| Date | November 2013 | Marks available | 2 | Reference code | 13N.1.sl.TZ0.15 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 15 | Adapted from | N/A |
Question
A computer virus spreads according to the exponential model
\[N = 200 \times {(1.9)^{0.85t}},{\text{ }}t \geqslant 0\]
where \(N\) is the number of computers infected, and \(t\) is the time, in hours, after the initial infection.
Calculate the number of computers infected after \(6\) hours.
Calculate the time for the number of infected computers to be greater than \({\text{1}}\,{\text{000}}\,{\text{000}}\).
Give your answer correct to the nearest hour.
Markscheme
\(200 \times {(1.9)^{0.85 \times 6}}\) (M1)
Note: Award (M1) for correct substitution into given formula.
\( = 5280\) (A1) (C2)
Note: Accept \(5281\) or \(5300\) but no other answer.
[2 marks]
\(1\,000\,000 < 200 \times {(1.9)^{0.85t}}\) (M1)(M1)
Note: Award (M1) for setting up the inequality (accept an equation), and (M1) for \({\text{1}}\,{\text{000}}\,{\text{000}}\) seen in the inequality or equation.
\(t = 15.6\) \((15.6113…)\) (A1)
\(16\) hours (A1)(ft) (C4)
Note: The final (A1)(ft) is for rounding up their answer to the nearest hour.
Award (C3) for an answer of \(15.6\) with no working.
Accept \({\text{1}}\,{\text{000}}\,{\text{001}}\) in an equation.
[4 marks]
Examiners report
This question was answered very well, although some candidates were not awarded the final mark because the answer was not an integer number of computers. In part (b), some candidates neglected to give their answer correct to the nearest hour and lost the final mark.
This question was answered very well, although some candidates were not awarded the final mark because the answer was not an integer number of computers. In part (b), some candidates neglected to give their answer correct to the nearest hour and lost the final mark.
