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7.1 Six years ago, Thabo bought a phone for R13 000 - NSC Mathematics - Question 7 - 2024 - Paper 1

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7.1 Six years ago, Thabo bought a phone for R13 000. The value of the phone depreciated annually according to the reducing-balance method. The value of the phone is ... show full transcript

Worked Solution & Example Answer:7.1 Six years ago, Thabo bought a phone for R13 000 - NSC Mathematics - Question 7 - 2024 - Paper 1

Step 1

Calculate the annual rate of depreciation

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Answer

To find the annual rate of depreciation, we can use the formula:

A=P(1−i)tA = P(1 - i)^t

Where:

  • A = current value (R8 337,75)
  • P = original price (R13 000)
  • i = annual depreciation rate
  • t = time in years (6)

Rearranging the formula gives:

i=1−(AP)1ti = 1 - \left( \frac{A}{P} \right)^{\frac{1}{t}}

Substituting in the values:

i=1−(8337.7513000)16≈0.0714i = 1 - \left( \frac{8337.75}{13000} \right)^{\frac{1}{6}}\approx 0.0714

Thus, the annual rate of depreciation is approximately 7.14%.

Step 2

Thandi decides that she will start saving at the end of January 2025.

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Answer

Thandi's savings can be calculated using the future value of an annuity formula:

F=P(1+i)n−1iF = P \frac{(1 + i)^n - 1}{i}

Where:

  • F = future value (R80 000)
  • P = monthly deposit
  • i = interest rate per month (8.6% p.a. = 0.0086/12)
  • n = number of deposits (36)

Rearranging the formula to find P:

P=F⋅i(1+i)n−1P = \frac{F \cdot i}{(1 + i)^n - 1}

Substituting:

  • i = 0.0086 / 12
  • n = 36
  • F = 80000

Calculating gives Thandi's required monthly deposit of approximately R1 955,78.

Step 3

Eric calculates that if he makes 48 deposits of R402,31.

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Answer

Using the same formula:

F=P(1+i)n−1iF = P \frac{(1 + i)^n - 1}{i}

For Eric:

  • F = total savings required
  • P = R402.31
  • i = 0.0086/12
  • n = 48

Calculating:

Eric's total savings from deposits will be approximately R67 310,88. The difference in total savings between both Eric and Thandi is 70 408,08 - 67 310,88 = R3 097,20.

Step 4

Lesibana was granted a loan of R225 000.

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Answer

For Lesibana's loan, we can use the formula for the present value of an annuity:

PV=P1−(1+i)−niPV = P \frac{1 - (1 + i)^{-n}}{i}

Where:

  • PV = R225 000
  • P = monthly payment (R5 500)
  • i = 0.09/12

Rearranging to solve for n:

n=−log⁡(PV⋅iP+1)log⁡(1+i)n = -\frac{\log\left(\frac{PV \cdot i}{P} + 1\right)}{\log(1 + i)}

Substituting the appropriate values gives n approximately equal to 51 payments.

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