At the start of 2018, the population of a town was 17 150 - OCR - GCSE Maths - Question 22 - 2019 - Paper 6

To find the value of $r$, we will use the population at the start of 2019:

$P = a r^1$

Thus, substituting in the known values:

$16807 = 17150 imes r$

Solving for $r$ gives us:

$r = \frac{16807}{17150} = 0.98$

In 2022, $t = 4$ (since it is 4 years after the start of 2018).

Using the formula:

$P = a r^t = 17150 imes (0.98)^4$

Calculating gives:

$P = 17150 imes 0.92236816 \approx 15818.68$

Thus, the population is predicted to be approximately 15818.68, which is less than 16000.

For the start of 2017, $t = -1$ (as it is one year before 2018).

Using the formula:

$P = a r^{-1} = 17150 imes (0.98)^{-1}$

Calculating gives:

$P = 17150 imes 1.020408163 \approx 17493$

Therefore, the population might have been approximately 17493.