In the year 2000 a shop sold 150 computers. Each year the shop sold 10 more computers than the year before, so that the shop sold 160 computers in 2001, 170 computers in 2002, and so on forming an arithmetic sequence. (a) Show that the shop sold 220 computers in 2007. (b) Calculate the total number of computers the shop sold from 2000 to 2013 inclusive. In the year 2000, the selling price of each computer was £900. The selling price fell by £20 each year, so that in 2001 the selling price was £880, in 2002 the selling price was £860, and so on forming an arithmetic sequence. (c) In a particular year, the selling price of each computer in £5 was equal to three times the number of computers the shop sold in that year. By forming and solving an equation, find the year in which this occurred.

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In the year 2000 a shop sold 150 computers - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Question 9

In the year 2000 a shop sold 150 computers. Each year the shop sold 10 more computers than the year before, so that the shop sold 160 computers in 2001, 170 computer... show full transcript

Worked Solution & Example Answer:In the year 2000 a shop sold 150 computers - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Step 1

Show that the shop sold 220 computers in 2007.

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Answer

To find the number of computers sold in 2007, we define the number of computers sold in any year as part of an arithmetic sequence:

Let:

The first term (2000) $a = 150$
The common difference $d = 10$

We can find the $n^{th}$ term using the formula: $a_n = a + (n-1)d$ In 2007, which is the 8th year since 2000, we have: $a_8 = 150 + (8-1) imes 10 = 150 + 70 = 220.$ Thus, the shop did sell 220 computers in 2007.

Step 2

Calculate the total number of computers the shop sold from 2000 to 2013 inclusive.

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Answer

To calculate the total number of computers sold from 2000 to 2013 (14 years), we need to find the sum of the arithmetic sequence.

Using the sum formula for an arithmetic series: $S_n = \frac{n}{2} \times (a + a_n)$ where:

$n = 14$ ,
$a = 150$ (computers sold in 2000),
$a_n = a + (n - 1)d = 150 + (14 - 1) \times 10 = 150 + 130 = 280$ .

Thus: $S_{14} = \frac{14}{2} \times (150 + 280) = 7 \times 430 = 3010.$ The total number of computers sold from 2000 to 2013 is 3010.

Step 3

In a particular year, the selling price of each computer in £5 was equal to three times the number of computers the shop sold in that year.

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Answer

Let the number of computers sold in that particular year be $x$ . The selling price in that year can be represented as: $ext{Selling Price} = 900 - 20(n-1),$ where $n$ is the year since 2000.

We also know that: $5 = 3x.$
From this, we have: $ext{Selling Price} = 900 - 20(n - 1) = 3x = 3 \times \frac{5}{3} = 5.$ Thus equating the two expressions: $900 - 20(n - 1) = 5.$ Solving for $n$ gives: $899 = 20(n - 1),\ n - 1 = \frac{899}{20},\ n = \frac{899}{20} + 1 = 45 + 1 = 46.$ This would yield the year 2000 + 46 = 2009. Therefore, the year when this occurred is 2009.

In the year 2000 a shop sold 150 computers - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Worked Solution & Example Answer:In the year 2000 a shop sold 150 computers - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Show that the shop sold 220 computers in 2007.

Calculate the total number of computers the shop sold from 2000 to 2013 inclusive.

In a particular year, the selling price of each computer in £5 was equal to three times the number of computers the shop sold in that year.

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