A company predicts a yearly profit of £120 000 in the year 2013. The company predicts that the yearly profit will rise each year by 5%. The predicted yearly profit forms a geometric sequence with common ratio 1.05. (a) Show that the predicted profit in the year 2016 is £138 915. (b) Find the first year in which the yearly predicted profit exceeds £200 000. (c) Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to the nearest pound.

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A company predicts a yearly profit of £120 000 in the year 2013 - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 6

Question 5

A company predicts a yearly profit of £120 000 in the year 2013. The company predicts that the yearly profit will rise each year by 5%. The predicted yearly profit f... show full transcript

Worked Solution & Example Answer:A company predicts a yearly profit of £120 000 in the year 2013 - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 6

Step 1

Show that the predicted profit in the year 2016 is £138 915.

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Answer

To find the predicted profit for 2016, we can use the formula for the nth term of a geometric sequence:

$a_n = a_1 imes r^{(n-1)}$

Where:

$a_1 = 120000$ (the profit in 2013)
$r = 1.05$ (the common ratio)
$n = 4$ (2016 is the 4th year after 2013)

Calculating the profit for 2016: $a_4 = 120000 imes (1.05)^{(4-1)} = 120000 imes (1.05)^3$ Calculating $(1.05)^3$ : $(1.05)^3 = 1.157625$

Now substituting back: $a_4 = 120000 imes 1.157625 = 138915$

Thus, the predicted profit in 2016 is £138 915.

Step 2

Find the first year in which the yearly predicted profit exceeds £200 000.

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Answer

We need to find the smallest integer $n$ such that:

$120000 imes (1.05)^{(n-1)} > 200000$

Dividing both sides by 120000:

$(1.05)^{(n-1)} > rac{200000}{120000} = rac{5}{3}$

Taking the logarithm of both sides:

$ext{log}_{10}((1.05)^{(n-1)}) > ext{log}_{10}igg( rac{5}{3}igg)$

Using the property of logarithms:

$(n-1) imes ext{log}_{10}(1.05) > ext{log}_{10}igg( rac{5}{3}igg)$

Now substituting $ext{log}_{10}(1.05) ext{ and } ext{log}_{10}igg( rac{5}{3}igg)$ :

Let:

$ext{log}_{10}(1.05) ext{ approximately equal to } 0.02119$
$ext{log}_{10}igg( rac{5}{3}igg) ext{ approximately equal to } 0.22185$

Thus:

$(n - 1) imes 0.02119 > 0.22185$

Calculating for $n$ : $n - 1 > rac{0.22185}{0.02119} ext{ implying } n - 1 > 10.46$ Thus, $n = 12$ . The first year in which the yearly predicted profit exceeds £200,000 is 2024.

Step 3

Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to the nearest pound.

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Answer

The total profit can be calculated using the formula for the sum of a geometric series:

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

Where:

$a = 120000$ (the first term)
$r = 1.05$ (the common ratio)
$n = 11$ (the number of years from 2013 to 2023 inclusive)

Substituting into the formula:

$S_{11} = 120000 rac{(1 - (1.05)^{11})}{1 - 1.05}$

Calculating $(1.05)^{11}$ : $(1.05)^{11} ext{ approximately equals to } 1.7137$

Therefore:

$S_{11} = 120000 rac{(1 - 1.7137)}{-0.05}$

Calculating this yields: $S_{11} = 120000 rac{-0.7137}{-0.05} = 120000 imes 14.274$

Calculating the total predicted profit: $S_{11} = 1712484 = 1704814$

Thus, the total predicted profit for the years 2013 to 2023 is approximately £1704814.

A company predicts a yearly profit of £120 000 in the year 2013 - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 6

Worked Solution & Example Answer:A company predicts a yearly profit of £120 000 in the year 2013 - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 6

Show that the predicted profit in the year 2016 is £138 915.

Find the first year in which the yearly predicted profit exceeds £200 000.

Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to the nearest pound.

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