PRACTICE PAPER (SET 2) MARK SCHEME
H
Date – Morning/Afternoon
GCSE (9-1) MATHEMATICS
J560/06 Paper 6 (Higher Tier)
PRACTICE PAPER (SET 2) MARK SCHEME
Duration: 1 hour 30 minutes
MAXIMUM MARK 100
DRAFT
This docum
...
PRACTICE PAPER (SET 2) MARK SCHEME
H
Date – Morning/Afternoon
GCSE (9-1) MATHEMATICS
J560/06 Paper 6 (Higher Tier)
PRACTICE PAPER (SET 2) MARK SCHEME
Duration: 1 hour 30 minutes
MAXIMUM MARK 100
DRAFT
This document consists of 12 pagesJ560/06 Mark Scheme Practice Paper (Set 2)
2
Subject-Specific Marking Instructions
1. M marks are for using a correct method and are not lost for purely numerical errors.
A marks are for an accurate answer and depend on preceding M (method) marks. Therefore M0 A1 cannot be awarded.
B marks are independent of M (method) marks and are for a correct final answer, a partially correct answer, or a correct intermediate stage.
SC marks are for special cases that are worthy of some credit.
2. Unless the answer and marks columns of the mark scheme specify M and A marks etc, or the mark scheme is ‘banded’, then if the correct
answer is clearly given and is not from wrong working full marks should be awarded.
Do not award the marks if the answer was obtained from an incorrect method, i.e. incorrect working is seen and the correct answer clearly
follows from it.
3. Where follow through (FT) is indicated in the mark scheme, marks can be awarded where the candidate’s work follows correctly from a
previous answer whether or not it was correct.
Figures or expressions that are being followed through are sometimes encompassed by single quotation marks after the word their for clarity,
e.g. FT 180 × (their ‘37’ + 16), or FT 300 – (their ‘52 + 72’). Answers to part questions which are being followed through are indicated by e.g.
FT 3 × their (a).
For questions with FT available you must ensure that you refer back to the relevant previous answer. You may find it easier to mark these
questions candidate by candidate rather than question by question.
4. Where dependent (dep) marks are indicated in the mark scheme, you must check that the candidate has met all the criteria specified for the
mark to be awarded.
5. The following abbreviations are commonly found in GCSE Mathematics mark schemes.
- figs 237, for example, means any answer with only these digits. You should ignore leading or trailing zeros and any decimal point e.g.
237000, 2.37, 2.370, 0.00237 would be acceptable but 23070 or 2374 would not.
- isw means ignore subsequent working after correct answer obtained and applies as a default.
- nfww means not from wrong working.
- oe means or equivalent.
- rot means rounded or truncated.
- seen means that you should award the mark if that number/expression is seen anywhere in the answer space, including the answer
line, even if it is not in the method leading to the final answer.
- soi means seen or implied.J560/06 Mark Scheme Practice Paper (Set 2)
3
6. In questions with no final answer line, make no deductions for wrong work after an acceptable answer (i.e. isw) unless the mark scheme says
otherwise, indicated by the instruction ‘mark final answer’.
7. In questions with a final answer line following working space,
(i) if the correct answer is seen in the body of working and the answer given on the answer line is a clear transcription error allow full marks
unless the mark scheme says ‘mark final answer’. Place the annotation next to the correct answer.
(ii) if the correct answer is seen in the body of working but the answer line is blank, allow full marks. Place the annotation next to the
correct answer.
(iii) if the correct answer is seen in the body of working but a completely different answer is seen on the answer line, then accuracy marks
for the answer are lost. Method marks could still be awarded. Use the M0, M1, M2 annotations as appropriate and place the annotation
next to the wrong answer.
8. In questions with a final answer line:
(i) If one answer is provided on the answer line, mark the method that leads to that answer.
(ii) If more than one answer is provided on the answer line and there is a single method provided, award method marks only.
(iii) If more than one answer is provided on the answer line and there is more than one method provided, award zero marks for the question
unless the candidate has clearly indicated which method is to be marked.
9. In questions with no final answer line:
(i) If a single response is provided, mark as usual.
(ii) If more than one response is provided, award zero marks for the question unless the candidate has clearly indicated which response is
to be marked.
10. When the data of a question is consistently misread in such a way as not to alter the nature or difficulty of the question, please follow the
candidate’s work and allow follow through for A and B marks. Deduct 1 mark from any A or B marks earned and record this by using the MR
annotation. M marks are not deducted for misreads.J560/06 Mark Scheme Practice Paper (Set 2)
4
11. Unless the question asks for an answer to a specific degree of accuracy, always mark at the greatest number of significant figures even if this
is rounded or truncated on the answer line. For example, an answer in the mark scheme is 15.75, which is seen in the working. The
candidate then rounds or truncates this to 15.8, 15 or 16 on the answer line. Allow full marks for the 15.75.
12. Ranges of answers given in the mark scheme are always inclusive.
13. For methods not provided for in the mark scheme give as far as possible equivalent marks for equivalent work. If in doubt, consult your Team
Leader.
14. Anything in the mark scheme which is in square brackets […] is not required for the mark to be earned, but if present it must be correct.J560/06 Mark Scheme Practice Paper (Set 2)
5
Question Answer Marks Part marks and guidance
1 (a) 1.58 final answer 2
1 AO1.2
1 AO1.3a
M1 for 1.57[7…] seen
or their answer seen to more than 2dp
corrected to 2dp
Both rounded and unrounded value
must be seen
(b) (i) (1 + n)3 = 272 = 729 1
1 AO2.2
(ii) 8 1
1 AO3.1a
2 (a) Any two from 1, 2, 3, 4, 6 1
1 AO2.1a
(b) Any valid explanation 1
1 AO2.4a
e.g. S is a factor
3 (a) Insufficient trials 1
1 AO2.5a
Any acceptable reason
(b) 11
8 1
3
1 AO1.3b
2 AO2.1b
B2 for two correct
or for one correct with total balls = 20
Or M1 for 66 20
120
or 20
120
47
or 20
120
7
4 (a) Final amount is less than initial
investment
1
1 AO3.4b
Or equivalent correct reason
(b) Used an incorrect multiplier for the
interest rate
1
1 AO3.4a
Or equivalent correct reasonJ560/06 Mark Scheme Practice Paper (Set 2)
6
Question Answer Marks Part marks and guidance
(c) 6498.40 or 6498.39 3
3 AO1.3a
M2 for 5800 × 1.0235
Or M1 for 5800 × 1.023n oe Where n ≥ 1, n ≠ 5
5 (a)
1.6 or
8 5
oe
3
3 AO1.3b
M1 for 2(3x – 4) = x or 6x – 8 = x
M1FT for 6x – x = 8
Alternative method:
M2 for 3 4
x 2
x
Or M1 for 0
2
3x 4 x
(b)
2
3
y
x
3
