General Certificate of Education
Advanced Subsidiary Examination
January 2010
Mathematics
MM1B
Unit Mechanics 1B
Friday 15 January 2010
1.30 pm to 3.00 pm
For this paper you must have:
*
an 8-page answer book
*
the blue AQA booklet of formulae and statistical tables.
You may use a graphics calculator.
Time allowed
*
1 hour 30 minutes
Instructions
*
Use black ink or black ball-point pen. Pencil should only be used for drawing.
*
Write the information required on the front of your answer book. The Examining Body for
this paper is AQA. The Paper Reference is MM1B.
*
Answer all questions.
*
Show all necessary working; otherwise marks for method may be lost.
*
The final answer to questions requiring the use of calculators should be given to three
significant figures, unless stated otherwise.
*
Take g ¼ 9.8 m s2 , unless stated otherwise.
Information
*
The marks for questions are shown in brackets.
*
The maximum mark for this paper is 75.
*
Unit Mechanics 1B has a written paper only.
Advice
*
Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P22645/Jan10/MM1B 6/6/6/
MM1B
2
Answer all questions.
1 Two particles, A and B, are travelling in the same direction along a straight line on a smooth
horizontal surface. Particle A has mass 3 kg and particle B has mass 7 kg. Particle A has a
speed of 20 m s1 and particle B has a speed of 10 m s1 , as shown in the diagram.
A
B
20 m s1
10 m s1
Particle A and particle B collide and coalesce to form a single particle. Find the speed of this
single particle after the collision.
(3 marks)
2 A sprinter accelerates from rest at a constant rate for the first 10 metres of a 100-metre race.
He takes 2.5 seconds to run the first 10 metres.
(a) Find the acceleration of the sprinter during the first 2.5 seconds of the race.
(3 marks)
(b) Show that the speed of the sprinter at the end of the first 2.5 seconds of the race is
(2 marks)
8 m s1 .
(c) The sprinter completes the 100-metre race, travelling the remaining 90 metres at a
constant speed of 8 m s1 . Find the total time taken for the sprinter to travel the
100 metres.
(3 marks)
(d) Calculate the average speed of the sprinter during the 100-metre race.
(2 marks)
3 A particle of mass 3 kg is on a smooth slope inclined at 60° to the horizontal. The particle is
held at rest by a force of T newtons parallel to the slope, as shown in the diagram.
T
60°
(a) Draw a diagram to show all the forces acting on the particle.
(1 mark)
(b) Show that the magnitude of the normal reaction acting on the particle is 14.7 newtons.
(2 marks)
(c) Find T.
P22645/Jan10/MM1B
(2 marks)
3
4 A ball is released from rest at a height of 15 metres above ground level.
(a) Find the speed of the ball when it hits the ground, assuming that no air resistance acts
on the ball.
(3 marks)
(b) In fact, air resistance does act on the ball. Assume that the air resistance force has a
constant magnitude of 0.9 newtons. The ball has a mass of 0.5 kg.
(i) Draw a diagram to show the forces acting on the ball, including the magnitudes of
the forces acting.
(1 mark)
(ii) Show that the acceleration of the ball is 8 m s2 .
(3 marks)
(iii) Find the speed at which the ball hits the ground.
(2 marks)
(iv) Explain why the assumption that the air resistance force is constant may not be
valid.
(1 mark)
5 The constant forces F1 ¼ ð8i þ 12jÞ newtons and F2 ¼ ð4i 4jÞ newtons act on a particle.
No other forces act on the particle.
(a) Find the resultant force acting on the particle.
(2 marks)
(b) Given that the mass of the particle is 4 kg, show that the acceleration of the particle is
ð3i þ 2jÞ m s2 .
(2 marks)
(c) At time t seconds, the velocity of the particle is v m s1 .
(i) When t ¼ 20 , v ¼ 40i þ 32j .
Show that v ¼ 20i 8j when t ¼ 0 .
(ii) Write down an expression for v at time t.
(iii) Find the times when the speed of the particle is 8 m s1 .
(3 marks)
(1 mark)
(6 marks)
Turn over for the next question
P22645/Jan10/MM1B
s
Turn over
4
6 A small train at an amusement park consists of an engine and two carriages connected to
each other by light horizontal rods, as shown in the diagram.
Carriage 2
Carriage 1
Engine
The engine has mass 2000 kg and each carriage has mass 500 kg.
