| Date | May 2018 | Marks available | 5 | Reference code | 18M.2.hl.TZ1.7 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
It is known that the number of fish in a given lake will decrease by 7% each year unless some new fish are added. At the end of each year, 250 new fish are added to the lake.
At the start of 2018, there are 2500 fish in the lake.
Show that there will be approximately 2645 fish in the lake at the start of 2020.
Find the approximate number of fish in the lake at the start of 2042.
Markscheme
EITHER
2019: 2500 × 0.93 + 250 = 2575 (M1)A1
2020: 2575 × 0.93 + 250 M1
OR
2020: 2500 × 0.932 + 250(0.93 + 1) M1M1A1
Note: Award M1 for starting with 2500, M1 for multiplying by 0.93 and adding 250 twice. A1 for correct expression. Can be shown in recursive form.
THEN
(= 2644.75) = 2645 AG
[3 marks]
2020: 2500 × 0.932 + 250(0.93 + 1)
2042: 2500 × 0.9324 + 250(0.9323 + 0.9322 + … + 1) (M1)(A1)
\( = 2500 \times {0.93^{24}} + 250\frac{{\left( {{{0.93}^{24}} - 1} \right)}}{{\left( {0.93 - 1} \right)}}\) (M1)(A1)
=3384 A1
Note: If recursive formula used, award M1 for un = 0.93 un−1 and u0 or u1 seen (can be awarded if seen in part (a)). Then award M1A1 for attempt to find u24 or u25 respectively (different term if other than 2500 used) (M1A0 if incorrect term is being found) and A2 for correct answer.
Note: Accept all answers that round to 3380.
[5 marks]
