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Date	November 2019	Marks available	3	Reference code	19N.1.SL.TZ0.T_8
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Calculate	Question number	T_8	Adapted from	N/A

Question

Siân invests $50\,000$ Australian dollars (AUD) into a savings account which pays a nominal annual interest rate of $5.6$ % compounded monthly.

Calculate the value of Siân’s investment after four years. Give your answer correct to two decimal places.

[3]

After the four-year period, Siân withdraws $40\,000$ AUD from her savings account and uses this money to buy a car. It is known that the car will depreciate at a rate of $18$ % per year.

The value of the car will be $2500$ AUD after $t$ years.

Find the value of $t$ .

[3]

Markscheme

$FV = 50\,000{\left( {1 + \frac{{5.6}}{{100 \times 12}}} \right)^{12 \times 4}}$ (M1)(A1)

Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

$N = 4$

$I\,\% = 5.6$

$PV = \left( \mp \right)50\,000$

$P{\text{/}}Y = 1$

$C{\text{/}}Y = 12$ (A1)(M1)

Note: Award (A1) for $C{\text{/}}Y = 12$ seen, (M1) for all other correct entries.

$N = 48$

$I\,\% = 5.6$

$PV = \left( \mp \right)50\,000$

$P{\text{/}}Y = 12$

$C{\text{/}}Y = 12$ (A1)(M1)

Note: Award (A1) for $C{\text{/}}Y = 12$ seen, (M1) for all other correct entries.

$\left( {FV = } \right)\,62\,520.97$ (AUD) (A1) (C3)

Note: The final (A1) is not awarded if the answer is not correct to 2 decimal places.

[3 marks]

$2500 = 40\,000{\left( {1 - \frac{{18}}{{100}}} \right)^t}$ OR $2500 = 40\,000{\left( {0.82} \right)^{t - 1}}$ (M1)(A1)

Note: Award (M1) for substitution into compound interest formula or term of a geometric sequence formula and equating to $2500$ , (A1) for correct substitution.

$I\,\% = - 18$

$PV = \left( \pm \right)40\,000$

$FV = \left( \mp \right)2500$

$P{\text{/}}Y = 1$

$C{\text{/}}Y = 1$ (A1)(M1)

Note: Award (A1) for $FV = \left( \mp \right)2500$ seen, (M1) for all other correct entries. $P V$ and $F V$ must have opposite signs.

$32800{\text{, }}26896{\text{, }}22054.72{\text{, }}18084.87{\text{, }}14829.59{\text{, }} \ldots$ (M1)

${t_{13}} = 3031.38$ and ${t_{14}} = 2485.73$ (A1)

Note: Award (M1) for a list of at least $5$ correct terms beginning with $32800$ , (A1) for identifying ${t_{13}}$ and ${t_{14}}$ .

$\left( {t = } \right)\,14.0\,\,\,\left( {13.9711 \ldots } \right)$ (A1) (C3)

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.4—Financial apps – compound interest, annual depreciation

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19N.1.SL.TZ0.T_8b:
After the four-year period, Siân withdraws $40\,000$ AUD from her savings account and uses this money to buy a car. It is known that the car will depreciate at a rate of $18$ % per year.

The value of the car will be $2500$ AUD after $t$ years.

Find the value of $t$ .
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the annual interest rate of the loan.
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