The adult population of a town is 25 000 at the end of Year 1 - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 3

The common ratio of the geometric sequence is:

$r = 1.03$

We start with the population growth model:

$P_n = P_1 \cdot r^{n-1}$

To find when it exceeds 40000:

$25000 \cdot 1.03^{N-1} > 40000$

Dividing both sides by 25000 gives:

$1.03^{N-1} > \frac{40000}{25000} = 1.6$

Taking the logarithm of both sides:

$(N - 1) \log 1.03 > \log 1.6$

Hence we have shown that $(N - 1) \log 1.03 > \log 1.6$.

From the previously derived inequality:

$(N - 1) \log 1.03 > \log 1.6$

We can isolate N:

$N - 1 > \frac{\log 1.6}{\log 1.03}$

Calculating:

$N > 1 + \frac{\log 1.6}{\log 1.03}$

Using approximate values:

$\log 1.6 \approx 0.2041 \quad \text{and} \quad \log 1.03 \approx 0.0128$

Then,

$N > 1 + \frac{0.2041}{0.0128} \approx 17.95$

Thus, rounding up, we find:

$N = 18$

To calculate the total amount given to the charity fund, we consider each year’s population.

For years 1 to 10, the populations will be:

• Year 1: 25,000
• Year 2: 25,750
• Year 3: 26,522.5
• Year 4: 27,318.5
• Year 5: 28,140.3
• Year 6: 28,988.7
• Year 7: 29,864.5
• Year 8: 30,768.2
• Year 9: 31,700.5
• Year 10: 32,662.3

Adding these amounts:

$25000 + 25750 + 26522.5 + 27318.5 + 28140.3 + 28988.7 + 29864.5 + 30768.2 + 31698 + 32662.3 \approx 287000$

So the total amount that will be given to the charity fund is approximately £287,000.