Respuesta :
Answer:
R 19734.32
Step-by-step explanation:
If an amount P is deposited at R% interest compounded annually for n years, the amount at the end of n years is given by the formula
[tex]P = P(1 + r)^t[/tex]
r = R/100
Note
Interest rate 8% = 8/100 =0.08 for computation
Interest rate 9% = 8/100 =0.09 for computation
Since the interest rate changes after 3 years, the first thing to do is to find out how much the value is after 3 years
[tex]V(3 years) = 6000(1 + 0.08)^3[/tex] = 6000 x1.08³ = 6000 x 1.259712 = 7558.27
After 3rd year interest changes to 9%(0..09) but an additional amount is deposited
Let's find the value after the 4th year at 9% compounded annually using the general formula with r = 0.09 and t =1
V (after 4th year) = 7558.27(1.09)¹ = 7558.27 x 1.09 = 8238.51
We are depositing 7000 at the end of the 4th year so the principal becomes 8238.51 + 7000 = 15,238.51
This new amount accrues interest at 9% for 3 more years (7-4)
So value of savings at end of 7th year is
V = 15238.51(1+0.09)³ = 15238.51 x 1.259712 = 19734.32
Answer:
R19734.32
Step-by-step explanation:
Annual Compound Interest Formula
[tex]\large \text{$ \sf A=P\left(1+r\right)^{t} $}[/tex]
where:
- A = final amount
- P = principal amount
- r = interest rate (in decimal form)
- t = time (in years)
Years 1-3
Given:
- P = R6000
- r = 8% = 0.08
- t = 3
Substitute the given values into the formula and solve for A:
[tex]\implies \sf A=6000(1+0.08)^3[/tex]
[tex]\implies \sf A=6000(1.08)^3[/tex]
[tex]\implies \sf A=6000(1.259712)[/tex]
[tex]\implies \sf A=7558.272[/tex]
Therefore, the amount in the account after 3 years was R7558.272.
Year 4
The interest rate changed to 9%.
Given:
- P = R7558.272
- r = 9% = 0.09
- t = 1
Substitute the given values into the formula and solve for A:
[tex]\implies \sf A=7558.272(1+0.09)^1[/tex]
[tex]\implies \sf A=7558.272(1.09)[/tex]
[tex]\implies \sf A=8238.51648[/tex]
Therefore, the value of savings at the end of the 4th year was R8238.51648.
Years 5-7
After four years, R7000 was added to the account.
Given:
- P = R8238.21648 + R7000 = R15238.21648
- r = 9% = 0.09
- t = 3
Substitute the given values into the formula and solve for A:
[tex]\implies \sf A=15238.21648(1+0.09)^3[/tex]
[tex]\implies \sf A=15238.21648(1.09)^3[/tex]
[tex]\implies \sf A=15238.21648(1.295029)[/tex]
[tex]\implies \sf A=19734.3207...[/tex]
Therefore, the value of the savings at the end of the seventh year is R19734.32.
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