A trading company made a profit of £50 000 in 2006 (Year 1) - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2
Question 10
A trading company made a profit of £50 000 in 2006 (Year 1).
A model for future trading predicts that profits will increase year by year in a geometric sequence wit... show full transcript
Worked Solution & Example Answer:A trading company made a profit of £50 000 in 2006 (Year 1) - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2
Step 1
Write down an expression for the predicted profit in Year n.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The expression for the predicted profit in Year n can be written as:
[ P_n = 50000r^{n-1} ]
Where ( P_n ) is the profit in Year n.
Step 2
Show that n > \( \frac{\log 4}{\log r} + 1 \).
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To show that the profit exceeds £200 000:
[ 50000r^{n-1} > 200000 ]
Dividing both sides by £50 000 gives:
[ r^{n-1} > 4 ]
Taking logarithms of both sides:
[ (n-1) \log r > \log 4 ]
Solving for n:
[ n-1 > \frac{\log 4}{\log r} ]
Thus,
[ n > \frac{\log 4}{\log r} + 1 ]
Step 3
find the year in which the profit made will first exceed £200 000.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using r = 1.09, we can calculate:
[ n > \frac{\log 4}{\log 1.09} + 1 ]
Calculating the logs:
(\log 4 \approx 0.60206)
(\log 1.09 \approx 0.03743)
Substituting these values in:
[ n > \frac{0.60206}{0.03743} + 1 \approx 17.0 ]
Therefore, the profit will first exceed £200 000 in Year 18 (2023).
Step 4
find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the total profits over 10 years, we can use the formula for the sum of a geometric series: