Mark Scheme (Results)

January 2023

Pearson Edexcel International Advanced Level

In Decision Mathematics (WDM11) Paper 01
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January 2023
Question Paper Log Number P63154A
Publications Code WDM11_01_MS_2301
All the material in this publication is copyright
© Pearson Education Ltd 2023
General Marking Guidance

• All candidates must receive the same treatment. Examiners

must mark the first candidate in exactly the same way as they
mark the last.
• Mark schemes should be applied positively. Candidates must
be rewarded for what they have shown they can do rather than
penalised for omissions.
• Examiners should mark according to the mark scheme not
according to their perception of where the grade boundaries
may lie.
• There is no ceiling on achievement. All marks on the mark
scheme should be used appropriately.
• All the marks on the mark scheme are designed to be awarded.
Examiners should always award full marks if deserved, i.e. if the
answer matches the mark scheme. Examiners should also be
prepared to award zero marks if the candidate’s response is not
worthy of credit according to the mark scheme.
• Where some judgement is required, mark schemes will provide
the principles by which marks will be awarded and
exemplification may be limited.
• When examiners are in doubt regarding the application of the
mark scheme to a candidate’s response, the team leader must
be consulted.
• Crossed out work should be marked UNLESS the candidate has
replaced it with an alternative response.
PEARSON EDEXCEL IAL MATHEMATICS

General Instructions for Marking

1. The total number of marks for this paper is 75.

2. The Edexcel Mathematics mark schemes use the following types of marks:

‘M’ marks
These are marks given for a correct method or an attempt at a correct method. In Mechanics they are
usually awarded for the application of some mechanical principle to produce an equation.
e.g. resolving in a particular direction, taking moments about a point, applying a suvat equation,
applying the conservation of momentum principle etc.
The following criteria are usually applied to the equation.

To earn the M mark, the equation

(i) should have the correct number of terms
(ii) be dimensionally correct i.e. all the terms need to be dimensionally correct
e.g. in a moments equation, every term must be a ‘force x distance’ term or ‘mass x distance’, if we allow
them to cancel ‘g’ s.
For a resolution, all terms that need to be resolved (multiplied by sin or cos) must be resolved to earn the
M mark.

M marks are sometimes dependent (DM) on previous M marks having been earned.
e.g. when two simultaneous equations have been set up by, for example, resolving in two directions and
there is then an M mark for solving the equations to find a particular quantity – this M mark is often
dependent on the two previous M marks having been earned.

‘A’ marks
These are dependent accuracy (or sometimes answer) marks and can only be awarded if the previous M
mark has been earned. e.g. M0 A1 is impossible.

‘B’ marks
These are independent accuracy marks where there is no method (e.g. often given for a comment or for
a graph).

A few of the A and B marks may be f.t. – follow through – marks.

3. General Abbreviations

These are some of the traditional marking abbreviations that will appear in the mark schemes.

• bod – benefit of doubt

• ft – follow through
• the symbol will be used for correct ft
• cao – correct answer only
• cso - correct solution only. There must be no errors in this part of the question to
obtain this mark
• isw – ignore subsequent working
• awrt – answers which round to
• SC – special case
• oe – or equivalent (and appropriate)
• dep – dependent
• indep – independent
• dp – decimal places
• sf – significant figures
• – The answer is printed on the paper
• – The second mark is dependent on gaining the first mark

4. All A marks are ‘correct answer only’ (cao), unless shown, for example as A1 ft to indicate that
previous wrong working is to be followed through. After a misread however, the subsequent
A marks are treated as A ft, but manifestly absurd answers should never be awarded A
marks.

5. For misreading which does not alter the character of a question or materially simplify it,
deduct two from any A or B marks gained, in that part of the question affected.

6. If a candidate makes more than one attempt at any question:

• If all but one attempt is crossed out, mark the attempt which is NOT crossed out.
• If either all attempts are crossed out or none are crossed out, mark all the attempts
and score the highest single attempt.

7. Ignore wrong working or incorrect statements following a correct answer.

Question
Scheme Marks
number
A–B–D–F–C–E–G–A M1
43 + 45 + 49 + 55 + 50 + 48 + 55 = 345 (m)
A1
1. (a)
A–B–D–F–G–E–C–A
43 + 45 + 49 + 55 + 48 + 50 + 52 = 342 (m) A1 (3)

(b) RMST weight = 237 (m) or arcs in RMST are BD, BE, BG, DF, CE B1
237 + 43 + 47 = 327 (m) M1 A1 (3)
(c) 327 optimal distance 342 B1ft (1)
(7 marks)
Notes for Question 1
a1M1: First four nodes correct for a nearest neighbour route starting at A – so must have at least
A – B – D – F – … or in terms of arcs (AB, BD, DF,…) or in terms of weights (43, 45, 49,…)
a1A1: One correct route (in terms of arcs or nodes but not just weights), must return to A and
corresponding correct length (units not required)
a2A1: Both routes correct (in terms of arcs or nodes but not just weights) and their corresponding correct
lengths (units not required)

b1B1: cao for RMST weight (237) or correct arcs (BD(45), BE(46), BG(47), DF(49), CE(50) only) or
45 + 46 + 47 + 49 + 50 stated
b1M1: Adding the two correct least weighted arcs (AB(43) and AD(47)) to their attempt at RMST weight
where their attempt at the RMST has a weight in the interval 224 RMST weight 250 (give bod if not
clear where their attempt at RMST weight comes from but if working shown then must be from summing
the weight of exactly five arcs). Allow unsimplified answers which imply the correct two arcs added to the
weight of the RMST (e.g. 45 + 46 + 47 + 49 + 50 + 43 + 47 is equivalent to the two smallest arcs (43, 47)
added to five arcs which sum to a value in the given interval).

If a candidate uses one of their NN routes from (a) and removes the arcs incident to A from this
route and adds on the 43 and 47 (e.g. you may see 45 + 49 + 55 + 50 + 48 + 43 + 47) then this can still
score M1 as they have added the weight of five arcs that form a spanning tree (not minimal but this
is their attempt - with a total in the required interval) and they have then added on the two correct
least weighted arcs

b1A1: CAO (327) – the correct answer of 327 with no working can score all three marks in this part

c1B1ft: Their numbers correctly used (their answer to (b) and their least value from (a)) with correct
inequalities (allow strict inequality for lower bound) so an answer of 327 – 342 is B0. Lower bound
must be less than upper bound. The LB must be 314 LB 340 and is dependent on scoring the M
mark in (b). The UB is dependent on the M mark in (a) and there must be two different values in (a) stated
(and they must have chosen the smaller of the two). Allow equivalent notation e.g. [314, 340] or (314,
340]

Due to the original question paper showing H in the first vertical column on the table instead of G,
use the following marking guidance.

In (a) allow use of H for G or a combination of Gs and Hs. The following are all examples that would be
acceptable for the correct route A – B – D – F – C – E – G – A in (a)
 • A–B–D–F–C–E–H–A
• A – B – D – F – C – E – G/H – A
• A–B–D–F–C–E–H–G–A
• A–B–D–F–C–E–G–H–G–A
• A – B – D – F – C – E – H or G – G – A
• AB, BD, DF, FC, CE, EH/G, G/HA
• AB, BD, DF, FC, CE, EG (or EH), HA (or GA)
• AB, BD, DF, FC, CE, EH, HG, GA
• AB, BD, DF, FC, CE, EH, GA

In general accept a cycle of the form A – B – D – F – C – E – X – A where X is G or H or any

combination of Gs and Hs (and similarly A – B – D – F – X – E – C – A for the second cycle or their
equivalents in terms of arcs).
Question
Scheme Marks
number

M1

A1 (BCDE)
2. (a)
A1 (HG)

A1ft (FJ)

Shortest path: A – D – G – F – J A1
Length: 28 (miles) A1ft (6)
(b) Shortest path: J – F – G – D – A – B – C – E – H B1
Length: 28 + 20 = 48 (miles) B1ft (2)
A(BC)E + FG = 15 + 3 = 18* M1
(c) A(DG)F + E(HF)G = 27 + 16 = 43 A1
A(D)G + E(H)F = 24 + 13 = 37 A1
Repeat arcs: AB, BC, CE, FG A1
Length: 193 + 18 = 211 (miles) A1ft (5)
EF (13) is the shortest link between two odd nodes excluding G
(d) M1
Repeat EF (13) since this is the shortest path excluding G
We finish at A A1
Length of route = 193 + 13 = 206 (miles) A1 (3)
(16 marks)
Notes for Question 2
In (a) it is important that all values at each node are checked very carefully – the order of the
working values must be correct for the corresponding A mark to be awarded e.g. at H the working
values must be 21 20 in that order (so 20 21 is incorrect)
It is also important that the order of labelling is checked carefully. The order of labelling must be a
strictly increasing sequence – so 1, 2, 3, 3, 4, … will be penalised once (see notes below) but 1, 2, 3, 5,
6, … is fine. Errors in the final values and working values are penalised before errors in the order of
labelling
a1M1: A larger value replaced by a smaller value at least twice in the working values at either E, F, H or J
a1A1: All values at B, C, D and E correct and the working values in the correct order
a2A1: All values at H and G correct and the working values in the correct order
a3A1ft: All values in F and J correct on the follow through and the working values in the correct order. To
follow through F check that the working values at F follow from the candidate’s final values for the nodes
that are directly attached to F (which are A, D, H, G (and J)). For example, if correct then the order of
labelling of nodes A, D, H and G are 1, 4, 6 and 7 respectively so the working values at F should come from
A, D, H and G in that order. The first working value at F should be 30 (from A), the second working value
at F should be their 29 (the Final value at D) + 21 (the weight of the arc DF), the third working value at F
should be their 20 (the Final value at H) + 8 (the weight of the arc HF) and the fourth working value at F
should be their 24 (the Final value at G) + 3 (the weight of the arc FG). Repeat the process for J (which will
have working values from B, H, G and F with the order of these nodes determined by the candidate’s order
of labelling at B, H, G and F)
a4A1: CAO (ADGFJ or AD, DG, GF, FJ but not JFGDA or equivalent from J to A)
a5A1ft: Follow through their final value at J only – if their answer is 28 but this is not the Final Value at J
then A0

b1B1: CAO for the route (JFGDABCEH or JF, FG, GD, DA, AB, BC, CE, EH )
b2B1ft: 48 or follow through their final value at J + their final value at H

c1M1: Three distinct pairings of the nodes A, E, F and G

c1A1: Any two rows correct including pairings and totals
c2A1: All three rows correct including pairings and totals
c3A1: CAO - correct arcs clearly stated and not just in their working as AB, BC, CE and FG (allow BA,
CB, etc.) – must be these arcs. Do not accept ABCE or AE via B and C
c4A1ft: Correct answer of 211 or follow through 193 + their least total from a choice of three

d1M1: Identifies the need to repeat one path of the three (AE, AF, EF) which does not include G (maybe
implicit) or listing of only these possible repeats – this mark is dependent on either scoring the M mark
in (c) or stating all three possible paths. If stating more than these three paths (AE, AF, EF) then it must
be clear from later working that they are only considering these three. As a minimum stating just one of
these three paths (or any combination of these three paths with no others) can score this mark (so, for
example, just stating AE and AF scores this mark) provided that they do not further imply that a
path including G should be repeated (as this would indicate that mentioning one (or more) of these
paths is for the purpose of not repeating it)
d1A1: Identifies EF as the least and A as the finishing point. They have to explicitly state that EF is the
least path that does not include G
d2A1: CAO (206)
Question
Scheme Marks
number
Bin 1: 1.8 1.4 1.6
Bin 2: 2.6 0.9 0.8 0.6
Bin 3: 2.8 1.2
3.(a) Bin 4: 3.1 M1 A1 A1 (3)
Bin 5: 2.4
(b)(i) 1.8 2.6 1.6 2.8 1.4 3.1 0.9 1.2 2.4 0.8 0.6 B1
(ii) Comparisons: 10 Swaps: 6 B1 B1 (3)
e.g. using middle right