The train moves along a straight horizontal track. A resistance force of magnitude
400 newtons acts on the engine, and resistance forces of magnitude 300 newtons act on each
carriage. The train is accelerating at 0.5 m s2 .
(a) Draw a diagram to show the horizontal forces acting on Carriage 2.
(1 mark)
(b) Show that the magnitude of the force that the rod exerts on Carriage 2 is 550 newtons.
(2 marks)
(c) Find the magnitude of the force that the rod attached to the engine exerts on Carriage 1.
(3 marks)
(d) A forward driving force of magnitude P newtons acts on the engine. Find P. (3 marks)
7 A ball is projected horizontally with speed V m s1 at a height of 5 metres above horizontal
ground. When the ball has travelled a horizontal distance of 15 metres, it hits the ground.
V m s1
5m
15 m
(a) Show that the time it takes for the ball to travel to the point where it hits the ground is
1.01 seconds, correct to three significant figures.
(3 marks)
(b) Find V .
(2 marks)
(c) Find the speed of the ball when it hits the ground.
(4 marks)
(d) Find the angle between the velocity of the ball and the horizontal when the ball hits the
ground. Give your answer to the nearest degree.
(3 marks)
(e) State two assumptions that you have made about the ball while it is moving. (2 marks)
P22645/Jan10/MM1B
5
8 A crate, of mass 200 kg, is initially at rest on a rough horizontal surface. A smooth ring is
attached to the crate. A light inextensible rope is passed through the ring, and each end of
the rope is attached to a tractor. The lower part of the rope is horizontal and the upper part
is at an angle of 20° to the horizontal, as shown in the diagram.
20°
When the tractor moves forward, the crate accelerates at 0.3 m s2 . The coefficient of
friction between the crate and the surface is 0.4 .
Assume that the tension, T newtons, is the same in both parts of the rope.
(a) Draw and label a diagram to show the forces acting on the crate.
(2 marks)
(b) Express the normal reaction between the surface and the crate in terms of T. (3 marks)
(c) Find T.
(5 marks)
END OF QUESTIONS
P22645/Jan10/MM1B
Version 1.0 0110
hij
General Certificate of Education
Mathematics 6360
MM1B
Mechanics 1B
Mark Scheme
2010 examination - January series
Mark schemes are prepared by the Principal Examiner and considered, together with the
relevant questions, by a panel of subject teachers. This mark scheme includes any
amendments made at the standardisation meeting attended by all examiners and is the scheme
which was used by them in this examination. The standardisation meeting ensures that the
mark scheme covers the candidates’ responses to questions and that every examiner
understands and applies it in the same correct way. As preparation for the standardisation
meeting each examiner analyses a number of candidates’ scripts: alternative answers not
already covered by the mark scheme are discussed at the meeting and legislated for. If, after
this meeting, examiners encounter unusual answers which have not been discussed at the
meeting they are required to refer these to the Principal Examiner.
It must be stressed that a mark scheme is a working document, in many cases further
developed and expanded on the basis of candidates’ reactions to a particular paper.
Assumptions about future mark schemes on the basis of one year’s document should be
avoided; whilst the guiding principles of assessment remain constant, details will change,
depending on the content of a particular examination paper.
Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk
Copyright © 2010 AQA and its licensors. All rights reserved.
COPYRIGHT
AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material
from this booklet for their own internal use, with the following important exception: AQA cannot give permission to
centres to photocopy any material that is acknowledged to a third party even for internal use within the centre.
Set and published by the Assessment and Qualifications Alliance.
The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334).
Registered address: AQA, Devas Street, Manchester M15 6EX
Dr Michael Cresswell Director General
MM1B - AQA GCE Mark Scheme 2010 January series
Key to mark scheme and abbreviations used in marking
M
m or dM
A
B
E
or ft or F
CAO
CSO
AWFW
AWRT
ACF
AG
SC
OE
A2,1
–x EE
NMS
PI
SCA
mark is for method
mark is dependent on one or more M marks and is for method
mark is dependent on M or m marks and is for accuracy
mark is independent of M or m marks and is for method and accuracy
mark is for explanation
follow through from previous
incorrect result
correct answer only
correct solution only
anything which falls within
anything which rounds to
any correct form
answer given
special case
or equivalent
2 or 1 (or 0) accuracy marks
deduct x marks for each error
no method shown
possibly implied
substantially correct approach
MC
MR
RA
FW
ISW
FIW
BOD
WR
FB
NOS
G
c
sf
dp
mis-copy
mis-read
required accuracy
further work
ignore subsequent work
from incorrect work
given benefit of doubt
work replaced by candidate
formulae book
not on scheme
graph
candidate
significant figure(s)
decimal place(s)
No Method Shown
Where the question specifically requires a particular method to be used, we must usually see evidence of use of this
method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate,
particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be
provided on the mark scheme.
Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can
be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates
showing no working is that incorrect answers, however close, earn no marks.
Where a question asks the candidate to state or write down a result, no method need be shown for full marks.
Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct
answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark
scheme, when it gains no marks.
Otherwise we require evidence of a correct method for any marks to be awarded.
3
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B
Q
1
Solution
7 × 10 + 3 × 20 = 10v
v=
130
= 13 ms-1
10
Marks
M1
A1
Total
Comments
M1: Three term equation for conservation
of momentum. Do not penalise inclusion
of negative signs. Must see a combined
mass of 10.
A1: Correct equation. Accept 3v + 7v in
place of 10v.
A1
3
A1: Correct speed.
Consistent use of mg instead m throughout
deduct 1 mark.
Total
2(a)
M1
A1
20
= 3.2 ms -2
2.52
A1
a=
(b)
3
1
10 = 0 × 2.5 + a × 2.52
2
M1
A1
1
10 = (0 + v) × 2.5
2
v = 8 ms -1
OR
1
10 = v × 2.5
2
v = 8 ms -1
OR
v 2 = (02 + )2 × 3.2 × 10
M1: Use of constant acceleration equation
to find a with u = 0 .
A1: Correct equation.
NOTE: If v is found first, do not award
any marks for part (a) until an equation to
find a is produced. This could be from
graphical method or from the use of
1
s = (u + v)t .
2
3
A1: Correct acceleration
2
M1: Use of constant acceleration equation
to find v with u = 0 .
A1: Correct speed from correct working.
NOTE: If v is found in part (a), with
correct working award full marks.
NOTE: Accept 3.2 × 2.5 = 8
−1
v = 8 ms
OR
v = (0+)3.2 × 2.5 = 8 ms -1 AG
(c)
(d)
t=
90
= 11.25 s
8
B1
Total Time = 2.5 + 11.25
= 13.75
= 13.8 s
M1
100
= 7.27 ms -1
13.75
M1
A1F
A1
Total
3
2
10
4
B1: Calculation of correct additional time.
Could be implied by later working.
M1: Addition of their time for the 90
metres and the 2.5 seconds.
A1: Correct total time. Accept 13.75.
NOTE: 22.5 + 2.5 = 25 scores B0M1A0
M1: Finding average speed. Must see 100
and their answer from part (c).
A1F: Follow through candidate’s time
from part (c), regardless of working in
part (c).
Allow 7.25 ms-1 from 13.8 seconds.
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B(cont)
Q
3(a)
Solution
Marks
B1
Total
1
R
T
Comments
B1: Correct force diagram with arrows
and sensible labels.
If R is shown as vertical award B0.
If F is included, award B0
3g or mg or W or 29.4
Accept a reflection of the diagram in a
vertical line.
Ignore components if shown with a
different notation, eg dotted lines.
(b)
(c)
( R = ) 3g cos 60°
M1
( R = )14.7
A1
AG
(T = ) 3g sin 60°
M1
(T = ) 25.5
A1
Total
2
M1: Resolving parallel to the slope. Must
see cos60° or sin30° or cos30° or sin60°
and 3g or 29.4.
2
5
5
M1: Resolving perpendicular to the slope.
Must see cos60° or sin30° or cos30° or
sin60° and 3g or 29.4.
3g
NOTE:
= 14.7 or equivalent without
2
the use of a trig term scores M0.
A1: Correct value from correct working.
NOTE: If candidates use g = 9.81, deduct
one mark here. If candidates obtain 14.7
from 14.715 they will have used g = 9.81.
Note: “R = ” does not need to be seen.
A1: Correct value. AWRT 25.5 or
truncation to 25.4.
NOTE: If candidates use g = 9.81 again,
do not penalise. Use of g = 9.81 gives
25.5 for the tension.
Note: “T = ” does not need to be seen.
