[PrinceUpsb]

Mark Scheme in Word Format - very thorough, many thanks to Empty Child for posting this! (Click below link)
http://www.thestudentroom.co.uk/atta...0&d=1339627392

Click Spoiler for Original Unofficial Mark Scheme (unfinished)

Spoiler:

Show

Pictures are for reminders of Qs as well as method explanation. Black writing = given, red writing = not given.
NB: Pictures are in no way accurately drawn, they are just to help - these are the questions in the exam, just not accurately drawn.
**Click on picture thumbnail for larger image

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Question 1: (Straight line, 3 different angles: 2x, 3x and 4x - calculate smallest angle)

2x + 3x + 4x = 180 Degrees
9x = 180 Degrees
x = 180 / 9
x = 20
2x = 20 * 2 = 40 Degrees

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Question 10: (Concert - standing area - amount of people)

(15*10)-((Pi*3^2)/2)= 135.862...

135.862.../0.3 = 452.876...

Answer = 452 as you cannot have 0.876 of a person and 453 people would require more space

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Question ?: (Find t - triangle - parallel lines)

Angle opposite w+40 must also be w+40.
Angles within parallel lines add to 180.
This means you had four angles around the point that were as follows: w+40, 108, w+40, 108.
Work out w: 2w+80=(360-108-108)=144
2w=64
w=32

t=180-(w+40)-(2*w)=44

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Question ?: (Perimeter of shape with triangles cut out)

Each side of the shape was a value of 5.2cm as an equilateral triangle was cut out. 8 sides means you did
5.2*8=41.6cm

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Question ?: (Find arc length AB)

The arc was from two points on an equilateral triangle, meaning the angle was 60 degrees.
(60/360)*16pi=8.377...
Answer = 8.38 (3sf)

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Question ?: (Direct proportion M to r^3)
M=200, r=8
M=k(r^3)
k=1.6

If M=3125, what is r
cbrt(3125/1.6)=12.5

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Question ?: (Quadratic Equation - quadratic formula)
3x^2+4x-10=0
Use the quadratic equation to get two values of x which were:
-2.61.. and 1.27..

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Question ?: (7 marks, area of shape, multiple steps)
There was a lot of info on this one and I haven't managed to remember it all. If anyone remembers all the values please post

But as far as I remember I got 36.01cm, as did a lot of other people on TSR, that's what I believe it is.

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Question ?: (Vector Questions)
a) 1.5s as the ratio was 3:2
b) -s+t+1.5s= t+0.5s

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Question ?: (x and y perimeter)
x=2.5y

2x+2y = perimeter

y=0.4x
Hence total perimeter expressed in terms of x was 2.8x

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Question ?: (bearings questions)

b)F cannot be greater than 180 as a bearing cannot be greater than 360 (in relation to the formula given)
c)The bearing from C->D is 342. What is the bearing of D->C. 180-(360-342)=162

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Question 4: (Enlargement of shape on cm grid)
12*2^2=48

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Question 3: (Concrete bases)
55% of 3.8 = 2.09
6 loads of 5 = 30 total
30/2.09=14.35
How many bases can they make? I would assume answer is just 14 as its the maximum whole number
Answer was 14

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Question ?: (volume of cuboid)
Values were x, 5x+1, 2x+3...
Length=width+7
5x+1=2x+10
5x=2x+9
3x=9
x=3
sub in 3 for x
Answer = 432

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Question ?: (Circle Theorems)

b) APD (?) is (?) because opposite angles in a cyclic quadrilateral add up to 180 (just explaining why will get you the mark)

For everyone inquiring about Grade Boundaries; as this is the first time this paper has been sat, there is no previous grade boundaries to give a rough indication.

It has been suggested, with the paper being out of 80 marks that the following grade boundaries may suit:
66 - A*
52 - A
41 - B
33 - C

**Please don't assume these are correct, they are just rough ideas posted on TSR

[taz180496]

no one post any answers as this will give foreigners 12hours behind us the advantage which will in return result the increase of the boundaries.

[AP1994]

yes please dont post anything and take that question with the answer?
dont wanna fail

[PrinceUpsb]

(Original post by AP1994)
yes please dont post anything and take that question with the answer?
dont wanna fail

(Original post by taz180496)
no one post any answers as this will give foreigners 12hours behind us the advantage which will in return result the increase of the boundaries.

This is AQA. Those rules only apply to Edexcel exams.

[07khan_A]

can u please post some more answers thanks

[07khan_A]

(Original post by PrinceUpsb)
//Under Construction
//Under Construction

Question 10: (Concert - standing area - amount of people)
(15*10)-((Pi*3^2)/2)= 135.862...

135.862.../0.3 = 452.876...

Answer = 452 as you cannot have 0.876 of a person and 453 people would require more space
----
Question ?: (Find t - triangle - parallel lines)

Angle opposite w+40 must also be w+40.
Angles within parallel lines add to 180.
This means you had four angles around the point that were as follows: w+40, 108, w+40, 108.
Work out w: 2w+80=(360-108-108)=144
2w=64
w=32

t=180-(w+40)-(2*w)=44
---

hi i took this paper today and got the same answer what did you get for the question with the quadrilateral made of 2 triangles - you had to calculate the area ??? thanks

[Maths94]

For question 15 where it was something like 3x+x2-10=0
What answers did use get and did you use trial and improvement?

[xxDanxx]

For Q10, the answer 135.... I think the paper said it had to be squared then / 0.3

Maximum number = area of standing area(squared) / 0.3

Giving an answer of around 61,528.27 (61,528 rounded)

[xxDanxx]

(Original post by Maths94)
For question 15 where it was something like 3x+x2-10=0
What answers did use get and did you use trial and improvement?

That wasn't the question it was

3x(Squared) + 4x - 10

You had to factorise by using the quadratic equation. Obtaining 2 different x values.

[Maths94]

Yeah that question. What was your final answer ?
I don't know why i got x = 0.90 ( guess really lol )
On the question where there is 2 parts (a) (b) you had to find R through M or something what did you get for them two ?

[PrinceUpsb]

(Original post by xxDanxx)
For Q10, the answer 135.... I think the paper said it had to be squared then / 0.3

Maximum number = area of standing area(squared) / 0.3

Giving an answer of around 61,528.27 (61,528 rounded)

I don't believe it did

[xxDanxx]

Can't remember my answer for R and M question but for my x values I think I had 1.27 and -2.6

OP:

Q:1

It was a straight line with 3 different angles upon it. 2x, 3x and 4x. You have to calculate smallest angle.

ANSWER:

2x + 3x + 4x = 180 Degrees
9x = 180 Degrees
x = 180 / 9
x = 20
2x = 20 * 2 = 40 Degrees

[Maths94]

Yeah for question 1 i got 40 Degrees

[xxDanxx]

There was a rotational question. Rotate a shape 90 degrees clockwise around point (0,1)

CBA drawing graph but that was question

[Antheran]

(Original post by xxDanxx)
For Q10, the answer 135.... I think the paper said it had to be squared then / 0.3

Maximum number = area of standing area(squared) / 0.3

Giving an answer of around 61,528.27 (61,528 rounded)

I was about to make the same mistake as you but then I realised that the area of the place is already in metres squared so squaring it again kind of suggests that you're not entirely sure what you're doing and is obviously wrong.

[dc747]

Am I the only one who thinks the question with the triangles cut from the square was very misleading? The diagram (though not drawn accurately I acknowledge) seemed to imply the sides of the triangle were not the same as the base, yet the question said they were. The discrepency between the diagram and the question was huge, so I worked out what the sides of the triangle were and used that for perimeter. In hindsight I guess I should have ignored the diagram

[Vernish]

(Original post by dc747)
Am I the only one who thinks the question with the triangles cut from the square was very misleading? The diagram (though not drawn accurately I acknowledge) seemed to imply the sides of the triangle were not the same as the base, yet the question said they were. The discrepency between the diagram and the question was huge, so I worked out what the sides of the triangle were and used that for perimeter. In hindsight I guess I should have ignored the diagram

Did that too.

[JamesWire]

They were, but as they were slanted they did not reach the top of the square.

[AdamskiUK]

(Original post by dc747)
Am I the only one who thinks the question with the triangles cut from the square was very misleading? The diagram (though not drawn accurately I acknowledge) seemed to imply the sides of the triangle were not the same as the base, yet the question said they were. The discrepency between the diagram and the question was huge, so I worked out what the sides of the triangle were and used that for perimeter. In hindsight I guess I should have ignored the diagram

I considered doing that, but it did say not drawn accurately (I think). I was tempted though.

[jacqueline789]

can you include how many marks each questions is worth on a mark scheme?
and grade boundaries?