Question / Working / Answer / Mark / Notes /
9
1 / (a) / reflected shape / 2 / B2 correct triangle drawn with vertices (4,4), (5,4) and (5,6)(B1 for a correct reflection in x = a)
(b) / rotation
centre (0,1)
90° anti-clockwise or 270o clockwise / 3 / B1 rotation
B1 about the centre (0,1)
B1 90o anticlockwise or 270° clockwise
NB If more than one transformation seen then B0
(c) / Rotation 90° clockwise centre (1,1) / 3 / B1 for rotation
B1 for 90° clockwise or 270° anticlockwise
B1 for (1,1)
(B0 for any combination of transformations)
2 / (a) / y = kx
10 = 600k
k = 10÷600 = / y = / 3 / M2 for 10 = k × 600 oe or 10 = oe or k =
(M1 for y=kx or y = or y 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 /
=13018867.92 = / 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
1MA0 Higher Tier – Practice Paper 2H (Set D) /
Question / Working / Answer / Mark / Notes /
4 / AB = 5
Or
AD = BC
OR
cos 54 =
BC = / 8.51 / 4 / B1 AB = 5
M1
M1
A1 8.5 – 8.51
OR
M1
M1
B1
A1 8.5 – 8.51
OR
B1 angle DCB = 54 or angle DBC = 36
M1 cos 54 =
M1 BC =
A1 8.5 – 8.51
NB Other methods such as tan + Pythagoras must be complete methods and will earn M2
5 / ÷ = ×
OR
3.33… ÷ 4.75 / or
0.70175(4386…) / 2 / 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
A1 for or 0.70175(4386…)
6 / m2 = / m = / 2 / M1 m2 = or or =or m =
A1 m = or m = ± or m = –
7 / (a) / 2a + 2t = 5t + 7
2a = 3t + 7
2a – 7 = 3t / / 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.
NB: for accept working to 2 dp: 0.67, 0.66, 2.33 or better
(b) / x =
y = -1 / 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
A1 cao for both x = and y = -1 oe
SC: B1 for x = or y = -1 oe
NB: for accept working to 2 dp: 0.67 or 0.66 or better
8 / ½ litre = 500 ml
500 = 42 h
h = 500 ÷ ( 42 ) / 9.95 / 5 / 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
9 / 168000 = 112% (of original price)
168000 ÷ 112 100 / 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
10 / (a) / – / / 2 / M1 – in co-ordinates or vectors or or
A1 cao
[SC If no marks then B1 or
(b) / M = (3, 6)
N = (4, 8) + ½ (6, –4) = (7, 6)
= –
OR
= ½
= +=
OR
½+ ½
= ½ + ½ / / 3 / B1 M = (3, 6)
M1 N = (4, 8) + ½ (6, –4) or (7, 6) or = –
A1 cao
OR
B1 = ½
M1 ft = + or
A1 cao
OR
B1 = ½ + ½
M1 ft = ½ + ½
A1 cao
11 / (a) / ½ 9.2 14.6 sin 64 / 60.4 / 2 / M1 ½ 9.2 14.6 sin 64
A1 60.3 – 60.4
(b) / AB2 = 9.22 + 14.62 – 2 × 9.214. 6cos 64
AB2 = 297.8 – 268.64 cos 64° = 297.8 - 268.64×0.43837..
AB2 = 297.8 – 117.76..
AB2 = 180.03
AB = √ / 13.4 / 3 / 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
12 / DB2 = 5.62 + 8.22 – 2 ×5.6 × 8.2cos78
DB2 = 79.505…
DB = 8.9165795..
=
DC = =8.9165..×0.6572..
=5.8198 / 5.82 / 6 / 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 =
M1 (=5.8198)
A1 for answer 5.80 to 5.83
If working in RAD or GRAD award method marks only.
RAD: DB=13.318.., DC=-9.98..
GRAD: DB=8.2152…, DC=5.0773…
13 / (a) / i CA = 2OA
ii BA=BO+OA= –b+a
iii BC=BO+OC= –b–a / 2a
a – b
–a – b / 3 / B1 for 2a oe
B1 for a – b oe
B1 for –a – b oe
(b) / 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 / 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)
14 / 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.
15 / (a) / eg = oe
eg = = / 35 - 41 / 2 / M1 for use of or drawing a right angled triangle against the line, or inverse expression of gradient eg 0.025, 1:40, or correct answer given as algebra (eg y=40, 40x)
A1 for 35 – 41 (units not required)
(b) / 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.
