| Date | May 2009 | Marks available | 4 | Reference code | 09M.1.sl.TZ1.12 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Calculate | Question number | 12 | Adapted from | N/A |
Question
An amount, C, of Australian Dollars (AUD) is invested for 5 years at 2.5 % yearly simple interest. The interest earned on this investment is 446.25 AUD.
5000 AUD is invested at a nominal annual interest rate of 2.5 % compounded half yearly.
Calculate the length of time in years for the interest on this investment to exceed 446.25 AUD.
Markscheme
\(446.25 = 5000{\left( {1 + \frac{{2.5}}{{2(100)}}} \right)^{2n}} - 5000\) (M1)(A1)
Notes: Award (M1) for substitution into compound interest formula. Award (A1) for correct values.
\(5446.25 = 5000{\left( {1 + \frac{{2.5}}{{2(100)}}} \right)^{2n}}\) (A1)
n = 3.44
n = 3.5 (A1)
OR
5446.25 = 5000(1.0125)2n (A1)(M1)(A1)
Notes: Award (A1) for 5446.25 seen.
Award (M1) for substitution into compound interest formula.
Award (A1) for correct values.
n = 3.44 years
3.5 years required (A1) (C4)
Notes: For incorrect substitution into compound interest formula award at most (M1)(A0)(A1)(A0).
Award (A3) for 3.44 seen without working.
Allow solution by lists. In this case
Award (A1) for half year rate 1.25 % seen.
(A1) for 5446.25 seen.
(M1) for at least 2 correct uses of multiplication by 1.0125
5000 × 1.0125 = 5062.5 and 5062.5 × 1.0125 = 5125.78125
(A1) n = 3.5
If yearly rate used then award (A0)(A1)(M1)(A0)
[4 marks]
Examiners report
This question was poorly answered by many of the candidates. Candidates confuse interest with principal in the formulas.
The idea of compounding periods and the implication for determining the level of interest is poorly understood. The correct answer is 3.5 years. Interpretation of compounding periods is expected.
