Notes

If correct they must use two slack, two surplus and two artificial variables. Accept alternative letters for these as long as the artificial variables are clearly identifiable

Please check suffices on \(s\) and \(a\) terms carefully – they may be in a different order – check that these are consistent

a1B1: Correctly re-writing the two \(\leqslant\) inequalities as equations with slack variables (can be implied by the corresponding two correct rows in Simplex tableau – our row 1 and 4)

a2B1: Correctly re-writing one of the \(\geqslant\) inequalities as an equation with surplus and artificial variables (can be implied by a correct corresponding row in Simplex tableau – our row 2 or 3)

a3B1: Correctly re-writing both \(\geqslant\) inequalities

a1M1: Forming an objective of the form \(P = 11x + ky - M(a_1 + a_2)\) and substituting for \(a_1\) and \(a_2\) (we must see the substitution but this does not need to be a correct expression for this mark)

a1A1: CAO for new objective (accept equivalent equation with terms in \(x\) and \(y\) collected) (M1 A1 may be implied by a correct objective row in the tableau)

a2M1: Any two rows correct on the ft from the candidate’s stated equations (ignore b.v. for this mark)

a2A1: CAO (including consistent b.v. column) – note that the candidate’s order in which the rows appear in the tableau (and choice of slack variable) may be different

A fully correct tableau implies all marks in (a) provided that there are no errors seen in the formation of the objective function

Notes

b1B1: Either using the correct values of \(x\) and \(y\) in the objective function or stating \(P = \frac{198}{7} + \frac{2}{7}k\)

b1M1: Considering at least one of the expressions (\(s_3\) or \(s_4\) columns) in the \(P\) row that involve \(k\) and compare with 0 (accept any correct inequality or equals)

b1A1: Correct range of values for \(k\) (condone missing 0 < ) but must have considered both possibilities and chosen 11/3 (may be implied by subsequent working) (allow \(k\) = 11/3 stated as the maximum value)

b2dM1: Dependent on previous M mark – using their \(k\) in given \(P\) which must come from a correct inequality or equation (they may choose a value of \(k\) from the correct range e.g. \(k = 3\))

b2A1: CAO - Correct value of \(P\) for their choice of \(k\)
(accept answers in the range \(\frac{198}{7} \lt P \leqslant \frac{88}{3}\) if \(k = 3\) \(P = 204/7\))

Notes

a1B1: Correct reason why Simplex cannot be used to solve the LP problem. Please mark positively and award if a correct statement is seen.
SC accept Because \(z \leqslant -3\) (and variables must be \(\geqslant 0\))

Notes

b1M1: Substituting given equations into all three given inequalities to form inequalities or equations with slack variables in terms of \(X\), \(Y\) and \(Z\) (condone at most two slips but see Special Case below)

b1A1: At least two of the four expressions simplified correctly (three inequalities and the new objective) (accept any equivalent rearrangement but the objective must be in terms of \(Q\) not \(P\)) (accept equations with slack variables)

b2A1: CAO (must be inequalities)

Special Case
(b) If they make a consistent error when substituting by using an incorrect expression for \(X\), \(Y\) or \(Z\) (e.g. substituting \(x = X + 1\) instead of \(x = X + 4\) in all terms) they may score M1 A1 (for at least two of their four expressions correct) A0

Notes

c1B1: CAO (must be in terms of \(Q\))

Notes

d1B1: CAO (Must include mention that \(Z\) can be increased but do not award if any incorrect statement seen)
Do not accept just a statement that the objective row contains a negative value and therefore it is not optimal

Notes

Note – accept correct recurring decimals in place of fractions or any equivalent fractions

ei1B1: Pivot row (\(Z\) row) correct including change of b.v. (ignore row ops)

ei1M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(s_1, s_2, s_3\) or value) correct

ei1A1: Row operations used correctly at least twice, i.e. two of the non-pivot rows or two of the non zero and one columns (\(s_1, s_2, s_3\) or value) – ignore row operations for this mark

ei2A1: CAO all values and row operations correctly stated including b.v column (allow alternative numbering of rows as long as this is clear. Condone use of \(r_2\) throughout. Do not accept in terms of b.v.) (Accept in terms of original r2 so r1 + 1/3r2, 4/9r2, r3 + 1/9r2, r4 + 5/9r2) (Row ops for pivot row may be written as r2 ÷ 9/4)

eii2M1: Stating (or implying) optimal values of \(X\), \(Y\), \(Z\) and \(Q\) (see special case)

eii3A1ft: CAO for \(x\), \(y\), \(z\) and \(P\) (the correct 4 values implies both marks) (follow through their values for \(X\), \(Y\), \(Z\) and \(Q\) from the tableau)

Special Case
(e) If they use the same incorrect substitution and do not state the values of \(X\), \(Y\), \(Z\) and \(Q\) they may score M1 A0 for the implied values of \(X\), \(Y\), \(Z\) and \(Q\) from their \(x\), \(y\), \(z\) and \(P\)

Notes

B1: CAO (correct column of \(x\) stated)

dB1: Dependent on first B1 Correct statement that \(x\) is now a basic variable (oe) Accept that \(x\) now appears in the first (bv) column or states \(x\) column has one 1 and rest 0 or \(s_2\) row has been replaced with \(x\)

Notes

B1: Either the constraint for the x-row or \(s_3\)- row correct (allow any equivalent form including non-integer coefficients but must be inequalities) Do not accept strict inequalities

M1: Eliminating \(s_2\) from the equation from the \(s_1\) row using the equation from the \(x\) row Correct constraint implies this mark

M1: Eliminating \(s_2\) from the equation from the \(P\) row using the equation from the \(x\) row Correct objective implies this mark

A1: \(P - 3x - y - 2z = 0\) or \(2x + z + s_1 = 50\) (allow any equivalent form including non-integer coefficients but must have been simplified to a single term in each variable) Correct constraint or objective implies the corresponding M mark

A1: Correct LP formulation (condone lack of ‘maximise’ and condone lack of the non-negative trivial constraints)

Note: It is possible to score M1 M0 A1 A0 or M0 M1 A1 A0

Alternatively – reproduces original tableau

B1: \(s_2\) row correct

M1: \(s_1\) row correct - Correct constraint implies this mark

M1: P row correct - Correct objective implies this mark

A1: fully correct tableau

A1: Correct LP formulation (condone lack of ‘maximise’ and condone lack of the non-negative trivial constraints)

Notes

(c) Note – accept correct recurring decimals in place of fractions

B1: Pivot row completely correct including change of b.v.

M1: All values in one of the non-pivot rows correct (so ignore b.v. column and ‘Row Ops’ column) or one of the ‘non zero and one’ columns (which are y, \(s_1\), \(s_2\) or Value) correct (must have pivoted on the correct value)

A1: cao all values including b.v. column – ignore ‘Row Ops’ column for this mark

B1: Correct row operations stated (allow alternative numbering of rows as long as this is clear. Condone use of \(\text{R}_1\) throughout)

Notes

(i) dB1: CAO (profit row still contains negative values (but not negative variables) – dependent on the M mark in (c) and a completed profit row in (c) (accept P row or objective row but not operations row or bottom row) Do not accept that not all values are positive

(ii) dB1ft: Follow through their values of \(z\), \(x\), and \(s_3\) – dependent on the M mark in (c) and a completed tableau in (c) ignore mention of any other variables including \(P\) All values must be positive

Notes

B1: CAO (e.g. correct indication that the (basic variable) column for \(y\) is not correct – see bold statement for the minimum acceptable) Accept there should not be a 1 in the objective row in the \(y\) column or that there should not be two 1s in the \(y\) column. If \(s_2\) column also mentioned then B0

Notes

(a) NOTE: if correct they must use one slack, two surplus and two artificial variables. Accept alternative letters for these as long as the artificial variables are clearly identifiable

B1: one correct equation or two correct inequalities (do not accept strict inequalities)

B1: two correct equations or three correct inequalities

B1: all three equations correct (please check suffices on \(s\) and \(a\) terms carefully – they may be in a different order)

M1: setting up the new objective which must be \(P = -x - y - M(a_1 + a_2)\) and an attempt to substitute for their \(a_1\) and \(a_2\) (accept any equivalent form, which may not be fully simplified) (accept the use of \(Q\) instead of \(P\) throughout)

M1: setting up initial tableau – all four rows complete with two correct rows (but ignore b.v. column for this mark) (Note the order of rows may be different from above) Check that the slack, surplus and artificial variables correspond to their equations

A1: CAO (any equivalent correct form, but the terms in the objective row must be simplified)

Notes

(i) B1: CAO (\(x = 130\))

(ii) M1: Substitute \(x = 130\) into candidate’s inequalities from (a) (at least 3 inequalities seen or both \(y \geqslant 30\) and \(y \leqslant 30\)) (condone \(y \leqslant 130\))