3 AO1.3a
M1 for y + 2 = 3x2
M1FT for 2
3
2
x
y
6 (a) 175 1
1 AO1.3a
(b) 28 to 31 with correct working 4
2 AO3.1d
1 AO3.2
1 AO3.3
M2 for 0.7 × their 175 oe and 0.85 × their
175 oe
Or M1 for 0.7 × their 175 oe or 0.85 ×
their 175 oe
AND
M1 for reading from graph using their
cardio interval
Implied by 122.5 or 123 and 148.75
or 149 seen
Using their max and min cardio rates
(c) Heart rate out of zone for about 4
minutes
1
1 AO2.4a
Or heart rate less than 87.5 during
the 50 minutes
7 69 4
1 AO1.3b
2 AO3.1d
1 AO3.3 M1 for n – 6 + n + 3n = 109 oe
M1FT for 5n = 109 + 6
A1 for n = 23
Allow equivalent part marks for use of
different person as starting point
Rearrangement of their equation to
isolate n termsJ560/06 Mark Scheme Practice Paper (Set 2)
7
Question Answer Marks Part marks and guidance
8 604.8 kg 4
1 AO1.1
1 AO1.3b
1 AO3.1d
1 AO3.2
B3 for answer 10.08 [kg]
OR
M1 for 400 × 400 × 28 soi
M1 for their volume ÷ 10003 soi
M1 for 2250 × their volume [× 60] soi
Volume calculation using consistent
units
Conversion of mm to m for all 3
dimensions done at any stage
Calculation of mass of 1 or 60 slabs
9 (a)
10
4
oe on first set of branches
5 9
,
4 9
,
6 9
,
3 9
on second set of
branches
2
2 AO2.3b
B1 for two or more correct probabilities
(b) 8
15
3
2 AO1.3b
1 AO2.3a
M2 for 6 4 4 6
10 9 10 9
oe
Or M1 for 6 4
10 9
oe or
4 6
10 9
oe
FT probabilities from their tree
diagram for method marks
10 (a) 3.20 × 106
4.04 × 105
2
1 AO1.2
1 AO2.3b
B1 for one correct
or for 3.195[3] × 106 and 4.042 × 105
or for 3.20 × 10m and 4.04 × 10n Where m and n are integers
(b) 3.54 × 107 1
1 AO1.3a
(c) 5.7[0] × 105 3
2 AO1.3b
1 AO2.3a
B1 for 2.455 × 106 or 1.885 × 106 seen
M1 for their (2.455 – 1.885) × 106 Using values in ranges 2.445 ≤ n ≤
2.455 and 1.885 ≤ n ≤ 1.895
11 (a) E
F
B1
B1
2 AO2.3aJ560/06 Mark Scheme Practice Paper (Set 2)
8
Question Answer Marks Part marks and guidance
(b) 1
1 AO2.3b
Clear intention
12 (a) 24 48 71 1
1 AO2.3b
(b) 2
2 AO2.3b
B1FT for at least 5 points plotted
correctly
If 0 scored, SC1 for translation of correct
curve
(c) States correct with comparison
showing approximately 18
employees over 55 and one quarter
of 80 = 20
2
1 AO2.1b
1 AO2.5a
FT reading from their cumulative
frequency curve at 55
B1 for one quarter of employees = 20
or for approximately 18 employees over
55 FT their curve
13 AM = MD given
BMA = CMD vertically opposite
BAM = CDM alternate angles
Triangles AMB, DMC congruent,
ASA
M1
M1
M1
A1
4 AO2.4b
After M0, B2 for two pairs of equal
angles and one pair of equal sides with
insufficient or no reasons
Or B1 for two pairs of equal angles
identified
Accept any correct proofJ560/06 Mark Scheme Practice Paper (Set 2)
9
Question Answer Marks Part marks and guidance
14 (a) x ≤ -3, x ≥ 4 3
