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On the 2nd day of January 2015 a company bought a new printer for R150 000 - NSC Mathematics - Question 6 - 2017 - Paper 1

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On the 2nd day of January 2015 a company bought a new printer for R150 000. The value of the printer decreases by 20% annually on the reducing-balance method. When ... show full transcript

Worked Solution & Example Answer:On the 2nd day of January 2015 a company bought a new printer for R150 000 - NSC Mathematics - Question 6 - 2017 - Paper 1

Step 1

Calculate the book value of the printer on the 2nd day of January 2017

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Answer

To calculate the book value after two years of depreciation of 20% per year, we use the formula for reducing balance:

A=P(1r)nA = P(1 - r)^n

Where:

  • AA is the book value.
  • PP is the initial price (R150,000).
  • rr is the depreciation rate (0.20).
  • nn is the number of years.

Calculate:

A=150000(10.2)2A = 150000(1 - 0.2)^2

=150000(0.8)2= 150000(0.8)^2

=150000(0.64)= 150000(0.64)

=R96,000= R96,000

Step 2

At the beginning of which year will the company have to replace the printer? Show ALL calculations.

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Answer

To find when to replace the printer, we stay with the reducing-balance method:

We need to find when the book value drops to R49,152:

49152=150000(10.2)n49152 = 150000(1 - 0.2)^n

Dividing both sides:

rac{49152}{150000} = (0.8)^n

Calculating the left-hand side:

0.32768=(0.8)n0.32768 = (0.8)^n

Taking logarithm on both sides:

extlog(0.32768)=nimesextlog(0.8) ext{log}(0.32768) = n imes ext{log}(0.8)

Solving for nn:

≈ 5$$ Thus, the printer will be replaced at the beginning of the year 2020.

Step 3

Calculate the amount that should be deposited every three months to have enough money to replace the printer on 2 January 2020.

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Answer

First, calculate the amount needed:

The new printer costs R280,000, and the old printer sells for R49,152. The balance required is:

28000049152=230848280000 - 49152 = 230848

Now, we use the future value formula for compound interest:

FV = P imes rac{(1 + i)^n - 1}{i}

Where:

  • FVFV is the future value (R230,848).
  • PP is the deposit amount.
  • ii is the interest rate per period (0.02125, for quarterly interest at 8.5%).
  • nn is the total number of deposits (20 quarters).

Rearranging to solve for P gives:

P = rac{FV imes i}{(1 + i)^n - 1}

By substituting the values:

≈ R9,383.26$$

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