To determine the accumulated amount in the savings account after 3 years with an annual interest rate of 18% compounded monthly, we need to use the formula for compound interest.
The compound interest formula is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given values:
- Principal ([tex]\( P \)[/tex]): R100
- Annual interest rate ([tex]\( r \)[/tex]): 18% or 0.18 (in decimal form)
- Number of compounding periods per year ([tex]\( n \)[/tex]): 12 (since the interest is compounded monthly)
- Time ([tex]\( t \)[/tex]): 3 years
Let's plug these values into the formula:
[tex]\[ A = 100 \left(1 + \frac{0.18}{12}\right)^{12 \times 3} \][/tex]
First, calculate the monthly interest rate ([tex]\( \frac{r}{n} \)[/tex]):
[tex]\[ \frac{0.18}{12} = 0.015 \][/tex]
Next, calculate the total number of compounding periods ([tex]\( nt \)[/tex]):
[tex]\[ 12 \times 3 = 36 \][/tex]
Now, use these values to find the accumulated amount ([tex]\( A \)[/tex]):
[tex]\[ A = 100 \left(1 + 0.015\right)^{36} \][/tex]
[tex]\[ A = 100 \left(1.015\right)^{36} \][/tex]
Now we need to calculate [tex]\( (1.015)^{36} \)[/tex]:
[tex]\[ (1.015)^{36} \approx 1.80967 \][/tex]
Finally, multiply this value by the principal ([tex]\( 100 \)[/tex]) to find [tex]\( A \)[/tex]:
[tex]\[ A = 100 \times 1.80967 \][/tex]
[tex]\[ A \approx 180.97 \][/tex]
So, the amount accumulated in the savings account after 3 years is approximately R180.97. Since the provided options are:
A) 200.91
B) None of these
C) 120.91
D) 180.91
E) 100.91
The closest answer to our calculation is:
D) 180.91
