Question
1
(a)
Working
(b)
(c)
2
(a)
y = kx
10 = 600k
2MB01 Higher Tier – Practice Paper 3H (Set D)
Answer
Mark
Notes
2
reflected shape
B2 correct triangle drawn with vertices (4,4), (5,4) and (5,6)
(B1 for a correct reflection in x = a)
rotation
centre (0,1)
90 anticlockwise or
270o
clockwise
3
Rotation 90°
clockwise
centre (1,1)
3
B1 rotation
B1 about the centre (0,1)
B1 90o anticlockwise or 270 clockwise
NB If more than one transformation seen then B0
B1 for rotation
B1 for 90° clockwise or 270° anticlockwise
B1 for (1,1)
(B0 for any combination of transformations)
3
y=
M2 for 10 = k × 600 oe or 10 =
k = 10÷600 =
(M1 for y=kx or y =
or y
oe or k =
x, k any letter or value)
A1 for y =
x oe
SC: B2 for 60y = x
NB: for 1/60 accept 0.016 to 0.017
3
1.3×107
3
M1 for time = distance ÷ speed expressed numerically.
M1 for 13000000 to 13100000 or digits 130188.. 130189.. or 1.3×10n
to 1.31×10n where n is a number other than 7, or absent, or digits
13(01…) ×10n
A1 1.3(0)×107 – 1.31×107
=13018867.92 =
1
Question
4
Working
5
AB = 5 sin 36 
AD
AD 
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
8.51
4
B1 AB = 5
Notes
5
sin36 sin 90
or

AD
5
AD
5
5 sin 90
M1 AD 
or AD 
sin 36
sin 36
M1 sin 36 
5
sin 36
A1 8.5 – 8.51
OR
Or
5
sin36 sin 90
or

BC
5
BC
5
5 sin 90
or BC 
M1 BC 
sin 36
sin 36
5
BC
5
BC 
sin 36
sin 36 
M1 sin 36 
AD = BC
B1 AD ' BC '
A1 8.5 – 8.51
OR
B1 angle DCB = 54 or angle DBC = 36
OR
5
BC
5
BC =
cos 54
5
BC
5
M1 BC =
cos 54
cos 54 =
M1 cos 54 =
A1 8.5 – 8.51
NB Other methods such as tan + Pythagoras must be complete methods
and will earn M2
5
2
÷
=
OR
3.33… ÷ 4.75
×
M1 for
oe and
oe or 3.33(…) and 4.75 or 40 ÷ 57 or 0.7, 0.70,
0.701, 0.702, 0.7017, 0.7018
or
0.70175(4386
…)
A1 for
2
or 0.70175(4386…)
Question
6
7
Working
m2 =
(a)
k
6
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
2
6m 2 k
k
k
2
M1
m
=
or
 or
m=
6m 2 = k or 6 m = k
6
6
6
k
k
k
A1 m =
or m = ±
or m = –
6
6
6
6
2a + 2t = 5t + 7
2a = 3t + 7
2a – 7 = 3t
Notes
3
M1 for expansion of bracket eg 2×a+2×t or divide all terms by 2
M1 for attempt at rearrangement of t term
eg –2t each side; 2a = 3t+? but with separate terms.
A1
oe but must have one term in t.
2
NB: for 3 accept working to 2 dp: 0.67, 0.66, 2.33 or better
(b)
3
M1 for correct process to eliminate either x or y (condone one
arithmetic error)
M1 (dep on 1st M1) for correct substitution of their found variable or
other acceptable method
x=
y = -1
A1 cao for both x =
SC: B1 for x =
NB: for
8
½ litre = 500 ml
500 =   42  h
h = 500 ÷ (   42 )
9.95
5
and y = -1
oe
or y = -1 oe
accept working to 2 dp: 0.67 or 0.66 or better
B1 ½ litre = 500ml or 500 seen
M1   42  h (= 50.2.×.h )or   42 (= 50.2..)
M1 “500” =   42  h oe
M1 (h =) “500” ÷ (   42 ) oe
A1 9.9 – 10.0
3
Question
9
10
(a)
Working
168000 = 112% (of
original price)
168000 ÷ 112  100
 4  2
   
8  –  4
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
Notes
150000
3
M1 168000 = 112% or 112 or 100 + 12 or 1.12 or 1 + 0.12 with an
intention to divide
M1 168000 ÷ 1.12 or 16800 ÷ 112  100
A1 cao
 2
 
