C1 June 2014 Q8
8. 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.
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.
| Scheme | Marks |
|---|---|
| Use \(n^{th}\text{ term} = a + (n - 1)d\) with \(d = 10\); \(a = 150\) and \(n = 8\), or \(a = 160\) and \(n = 7\), or \(a = 170\) and \(n = 6\) : \(= 150 + 7\times 10\) or \(160 + 6\times 10\) or \(170 + 5\times 10\) | M1 |
| = 220* (Or gives clear list – see note) | A1* |
| (2) |
Notes
M1: Attempt to use \(n^{th}\text{ term} = a + (n - 1)d\) with \(d = 10\), and correct combination of \(a\) and \(n\) i.e. \(a = 150\) and \(n = 8\) or \(a = 160\) and \(n = 7\), or \(a = 170\) and \(n = 6\)
A1 *: Shows that 220 computers are sold in 2007 with no errors
Note that this is a given solution, so needed \(150 + 7\times 10\) or \(160 + 6\times 10\) or \(170 + 5\times 10\) or equivalent.
Accept a correct list showing all values and years for both marks Just 150,160,170,180,190,200,210,220 is M1A0
Need some reference to years as well as the list of numbers of computers for A1.
Or
| Scheme | Marks |
|---|---|
| If answer 220 is assumed and \(150 + (n - 1)10 = 220\) or variation is solved for \(n =\) | M1 |
| Then \(n = 8\), so 2007 is the year (must conclude the year) | A1* |
| (2) |
| Scheme | Marks |
|---|---|
| Use \(S_n = \dfrac{n}{2}\left\{2a + (n - 1)10\right\}\) Or \(S_n = \dfrac{n}{2}\left\{a + l\right\}\) and \(l = a + (n - 1)10\) | M1 |
| \(= 7(300 + 13\times 10)\) or \(7(150 + 280)\) | A1 |
| \(= 7\times 430\) \(= 3010\) | A1 |
| (3) |
Notes
M1: Attempts to use \(S_n = \dfrac{n}{2}\left\{2a + (n - 1)d\right\}\) with \(d = 10\), and correct combination of \(a\) and \(n\) i.e. \(a = 150\) and \(n = 14\), or \(a = 160\) and \(n = 13\), or \(a = 170\) and \(n = 12\)
A1: Uses \(S_n = \dfrac{n}{2}\left\{2a + (n - 1)d\right\}\) with \(a = 150\), \(d = 10\) and \(n = 14\) [N.B. \(S_n = \dfrac{n}{2}\left\{a + l\right\}\) needs \(l = a + (n - 1)d\) as well
NB A0 for \(a = 160\) and \(n = 13\) or \(a = 170\) and \(n = 12\) unless they then add the first, or first two terms respectively.
A1: Cao 3010 . This answer (with no working) implies correct method M1A1A1.
Special case: If a complete list 150+160+170+180+190+200+210+220+230+240+250+260+270+280 is seen, then there is an error finding the sum then score M1A1A0, but incomplete or wrong lists score M0A0A0
| Scheme | Marks |
|---|---|
| Cost in year \(n = 900 + (n - 1)\times -20\) Sales in year \(n = 150 + (n - 1)\times 10\) | M1 |
| Cost \(= 3\times\)Sales \(\Rightarrow 900 + (n - 1)\times -20 = 3\times\left(150 + (n - 1)\times 10\right)\) \(900 - 20n + 20 = 450 + 30n - 30\) | M1 |
| \(500 = 50n\) \(n = 10\) | M1 |
| Year is 2009 | A1 |
| As \(n\) is not defined they may work correctly from another base year to get the answer 2009 and their \(n\) may not equal 10. If doubtful – send to review. | |
| (4) | |
| (9 marks) |
Notes
M1: Writes down an expression for the cost \(= 900 + (n - 1)\times -20\) or writes \(900 + (n - 1)d\) and states \(d = -20\)
Allow \(900 + n\times -20\). Allow recovery from invisible brackets.
M1: Attempts to write down an equation in \(n\) for statement ‘cost \(= 3\times\)sales’
\(900 + (n - 1)\times -20 = 3\times\left(150 + (n - 1)\times 10\right)\). Accept the 3 on the wrong side and allow use of 20 instead of \(-20\) and allow \(n\) (consistently) instead of \(n - 1\) for this mark. Ignore £ signs in equation.
M1: Solves the correct linear equation in \(n\) to achieve \(n = 10\) (for those using \(n - 1\)) or \(n = 9\) (for those using \(n\)). Ignore £ signs.
A1: Cso Year 2009 (A0 for the answer Year 10 if 2009 is not given )
Special case. Just answer or trial and improvement with no equation leading to answer scores SC 0,0,1,1
Equations satisfying the method mark descriptors followed by trial and improvement could get all four marks