A trading company made a profit of £50 000 in 2006 (Year 1) - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2

To show that ( 50000 r^{n-1} > 200000 ), we divide both sides by 50000:

[ r^{n-1} > 4 ]

Next, we take the logarithm of both sides:

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

Dividing by ( log r ):

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

Thus,

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

Using ( r = 1.09 ), we apply the inequality:

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

This simplifies to:

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

Taking logarithm on both sides:

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

Calculating:

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

This leads to:

[ n > 1 + \frac{log 4}{log(1.09)} \approx 1 + 17.086 = 18.086 ]

Thus, the profit will first exceed £200 000 in Year 18, which corresponds to the year 2023.

The total profit from Year 1 to Year 10 can be calculated using the formula for the sum of a geometric series:

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

Where:

• ( a = 50000 )
• ( r = 1.09 )
• ( n = 10 )

Substituting the values:

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

Calculating ( (1.09)^{10} \approx 2.367 ):

[ S_{10} = 50000 \frac{1 - 2.367}{-0.09} \approx 50000 \frac{1.367}{0.09} \approx 760000 ]

Thus, the total profits over the 10 years is approximately £760 000.