UNOFFICIAL MARKSCHEME Edexcel Maths Calculator paper 09/06/2016

How many marks will i get for these questions. Its driving me nuts just thinking about it.



Pauls car question

My answer:15000(1-23/100)^1=11500 <==first year
1550(1-18/100)^2 = 7766.22 <=== 2 years added on

1/2 of 15000 = 7500

I then added the 7766.22 and 11500 by mistake.(4 mark question)


Q.colins drive
My answer:

3.50x5 = £17.50
18*5=90
90*2=180
180/45.2=3.982300885 - 2DP = 3.98
1836.9*18.1488 = 2484.57072p

p to £ x100 = £24.85

I for got to add the £17.50 at the end(5Mark question)

For the first one marked harshly 2 and marked leniently 3 however I think you're likely to get 3?

For the last one I think you got 4/5 because you only missed out one step.

Can I quickly pm you?


Posted from TSR Mobile

Your actually wrong the answer is in fact, 3.7.

Stop embarrassing yourself, it is a grade C question. When you put 3.7 into the equation, the answer is too low. When you put 3.8 into the equation, the answer is too high. When you put 3.75 into the equation, it is too low. This means that x is between 3.75 and 3.8. This means that no matter what number x is between 3.75 and 3.8, they will always round up to 3.8 to 1 decimal place. Stop answering back when you're not sure about the answer and learn from people who know the correct answer.

Lol your deluded my friend, go on this past test paper, question 11 on this link.
http://www.mathsgenie.co.uk/papers/2hjune2012ans.pdf

The answer on this past test paper trial and improvement question is 4.6 wanna know why?
4.7- 75.623-72 = 3.623
4.6- 72-69.736 = 2.264

If you take away you can see that 4.6 is closer, plus 4.65 is obviously closer to 4.7 then 4.6. See you dont need to do any fancy rounding stuff, you know it's 3.7 but you don't want to admit it, 3.7 is much closer to 34 than 3.8.

In every single trial and improvement question you have to round up if the answer is closer to the original number and is at least half way. The answer is 38. Look on past exam questions.


Posted from TSR Mobile

Here is the paper 2 GCSE June 2015 edexcel trial and improvement question:

"The equation x³ – x² = 30 has a solution between 3 and 4
Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place.
You must show ALL your working
(Total for Question 2 is 4 marks)"

And here is the official mark scheme:
"B2 for a trial 3.4 ≤ x ≤ 3.5 evaluated correctly
(B1 for a trial evaluated correctly for 3 ≤ x ≤ 4 )
B1 for a different trial evaluated correctly for 3.45≤ x < 3.5
B1 (dep on at least one previous B1) for 3.5
[Note: Trials should be evaluated to at least accuracy shown in table, truncated or rounded]
[NB No working scores 0 marks]"

This is the table of solutions for the function and values that candidates may have used:
x x³ – x²
3.1 20.(181)
3.2 22.(528)
3.3 25.(047)
3.4 27.(744)
3.5 30.(625)
3.6 33.(696)
3.7 36.(963)
3.8 40.(432)
3.9 44(109)
3.45 29.1(61125)
3.46 29.4(50136)
3.47 29.7(41023)
3.48 30.0(33792)
3.49 30.3(28449)

To summarise, based on previous official markschemes,the mark scheme this year is likely to be as follows:
B2 for correctly evaluated trial with 3.7 ≤ x ≤ 3.8
B1 for different correctly evaluated trial with 3.75 ≤ x < 3.8
B1 for 3.8

The correct answer is 3.8, and if you have not been taught to use the halfway point (in this case 3.75) then you are being misinformed. As a year 11 teacher, I had a class of very able students, most of which obtained a A or B in their Feb mock (3 of them an A*) and I know they can look at the first point (worth B2) to decide the accurate value but I always emphasise that they must trial the midpoint. I'd be doing them a disservice if I didn't!