2.6 1.8 2.8 1.6 3.1 1.4 1.2 2.4 0.9 0.8 0.6 pivot 1.4 M1
2.6 1.8 2.8 1.6 3.1 2.4 1.4 1.2 0.9 0.8 0.6 pivots 1.6 0.8
2.6 1.8 2.8 3.1 2.4 1.6 1.4 1.2 0.9 0.8 0.6 pivots 2.8 0.9 (0.6) A1
3.1 2.8 2.6 1.8 2.4 1.6 1.4 1.2 0.9 0.8 0.6 pivot(s) (3.1) 1.8 (1.2) A1ft
3.1 2.8 2.6 2.4 1.8 1.6 1.4 1.2 0.9 0.8 0.6 pivot 2.4
A1
3.1 2.8 2.6 2.4 1.8 1.6 1.4 1.2 0.9 0.8 0.6 (sort complete)
(c)
e.g. using middle left

2.6 1.8 2.8 1.6 3.1 1.4 1.2 2.4 0.9 0.8 0.6 pivot 1.4
2.6 1.8 2.8 1.6 3.1 2.4 1.4 1.2 0.9 0.8 0.6 pivots 2.8 0.9
3.1 2.8 2.6 1.8 1.6 2.4 1.4 1.2 0.9 0.8 0.6 pivots (3.1) 1.8 (1.2) 0.8
3.1 2.8 2.6 2.4 1.8 1.6 1.4 1.2 0.9 0.8 0.6 pivot(s) 2.6 (1.6) (0.6)
3.1 2.8 2.6 2.4 1.8 1.6 1.4 1.2 0.9 0.8 0.6 (sort complete)

Bin 1: 3.1 1.8

Bin 2: 2.8 1.6 0.6
(d) Bin 3: 2.6 2.4 M1 A1 A1 (3)
Bin 4: 1.4 1.2 0.9 0.8

(13 marks)
Notes for Question 3
PLEASE NOTE NO MISREADS IN THIS QUESTION – MARK ACCORDING TO THE SCHEME
AND THE SPECIAL CASE IN PART (c) AND THE GUIDANCE FOR THE M MARK IN (d)

a1M1: The correct first four items placed correctly (the bold values) and at least eight values placed in bins
(allow repeated values). Condone cumulative totals for M1 only
a1A1: First eight values placed correctly (the bold and underlined values) with all eleven correct values
only placed in bins. This mark cannot be awarded if any repeated values or incorrect values are seen (even
if the first eight values have been placed correctly)
a2A1: CSO – no additional or repeated values (dependent on both previous marks)

bi1B1: CAO – isw after one complete pass. Please check these carefully as some candidates show all the
swaps and comparisons in the first pass and some show more than one complete pass. As a guide consider
the placement of the 0.8 (when this is the second value from the right-hand side this will indicate the
completion of the first pass)
bii1B1: Comparisons correct (10)
bii2B1: Swaps correct (6)
If the comparisons and swaps are not labelled then assume that the first number seen is the comparisons and
the second number is the swaps (so seeing after the 1st pass of bubble sort the numbers 10, 6 then award
both the second and third B marks in this part). If all they state is 6 then 10 then give SC B1 B0 for the final
two marks in this part

c1M1: Quick sort, pivot, p, chosen (must be choosing middle left or right – choosing first/last item as the
pivot is M0). After the first pass the list must read (values greater than the pivot), pivot, (values less than the
pivot). If only choosing one pivot per iteration then M1 only
c1A1: First pass correct and choosing consistent pivots for the second pass for this mark
c2A1ft: Second and third pass correct following through from their first pass and choice of pivots for the
second pass these pivots for the second pass must be consistent (either both middle left or both middle right)
c3A1: CSO (correct solution only – all previous marks in this part must have been awarded) including a
fifth pass shown (not just saying ‘sort complete’ after the fourth pass) for ‘middle right’ or a fourth pass
shown for ‘middle left’