A1: CAO (\(y = 30\)) we must see both \(y \geqslant 30\) and \(y \leqslant 30\) explicitly stated for this mark (must not follow from any incorrect working)

Notes

B1: Correct objective function \((8x + 20y + 40z)\) plus ‘maximise’ or ‘max’ but not ‘maximum’ (and ‘\(P =\)’ is not required)

M1: Either one correct inequality (need not be simplified) or both \(4x + 8y + 15z \leqslant k_1\) and \(x + 5y + 12z \geqslant k_2\) where \(k_1, k_2 \gt 0\)

A1: Both correct (\(4x + 8y + 15z \leqslant 9000,\ x + 5y + 12z \geqslant 3600\)) – allow equivalent answers (provided 4 terms only and integer coefficients e.g. \(2x + 10y + 24z - 7200 \geqslant 0\))

B1: CAO (\(x + y + z \geqslant 1600\) oe provided 4 terms only and integer coefficients)

B1: CAO (\(3y \leqslant x\) oe provided 2 terms only and integer coefficients)

Notes

(b) Note in this part that the numbering of the suffices for the slack, surplus and artificial variables seen in both the candidate’s equations and Simplex tableau will most likely be different from what is seen in the MS (e.g. \(4x + 8y + 15z + s_3 = 9000\) is correct). The correct values in the rows of the tableau do NOT imply the first four marks (the question explicitly asked for the constraints as equations and the rows for the two objectives to be explicitly stated)

B1: Any two correct inequalities converted into equations correctly (condone the same letter, say \(s_3\), being used twice) – any equivalent forms of the correct equations are acceptable (e.g. variables do not need to be on the same side)

B1: All four correct equations (using distinct slack, surplus and artificial variables) – any equivalent forms of the correct equations are acceptable

M1: Using \(I = -(a_1 + a_2)\) with their expressions for \(a_1\) and \(a_2\) (allow slips in forming \(I\) from their two expressions for the two artificial variables) but must be a clear intention to calculate \(I = -(a_1 + a_2)\)) – must be exactly two artificial variables in their expression for \(I\) for this mark

A1: CAO for \(I\) and \(P\) (must be stated as \(P - 8x - 20y - 40z = 0\) (A0 if = 0 missing) and \(I - 2x - 6y - 13z + s_2 + s_3 = -5200\) - so variables on one side and constant on the other)

M1: setting up initial tableau – all six rows complete (with no blanks) and two correct rows (but ignore b.v. column for this mark)

A1: CAO (any equivalent correct form) - note that the candidate’s order in which the rows appear in the tableau (and choice of letter to represent the slack, surplus and artificial variables) may be different – check to ensure that the basic variable column is consistent with their choice of lettering for the slack and artificial variables

Notes

B1: Pivot (top) row completely correct including change of b.v. (but not ‘Row Ops’ column)

M1: All values in one of the non-pivot rows correct (so ignore b.v. column and ‘Row Ops’ column) or one of the ‘non zero and one’ columns (which are \(y, s_1, s_2\) or Value) correct (must have pivoted on the correct value)

A1: Row operations used correctly at least twice, i.e. two of the ‘non zero and one’ columns (\(s_1, s_2\), \(y\) or Value) correct

A1: CAO all values including b.v. column – ignore ‘Row Ops’ column for this mark

B1: Correct row operations stated – alternatives are \(\dfrac{1}{3}\text{r}_1,\ \text{r}_2 - \dfrac{1}{33}\text{r}_1,\ \text{r}_3 + \dfrac{12}{33}\text{r}_1,\ \text{r}_4 + \dfrac{12}{33}\text{r}_1,\ \text{r}_5 + \dfrac{56}{33}\text{r}_1\)

Notes

(d)(i)

B1: CAO (20 500)

(ii)

B1: CAO (70 only)

Notes

B1: CAO – must explicitly mention the fact that it is possible that not all the books will be sold. Ignore reasons for why, provided they do not relate to the publisher producing less than the optimal number of books

Notes

(a) M1: Correctly re-writing any two inequalities as equations with slack variables (can be implied by two correct rows in the Simplex tableau ignoring b.v. column). Or correctly stating all four constraints as inequalities

A1: Correctly re-writing all inequalities as equations with slack variables (can be implied by the four correct constraint rows in the Simplex tableau ignoring b.v. column)

B1: Correctly re-writes objective function (can be implied by correct row in tableau)

M1: Any two rows correct including consistent b.v. column entries or any three rows correct (ignoring b.v. column)

A1: cao (including consistent b.v. column) – note that the candidate’s order in which the rows appear in the tableau (and choice of letter to represent the slack variable) may be different. A correct tableau implies full marks in this part

Notes

(b) B1: cao for \(x\), \(y\), \(s_1, s_2, s_3\) and \(s_4\) only (ignore any mention of \(P\))

Notes

(c) B1: Pivot row completely correct including change of b.v.

