| Date | November 2012 | Marks available | 3 | Reference code | 12N.1.sl.TZ0.14 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 14 | Adapted from | N/A |
Question
Jackson invested 12 000 Australian dollars (AUD) in a bank that offered simple interest at an annual interest rate of r %. The value of Jackson’s investment doubled after 20 years.
Maddison invests 15 000 AUD in a bank that offers compound interest at a nominal annual interest rate of 4.44 %, compounded quarterly.
Calculate the number of years that it will take for Maddison’s investment to triple in value.
Markscheme
\(45000 = 15000{\left( {1 + \frac{{4.44}}{{400}}} \right)^{4n}}\) (M1)(A1)
Note: Award (M1) for substituted compound interest formula, (A1) for a correctly substituted formula and correctly equated to 45 000.
OR
\(3= {\left( {1 + \frac{{4.44}}{{400}}} \right)^{4n}}\) (M1)(A1)
Note: Award (M1) for substituted compound interest formula, (A1) for a correctly substituted formula and correctly equated to 3.
n = 25 years (A1) (C3)
Notes: Award (A1)(M0)(A0) if 24.9 or 24.88 seen as a final answer, with no working seen. Award, at most, (A1)(M1)(A0) if working is seen and a final answer of 24.9 or 24.88 is given.
[3 marks]
Examiners report
In part (b), writing down any substituted form of the compound interest formula led to a significant number of candidates scoring at least 1 mark for method here. Often, however, the formula was incorrectly substituted or was not correctly equated to 45 000 and subsequent marks were then lost.