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B(cont)
Q
Solution
4(a) v 2 = 02 + 2 × 9.8 × 15
v 2 = 294
v = 17.1 ms-1
(b)(i)
Marks
M1
A1
Total
A1
3
B1
1
0.9
4.9 or 0.5g
(b)(ii)
Comments
M1: Use of constant acceleration equation
to find v with u = 0 and a = ± 9.8 .
A1: Correct equation
A1: Correct speed from correct working.
Accept AWRT 17.1. Accept 17.15.
Accept 7 6
Note: If g = 9.81 is used for the first time
deduct one mark. Should get 17.2 ms-1
from g = 9.81.
4.9 − 0.9 = 0.5a
( a =)
4
= 8 ms -2
0.5
M1B1
AG
A1
3
B1: Correct diagram, with arrows and
labels. Must see 0.9 and 4.9 or 0.5g (or
4.905 if working with g = 9.81).
M1: Uses 0.5a.
B1: Explicit statement of “4.9 – 0.9” or
“mg – 0.9” or “0.5g – 0.9”.
A1: Correct acceleration from correct
working. Can be awarded without the B1
mark.
4.9( or 0.5 g ) − 0.9
4
Must see
or
0.5
0.5
or 4 = 0.5a
Note: If g = 9.81 is used candidates will
get 8.01 ms-2. Deduct 1 mark if 8.01 is
seen.
Examples:
4.9 = 0.5a + 0.9
a =8
M1B0A0
4 = 0.5a
M1B0A1
a =8
If candidates only write
4
a=
= 8 award M0B0A0.
0.5
(b)(iii)
v 2 = 02 + 2 × 8 × 15
M1
v = 15.5 ms-1
A1
(b)(iv) The air resistance force will not be
constant, but changes as the speed of the
ball changes (or changes as the ball
accelerates).
B1
Total
2
M1: Use of constant acceleration equation
to find v with u = 0 and a = ±8 .
A1: Correct speed from correct working.
Accept AWRT 15.5 or truncated to 15.4.
Accept 4 15 .
1
B1: Correct explanation, key words in
bold.
Do not award mark for statements that
imply that the acceleration causes the air
resistance to change.
10
6
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B(cont)
Q
Solution
5(a) (8i + 12 j) + (4i − 4 j) = 12i + 8 j
(b)
4a = 12i + 8 j or (a =)
( a = ) 3i + 2 j
(c)(i)
Marks
M1A1
M1
12i + 8 j
4
AG
A1
= −20i − 8 j
2
B1
M1
40i + 32 j = v + (3i + 2 j) × 20
40i + 32 j = v + 60i + 40 j
v = (40i + 32 j) − (60i + 40 j)
Total
2
M1: Use of Newton’s second law with 4a
and their answer to part (a).
A1: Correct acceleration from correct
equation.
B1: Seeing 60i + 40j or (3i + 2j) × 20
M1: Use of constant acceleration equation
with t = 20 and a = 3i ± 2 j .
AG
A1
Comments
M1: Adding forces to find resultant.
A1: Correct resultant force.
3
A1: Correct velocity from correct
working, with one of the intermediate
lines of working (or equivalent) shown.
Note: Candidates may use u instead of v
in their working.
Example:
Starting with
v = 40i + 32 j + (3i + 2 j) × 20
Scores B1M1A0.
Note on Verification Method:
v = ( −20i − 8 j) + (3i + 2 j) × 20 B1M1
= ( −20 + 60)i + (−8 + 40) j
= 40i + 32 j
A1
Similarly, verification to confirm
acceleration from the two velocities is
acceptable.
(c)(ii)
( v = ) (−20i − 8 j) + (3i + 2 j)t
B1
(c)(iii)
( v =) (3t − 20)i + (2t − 8) j
M1
(3t − 20) 2 + (2t − 8) 2 = 82
dM1
A1
A1
dM1
13t 2 − 152t + 400 = 0
t=
152 ± 1522 − 4 × 13 × 400
2 × 13
t = 4 or t = 7.69
A1
Total
1
M1: Velocity vector seen split into
components. Condone omission of i and j
Note: This can be implied by later
working, such as the second line of this
solution.
dM1: Equation based on speed of 8.
A1: Correct unsimplified equation.
A1: Simplified quadratic equation
dM1: Solving quadratic equation, to
obtain two solutions.
6
14
7
B1: Correct velocity vector.
Note “v = ” does not need to be seen.