Here is the paper 2 GCSE June 2015 edexcel trial and improvement question:

"The equation x³ – x² = 30 has a solution between 3 and 4
Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place.
You must show ALL your working
(Total for Question 2 is 4 marks)"

And here is the official mark scheme:
"B2 for a trial 3.4 ≤ x ≤ 3.5 evaluated correctly
(B1 for a trial evaluated correctly for 3 ≤ x ≤ 4 )
B1 for a different trial evaluated correctly for 3.45≤ x < 3.5
B1 (dep on at least one previous B1) for 3.5
[Note: Trials should be evaluated to at least accuracy shown in table, truncated or rounded]
[NB No working scores 0 marks]"

This is the table of solutions for the function and values that candidates may have used:
x x³ – x²
3.1 20.(181)
3.2 22.(528)
3.3 25.(047)
3.4 27.(744)
3.5 30.(625)
3.6 33.(696)
3.7 36.(963)
3.8 40.(432)
3.9 44(109)
3.45 29.1(61125)
3.46 29.4(50136)
3.47 29.7(41023)
3.48 30.0(33792)
3.49 30.3(28449)

To summarise, based on previous official markschemes,the mark scheme this year is likely to be as follows:
B2 for correctly evaluated trial with 3.7 ≤ x ≤ 3.8
B1 for different correctly evaluated trial with 3.75 ≤ x < 3.8
B1 for 3.8

The correct answer is 3.8, and if you have not been taught to use the halfway point (in this case 3.75) then you are being misinformed. As a year 11 teacher, I had a class of very able students, most of which obtained a A or B in their Feb mock (3 of them an A*) and I know they can look at the first point (worth B2) to decide the accurate value but I always emphasise that they must trial the midpoint. I'd be doing them a disservice if I didn't!

Here is the paper 2 GCSE June 2015 edexcel trial and improvement question:

"The equation x³ – x² = 30 has a solution between 3 and 4
Use a trial and improvement method to find this solution.
Give your answer correct to one decimal place.
You must show ALL your working
(Total for Question 2 is 4 marks)"

And here is the official mark scheme:
"B2 for a trial 3.4 ≤ x ≤ 3.5 evaluated correctly
(B1 for a trial evaluated correctly for 3 ≤ x ≤ 4 )
B1 for a different trial evaluated correctly for 3.45≤ x < 3.5
B1 (dep on at least one previous B1) for 3.5
[Note: Trials should be evaluated to at least accuracy shown in table, truncated or rounded]
[NB No working scores 0 marks]"

This is the table of solutions for the function and values that candidates may have used:
x x³ – x²
3.1 20.(181)
3.2 22.(528)
3.3 25.(047)
3.4 27.(744)
3.5 30.(625)
3.6 33.(696)
3.7 36.(963)
3.8 40.(432)
3.9 44(109)
3.45 29.1(61125)
3.46 29.4(50136)
3.47 29.7(41023)
3.48 30.0(33792)
3.49 30.3(28449)

To summarise, based on previous official markschemes,the mark scheme this year is likely to be as follows:
B2 for correctly evaluated trial with 3.7 ≤ x ≤ 3.8
B1 for different correctly evaluated trial with 3.75 ≤ x < 3.8
B1 for 3.8

The correct answer is 3.8, and if you have not been taught to use the halfway point (in this case 3.75) then you are being misinformed. As a year 11 teacher, I had a class of very able students, most of which obtained a A or B in their Feb mock (3 of them an A*) and I know they can look at the first point (worth B2) to decide the accurate value but I always emphasise that they must trial the midpoint. I'd be doing them a disservice if I didn't!

I cannot believe how much time and space has been devoted to such a simple question (Trial and Improvement) and all over 1 mark. Thank God it's been taken off the specification for next year !!!

For paper 1 I got around 76 and for paper 2 I got around 96, what grade do you think that will be?

I'm sure you're just being too harsh on yourself!! What are you aiming for?

