| Date | November 2009 | Marks available | 2 | Reference code | 09N.1.sl.TZ0.12 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 12 | Adapted from | N/A |
Question
The population of big cats in Africa is increasing at a rate of 5 % per year. At the beginning of 2004 the population was \(10\,000\).
Write down the population of big cats at the beginning of 2005.
Find the population of big cats at the beginning of 2010.
Find the number of years, from the beginning of 2004, it will take the population of big cats to exceed \(50\,000\).
Markscheme
\(10\,000 \times 1.05\)
\( = 10\,500\) (A1) (C1)
[1 mark]
\(10\,000 \times {1.05^6}\) (M1)
Note: Award (M1) for correct substitution into correct formula.
\( = 13\,400\) (A1) (C2)
[2 marks]
\(50\,000 = 10\,000 \times 1.05''\) (M1)(A1)
Note: Award (M1) for \(10\,000 \times 1.05''\) or equivalent, (A1) for \(50\,000\)
\(n = 33.0\) (Accept 33) (A1) (C3)
[3 marks]
Examiners report
This question was well answered by many candidates, particularly part (a).
This question was well answered by many candidates, particularly part (a). However, a significant number of students lost a mark for rounding up rather than down in part (b).
This question was well answered by many candidates, particularly part (a). However, a significant number of students lost a mark for rounding up rather than down in part (b). Part (c) proved to be the most difficult both for generating the equation and for solving it.
