(COMPLETE ANSWERS)
Semester 1 2025 - DUE
April 2025
For assistance contact
Email:
, Question 1: An interest rate of 17,5% per year, compounded quarterly, is equivalent to a
continuous compounding rate of:
The nominal annual interest rate is i(m)=17.5%=0.175, and the number of compounding periods
per year is m=4. The effective interest rate per compounding period is i=mi(m)=40.175=0.04375.
The accumulation factor for one year with quarterly compounding is (1+i)m=(1+0.04375)4.
Let r be the equivalent continuous compounding rate. The accumulation factor for one year with
continuous compounding is er.
We set the two accumulation factors equal to each other: er=(1+0.04375)4 er=(1.04375)4
er≈1.18449
To find r, we take the natural logarithm of both sides: r=ln(1.18449) r≈0.16987
Converting this to a percentage and rounding to three decimal places: r≈16.987%
Looking at the options provided, none of them exactly match. Let's re-calculate with more
precision: (1.04375)4=1.184490015625 r=ln(1.184490015625)≈0.16987005 r≈16.987%
There seems to be a discrepancy with the provided options. Let's double-check the calculation.
er=(1+40.175)4=(1.04375)4 r=4ln(1.04375)=4×0.042806≈0.171224 r≈17.122%
Rounding to three decimal places gives 17.122%. The closest option is [1] 17,128%.
Question 2: An amount borrowed at 29% interest per year, compounded continuously, has
accumulated to R38 279,20 after four years. The initial amount borrowed was:
The formula for continuous compounding is A=Pert, where A is the accumulated amount, P is
the principal (initial amount), r is the continuous interest rate, and t is the time in years.
Given: A=R38279.20, r=29%=0.29, t=4 years. We need to find P.
38279.20=Pe0.29×4 38279.20=Pe1.16
Calculate e1.16: e1.16≈3.19451
Now, solve for P: P=3.1945138279.20 P≈12000.00
The initial amount borrowed was approximately [2] R12 000,00.
Question 3: The effective rate for a continuous compounding rate of 17,5% per year, is: