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Calculate the future value of an investment using simple interest and compound interest (annually and quarterly).

Mathematics

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Answers

5 parts
a

N812500.00\text{N}812500.00

b(i)

N2034265.48\text{N}2034265.48

b(ii)

N2357251.75\text{N}2357251.75

b(iii)

N2189247.50\text{N}2189247.50

b(iv)

N2371933.75\text{N}2371933.75

Step-by-step solution

16 steps

Here's the solution to question 1.

Given:
Principal (PP) = N250000
Interest rate (rr) = 15% = 0.15
Time (tt) = 15 years

Part a· 4 steps

Simple interest

The formula for simple interest is A=P(1+rt)A = P(1 + rt).

  1. Substitute the given values into the formula.

    A=250000(1+0.15×15)A = 250000(1 + 0.15 \times 15)
  2. Calculate the product of rr and tt.

    A=250000(1+2.25)A = 250000(1 + 2.25)
  3. Add the values inside the parenthesis.

    A=250000(3.25)A = 250000(3.25)
  4. Multiply to find the final amount.

    A=812500A = 812500

    The amount after 15 years with simple interest is N812500.00\boxed{\text{N}812500.00}.

Part b(i)

N2034265.48\text{N}2034265.48

annually

Interest compounded

The formula for compound interest compounded nn times per year is A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}.
The formula for compound interest compounded continuously is A=PertA = Pe^{rt}.

For annual compounding, n=1n=1.

  1. Substitute the values into the formula.

    A=250000(1+0.151)1×15A = 250000\left(1 + \frac{0.15}{1}\right)^{1 \times 15} A=250000(1.15)15A = 250000(1.15)^{15}
  2. Calculate (1.15)15(1.15)^{15}.

    A=250000×8.1370619A = 250000 \times 8.1370619
  3. Multiply to find the final amount.

    A=2034265.475A = 2034265.475

    The amount after 15 years compounded annually is N2034265.48\boxed{\text{N}2034265.48}.

Part b(ii)

N2357251.75\text{N}2357251.75

quarterly

For quarterly compounding, n=4n=4.

  1. Substitute the values into the formula.

    A=250000(1+0.154)4×15A = 250000\left(1 + \frac{0.15}{4}\right)^{4 \times 15} A=250000(1+0.0375)60A = 250000(1 + 0.0375)^{60} A=250000(1.0375)60A = 250000(1.0375)^{60}
  2. Calculate (1.0375)60(1.0375)^{60}.

    A=250000×9.429007A = 250000 \times 9.429007
  3. Multiply to find the final amount.

    A=2357251.75A = 2357251.75

    The amount after 15 years compounded quarterly is N2357251.75\boxed{\text{N}2357251.75}.

Part b(iii)

N2189247.50\text{N}2189247.50

semi-annually

For semi-annual compounding, n=2n=2.

  1. Substitute the values into the formula.

    A=250000(1+0.152)2×15A = 250000\left(1 + \frac{0.15}{2}\right)^{2 \times 15} A=250000(1+0.075)30A = 250000(1 + 0.075)^{30} A=250000(1.075)30A = 250000(1.075)^{30}
  2. Calculate (1.075)30(1.075)^{30}.

    A=250000×8.75699A = 250000 \times 8.75699
  3. Multiply to find the final amount.

    A=2189247.5A = 2189247.5

    The amount after 15 years compounded semi-annually is N2189247.50\boxed{\text{N}2189247.50}.

Part b(iv)

N2371933.75\text{N}2371933.75

continuously

For continuous compounding, the formula is A=PertA = Pe^{rt}.

  1. Substitute the values into the formula.

    A=250000×e0.15×15A = 250000 \times e^{0.15 \times 15} A=250000×e2.25A = 250000 \times e^{2.25}
  2. Calculate e2.25e^{2.25}.

    A=250000×9.487735A = 250000 \times 9.487735
  3. Multiply to find the final amount.

    A=2371933.75A = 2371933.75

    The amount after 15 years compounded continuously is N2371933.75\boxed{\text{N}2371933.75}.

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