M1: All values in one of the non-pivot rows correct (so ignore b.v. column and ‘Row Ops’ column) or one of the ‘non zero and one’ columns (which are \(s_1, s_2\) or Value) correct (must have pivoted on the correct value)

A1: Row operations used correctly at least twice, i.e. two of the ‘non zero and one’ columns (\(s_1, s_2\) or Value) correct

A1: cao all values including b.v. column – ignore ‘Row Ops’ column for this mark

B1: Correct row operations stated
(alternatives row operations are 5r1;   r2 + 2r1;   r3 – 12r1;   r4 – r1;   r5 – \((k - 2)\)r1)

M1: Their optimal value (as a linear expression in \(k\)) stated correctly following their third iteration (must have pivoted on a positive value from the \(s_4\) column and completed the bottom row of the tableau). Condone this expression (\(4k + 4\) if correct) being stated as part of an equation/inequality (or as part of their final answer) – sight of this expression (but must be seen outside of the tableau) scores this mark

B1: Correctly inferring that \(k \lt 2\) either from \(-\tfrac{2}{5} + \tfrac{1}{5}k \lt 0\) or from \(4k + 4 \gt 5k + 2\) – just stating \(k \lt 2\) without it being clear where this comes from is B0

dM1: Considering (at least) two of their linear expressions in \(k\) from their objective row (not including the Value column) after the third iteration \(\geqslant 0\) (dependent on the previous M mark) – note that working may be minimal here so please follow through their expressions in \(k\) from their objective row (so if correct, stating \(k \leqslant 2\) and \(k \geqslant 1\) implies this mark)

A1: cao for the range of values for \(P\) - this mark is dependent on a correct objective row in the tableau and the previous three marks in this part

Notes

(a) B1: One correct non-trivial inequality (allow strict inequality provided direction of inequality sign is correct) – equations with slack variables etc. scores no marks unless replaced with correct inequalities

B1: All three non-trivial inequalities correct

M1: Either expression stated correctly (allow equal to (or an inequality with) any letter e.g. \(P = -2x + 3y + z\) but not equal to a value e.g. = 0) – ignore any mention of maximum/minimum for this mark

A1: Both expressions correct including max/min correctly matched with each expression (allow equal to any letter only) – do not isw if they continue and place their expression(s) equal to a value(s)

Notes

(b) B1: Correct reasoning that the pivot is a value from the \(x\)-column – as a minimum must state that the \(2 - 4M\) is the only negative (condone most negative) in the objective row (allow profit row or \(P\) row, condone ‘bottom row’)

B1: Correct justification of why the 3 in the \(a_2\) row or the 3 in the \(x\) column is the pivot – so must state the correct pivot in a clear unambiguous way (so just saying the pivot is ‘the 3’ is B0) and comparing or stating that \(\dfrac{12}{3}\) or 4 is less than/least positive for both \(\dfrac{8}{1}\) or 8 and \(\dfrac{13}{2}\) or 6.5 – must see all three values so do check the table for possibly stating the \(\theta\) values there. However, just stating that the 3 is the pivot because it is the smallest \(\theta\) value (without seeing anywhere these \(\theta\) values) is B0

Notes

(a) B1: cao (\(x + y + z \leqslant 39\))

M1: \(\tfrac{2}{5}(x + y + z) \,\square\, x\) where \(\square\) is any inequality or equals

A1: cao

B1: cao (\(x + z \geqslant 28\))

B1: Correct objective function (\(P = y + z\)) plus ‘maximise’ or ‘max’ but not ‘maximum’

Notes

(b) M1: One \(\leqslant\) constraint re-formulated as an equation using slack variables – dependent on either the first B mark in (a) or the M mark in (a)

A1: cao (both \(\leqslant\) constraints)

B1: \(\geqslant\) constraint re-formulated as an equation using one surplus and one artificial variable

M1: Formulates second objective with \(I = -a_1\) and their expression for \(a_1\)

M1: Setting up the initial tableau – all five rows complete with two correct rows (but ignore b.v. column for this mark)

A1: cao (any equivalent correct form)

Notes

(c) B1: Correct reasoning that the pivot is a value from the \(z\)-column – condone any mention of negative value in \(P\) row

B1: Correct justification of why the 5 in the \(s_2\) row is the next pivot – so must compare or state that 12.4 is less than 28 (not sufficient to just say that 12.4 (oe) is the least)

Notes

(d) B1: Pivot row correct including change of b.v.

