| Date | November 2014 | Marks available | 2 | Reference code | 14N.1.sl.TZ0.13 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 13 | Adapted from | N/A |
Question
A potato is placed in an oven heated to a temperature of 200°C.
The temperature of the potato, in °C, is modelled by the function \(p(t) = 200 - 190{(0.97)^t}\), where \(t\) is the time, in minutes, that the potato has been in the oven.
Write down the temperature of the potato at the moment it is placed in the oven.
Find the temperature of the potato half an hour after it has been placed in the oven.
After the potato has been in the oven for \(k\) minutes, its temperature is 40°C.
Find the value of \(k\).
Markscheme
For parts (a) and (b) only, the first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
\(200 - 190{(0.97)^0}\) (M1)
Note: Award (M1) for correct substitution.
\( = 10\,^\circ {\text{C}}\) (A1) (C2)
Note: Units are required.
For parts (a) and (b) only, the first time a correct answer has incorrect or missing units, the final (A1) is not awarded.
\(200 - 190{(0.97)^{30}}\) (M1)
Note: Award (M1) for correct substitution.
\( = 124^\circ {\text{C }}(123.808 \ldots ^\circ {\text{C}})\) (A1) (C2)
Note: Units are required, unless already omitted in part (a).
\(200 - 190{(0.97)^k} = 40\) (M1)
Note: Award (M1) for correct substitution.
\(k = 5.64{\text{ (minutes) }}(5.64198 \ldots )\) (A1) (C2)
