M2. (A) Correct Stem

M1. (a) 5.285(7 ...)

(b) 5.3

B1ft

[2]

M2. (a) Correct stem

3, 4, 5, 6, placed vertically

Leaves (allow one error)
7 1 3 7
0 3 8 4 4 7 3 8 1
2 8 4 9 2 0
0 0 0 0

Ordered leaves
1 3 7 7
0 1 3 3 4 4 7 8 8
0 2 2 4 8 9
0 0 0 0

Key unambiguously stated

eg, 4 | 0 means 40 oe

(b) (i) 60

Allow 1 hour if unit corrected

(ii) These students used all the time up

Allow finished right at the end or only just
enough time/did not finish or exam was one hour oe

(ii) 58 – 41

[9]

M3. (a) 41

(b) 180 – 67

71 + their 41 oe 360 – 41 – 67 – 139

113

[3]

M4. –2, 1, 6

–1 each error or emission.

Ignore extra terms

12 – 3, 22 – 3, 32 – 3 is B1

[2]

M5. (a) 18 and – 12 seen

(b) 22x or 22 × x or x ×22

or 500 – x22

500 – 22x oe

SC1 5 – 0.22x oe

[4]

M6. (a) Either or

A1 ft

(b)× 24

[5]

M7. Allow embedded solutions, but if contradicted M marks only

4 × 142 + 5 × 146 + 8 × 150 + 7 × 154 + 5 × 158 + 1 × 162

where x is midpoint or end point or

Values ± 0.5

For at least 2 multiplications and additions seen

Their 4528 ÷ 30

M1 dep

150.9(3...)

151 with working

[3]

M8. (a) Two points calculated or plotted

B1 For each point or (–1, –5)

B1 Line through (0, –3) (0, –3)

B1 Line gradient 2 (1, –1)

(2, 1)

(3, 3)

(4, 5)

Straight line drawn

(b) Attempt to read off at y = 4.5 or 2x = 7.5

or 4.5 as y coordinate

3.75

ft Their graph

± 1 mm (square)

B1 ft

[5]

M9. (a) BC2 = 192 – 92(= 280)

x2 + 92 = 192

BC = √280

For squaring, subtracting and evidence of square rooting

DM1

BC = 17 or 16.7(....)

17 with no working gets 3

(b) Sight of tangent

or
Angle = tan–l (1 ÷ 24)

tan –1 (0.458)
M2 for any complete correct method
Sin = 11/√697 or 11/26.4
Cos = 24/√697 or 24/26.4

DM1

25 or 24.6(....)

25 with no working gets 3
Radians 0.43 gradians 27.35
Penalise on first occurrence only.

[6]

M10. 700 × 1.12 – 700

or 700 × 0.1 or 70 or 700 × 1.1 or 770
or 700 × 1.12
or 847 or 140

147(.00)

[2]

M11. Trial between 2 and 3

Trial between 2.3 and 2.4 inclusive that “bracket” the answer

Trial at 2.35 or 2.36 or 2.37 and 2.4 stated as answer

DB1

In this question final answer on its own will not get any marks
Working must be seen.
All trials must be correctly evaluated either rounded or truncated to a degree of accuracy that allows comparison.

[3]

M12. Large is 3 times standard

Or standard is large

Price per biscuit 9...p and 8...p

Which costs 3 × £1.09
= £3.27 Large is better

Which is £1.05 or £1.06
Large is better

[2]

M13. (a) 80 × 1.75

accept 80 × 1.45 and 80 × 105

140

(b) {190 – (their 140)} ÷ (2.25 – 1.75)

Or (their 50) ÷ 0.

Allow (their 50) in 30 minutes

100

ft from their (a)

A1 ft

[4]

M14. (a) 6 right, 6 down

or as vector

(b)y = x

90° or 270° anti-clockwise or clockwise
about (–3, –3) or (3, 3)

Centre (0, 0) or origin

Check alternative fully correct for 2 marks,
2 parts correct for 1 mark.

[4]

M15. (a)× (7 + 11) × 5

(b) Their 45 × 16

or 720

19.3 × their 720

13896

13.896

ft if both Ms awarded

A1 ft

[6]

M16. (a) (i) 8, 38, 62, 75, 80

Rest of question must be from an increasing
cumulative frequency diagram (not linear)

(ii) Upper class boundaries used

±square

Their correct heights

±square

Ignore (20, 0)

Ignore curve before (30, 8)

B1 ft

Straight lines or smooth curve connecting points

±square

Ignore curve before (30, 8)

(b) Locating and subtracting quartiles ie “49” – “35”
If no working check graph

From 60, 20 or their quartiles

eg 17.5, 52.5 or methods

= “14”

A1 ft

[5]

M17. (a) The views fairly represent the different classes of passengers

(b)(= 0.8) or (= 0.2)

M1 dep

Their 0.8 × 40 or their 0.2 × 40

32 and 8

SC2 Answers wrong way round

[5]

M18. 3x2 = x + 2

y = 3(y – 2)2

3x2 – x – 2 = 0

3y2 – 13y + 12 = 0

(3x + 2)(x – 1) = 0
or (x –)2 = ±√() or ±

x =

(3y – 4)(y – 3) = 0 (Reverse A1 s below)
Must be for factorising a quadratic.
x (or y) terms must have product equal to square term and
number terms must have a product equal to ± constant term.
If completing the square or formula used must be to at least the
stage shown for Method mark. or (y –)2 = ±√() or ±

y =

x = 1 and –

Need both

y = 3 and

Must match appropriate values of y with x
Must use y = x + 2, or x = y – 2. Answers without any working is B1, otherwise answers must be supported by an algebraic method. Graphical method is M0.
Special case: x = 1, y = 3 without working B1. (Can be guessed). NB only award this if no other marks awarded.

A1 ft

[5]

M19. Sight of 5250 or 5350

Sight of 95 or 105

Their correct combination
ie.

= 50

Accept 49, with explanation that 50 would be right on the limit,
hence 49 is the maximum

[4]

M20. (a) Using frequency densities

Correct frequency densities

Height & widths plotted correctly

At least 2 correctLook for consistent use of ‘key’
4 out of 5 correct ... look for evidence on graph
± ½ square for their vertical scale

(b) Good attempt to ‘halve’ the area

735 minutes

Or halving the frequency + attempt at calculation
(could be a cum. freq. calculation)

[5]

M21. (x + 5)2 + (x – 2)2 = 102

x2 + 25 + x2 ± 4 = 100 implies M1

x2 + 10x + 25 + x2 – 4x + 4 = 100

SC1 x2 + 10 x + 25 + x2 – 4 x + 4 = 10

2x2 + 6x – 71 = 0

For rearranging into a 3 term quadratic = 0
or going to cts straight away.

M1 dep

Use of quadratic formula or completing the square to solve

Any evidence of formula or cts gets M1

Allow use of graphical calculator

M1 dep

4.6, 4.64, 4.65

4.64(4..) for T & I

[5]

M22. (a)

Sin B = 0.9679(1...)

B = 75.4(...)

(b)x2 = 222 + 232 – 2 × 22 × 23 × cos 48

x2 = 335.8(...)

x = 18.32(....)

ft only if an error made in calculation of x2
but not on (222 + 232 – 2 × 22 × 23 ( = 1)) cos 48

(= 0.669 = 0.818)

A1 ft

18 or 18.3

Independent mark. Award if value > 3sf seen
or calculation seen.

B1 ft

[7]

M23. Graph A is y = (x – 3)2

Graph B is y (x + 3)2

Graph C is y = –x2

Graph D is y = 3 – x2

[4]

Page 1