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Calculate the accumulated amount in the following situations: (a) A loan of R 300 at a rate of 8% for 1 year. (b) An investment of R 2 250 at a rate of 12,5% p.a. for 6 years.

Mathematics

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2 parts
a

R 324\text{R } 324

b

R 3 937.50\text{R } 3\,937.50

18.29 years18.29 \text{ years}

16.25%16.25\%

Step-by-step solution

13 steps

Here are the solutions to the problems:

1. Calculate the balance accumulated:

Step 1: Identify the given values for principal (PP), interest rate (rr), and time (tt).

P=R 3 500P = \text{R } 3\,500 r=7.5%=0.075r = 7.5\% = 0.075 t=2 yearst = 2 \text{ years}

Step 2: Use the simple interest formula for the accumulated amount, A=P(1+r⋅t)A = P(1 + r \cdot t).

A=3500(1+0.075×2)A = 3500(1 + 0.075 \times 2)

Step 3: Calculate the accumulated amount.

A=3500(1+0.15)A = 3500(1 + 0.15) A=3500(1.15)A = 3500(1.15) A=R 4 025A = \text{R } 4\,025

The balance accumulated is R 4 025\boxed{\text{R } 4\,025}.

2. Calculate the accumulated amount:

Part a· 3 steps

A loan of R 300 at a rate of 8% for 1 year.

  1. Identify the given values.

    P=R 300P = \text{R } 300 r=8%=0.08r = 8\% = 0.08 t=1 yeart = 1 \text{ year}

  2. Use the formula A=P(1+r⋅t)A = P(1 + r \cdot t).

    A=300(1+0.08×1)A = 300(1 + 0.08 \times 1)

  3. Calculate the accumulated amount.

    A=300(1+0.08)A = 300(1 + 0.08) A=300(1.08)A = 300(1.08) A=R 324A = \text{R } 324

    The accumulated amount is R 324\boxed{\text{R } 324}.

Part b

R 3 937.50\text{R } 3\,937.50

An investment of R 2 250 at a rate of 12.5% p.a. for 6 years.

  1. Identify the given values.

    P=R 2 250P = \text{R } 2\,250 r=12.5%=0.125r = 12.5\% = 0.125 t=6 yearst = 6 \text{ years}

  2. Use the formula A=P(1+r⋅t)A = P(1 + r \cdot t).

    A=2250(1+0.125×6)A = 2250(1 + 0.125 \times 6)

  3. Calculate the accumulated amount.

    A=2250(1+0.75)A = 2250(1 + 0.75) A=2250(1.75)A = 2250(1.75) A=R 3 937.50A = \text{R } 3\,937.50

    The accumulated amount is R 3 937.50\boxed{\text{R } 3\,937.50}.

    3. Calculate the number of years:

  4. Identify the given values for principal (PP), accumulated amount (AA), and interest rate (rr).

    P=R 1 000P = \text{R } 1\,000 A=R 2 500A = \text{R } 2\,500 r=8.2%=0.082r = 8.2\% = 0.082

  5. Use the simple interest formula A=P(1+r⋅t)A = P(1 + r \cdot t) and substitute the known values.

    2500=1000(1+0.082⋅t)2500 = 1000(1 + 0.082 \cdot t)

  6. Solve for tt.

    25001000=1+0.082⋅t\frac{2500}{1000} = 1 + 0.082 \cdot t 2.5=1+0.082⋅t2.5 = 1 + 0.082 \cdot t 2.5−1=0.082⋅t2.5 - 1 = 0.082 \cdot t 1.5=0.082⋅t1.5 = 0.082 \cdot t t=1.50.082t = \frac{1.5}{0.082} t≈18.29 yearst \approx 18.29 \text{ years}

    The number of years is approximately 18.29 years\boxed{18.29 \text{ years}}.

    4. Calculate the interest rate:

  7. Identify the given values for principal (PP), accumulated amount (AA), and time (tt).

    The time period is from the son's 5th birthday to his 21st birthday.

    P=R 5 000P = \text{R } 5\,000 A=R 18 000A = \text{R } 18\,000 t=21−5=16 yearst = 21 - 5 = 16 \text{ years}

  8. Use the simple interest formula A=P(1+r⋅t)A = P(1 + r \cdot t) and substitute the known values.

    18000=5000(1+r×16)18000 = 5000(1 + r \times 16)

  9. Solve for rr.

    180005000=1+16r\frac{18000}{5000} = 1 + 16r 3.6=1+16r3.6 = 1 + 16r 3.6−1=16r3.6 - 1 = 16r 2.6=16r2.6 = 16r r=2.616r = \frac{2.6}{16} r=0.1625r = 0.1625

  10. Convert the rate to a percentage.

    r=0.1625×100%=16.25%r = 0.1625 \times 100\% = 16.25\%

    The interest rate was 16.25%\boxed{16.25\%}.

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