A1: Both correct solutions. Accept
100
AWRT 7.7 or 7.6 or
.
13
Note: Using calculator to solve quadratic
is acceptable.
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B (cont)
Q
6(a)
Solution
300
(b)
T1 − 300 = 500 × 0.5
(T1 =) 300 + 250
= 550 N
T1
Marks
Total
B1
1
M1
AG
A1
2
Comments
B1: Force diagram with two arrows clearly
in opposite directions. Must see 300 and
one other label (a letter) or 550.
Do not penalise if vertical forces included,
even if wrong.
M1: Three term equation of motion.
A1: Correct force from correct working.
Examples:
T1 = 300 + 250 = 550 N
scores M0A0
T1 − 300 = 250
T1 = 550 N
scores M1A1
T1 = 300 + 500 × 0.5 = 550 N
scores M1A1
Just 300 + 500 × 0.5 = 550 N
scores M0A0
700 + T1 = 2500 × 0.5
T1 = 550
scores M0A0
(c)
M1A1
T2 − 550 − 300 = 500 × 0.5
A1
T2 = 550 + 300 + 250 = 1100 N
OR
T2 − 600 = 1000 × 0.5
(d)
3
(M1)
(A1)
T2 = 600 + 500 = 1100 N
(A1)
P − 1100 − 400 = 2000 × 0.5
M1
A1F
P = 1100 + 400 + 1000 = 2500
OR
P − 1000 = 3000 × 0.5
A1F
(3)
3
(M1)
(A1F)
P = 1000 + 1500 = 2500
(A1F)
Total
(3)
9
8
M1: Four term equation of motion for
Carriage 1 including 550 and 300 with
mass 500
A1: Correct equation.
A1: Correct force
M1: Three term equation of motion for
Carriages 1 and 2 together including 300
twice or 600 with mass 1000.
A1: Correct equation.
Accept T2 = 600 + 500 or similar.
A1: Correct tension
M1: Four term equation of motion for
engine with mass 2000, a force of 400 and
their answer to part (c).
A1F: Correct equation.
A1F: Correct force
M1: Three term equation of motion for
whole train with mass 3000 and 1000 (OE)
force.
A1F: Correct equation.
A1F: Correct force
Follow through from incorrect T2 in part
(c).
Don’t penalise candidates who use a letter
other than P.
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B (cont)
Q
7(a)
1
5 = × 9.8t 2
2
t=
(b)
(c)
Solution
5
= 1.01 s AG
4.9
5
4.9
4.9
V = 15
= 14.8
5
15 = V ×
vV = ±9.8 ×
5
( = ±9.899 )
4.9
Marks
M1
A1
Total
A1
3
M1
A1
M1: Using distance = speed×time OE
2
or
vV = 2 × 9.8 × 5 = 9.899
(d)
dM1
A1F
9.899
14.8
or
14.8
9.899
M1
A1F
A1F
tan α =
α = 34°
9.899
14.8
or
17.8
17.8
α = 34°
sin α =
cos α =
14.8
9.899
or
17.8
17.8
α = 34°
(e) Particle
Experiences no air resistance or no wind
or only gravity or no other forces acting
or no spin.
Total
A1: Correct speed.
Accept AWRT 14.8 or 14.9.
Note: If g = 9.81 is used for the first time
deduct one mark. Should get 14.9 ms-1
from g = 9.81.
M1: Calculating vertical component of
velocity.
A1: Correct value. Accept 9.9 or similar
M1A1
v = 9.8992 + 14.82 = 17.8 ms -1
Comments
M1: Equation based on vertical motion
with no velocity component, with ±5 and
±9.8
A1: Correct equation
A1: Correct time from correct working.
Must see square root or t 2 = 1.02 OE
Note: If g = 9.81 is used for the first time
deduct one mark. Should still get 1.01
seconds.
4
3
dM1: Finding magnitude (with addition
not subtraction of squares inside the square
root).
A1: Correct speed.
Accept AWRT 17.8 or AWRT 17.9.
Note: If g = 9.81 is used for the first time
deduct one mark. Should get 17.9 ms-1
from g = 9.81
M1: Use of one of trig equations shown.
A1F: Anything which rounds to 34° or 56°
A1F: 34° CAO (33° scores M1A1A0)
Only follow through if all method marks in
(b) and (c) have been awarded (except the
dM if tan used) .
