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# UNOFFICIAL MARKSCHEME Edexcel Maths Calculator paper 09/06/2016

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1. (Original post by Abzzzz)
Trial and improvement was 3.7
You mean 3.8
2. I didn't do a curve on the graph as it was up and down numbers if you get what I mean, rather than increasing. rip A grade I was really hoping for
3. (Original post by Abzzzz)
Trial and improvement was 3.7
It was 3.8 lololol
4. (Original post by Abzzzz)
Trial and improvement was 3.7
i got 3.8?
5. (Original post by Disfuligee)
15000*0.77*0.82*0.82=7766.22

This is the correct answer so Paul is wrong
Oh so I was wrong in my working

Do you think I'll get any marks from that?
6. (Original post by robolliffe)
yeah i did and I'm very confident i am correct
Were we supposed to round it? I got 28.94
7. (Original post by 9091student)
i got 3.8?
You are right
8. (Original post by Purpleunicorn197)
Hi I'm going to start and unofficial markscheme so add your answers
Reps are not necessary but would be appreciated

It would be really useful if you could remember the question numbers
1. Order numbers correctly and include a key
b) 9/20=
0.45
2a. 5a*2b*3c =
30abc
b. Factorise fully 3y +6 =
3(y+2)
c. Expand x(x-3)=
x^2-3x
3.
x=9.25
4. Smallest packet was best value for money because it was cheapest per bag of crisps
5. 4 times as many red than green: There were 0.4 blue and 0.15 yellow.
0.4+0.15+4x+x=1
0.55+5x=1
5x=0.45
x=0.09
4x=0.36
Green- 0.09 Red- 0.36
6a. 20.3
b. 68.04kg
c. 2.61m
7. 3.8
8a. 4:3
b. 480 pixels
9.x= 124 degrees
10. 7.92
11.£42.29
12. x=144 degrees
13.
-2 -1 0 1 2
-1 3 1 -1 3
Draw curve
14. 3 Bars
15. No, because 7766.22 was greater than 7500 pounds
16. 128g
b.11g
17. 30.1
18. 10a^5b^4
b. y= 5p^2-x
19.11 males
20. x= 0.29 or x= -2.29
21. 29.25
22. F.d- 1.6, 3.2, 2.4, 1.8, 0.3
23.28.9
24a. -1/3x+4
b. 2x/(x+1)(x-1)
25. 80.4
I got 20 for question 19 is that incorrect 128/50 and then divide the answer by 51???
9. (Original post by yoyojames1234567)
It was 10 u cant have .7 of a person
They usually give 10 or 11, depending on how you round it
10. (Original post by EdanBrooke)
I didn't do a curve on the graph as it was up and down numbers if you get what I mean, rather than increasing. rip A grade I was really hoping for
It was an up and down graph 😁 Just a curvy one 😂😂
11. (Original post by robolliffe)
yeah i did and I'm very confident i am correct
same
12. (Original post by 9091student)
i got 3.8?
Was 3.8
13. Can someone please upload a completed paper which has solutions in a pdf format, big shout out and thanks to whoever does! Think i got about 93!
14. (Original post by Abzzzz)
Trial and improvement was 3.7
No it wasnt it was 3.8

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15. What was he 126g and the 11g question for?
16. (Original post by DOMAR34)
How'd you work it out? Scared to do my working on here incase I get roasted but here goes:

====
Paul buys a new car for \$(American computer, insert pounds)15,000

The magazine said new cars lose over half their value after 3 years. Is this true?

100-x% (X is the percentage depreciation after the first year, don't know value)

X% So X/100*15,000 = x* (asterik is for future notation)

100-x% (X is the percentage depreciation for the next years after the first, don't know value)

X% so X/100*x*(from previous answer) = 7392 <---- the answer I got.

Yes, the car lost over half its value as (15000/2 = 7500 so 7500-7392 = 108).
The value decreased by 23% the first year, so by the end of Year 1, it was worth £11550. (15000x0.77)
It decreased by 18% each year for the next two years, so at the end of Year 2, its value was £9471 (11550x0.82).
At the end of year 3, its value was £7766.22 (9471x0.82), which is more than half of £15000, so it has not lost half of its value in the first three years.
17. (Original post by Sullivan0)
I got 20 for question 19 is that incorrect 128/50 and then divide the answer by 51???
It would be 50/128, the number should be below 1
18. (Original post by Edgerulezs)
It was an up and down graph 😁 Just a curvy one 😂😂
Yeah didn't do curves. Damn.
19. (Original post by SweetStarlight14)
The value decreased by 23% the first year, so by the end of Year 1, it was worth £11550. (15000x0.77)
It decreased by 18% each year for the next two years, so at the end of Year 2, its value was £9471 (11550x0.82).
At the end of year 3, its value was £7766.22 (9471x0.82), which is more than half of £15000, so it has not lost half of its value in the first three years.
Will I get 0 marks for that answer or maybe 1-2?
20. What number was the circumference perimeter question?

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