Higher June 2019 Paper 6 Q22
22 At the start of 2018, the population of a town was 17 150.
At the start of 2019, the population of the town was 16 807.
It is assumed that the population of the town is given by the formula
\[P = ar^t\]where \(P\) is the population of the town \(t\) years after the start of 2018.
(a) Write down the value of \(a\). [1]
(b) Show that \(r = 0.98\). [1]
(c) Show that the population is predicted to be less than 16 000 at the start of 2022. [2]
(d) Use the formula to work out what the population might have been at the start of 2017. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17 150 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 16 807 ÷ 17 150 = 0.98 | 1 | Condone: 17150 × [0].98 = 16807 16807 ÷ [0].98 = 17150 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 15 818 to 15 819 | 2 | M1 for \(17150 \times 0.98^4\) or their (a) \(\times\ 0.98^4\) or for \(16807 \times 0.98^3\) and A1FT from their (a) \(\times\ 0.98^4\) correctly evaluated Alternative methods using division M1 for \(16000 \div 0.98^4\) A1 for 17300 to 17350 is greater than 17150 OR M1 for \(16000 \div 0.98^3\) A1 for 16900 to 17000 is greater than 16807 | FT from their (a), and only if method shown Accept “[population in] 2018” for 17150 Accept “[population in] 2019” for 16807 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17 500 nfww | 2 | M1 for \(17150 \times 0.98^{-1}\) oe or their (a) \(\times\ 0.98^{-1}\) oe or \(16807 \times 0.98^{-2}\) oe | NB: M1 for \(0.98^{-1}\) = 1.02[04…] and 17150 × 1.02[04…] but M0 for 17150 × 1.02 = 17493 |