Hi I'm going to start and unofficial markscheme so add your answers
Reps are not necessary but would be appreciated

Sorry for the hold up with the questions
1. Order numbers correctly and include a key (Stem and leaf) (3 marks)
b) 9/20=
0.45 (2 marks)
2a. 5a*2b*3c =
30abc (1 mark)
b. Factorise fully 3y +6 =
3(y+2) (1 mark)
c. Expand x(x-3)=
x^2-3x (1 mark)
3. Perimeter of a rectangle with side lengths x and x+4 is 45
x+x+x+4+x+4= 4x+8
4x+8=45
4x=37
x=9.25 (3 marks)
4. Smallest packet was best value for money because it was cheapest per bag of crisps (4 marks)
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 (3 marks)
6a. Body mass index= m/h^2
Mass= 50kg
Height= 1.57m
20.3 (2 marks)
b. What mass will result in a BMI of 21 with a height of 1.8m
68.04kg (2 marks)
c. Tom's skis can be a maximum of 145% of his height what is their max. tenght (Tom is 1.8m)
2.61m (3 marks)
7. Trial and error. x^3-5x=34 with a number between 3 and 4
x^3-5x=34
3.7 gives you 32.153 which is too small
3.8 gives you 35.872 which is too big
So you go in the middle at 3.75 and if it is too big then its you go with the smaller option and viceversa so 3.75 gives you 33.984375 which is too small so you go for the larger option of 3.8
3.8 (4 marks)
8a. Photographs with of width 720 pixels and a height of 540 pixels what is the ratio width:height in its simplest form?
4:3 (2 marks)
b. The person now wants the ratio to be 2:3, without changing the width of the picture, what is the new height?
480 pixels (2 marks)
9.x= 124 degrees (4 marks)
10. Mean of grouper freq. table 396(freq.*midpoint total)/50 (freq. total)
7.92 (4 marks)
11. Colin's drive to work
£42.29 (5 marks)
12. The middle angle of a pentagon
x=144 degrees (4 marks)
13.
-2 -1 0 1 2
-1 3 1 -1 3 (2 marks)
Draw curve (2 marks)
14. steel bar with length of 120cm and a cross sectional area of 72cm^2 and a density of 8g/cm^3. A trolley can carry 250kg so how many bars can it carry?
3 Bars (5 marks)
15. Pauls buys a car worth 15,000 pounds and the value depreciates by 23% in the first year and 18% after the first year. Has Paul's car lost half of its value in theree years?
No, because 7766.22 was greater than 7500 pounds (4 marks)
16. Median on cumulative freq. graph
128g (1 mark)
b. Use graph to find interquartile range
11g (2 marks)
17. Triangle with sides 15.6cm (opp), 18(hyp). Angle is split into 2, 30 degrees and x degrees. the whole angle is 60.1 so
x=30.1 (4 marks)
18. 10a^5b^4 (2 marks)
b. y= 5p^2-x (3 marks)
19. 51 out of the 234 were maths in yr 12 so 50 * 51/234
11 males (2 marks)
20. 3x^2 + 6x -2
x= 0.29 or x= -2.29 (3 marks)
21. v=14 to 2 s.f u= 8.7 to 2 s.f
What is the upper bound of I when I= 5(v-u)
29.25 (3 marks)
22. Histogram -F.d- 1.6, 3.2, 2.4, 1.8, 0.3 (3 marks)
23. Perimeter of the shaded part of the sector.
Measurements were 4+4+5/2pi+25/6pi
28.9 (3 marks)
24a. Simplify (3-x)/3x^2-5x-12
-1/3x+4 (2 marks)
b. Write are one fraction in its simplest for (x/x-1)- (x/x+1)
2x/(x+1)(x-1) (3 marks)
25. Triangle with side lengths 7 and 8 finding the top angle using 1/2absinC , sine rule and cosine rule
80.4 (6 marks)

Made the same mistake on the 2015 mock, ahhh this time i put the key in first before doing anything.

Don't worry about it, everyone's bound to make silly mistakes, its very hard to avoid them. Especially under pressure of exams.

How much do you think it would be for an A?