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Niamh has saved to buy a car - Leaving Cert Mathematics - Question 4 - 2013

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Niamh has saved to buy a car. She saved an equal amount at the beginning of each month in an account that earned an annual equivalent rate (AER) of 4%. (i) Show tha... show full transcript

Worked Solution & Example Answer:Niamh has saved to buy a car - Leaving Cert Mathematics - Question 4 - 2013

Step 1

(i) Show that the rate of interest, compounded monthly, which is equivalent to an AER of 4% is 0.327%, correct to 3 decimal places.

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Answer

To find the monthly interest rate that corresponds to an annual equivalent rate (AER) of 4%, we can use the formula:

(1+i)12=1+0.04(1 + i)^{12} = 1 + 0.04

Taking the 12th root:

1+i=(1.04)1/121 + i = (1.04)^{1/12}

Calculating this gives:

\approx 0.003273$$ To express this as a percentage: $$i \times 100 \approx 0.327\%$$ Thus, the monthly interest rate is approximately 0.327%, correct to 3 decimal places.

Step 2

(ii) Niamh has €15,000 in the account at the end of 36 months. How much has she saved each month, correct to the nearest euro?

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Answer

The future value of the savings can be calculated using the formula for the future value of an annuity:

P=A×(1+i)n1iP = A \times \frac{(1 + i)^n - 1}{i}

Where:

  • PP is the future value (€15,000)
  • AA is the monthly savings amount
  • ii is the monthly interest rate (0.003273)
  • nn is the number of months (36)

Rearranging for AA gives us:

A=P×i(1+i)n1A = P \times \frac{i}{(1 + i)^n - 1}

Substituting the values:

A15000×0.003273(1+0.003273)361A \approx 15000 \times \frac{0.003273}{(1 + 0.003273)^{36} - 1}

Calculating this yields:

A392.02A \approx 392.02

Therefore, Niamh has saved approximately €392 each month, rounded to the nearest euro.

Step 3

Conall borrowed to buy a car. He borrowed €15,000 at a monthly interest rate of 0.866%. He made 36 equal monthly payments to repay the entire loan. How much, to the nearest euro, was each of his monthly payments?

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Answer

We can use the formula for calculating monthly payments on a loan:

A=P×i(1+i)n(1+i)n1A = P \times \frac{i(1 + i)^n}{(1 + i)^n - 1}

Where:

  • PP = €15,000 (loan amount)
  • ii = 0.00866 (monthly interest rate)
  • nn = 36 (number of payments)

Substituting these values into the formula:

A=15000×0.00866(1+0.00866)36(1+0.00866)361A = 15000 \times \frac{0.00866(1 + 0.00866)^{36}}{(1 + 0.00866)^{36} - 1}

Calculating this results in:

A487A \approx 487

Thus, each of Conall's monthly payments is approximately €487.

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