SC for (c): If using the original list or an incorrect list from the start of (c), or after the first pass, with
only one error (an error is either one missing number, one extra number, two numbers transposed or
one incorrect number) then they can score at most M1A0A1ftA0. If the candidate sorts into ascending
order they can score M1 as per the main scheme (but with the values either side of the pivot
reversed), A1 for a fully correct sort then A0 A0 even if the list is reversed at the end (so 2 marks
max.).

d1M1: Their five largest items placed correctly and at least eight values placed in bins (if correct this will
be the bold items but must check their packing if any of their five largest values are incorrect – note that
the maximum weight of a bin is 5). Condone cumulative totals for M1 only. First-fit increasing scores no
marks in this part. If no sort seen in (c) then mark (d) assuming the correct ordered list is being used
d1A1: First eight values placed correctly (the bold and underlined values) with all eleven correct values
only placed in bins. This mark cannot be awarded if any repeated values or incorrect values are seen (even
if the first eight values have been placed correctly)
d2A1: CSO – no additional or repeated values (dependent on both previous marks)
Question
Scheme Marks
number

Activity Immediately Activity Immediately Activity Immediately

preceded by preceded by preceded by
A - G A, B, E M D, G
4.(a) B - H A, B, E N H, K B2, 1, 0 (2)
C - I A, B, E P H, K
D A J A, B, E, F Q H, I, J, K
E C K D, G R P, Q
F C L D, G

M1

A1
(b)
M1

A1 (4)

(c) Critical activities: C, E, G, K and N B1 (1)

(d) Float on J = 35 – 22 – 5 = 8 B1ft (1)
133
(e) Lower bound is = 3.0930... = 4 B1 (1)
43
e.g.

M1

A1
(f)
A1

A1 (4)

(13 marks)
Notes for Question 4
a1B1: Any four (of the eight blank) rows correct
a2B1: All eight rows correct

b1M1: All top boxes complete, values generally increasing in the direction of the arrows (so generally
going from ‘left to right’ across the network), condone one ‘rogue’ value (if values do not increase in the
direction of the arrows then if one value is ignored and the remaining values do increase in the direction of
the arrows then this is considered to be a single rogue value). Note that all values in the top boxes could be
incorrect but it can still score the M mark if the values are increasing in the way stated above
b1A1: CAO – all values correct in the top boxes
b2M1: All bottom boxes complete (but condone a blank box for the late event time at the end event node
for the M mark only). Values generally decreasing in the opposite direction of the arrows (so generally
going from ‘right to left’ across the network), condone one ‘rogue’ (as described above in b1M1)
b2A1: CAO – all values correct in the bottom boxes

c1B1: CAO (C, E, G, K and N only)

d1B1ft: correct calculation seen for their J (provided total float is non-negative). Correct answer or the
correct answer following through the event times for J with no working seen scores B0 – must see all three
numbers in their calculation

e1B1: CSO – 4 together with either a correct calculation seen or an awrt 3.1. An answer of 4 with no
working scores B0. If working seen then it must be correct

f1M1: Not a cascade (Gantt) chart. 5 ‘workers’ used at most and at least 8 new (11 in total) activities placed
f1A1: 4 workers. All 14 new (17 in total) activities present (just once). Condone at most three errors. An
activity can give rise to at most three errors; one on duration, one on time interval and only one on IPA
f2A1: 4 workers. All 14 new (17 in total) activities present (just once). Condone one error either
precedence or time interval or activity length; An activity can give rise to at most three errors; one on
duration, one on time interval and only one on IPA
f3A1: 4 workers. All 14 new (17 in total) activities present (just once). No errors.

