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1. Does anyone know how to anwer this question
9. A store begins to stock a new range of DVD players and achieves sales of £1500
of these products during the first month.
In a model it is assumed that sales will decrease by £x in each subsequent month,
so that sales of £(1500 − x) and £(1500 − 2x) will be achieved in the second and
third months respectively.
Given that sales total £8100 during the first six months, use the model to
(a) find the value of x, (4)
(b) find the expected value of sales in the eighth month, (2)
(c) show that the expected total of sales in pounds during the first n months is
given by kn(51 − n), where k is an integer to be found. (3)
(d) Explain why this model cannot be valid over a long period of time. (1)

I can do this question - BUT in the mark scheme it says x = 60 i also have x = to 60.

But it makes more sense if x = -60 as some people do have in this forum.

I am just not sure which is correct....
2. The correct answer is x = 60 because the model states that each month the sales DECREASE by £(1500 - x). If x = -60 this would mean that the sales for each month would be £(1500 - -60), which is the same as £(1500 + 60) because two negatives make a positive. Therefore this means that the value of x has to be positive in order for sales to decrease each month. Hope this helps
3. (Original post by christinajane)

But it makes more sense if x = -60 as some people do have in this forum.

I am just not sure which is correct....
x=60 is correct, anybody who says otherwise is wrong.
4. Yeah I thought x = 60

Its just for part c

if x = 60 then wouldnt the equation be 30n(49-n)

Hmmm

Im getting confused by these negative signs haha!

It makes sense what you said about the -60 bit I get that - just for part c I cant get it to match what they want
5. (Original post by christinajane)
Yeah I thought x = 60

Its just for part c

if x = 60 then wouldnt the equation be 30n(49-n)

Hmmm

Im getting confused by these negative signs haha!

It makes sense what you said about the -60 bit I get that - just for part c I cant get it to match what they want
It might make more sense if you did it like this:

(1500-x) + (1500 - 2x) + (1500-3x) + ... ( 1500-nx)

= (1500 + 1500 + 1500 +... + 1500) + (-x - 2x - 3x - 4x -... - nx)
=1500n - (x + 2x + 3x + 4x + 5x + ... + nx)
=...
To be honest, I could even factor an x out if I wanted to:

1500n - x(1 + 2 + 3 + .. + n)
= 1500n - x(n/2 (n+1) )
Can you see how I factored the negative sign out of the bracket so I'm left with 1500n - (blah)?
6. Ahh yeah I get you..

Sometimes I use that method when I have a negative quadratic and it normally seems to work

Thank you once again everyone and especially zacken!
7. (Original post by christinajane)
Ahh yeah I get you..

Sometimes I use that method when I have a negative quadratic and it normally seems to work

Thank you once again everyone and especially zacken!
No worries.

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