M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(s_1, s_2\) or value) correct (from their choice of pivot)

A1: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(s_1, s_2\) or value)

A1: For all values and row operations correctly stated – do not penalise lack of correct b.v. in pivot row twice. Condone blank Row Ops in the first row only

A1: Correct allocation of training times – must be in context (so not just in terms of \(x\), \(y\) and \(z\))

Notes

(a) B1: CAO – including maximise (or max)

B1: CAO (oe)

B1: CAO (oe)

B1: CAO (oe)

Notes

(b) M1: Setting up the new objective and substituting for \(a_1\) and \(a_2\)

A1: Correct values substituted into Table 1

Notes

(c) B1: CAO – mention that \(A = 0\)

B1: At least three values stated correctly

B1: All six values correct (ignore values stated for \(a_1, a_2\) and \(P\))

Notes

(d) M1: Correct pivot located, attempt to divide row

A1: Pivot row correct including change of b.v.

M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(y\), \(s_1, s_3\) or value) correct following through their choice of pivot from column \(s_2\) or \(s_3\)

A1: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(y\), \(s_1, s_3\) or value) correct

A1: CAO all values and row operations correctly stated – allow if row operations given in terms of old row 1 – ignore b.v. column for this mark

Notes

(e)(i) B1: Correct reasoning of why solution is optimal or using \(P = 69 - \frac{15}{7}y - \frac{11}{7}s_1 - \frac{3}{7}s_3\) and mentioning increasing \(y, s_1, s_3\) would decrease \(P\) (oe)

(e)(ii) B1ft: their value of \(P\) – dependent on both M marks in (d)

(e)(iii) B1ft: their values of the basic variables only – dependent on both M marks in (d)

Notes

(a) B1: Any two correct (accept strict inequalities) – accept equivalent inequalities

B1: CAO (accept equivalent inequalities but inequalities must not be strict)

Notes

(b) B1: One correct vertex (must be exact)

B1: All three correct vertices (must be exact)

M1: Testing all three of their vertices in the correct objective function

A1: Correct three values of \(P\) and correct optimal vertex either stated or clearly indicated on the graph

Notes

(c) M1: Their optimal point from (b) evaluated in \(Q\) compared to their \(\left(\frac{60}{37}, \frac{150}{37}\right)\) evaluated in \(Q\) (with correct inequality)

A1: \(a \lt \frac{5}{2}\)

M1: Their optimal point from (b) evaluated in \(Q\) compared to their \(\left(\frac{2}{3}, \frac{5}{3}\right)\) evaluated in \(Q\) (with correct inequality)

A1: \(a \gt -3\)

Notes

(a) B1: CAO – allow any equivalent form of \(x + y \geqslant 60\) - do not condone strict inequality

B1: CAO – allow any equivalent form of \(y \leqslant \tfrac{4}{5}(x + y)\) (but not \(y \leqslant 80\%(x + y)\) only) and need not be simplified - do not condone strict inequality – isw if correct answer is incorrectly simplified

Notes

In (b), lines must be long enough to define the correct feasible region and would pass if extended through one small square of the points stated:

\(x + y = 60\) must pass within one small square of its intersection with the axes – (0, 60) and (60, 0)
\(y + 3x = 140\) must pass within one small square of its intersection with the axes – (0, 140) and \(\left(\frac{140}{3}, 0\right)\) (so at 46.666…, 0)
\(4y + x = 80\) must pass within one small square of its intersection with the axes – (0, 20) and (80, 0)
\(y = 4x\) must pass within one small square of (0, 0) and (25, 100)

In (b) condone for full marks lines which are drawn as dashed rather than solid

(b) B1: 2 lines drawn correctly

B1: 3 lines drawn correctly

B1: 4 lines drawn correctly

B1: Region, \(R\), correctly labelled – not just implied by shading – dependent on scoring the first three marks in this part

