \frac{6198.6 - 6000}{6000} = \frac{198.6}{6000} = 0.0331$. The effective interest rate during the 4th year is 3.31%.