DSC1630
EXAM PACK
,DSC1630 - INTRODUCTORY FINANCIAL MATHEMATICS
ASSIGNMENT 02 SEMESTER 1 2025
EXAM ANSWERS
Question 1 [4]
Determine the equivalent continuous compounding rate, to two decimal places, of
an interest rate of 17,5% per year, compounded quarterly.
The equivalent continuous rate is found using:
\[ e^\delta=\left(1+\frac{r}{m}\right)^m \]
Therefore,
\[ \delta=m\ln\left(1+\frac{r}{m}\right) \]
Given:
\(r=0.175\)
\(m=4\)
\[ \delta=4\ln\left(1+\frac{0.175}{4}\right) \]\[ =4\ln(1.04375) \]\[ =4(0.042819997) \]\[
=0.17128 \]
As a percentage:
\[ 0.17128\times100=17.128\% \]
Therefore, to two decimal places:
\[ \boxed{\delta=17.13\%\text{ per year}} \]
Answer: 17,13% per year.
,Question 2 [5]
If R35 000 accumulates to R48 320 at a continuous compounding rate of 8,6% per
year, determine the term under consideration in years to two decimal places.
For continuous compounding:
\[ A=Pe^{rt} \]
Given:
\(P=R35\,000\)
\(A=R48\,320\)
\(r=0.086\)
Substitute:
\[ 48\,320=35\,000e^{0.086t} \]
Divide both sides by 35 000:
\[ \frac{48\,320}{35\,000}=e^{0.086t} \]\[ 1.380571=e^{0.086t} \]
Take natural logarithms:
\[ \ln(1.380571)=0.086t \]
Therefore:
\[ t=\frac{\ln(1.380571)}{0.086} \]\[ t=3.74997 \]
To two decimal places:
\[ \boxed{t=3.75\text{ years}} \]
Answer: 3,75 years.
, Question 3 [5]
Calculate the accumulated amount after eight years of weekly payments of R1 900
each into an account earning 9,7% interest per year, compounded weekly. Draw
an appropriate timeline showing all the given values.
This is an ordinary annuity because the payments are made weekly.
Given:
Weekly payment: \(PMT=R1\,900\)
Annual interest rate: \(r=9.7\%=0.097\)
Compounding periods per year: \(52\)
Term: \(8\) years
Number of payments:
\[ n=8(52)=416 \]
Weekly interest rate:
\[ i=\frac{0.097}{52}=0.00186538 \]
The future value of an ordinary annuity is:
\[ FV=PMT\left[\frac{(1+i)^n-1}{i}\right] \]
Substitute:
\[ FV=1\,900\left[ \frac{(1+0.00186538)^{416}-1} {0.00186538} \right] \]\[ FV=1\,900\left[
\frac{(1.00186538)^{416}-1} {0.00186538} \right] \]\[ FV\approx R1\,192\,927.24 \]
Therefore:
EXAM PACK
,DSC1630 - INTRODUCTORY FINANCIAL MATHEMATICS
ASSIGNMENT 02 SEMESTER 1 2025
EXAM ANSWERS
Question 1 [4]
Determine the equivalent continuous compounding rate, to two decimal places, of
an interest rate of 17,5% per year, compounded quarterly.
The equivalent continuous rate is found using:
\[ e^\delta=\left(1+\frac{r}{m}\right)^m \]
Therefore,
\[ \delta=m\ln\left(1+\frac{r}{m}\right) \]
Given:
\(r=0.175\)
\(m=4\)
\[ \delta=4\ln\left(1+\frac{0.175}{4}\right) \]\[ =4\ln(1.04375) \]\[ =4(0.042819997) \]\[
=0.17128 \]
As a percentage:
\[ 0.17128\times100=17.128\% \]
Therefore, to two decimal places:
\[ \boxed{\delta=17.13\%\text{ per year}} \]
Answer: 17,13% per year.
,Question 2 [5]
If R35 000 accumulates to R48 320 at a continuous compounding rate of 8,6% per
year, determine the term under consideration in years to two decimal places.
For continuous compounding:
\[ A=Pe^{rt} \]
Given:
\(P=R35\,000\)
\(A=R48\,320\)
\(r=0.086\)
Substitute:
\[ 48\,320=35\,000e^{0.086t} \]
Divide both sides by 35 000:
\[ \frac{48\,320}{35\,000}=e^{0.086t} \]\[ 1.380571=e^{0.086t} \]
Take natural logarithms:
\[ \ln(1.380571)=0.086t \]
Therefore:
\[ t=\frac{\ln(1.380571)}{0.086} \]\[ t=3.74997 \]
To two decimal places:
\[ \boxed{t=3.75\text{ years}} \]
Answer: 3,75 years.
, Question 3 [5]
Calculate the accumulated amount after eight years of weekly payments of R1 900
each into an account earning 9,7% interest per year, compounded weekly. Draw
an appropriate timeline showing all the given values.
This is an ordinary annuity because the payments are made weekly.
Given:
Weekly payment: \(PMT=R1\,900\)
Annual interest rate: \(r=9.7\%=0.097\)
Compounding periods per year: \(52\)
Term: \(8\) years
Number of payments:
\[ n=8(52)=416 \]
Weekly interest rate:
\[ i=\frac{0.097}{52}=0.00186538 \]
The future value of an ordinary annuity is:
\[ FV=PMT\left[\frac{(1+i)^n-1}{i}\right] \]
Substitute:
\[ FV=1\,900\left[ \frac{(1+0.00186538)^{416}-1} {0.00186538} \right] \]\[ FV=1\,900\left[
\frac{(1.00186538)^{416}-1} {0.00186538} \right] \]\[ FV\approx R1\,192\,927.24 \]
Therefore:
