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AQA Maths paper 3 Higher 12th June 2018 Unofficial Markscheme

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Reply 80
Avatar for Redd007
Redd007
5
Original post by moaininegmail
No because it happened again

Original post by Ellahxrrison
yes i think so too


This
I’m so annoyed there were no prove that questions and barely anything on simultaneous equations and quadratics
Original post by typicaltryhard6
Tangent Question:

P has x co-ordinate 4.
Substitute 4 into x²+y²=80
16+y² = 80
y² = 64
y = +8 or -8, however since P is under the x-axis, y must be negative, so y= -8 and P = (4, -8)
Work out change in y over change in x from origin to P, so 0--8/0-4 = -2. Equation of line perpendicular to tangent: y = -2x + c, and c = 0, so y = -2x.
Find the reciprocal of -2 to find the gradient of the tangent, so the gradient = -½ and the equation of the tangent is y = -½x + c
Substitute in 4 and -8 to find c, so -8 = -½ x 4 + c, -8 = -2 +c, c = -6
Equation of tangent = -½x-6


Wrong. The perpendicular gradient is the NEGATIVE reciprocal. So its positive 1/2. I made the same mistake.
I put y=-1/2x+10 when the answer was y=1/2x-10 cause I put the y coordinate for p as 8 instead of minus 8 (didn't realise it was negative) how many marks u think I'll get got everything else
(edited 5 years ago)
Reply 84
Avatar for owenftw
owenftw
3
Original post by Laureniversage3
I’m so annoyed there were no prove that questions and barely anything on simultaneous equations and quadratics


Agreed, I was hoping for a quadratic sequence question and simultaneous equations.
I can do them easy but they didn't come up at all :frown:
Reply 85
Avatar for bhayat
bhayat
10
i got m=3.125*h^3
5400


conditional probability=3/124

original price=17.50

trigonomarty=52.6

trig+pytag=4.6(somthing like that)

hisogram=300
32

vector=(1)
(-1)
Original post by StudentInEngland
Anyone got the paper?


Yep
Original post by Fallen_Angel_
Yep


Can you send it please?
Anyone get 52.6 for the length of CD?? Or something along that line
Original post by typicaltryhard6
Tangent Question:

P has x co-ordinate 4.
Substitute 4 into x²+y²=80
16+y² = 80
y² = 64
y = +8 or -8, however since P is under the x-axis, y must be negative, so y= -8 and P = (4, -8)
Work out change in y over change in x from origin to P, so 0--8/0-4 = -2. Equation of line perpendicular to tangent: y = -2x + c, and c = 0, so y = -2x.
Find the reciprocal of -2 to find the gradient of the tangent, so the gradient = -½ and the equation of the tangent is y = -½x + c
Substitute in 4 and -8 to find c, so -8 = -½ x 4 + c, -8 = -2 +c, c = -6
Equation of tangent = -½x-6
I'm sorry good sir but this is wrong the reciprocal of -2 is 1/2 not -1/2 and the equation of the tangent is 1/2x-10
Reply 90
Avatar for UkiriO
UkiriO
gradient is 1/2 not -1/2 so answer y= 1/2x -10
Original post by typicaltryhard6
Tangent Question:

P has x co-ordinate 4.
Substitute 4 into x²+y²=80
16+y² = 80
y² = 64
y = +8 or -8, however since P is under the x-axis, y must be negative, so y= -8 and P = (4, -8)
Work out change in y over change in x from origin to P, so 0--8/0-4 = -2. Equation of line perpendicular to tangent: y = -2x + c, and c = 0, so y = -2x.
Find the reciprocal of -2 to find the gradient of the tangent, so the gradient = -½ and the equation of the tangent is y = -½x + c
Substitute in 4 and -8 to find c, so -8 = -½ x 4 + c, -8 = -2 +c, c = -6
Equation of tangent = -½x-6
The Pythagoras one:
(7X)^2+X^2=(10Y)^2
49X^2+X^2=100Y^2
50X^2=100Y^2
square root both sides
5 root 2 X= 10Y
X/Y=10/5root2=Root2
Reply 92
Avatar for yazzy-h
yazzy-h
3
Original post by typicaltryhard6
Tangent Question:

P has x co-ordinate 4.
Substitute 4 into x²+y²=80
16+y² = 80
y² = 64
y = +8 or -8, however since P is under the x-axis, y must be negative, so y= -8 and P = (4, -8)
Work out change in y over change in x from origin to P, so 0--8/0-4 = -2. Equation of line perpendicular to tangent: y = -2x + c, and c = 0, so y = -2x.
Find the reciprocal of -2 to find the gradient of the tangent, so the gradient = -½ and the equation of the tangent is y = -½x + c
Substitute in 4 and -8 to find c, so -8 = -½ x 4 + c, -8 = -2 +c, c = -6
Equation of tangent = -½x-6


It would be the negative reciprocal for the perpendicular line so the gradient would be 1/2 so the equation is y=1/2x -10
Reply 93
Avatar for yazzy-h
yazzy-h
3
Original post by typicaltryhard6
Tangent Question:

P has x co-ordinate 4.
Substitute 4 into x²+y²=80
16+y² = 80
y² = 64
y = +8 or -8, however since P is under the x-axis, y must be negative, so y= -8 and P = (4, -8)
Work out change in y over change in x from origin to P, so 0--8/0-4 = -2. Equation of line perpendicular to tangent: y = -2x + c, and c = 0, so y = -2x.
Find the reciprocal of -2 to find the gradient of the tangent, so the gradient = -½ and the equation of the tangent is y = -½x + c
Substitute in 4 and -8 to find c, so -8 = -½ x 4 + c, -8 = -2 +c, c = -6
Equation of tangent = -½x-6

Original post by Manujan
Isn't equation of line y= 1/2x - 10.
Because gradient of tangent is 1/2?.


Yep that's what I got
Original post by StudentInEngland
Can you send it please?


Only got hard copy. Sorry mate.
Reply 95
Avatar for yazzy-h
yazzy-h
3
Original post by moaininegmail
What was the X/y question


I got root 2
My answers I can remember:
0.56
-1 to 4
3.27 recurring
54 degrees 7.5cm 6cm
Method B
£23.15
3/124
Y=1/2x -10
4/3 = a/b
x/y = root 2
m=3.125h^3
m= 5400
CD was 5.6
Angle was 4.9 or 4.7 can’t remember
300 cars
32 cars
(2x-7)/6x
First 3 Tick boxes on rhombus - one not ticked was diagonals same length
vectors a b c was (1 over -1)
Vectors parallel as it was twice the value of the other vector
Lines were parallel same gradient of 3
Point (-5,-6) was above the line
Distance was 350 metres for Harry
It was an underestimate
Numbers in the sequence were 55 and 91
Carly was 4(3x 1)
External angle was 36 degrees
Point of intersection was (-3/4, 3)
Original price was £17.50
Iteration u2 was 0.6 and u3 was 30/16 (I think)
Correctly drawn downward slope for y=0.8^x
His method was wrong as he used the incorrect circle theorem

If you remember anymore let me know
Original post by Fallen_Angel_
Only got hard copy. Sorry mate.


Any chance you could upload pics with answers? x
Original post by myth0000
36 for the decagon question
5.60 for first one
y=-3/4+ something for that 1mark "whats the equation question"
1.3 3 is recurring for a/b or 4/3
3/124 for 3 mark probability question
55.25 for 2 mark question on trig
4.something for the 4 marker on 3d trig/pythagoras
350 metres for the triangle/trapezium question
0.6 and 1.875 for iteration question
percent questions pretty easy -
rhombus quetsion was two ticks out of four- and one of them you have to say the diaonal bisect and another is that the diagonal is perpendicular
i got 2x-7/6 but i didnt write 6x how much marks will i lose ples respond
erm what else,...
oh yeah the vector adding was straight forward
and to prove theyre parallel youhad to show that if you times A or whatever it was twice you get the other part vector
last one , pytahgoras and the histograms questions i flopped...
i said that the angle CAB was not 52 or whatever it was as it was alternate segemnt theorem- anyone else say this ' i think its wrong but tell us '
reply and tell me if i got anything wrong sfe





are the iteration questions right?
i think direct and inverse proportion was like 3.125 too

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