3 AO1.3b
M1 for (x – 4)(x + 3)
A1 for solutions -3 and 4 seen
(b) -2 4
2 AO2.1a
2 AO3.1b
M1 for 2y = x + 4 drawn
M1 for x + y = 5 drawn
M1FT for correct region/points identified
on graph
15 (a)
6 3
x y
2
2 AO1.3a M1 for
3
2
x y
or
12 3
6 6
x y
x y
(b) (i) 2x
y
1
1 AO1.3a
(ii) 3 14 2
( 2)( 3)
x x
x x
or
(3 14)
( 2)( 3)
x x
x x
or
2
(3 14)
6
x x
x x
3
3 AO1.3b
M1 for numerator 4x(x + 3) – x(x – 2) oe
M1 for denominator (x – 2)(x + 3) oe
16 (a) 1.44 3
3 AO1.3a
M2 for 9 × 42 = y × 102 oe
Or M1 for 9 × 42 or y = k2
x
soi
(b) 56.25 3
1 AO1.3a
1 AO3.1a
1 AO3.3
M2 for 1.5625 or
2
1
0.8
soi
Or M1 for 0.82 soi
Alternative method:
M1 for calculation of values of y for
their x and their 0.8x
M1 for calculation of percentage
increase in y valuesJ560/06 Mark Scheme Practice Paper (Set 2)
10
Question Answer Marks Part marks and guidance
17 (a)
Gradient AB = 4 2 1
7 3 2
Gradient BC = 2 4 2
10 7
Product of gradients = 2 1
1 2
Perpendicular because product of
gradients is -1
M1
M1
A1
B1
2 AO2.4b
2 AO3.1b
Calculation of gradient of AB
Calculation of gradient of BC
(b) 8.06[2…] or 65 4
2 AO1.3b
1 AO3.1a
1 AO3.2
B1 for identifying AC as hypotenuse
M2 for (10 3) (2 2) 2 2
Or M1 for attempt to use Pythagoras
18 (a) 3 hours 12 minutes 6
2 AO1.3b
2 AO3.1d
1 AO3.2
1 AO3.3
M2 for [AC2]
= 2.82 + 6.22 – 2 × 2.8 × 6.2 × cos 95
Or M1 for attempt to use cosine rule
AND
A1 for [AC =] 7.02
M2 for (2.8 + 6.2 + their 7.02) ÷ 5
Or M1 for attempt at their distance ÷ 5
(b) Any sensible assumption about
distance, speed or time
Any sensible explanation
1 1
2 AO3.5
e.g. he walks in a straight line
e.g. he doesn’t have a rest
e.g. underestimate so time would be
longer
19 (a) 4a – 3b 2
1 AO1.3b
1 AO2.3b
B1 for AD = 3a or BC = 2b soi Allow OD 4a or OC 3b for B1J560/06 Mark Scheme Practice Paper (Set 2)
11
Question Answer Marks Part marks and guidance
(b) AD = CE = 3a
AC = DE = 3b – a
Opposite sides equal and parallel
hence ACED is a parallelogram
M2
M2
A1
1 AO2.4a
3 AO3.1b
1 AO3.3
M1 for AD = 3a or CE = 3a
M1 for AC = 3b – a or DE = 3b – aJ560/06 Practice Paper (Set 2)
12
Assessment Objectives (AO) Grid
Question AO1 AO2 AO3 Total
1(a) 2 0 0 2
1(b)(i) 0 1 0 1
1(b)(ii) 0 0 1 1
2(a) 0 1 0 1
2(b) 0 1 0 1
3(a) 0 1 0 1
3(b) 1 2 0 3
4(a) 0 0 1 1
4(b) 0 0 1 1
4(c) 3 0 0 3
5(a) 3 0 0 3
5(b) 3 0 0 3
6(a) 1 0 0 1
6(b) 0 0 4 4
6(c) 0 1 0 1
7 1 0 3 4
8 2 0 2 4
9(a) 0 2 0 2
9(b) 2 1 0 3
10(a) 1 1 0 2
10(b) 1 0 0 1
10(c) 2 1 0 3
11(a) 0 2 0 2
11(b) 0 1 0 1
12(a) 0 1 0 1
12(b) 0 2 0 2
12(c) 0 2 0 2
13 0 4 0 4
14(a) 3 0 0 3
14(b) 0 2 2 4
15(a) 2 0 0 2
15(b)(i) 1 0 0 1
15(b)(ii) 3 0 0 3
16(a) 3 0 0 3
16(b) 1 0 2 3
17(a) 0 2 2 4
17(b) 2 0 2 4
18(a) 2 0 4 6
18(b) 0 0 2 2
19(a) 1 1 0 2
19(b) 0 1 4 5
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