The table below is helpful in checking an activities duration, time interval and IPA

Activity Duration Time interval IPA

D 12 9 – 23 A
E 3 9 – 12 C
F 13 9 – 30 C
G 11 12 – 23 A, B, E
H 8 12 – 30 A, B, E
I 12 12 – 35 A, B, E
J 5 22 – 35 A, B, E, F
K 7 23 – 30 D, G
L 6 23 – 43 D, G
M 6 23 – 43 D, G
N 13 30 – 43 H, K
P 7 30 – 40 H, K
Q 5 30 – 40 H, I, J, K
R 3 37 – 43 P, Q
Question
Scheme Marks
number
E.g. You cannot have a graph with an odd number of odd vertices

1+ 2 + 2 + 3 + 3 + 4 + 4 + 6
E.g. = 12.5 which is not an integer and so therefore
5.(a) 2
not possible to have a graph with the given vertex orders B1 (1)
As vertex C appears more than once A – C – D – E – C – B – F is not an
(b) B2, 1, 0 (2)
example of a path on T
M1; A1; A1
(c) AC, AB, CD; DH, DG; CF, DE
(3)

(d) B1 (1)

(e) 21 < x < 25 B2, 1, 0 (2)

(9 marks)
Notes for Question 5
a1B1: CAO – common examples that score B1:

• Cannot have (a graph with an) odd number of odd vertices

• Cannot have a graph with three odd vertices
• The sum of the degrees/order (of the vertices) is 25 which is not even therefore not possible (but not
just for obtaining 25 and saying ‘impossible’). The 25 must be linked either in words to the ‘sum of
the degrees/order’ or explicitly showing 1+ 2 + 2 + 3 + 3 + 4 + 4 + 6 = 25 so just ’25 is not even’
scores B0
• The sum of the degrees/order (of the vertices) is 25 which is odd therefore not possible (with
equivalent justification of the 25 as in the previous bullet-point)
1+ 2 + 2 + 3 + 3 + 4 + 4 + 6
• = 12.5 which is not an integer so therefore impossible. They do not have to
2
explain that they are using the result that vertex degrees 2(no of arcs) but they must explain
why a value of 12.5 leads to the required graph not being possible. A value of 12.5 with no working
(or explanation) scores B0
In (a) do not condone clearly incorrect technical language e.g. using ‘arc’ when it should be ‘vertex’

b1B1: No + attempt at a reason which includes either the mention of a cycle/circle/loop etc. or the
repeating of a vertex/node/point etc. is sufficient for this mark (condone incorrect technical language) – give
bod (but ‘no because there is a repeated arc’ only scores B0 unless we also see mention of a repeated vertex
(oe) as well)
b2DB1: No + correct reason (dependent on first B mark in (b)) – no bod – must refer to C appearing twice
(not just that a vertex is repeated) or that it contains the cycle C – D – E – C (not just that it contains a
cycle). All technical language must be correct if used for this mark and do not isw any incorrect
reasoning

The minimum acceptable answer for both marks in this part is, ‘it is not a path as C appears twice’

c1M1: Prim’s – first three arcs correctly chosen in order (AC, AB, CD) or first four nodes (A, C, B, D)
correctly chosen in order. If any explicit rejections seen at some point then M1 (max) only. Order of
nodes may be seen at the top of a matrix/table {1, 3, 2, 4, -, -, -, -}. However, do not accept a list of weights
only (as the weights in the network are not unique)
c1A1: First five arcs correctly chosen in order (AC, AB, CD, DH, DG) or
all eight nodes {A, C, B, D, H, G, F, E} correctly chosen in order. Order of nodes may be seen at the top of
a matrix so for the first two marks accept {1, 3, 2, 4, 8, 7, 6, 5} (no missing numbers). However, do not
accept a list of weights only (as the weights in the network are not unique)
c2A1: CSO – all arcs correctly stated and chosen in the correct order (with no additional incorrect arcs).
They must be considering arcs for this final mark (do not accept a list of the weights of each arc, nodes or
numbers across the top of the matrix unless the correct list of arcs (in the correct order) is also seen)