 4
2   x
   
y
4
in co-ordinates or vectors or   or  
2
M1
–
A1 cao
 4
 
2
[SC If no marks then B1   or
(b)
M = (3, 6)
N = (4, 8) + ½ (6, –4) = (7,
6)
 4
 
0
3
B1 M = (3, 6)
M1 N = (4, 8) + ½ (6, –4) or (7, 6) or
A1 cao
OR
7 3
   
6
6
= – 
OR
B1
=½
6   2 8 
     
4
4
0
=   + = 
M1 ft
A1 cao
OR
OR
½
 2 
 
 4 
+½
B1
 2
6 
 
 
4
4
= ½   + ½ 
M1 ft
A1 cao
4
=½
6 
 
4
=  +
= ½
 '2' 
8 
 
 
 '4'  or  0 
+½
 '2' 
6 
 
 
4
'4'
= ½   + ½ 
7
 
6
= –
3
 
6
Question
11 (a)
(b)
Working

½ 9.2  14.6  sin 64
AB2 = 9.22 + 14.62 – 2 ×
9.2  14. 6  cos 64
AB2 = 297.8 – 268.64 cos
64° = 297.8 268.64×0.43837..
AB2 = 297.8 – 117.76..
AB2 = 180.03
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
60.4
2
M1 ½  9.2  14.6  sin 64
A1 60.3 – 60.4
13.4
3
5.82
6
Notes
M1 9.22 + 14.62 – 2  9.2  14. 6  cos 64°
M1 (dep) for correct order of evaluation e.g. 297.(8) – 117.(7..)
A1 13.4 – 13.42
AB = √ 180.03
12
DB2 = 5.62 + 8.22 – 2
×5.6 × 8.2cos78
DB2 = 79.505…
DB = 8.9165795..
M1 Cosine rule: DB2=5.62+8.22–2×5.6×8.2×cos78
M1 √79.505… (=8.9165795..)
A1 for DB = 8.90 to 8.92
M1
=
=
DC =
M1
A1 for answer 5.80 to 5.83
=8.9165..×0.6572..
=5.8198
(=5.8198)
If working in RAD or GRAD award method marks only.
RAD: DB=13.318.., DC=-9.98..
GRAD: DB=8.2152…, DC=5.0773…
5
Question
13 (a)
(b)
Working
i CA = 2OA
ii BA=BO+OA= –b+a
iii BC=BO+OC= –b–a
i AX = AO + OX
= – a + 2a – b = a – b
ii AX = BA so AX is
parallel to BA; A is on
both AX and BA, so B, A,
X are all on a straight line
14
15
(a)
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
2a
3
B1 for 2a oe
a–b
B1 for a – b oe
–a – b
B1 for –a – b oe
a–b
explanation
3
M1 for AX = AO + OX
A1 for a – b oe
B1 (dep on M1) for explanation eg BX = BO + OX = –b + 2a – b = 2(a
– b)
Region
identified
4
M1 for any two of the lines y=1, y=2x–2, y=6–x
M1 for any two of the lines y=1, y=2x–2, y=6–x with at least one
showing shading (in or out)
M1 for any two of the lines y=1, y=2x–2, y=6–x with at least two
showing consistent and correct shading (in or out)
A1 lines drawn, and correct region identified by either shading in, or
shading out; the letter R is not required. Accept without shading only
with the correct region indicated by R.
35 - 41
2
eg =
M1 for use of
(b)
or drawing a right angled triangle against the
line, or inverse expression of gradient eg 0.025, 1:40,
answer given as algebra (eg y=40, 40x)
A1 for 35 – 41 (units not required)
oe
eg =
Notes
or correct
=
Interpretation
1
B1 ft from part (a) or equivalent written explanation placing the
gradient into the correct context, linking rate eg distance per gallon,
mpg, constant rate of use of petrol, constant rate.
6
Question
16 (a)
(b)
Working
2
2
1:2 or 2 :1
80 × 22 = 80 × 4 =
1:23 or 23:1
171700 × 23
= 171700 × 8 =
OR
ha=
=102.47589
hb = ha×2=204.95..
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
Notes
320
2
M1 for sight of 1:22 or 22:1 or 22 or for ratio of area or 80×4 or
identification of 4 as the scale factor
A1 cao
1 373 600
3
M1for sight of 1:23 or 23:1 or 23 or for ratio of volumes or
identification of 8 as the scale factor
M1 for 23 × 171700
A1 cao
OR
M1 for complete calculation to find the height of A (=102.47589..)
M1 (dep) for ha×2 and used to find volb
A1 cao
volb= π×802×204.95..
17
SF = (x2 – 1) ÷ 2 (x – 1)
= (x – 1)  (x + 1) ÷ 2 (x
– 1)
= ½ (x + 1)
Area DEF = 4 
1

 2  x  1 
M1 (x2 – 1) ÷ 2 (x – 1) or SF × 2 (x – 1) = (x2 – 1)
M1 ½ (x + 1) or (x – 1)  (x + 1) ÷ 2 (x – 1)
4
2
 x2 1 
 x 1


