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R6000 is deposited into a savings account. Four years later, R7000 is added to the savings. The interest rate for the first three years is 8% per annum compounded annually. Thereafter, the interest rate changes to 9% per annum compounded annually. Calculate the value of the savings at the end of the seventh year.​

Respuesta :

rvkacademic

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

semsee45

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.

Learn more about compound interest here:

https://brainly.com/question/28263709

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