In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A company made a profit of £20000 in its first year of trading, Year 1 A model for future trading predicts that the yearly profit will increase by 8% each year, so that the yearly profits will form a geometric sequence. According to the model, (a) show that the profit for Year 3 will be £23 328 (b) find the first year when the yearly profit will exceed £65 000 (c) find the total profit for the first 20 years of trading, giving your answer to the nearest £1000.

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In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

Question 7

In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A company made a profit of £200... show full transcript

Worked Solution & Example Answer:In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

Step 1

show that the profit for Year 3 will be £23 328

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Answer

To find the profit for Year 3, we can use the geometric sequence formula:

$u_n = u_1 imes r^{(n-1)}$

where:

$u_1 = £20000$ (the profit in Year 1)
$r = 1 + 0.08 = 1.08$ (the growth factor, since the profit increases by 8% each year)
$n = 3$ (the year we are calculating for)

Therefore, we calculate:

$u_3 = 20000 imes (1.08)^{(3-1)} = 20000 imes (1.08)^2$

First, compute $(1.08)^2$ :

$(1.08)^2 = 1.1664$

Now substitute this back into the equation:

$u_3 = 20000 imes 1.1664 = 23328$

Thus, the profit for Year 3 is indeed £23 328.

Step 2

find the first year when the yearly profit will exceed £65 000

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Answer

We need to solve for $n$ in the following inequality, based on the geometric sequence model:

$20000 imes (1.08)^{(n-1)} > 65000$

First, divide both sides by 20000:

$(1.08)^{(n-1)} > rac{65000}{20000} \rightarrow (1.08)^{(n-1)} > 3.25$

Next, take the logarithm of both sides:

$ext{log}((1.08)^{(n-1)}) > ext{log}(3.25)$

Using the power property of logarithms:

$(n-1) imes ext{log}(1.08) > ext{log}(3.25)$

Now, isolate $n$ :

$n - 1 > rac{ ext{log}(3.25)}{ ext{log}(1.08)}$

Calculating the right side gives:

ightarrow n - 1 > 15.37 $$ Thus, $$ n > 16.37 $$ Since $n$ must be a whole number, we round up to get: $$ n = 17 $$ Therefore, the first year when the yearly profit exceeds £65 000 is Year 17.

Step 3

find the total profit for the first 20 years of trading, giving your answer to the nearest £1000

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Answer

To find the total profit for the first 20 years, we use the formula for the sum of a geometric series:

$S_n = rac{u_1 (1 - r^n)}{1 - r}$

where:

$n = 20$ (the number of years)
$u_1 = £20000$ (the profit in Year 1)
$r = 1.08$ (growth factor)

Substituting these values, we can calculate:

$S_{20} = rac{20000 (1 - (1.08)^{20})}{1 - 1.08}$

Calculating $(1.08)^{20}$ :

$(1.08)^{20} = 4.6651$

Now substitute back:

$S_{20} = rac{20000 (1 - 4.6651)}{-0.08} = rac{20000 imes -3.6651}{-0.08} = rac{20000 imes 3.6651}{0.08}$

Calculating this results in:

$S_{20} = 91626.25$

Rounding to the nearest £1000 gives:

Total profit for the first 20 years is approximately £91,000.

In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

Worked Solution & Example Answer:In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

show that the profit for Year 3 will be £23 328

find the first year when the yearly profit will exceed £65 000

find the total profit for the first 20 years of trading, giving your answer to the nearest £1000

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