(M1)
(A1F)
(A1F)
(M1)
(A1F)
(A1F)
B1
B1
2
14
9
B1: Particle assumption
B1: Other assumption.
Ignore any other assumptions.
MM1B - AQA GCE Mark Scheme 2010 January series
MM1B (cont)
Q
8(a)
Solution
R
T
T
F
Marks
Total
B1
B1
2
Ignore components if shown with a
different notation, eg dotted lines.
mg or W or 200g or 1960
(b)
R + T sin 20° = 1960
OR
R + T sin 20° = 200g
M1A1
( R = )1960 − T sin 20°
A1
M1: Resolving vertically with three terms.
Must include sin20° or cos20° or sin 70°
or cos70° with T and 200g or 1960.
A1: Correct equation.
3
OR
( R = ) 200 g − T sin 20°
(c)
T cos 20° + T − F = 200 × 0.3
M1A1
T cos 20° + T − 0.4(1960 − T sin 20°)
= 200 × 0.3
T=
Total
TOTAL
5
10
75
10
A1: Correct expression for the normal
reaction.
Note: If g = 9.81 is used for the first time
deduct one mark. Should get 1962 instead
of 1960.
M1: Four term equation of motion. Must
include sin20° or cos20° or sin 70° or
cos70° with T and a second T term with
no trig.
A1: Correct equation
M1: Use of friction law with their
expression for R, provided that R has two
terms.
Note that this mark does not depend on
any previous marks.
M1
dM1
A1
60 + 784
= 406
cos 20° + 1 + 0.4sin 20°
Comments
B1: F, R and mg (or equivalent) with
arrows and labels.
B1: Two equal tension forces with arrows
and labels.
Example
If Candidate gives 1960 as answer to part
(b), then:
F = 0.4 × 1960 = 784
scores M0 here
dM1: Solving for T.
Note: This mark requires both of the
previous M marks.
A1: Correct tension.
Accept AWFW 406 to 407.
Note: If g = 9.81 is used should get 407
instead of 406.
klm
Scaled mark component grade boundaries - January 2010 exams
A2 units (legacy)
Component
Code
Component Title
Maximum
Scaled Mark
A
Scaled Mark Grade Boundaries
B
C
D
E
HS6U
GCE HISTORY UNIT 6U
40
32
30
27
24
22
HOA5
HOA6
GCE HISTORY OF ART UNIT 5
GCE HISTORY OF ART UNIT 6
40
40
29
29
25
25
22
22
19
19
16
16
HEC5
HEC6
HEC7
GCE HOME ECONOMICS UNIT 5
GCE HOME ECONOMICS UNIT 6
GCE HOME ECONOMICS UNIT 7
52
50
50
39
34
ICT4
ICT5
ICT6
GCE INFO AND COMM TECH UNIT 4
GCE INFO AND COMM TECH UNIT 5
GCE INFO AND COMM TECH UNIT 6
90
90
90
64
68
59
58
62
51
52
57
43
3
46
52
36
3
40
47
29
GCE LAW UNIT 4
GCE LAW UNIT 5
GCE LAW UNIT 6
85
85
70
60
57
47
55
52
43
50
48
39
45
44
35
40
40
31
GCE MATHEMATICS UNIT D01
GCE MATHEMATICS UNIT D02
GCE MATHEMATICS UNIT FP1
GCE MATHEMATICS UNIT FP2
GCE MATHEMATICS UNIT FP3
GCE MATHEMATICS UNIT FP4
GCE MATHEMATICS UNIT M1A WRITTEN
GCE MATHEMATICS UNIT M1A CWK
GCE MATHEMATICS UNIT M1B
GCE MATHEMATICS UNIT M2B
GCE MATHEMATICS UNIT PC1
GCE MATHEMATICS UNIT PC2
75
75
75
75
75
75
75
25
75
75
75
75
63
57
61
61
63
61
59
20
55
56
62
58
56
49
54
53
55
52
51
17
48
48
54
50
49
41
47
45
47
43
43
14
41
40
47
43
42
34
40
38
39
35
36
12
34
32
40
36
35
27
33
31
32
27
29
10
28
25
33
29
LAW4
LAW5
LAW6
MD01
MD02
MFP1
MFP2
MFP3
MFP4
MM1A/W
MM1A/C
MM1B
MM2B
MPC1
MPC2
(no candidates were entered for this component)
35
31
28
25
30
26
23
20