Misread in (c): Starting at a node other than A scores M1 only – must have the first three arcs (or four
nodes) correct (and in the correct order) – condone any rejections seen for this mark

d1B1: CAO (ignore weights on arcs even if incorrect)

e1B1: x  25 or x 25 or x  24 or x 24 or equivalent notation e.g. (… , 25)

e2B1: CAO ( 21  x  25 ) or equvialent notation (e.g. (21, 25))
Question
Scheme Marks
number
6. (a) Minimise (P =) 300x + 400y + 400z B1
Subject to:
275 x + 200 y + 100 z 5500 ( 11x + 8 y + 4 z 220) B1
5 x + 2 y + 3z 70 B1
x y z
+ + 1  4 x + 3 y + 2 z 60 M1 A1
15 20 30
x + y + z 18
B1 (6)
( x, y , z 0)
e.g Martin makes apple cakes and chocolate cakes in the ratio of 2:1
e.g. for every one chocolate cake that Martin makes he has to make 2 apple
cakes (assuming that he makes chocolate/apple cakes)
(b) B1 (1)
e.g. Martin should make twice as many apple cakes as chocolate cakes
e.g. The number of apple cakes that Martin makes should be double the number
of chocolate cakes he makes

B1

B1
(c)
B1

B1 (4)

(d) Drawing an objective line accept reciprocal gradient M1

Correct objective line A1
Martin should make 9 carrot cakes, 6 apple cakes and 3 chocolate cakes A1
Minimum amount of sugar is 6300 grams (or 6.3 kg) A1 (4)
(e) Martin has 1525 grams of flour remaining B1
Martin also has 4 eggs remaining B1 (2)
(17 marks)
 Notes on Question 6
a1B1: CAO – expression correct (300x + 400y + 400z or 0.3x +0.4y + 0.4z only) and ‘minimise’ or ‘min’
but not ‘minimum’. ISW if either of these two expressions are seen and then ‘simplified’

a2B1: CAO ( 275 x + 200 y + 100 z 5500 ) oe but must be four terms only with integer coefficients e.g.
11x + 8 y 220 − 4 z
a3B1: CAO ( 5 x + 2 y + 3z 70 ) oe but must be four terms only with integer coefficents
x y z
a1M1: Correct method + + • 1 where • is any inequality symbol or =
15 20 30
a1A1: CAO ( 4 x + 3 y + 2 z 60 ) oe must be four terms only with integer coefficients
a4B1: CAO ( x + y + z 18 ) oe must be four terms only with integer coefficients

b1B1: CAO but give bod if intention is correct. Some correct examples include:

• (Martin makes) apple (cakes) and chocolate (cakes) in the ratio of 2:1
• For every one chocolate (cake that Martin makes he has to) make 2 apple (cakes)
• (Martin should make) twice as many apple (cakes) as chocolate (cakes)
• The number of apple cakes (that Martin makes should be) double the number of chocolate cakes (he
makes)
Please check these carefully for those candidates that imply incorrectly that Martin should make two
chocolate cakes for every one apple cake. Furthermore, do not condone an answer that implies an inequality
(e.g. use of words such as, ‘at least’, ‘at most’, etc.)

The lines in (c) must define the correct FR and if extended would pass within a small square of their point
of intersection with the axes

c1B1: Any two lines correctly drawn

c2B1: Any three lines correctly drawn
c3B1: All four lines correctly drawn
c4B1: Correct R labelled – not just implied by shading – dependent on scoring the first three marks in this
part and all four lines being drawn from axis to axis (within one small square)

d1M1: Drawing the correct objective line (gradient –0.5) or its reciprocal (gradient –2) on the graph. Line
must be correct to within one small square if extended from axis to axis. If line is shorter than (0, 1) to (2, 0)
(or for the reciprocal (0, 2) to (1, 0)) then M0
d1A1: Correct objective line – same condition that the line must be correct to within one small square if
extended from axis to axis and be no shorter than the line from (0, 1) to (2, 0)

The final 4 marks are all dependent on the first three B marks in (c), the first two marks in part (d)
and they must not have implied an incorrect R in (c) (but give bod if region not labelled in (c) or if the
lines did not go axis to axis in (c) as this was penalised with the final mark in (c))

d2A1: CAO (in context) – as a minimuim accept 9 carrot, 6 apple and 3 chocolate
d3A1: CAO (6300 or 6.3) – no units required but if stated then must be correct – so 6300 kg is A0

e1B1: CAO (1525 (grams) or 1.525 (kg) of flour) – no units required but if stated then must be correct – so
1525 kg is A0
e2B1: CAO 4 (eggs)
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