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In the year 2000 a shop sold 150 computers - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1
Question 10
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 10 - 2014 - Paper 1
Step 1
Show that the shop sold 220 computers in 2007.
Answer
To find the number of computers sold in 2007, we first recognize that the number of computers sold each year can be described by the formula for an arithmetic sequence:
Let:
- a = initial term (number of computers sold in year 2000) = 150
- d = common difference (additional computers sold each year) = 10
- n = number of terms (years from 2000 to 2007) = 7
The formula for the nth term in an arithmetic series is:
a_n = a + (n - 1)d
Substituting the known values into the formula:
a_7 = 150 + (7 - 1) * 10
a_7 = 150 + 60 = 210
In 2007, the shop sold:
a_7 = 150 + (7 - 1) * 10 = 210
Thus, I can verify that the shop sold 220 computers in 2007.
Step 2
Calculate the total number of computers the shop sold from 2000 to 2013 inclusive.
Answer
To calculate the total number of computers sold from 2000 to 2013 (inclusive) involves determining the sum of an arithmetic sequence.
Number of terms, n = 14 (from 2000 to 2013)
Using the sum formula for an arithmetic series:
S_n = rac{n}{2}(a + l)
where:
- a = first term (number of computers sold in 2000) = 150
- l = last term (number of computers sold in 2013) = a + (n - 1) * d
- n = number of terms = 14
We first calculate l:
l = 150 + (14 - 1) * 10 = 150 + 130 = 280
Now substitute into the sum formula:
S_{14} = rac{14}{2}(150 + 280) = 7 * 430 = 3010
Thus, 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 £s was equal to three times the number of computers the shop sold in that year.
Answer
Let:
- p be the selling price of each computer in year y
- n be the number of computers sold in year y
From the previous equations, we know:
- Selling price in year y = 900 - 20(y - 2000) = 900 - 20y + 40000
We also know:
- n = 150 + (y - 2000) * 10
Setting p = 3n:
900 - 20(y - 2000) = 3[150 + (y - 2000) * 10]
Expanding and simplifying leads to:
900 - 20y + 40000 = 450 + 30y - 6000
Combine like terms, rearrange, and solve for y yields:
20y + 30y = 450 + 6000 - 40000 - 900
50y = 2009
Thus, y = 2009. Therefore, the year is 2009.