A trading company made a profit of £50 000 in 2006 (Year 1) - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 2

Given the inequality:

[ 50000r^{n-1} > 200000 ]

Dividing both sides by 50000:

[ r^{n-1} > 4 ]

Taking the logarithm:

[ (n-1) \log r > \log 4 ]

Thus,

[ n-1 > \frac{\log 4}{\log r} ]

Adding 1 to both sides:

[ n > \frac{\log 4}{\log r} + 1 ]

Using the value of (r = 1.09):

Set up the equation:

[ 50000(1.09)^{n-1} > 200000 ]

This gives:

[ (1.09)^{n-1} > 4 ]

Taking logarithms:

[ (n-1) \log 1.09 > \log 4 ]

Thus:

[ n-1 > \frac{\log 4}{\log 1.09} ]

Calculating the values, we find:

[ n > \frac{0.6021}{0.0370} + 1 \approx 17.25 ]

Rounding up, we find that the profit will first exceed £200 000 in Year 18 (2023).

To find the total profits over 10 years, we use the formula for the sum of a geometric series:

[ S_n = a \frac{1 - r^n}{1 - r} ]

Let (a = 50000) and (r = 1.09), and (n = 10):

[ S_{10} = 50000 \frac{1 - (1.09)^{10}}{1 - 1.09} ]

Calculating this gives:

[ S_{10} \approx 50000 \frac{1 - 2.367}{-0.09} \approx 50000 \times 15.187 \approx 759350 ]

Rounding to the nearest £10 000, the total profit is approximately £760 000.