4

4

2( x  1) 
2



M1
or
2
2
C1 fully correct convincing process
OR
M1 (x2 + 2x + 1) ÷ 4
= (x + 1)2
= x2 + 2x + 1
x
2
 2x  1  4
( x  1)( x  1)  4 or (x + 1) ÷ 2
M1
or
M1 2 (x – 1)  (x + 1) ÷ 2
C1 fully correct convincing process
7
Question
18
Working
2
x + (2x + 5)2 = 25
x2 + 4x2 + 20x + 25 = 25
5x2 + 20x = 0
5x ( x + 4) = 0
x = 0, x = –4
y=2×0+5
y = 2 × –4 + 5
1MA0 Higher Tier – Practice Paper 2H (Set D)
Answer
Mark
Notes
2
x = 0,
6
M1 x + (2x + 5)2 ( = 25)
y=5
A1 x2 + 4x2 + 10x + 10x + 25 (= 25)
or
M1 Use of factorisation or correct substitution into quadratic formula or
x = –4,
completing the square to solve an equation of the form
y = –3
ax 2  bx  c  0, a  0
A1 x = 0, x = –4
M1 substitution of an x value into an original equation
A1 y = 5, y = –3 correctly matched to x values
SC (If M0M0M0 then B1 for one pair (x, y) of correct answers)
8
New
Qn
Question
Number
Paper
Date
1a
1b
1c
2
3
4
Q10a
Q10b
Q06
Q13a
Q12
Q14
1211 3H
1211 3H
1206 3H
1206 3H
1206 3H
1211 3H
5
Q03
1206 3H
6
Q15
1211 3H
7a
Q16a
1206 3H
7b
Q16b
1206 3H
8
Q12
1211 3H
9
Q13
1211 3H
10a
Q18a
1211 3H
10b
Q18b
1211 3H
11a
Q19a
1211 3H
11b
12
13a
13b
Q19b
Q18
Q15a
Q15b
1211 3H
1206 3H
1206 3H
1206 3H
14
Q14b
1206 3H
15a
Q08a
1206 3H
15b
Q08b
1206 3H
16a
Q17a
1206 3H
16b
Q17b
1206 3H
2
3
3
3
3
4
1.32
1.89
1.85
1.83
1.75
2.21
66
63
62
61
58
55
Percentage
scoring full
marks
61.1
39.9
39.5
44.0
36.0
41.9
2
0.99
50
41.3
2
0.92
46
41.6
3
1.87
62
45.4
3
1.34
45
26.9
5
1.98
40
31.0
3
1.07
36
34.3
2
1.3
65
3
1.04
35
2
1.17
59
3
6
3
3
1.05
2.02
1.77
0.96
35
34
59
32
24.5
44.8
6.2
4
1.21
30
13.9
2
1.08
54
1
0.22
22
2
0.33
17
3
0.86
29
Maximum
score
Skill tested
Reflect a shape in a given line
Describe a rotation of a 2-D shape
Describe a rotation of a 2-D shape,
Set up and use equations to solve problems
Calculate with standard form
Use the trigonometric ratios to solve 2-D and 3-D problems
Add, subtract, multiply and divide whole numbers, integers,
negative numbers, fractions and decimals, and numbers in
index form
Substitute numbers into a formula and change the subject of
a formula
Change the subject of a formula where the subject appears
on both sides
Use elimination to solve two simultaneous equations
Solve problems involving more complex shapes and solids,
including segments of circles and frustums of cones
Use percentages to solve problems
Calculate, and represent graphically the difference of two
vectors
Solve geometrical problems in 2-D using vector methods
Calculate the area of a triangle given the length of two sides
and the included angle
Use the sine and cosine rules to solve 2-D and 3-D problems
Use the sine and cosine rules to solve 2-D and 3-D problems
Calculate the resultant of two vectors
Solve geometrical problems in 2-D using vector methods
Solve linear inequalities in one or two variables, and
represent the solution set on a number line or coordinate
grid
Find the gradient of a straight line from a graph
Use gradients to see how one variable changes in relation to
another
Use the effect of enlargement for perimeter, area and
volume of shapes and solids
Use the effect of enlargement for perimeter, area and
volume of shapes and solids
9
Mean
Score
Mean
Percentage
26.5
27.7
12.7
11.5
17
Q21
1211 3H
18
Q20
1211 3H
Solve a pair of simultaneous equations, one of which is
linear and the other quadratic
Use simple examples of the relationship between
enlargement and areas and volumes of simple shapes and
solids
10
6
1.02
17
8.5
4
0.24
6
1.7
80
33.29
41.6