16 / (a) / 1:22 or 22:1
80 × 22 = 80 × 4 = / 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
(b) / 1:23 or 23:1
171700 × 23
= 171700 × 8 =
OR
ha=
=102.47589
hb = ha×2=204.95..
volb= π×802×204.95.. / 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
17 / SF = (x2 – 1) ÷ 2 (x – 1)
= (x – 1) (x + 1) ÷ 2 (x – 1)
= ½ (x + 1)
Area DEF = 4
= (x + 1)2
= x2 + 2x + 1 / 4 / 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)
M1 or
C1 fully correct convincing process
OR
M1 (x2 + 2x + 1) ÷ 4
M1 or or (x + 1) ÷ 2
M1 2 (x – 1) (x + 1) ÷ 2
C1 fully correct convincing process
18 / x2 + (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 / x = 0,
y = 5
or
x = –4,
y = –3 / 6 / M1 x2 + (2x + 5)2 ( = 25)
A1 x2 + 4x2 + 10x + 10x + 25 (= 25)
M1 Use of factorisation or correct substitution into quadratic formula or completing the square to solve an equation of the form
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)
New
Qn / Question
Number / Paper
Date / Skill tested / Maximum
score / Mean
Score / Mean
Percentage / Percentage
scoring full marks
1a / Q10a / 1211 3H / Reflect a shape in a given line / 2 / 1.32 / 66 / 61.1
1b / Q10b / 1211 3H / Describe a rotation of a 2-D shape / 3 / 1.89 / 63 / 39.9
1c / Q06 / 1206 3H / Describe a rotation of a 2-D shape, / 3 / 1.85 / 62 / 39.5
2 / Q13a / 1206 3H / Set up and use equations to solve problems / 3 / 1.83 / 61 / 44.0
3 / Q12 / 1206 3H / Calculate with standard form / 3 / 1.75 / 58 / 36.0
4 / Q14 / 1211 3H / Use the trigonometric ratios to solve 2-D and 3-D problems / 4 / 2.21 / 55 / 41.9
5 / Q03 / 1206 3H / Add, subtract, multiply and divide whole numbers, integers, negative numbers, fractions and decimals, and numbers in index form / 2 / 0.99 / 50 / 41.3
6 / Q15 / 1211 3H / Substitute numbers into a formula and change the subject of a formula / 2 / 0.92 / 46 / 41.6
7a / Q16a / 1206 3H / Change the subject of a formula where the subject appears on both sides / 3 / 1.87 / 62 / 45.4
7b / Q16b / 1206 3H / Use elimination to solve two simultaneous equations / 3 / 1.34 / 45 / 26.9
8 / Q12 / 1211 3H / Solve problems involving more complex shapes and solids, including segments of circles and frustums of cones / 5 / 1.98 / 40 / 31.0
9 / Q13 / 1211 3H / Use percentages to solve problems / 3 / 1.07 / 36 / 34.3
10a / Q18a / 1211 3H / Calculate, and represent graphically the difference of two vectors / 2 / 1.3 / 65 / 26.5
10b / Q18b / 1211 3H / Solve geometrical problems in 2-D using vector methods / 3 / 1.04 / 35
11a / Q19a / 1211 3H / Calculate the area of a triangle given the length of two sides and the included angle / 2 / 1.17 / 59 / 27.7
11b / Q19b / 1211 3H / Use the sine and cosine rules to solve 2-D and 3-D problems / 3 / 1.05 / 35
12 / Q18 / 1206 3H / Use the sine and cosine rules to solve 2-D and 3-D problems / 6 / 2.02 / 34 / 24.5
13a / Q15a / 1206 3H / Calculate the resultant of two vectors / 3 / 1.77 / 59 / 44.8
13b / Q15b / 1206 3H / Solve geometrical problems in 2-D using vector methods / 3 / 0.96 / 32 / 6.2
14 / Q14b / 1206 3H / Solve linear inequalities in one or two variables, and represent the solution set on a number line or coordinate grid / 4 / 1.21 / 30 / 13.9
15a / Q08a / 1206 3H / Find the gradient of a straight line from a graph / 2 / 1.08 / 54 / 12.7
15b / Q08b / 1206 3H / Use gradients to see how one variable changes in relation to another / 1 / 0.22 / 22
16a / Q17a / 1206 3H / Use the effect of enlargement for perimeter, area and volume of shapes and solids / 2 / 0.33 / 17 / 11.5
16b / Q17b / 1206 3H / Use the effect of enlargement for perimeter, area and volume of shapes and solids / 3 / 0.86 / 29
17 / Q21 / 1211 3H / Solve a pair of simultaneous equations, one of which is linear and the other quadratic / 6 / 1.02 / 17 / 8.5
18 / Q20 / 1211 3H / Use simple examples of the relationship between enlargement and areas and volumes of simple shapes and solids / 4 / 0.24 / 6 / 1.7
80 / 33.29 / 41.6
9