Notes

(c) M1: Drawing the correct objective line (with gradient – 5) or its reciprocal (with gradient \(-\frac{1}{5}\)). Line must be correct to within one small square if extended from axis to axis. If lines shorter than (5, 0) to (0, 25) or (0, 5) to (25, 0) then M0. Or point testing at least two exact coordinates of their \(R\) using their objective function which must be of the form \(k(5x + y)\) or \(k(x + 5y)\) for some positive real value \(k\)

A1: Correct objective line – condone lack of labelling of the objective line. Or point testing at least two of the correct exact coordinates which are (20, 80), (40, 20), (80, 0) and \(\left(\frac{160}{3}, \frac{20}{3}\right)\) using a correct objective function of the form \(k(5x + y)\)

A1: Correct number of watches – must be in context (and not just in terms of \(x\) and \(y\)) – dependent on a correct feasible region in (b) (so must have scored the first three marks in (b) but may not have labelled the FR as \(R\))

Notes

Condone use of \(\boldsymbol{x}\) for \(\boldsymbol{a}\) and \(\boldsymbol{y}\) for \(\boldsymbol{d}\) in part (d)

(d) B1ft: A ‘correct’ equation (e.g. \(20a + 80d = 4\,455\)) involving their optimal point from (c) (accept any values even if non-integer) and 4455 –- note that for those who have done point testing in (c) the calculation 4455 / (their value for \(P\)) where \(P = 5x + y\) or \(x + 5y\) using their optimal point implies this mark

B1: CAO on the relationship between the costs of the two types of watches (\(a = 5d\)) – this mark may be implied e.g. \(20(5d) + 80d = 4455\) would score the first two marks in this part – note that for those who have done point testing in (c) the calculation 4455 / (their value for \(P\)) where \(P = 5x + y\) using their optimal point implies this mark e.g. just seeing 4455 / 180 is the first two marks in this part

dB1: CAO (dependent on first two B marks) – this mark is dependent on having the correct optimal point (20, 80) and is dependent on a correct feasible region in (b) (so must have scored at least the first three marks in (b)) – allow for \(a = 123.75\) and \(d = 24.75\) (so does not need to be in context or units) – the correct answers with no working scores no marks in this part (however, note that 4455 / 180 is the minimum amount of working that is acceptable)

Notes

(a) B1: CAO – e.g. not all of the constraints are \(\leqslant\), the origin is not a (basic feasible) solution of the LP

Notes

(b) B1: CAO \(3x + y + 2z + s_1 = 30\) (may be seen in the simplex tableau – allow any \(s_i\) (or \(s\)) for \(s_1\))

B1: CAO \(x - y + z - s_2 + a_1 = 8\) (may be seen in the simplex tableau – allow any consistent \(s_i\) for \(s_2\) (or \(t\) say) but not the same \(s_i\) as in the previous mark and allow any \(a_i\) for \(a_1\))

B1: CAO \(4y + 2z - s_3 + a_2 = 15\) (may be seen in the simplex tableau – same conditions as above)

M1: setting up the new objective which must be \(P = 2x + 2y - z - M(a_1 + a_2)\) and substituting for their \(a_1\) and \(a_2\) (if no working then the correct objective line in the tableau implies this mark)

A1: CAO \(P - (2 + M)x - (2 + 3M)y - (-1 + 3M)z + Ms_2 + Ms_3 = -23M\) (any equivalent form – need not be factorised and does not need to be re-arranged into this form - if no working then the correct objective line in the tableau implies this mark)

M1: setting up initial tableau – all four rows complete with two correct rows (but ignore b.v. column for this mark)

A1: CAO (any equivalent correct form)

Notes

(c) B1: CAO \(s_1 = 26.25,\ a_1 = 11.75,\ y = 3.75,\ x = z = s_2 = s_3 = a_2 = 0\) (ignore expression for \(P\) if given)

Notes

(d) B1: correct reasoning of why the solution is not feasible e.g. \(a_1\) is not zero but B0 for just stating that the artificial variable is non-zero (so must see either \(a_1\) or 11.75 being stated as non-zero)

Notes

(e) B1: correct reasoning of why the pivot comes from a value from the \(z\)-column so must say that the most negative value (in the objective row) is \(2 - 1.5M\) (or this expression clearly implied)

dB1: correct justification of why the 0.5 in the third row is the next pivot (dependent on previous B mark) – so must compare or state that \(\dfrac{3.75}{0.5}\) or 7.5 is less than both \(\dfrac{26.25}{1.5}\) or 17.5 and \(\dfrac{11.75}{1.5}\) or 7.8(3333….) – just stating that the 0.5 in the third row is the next pivot without reasoning is no marks in this part

Notes

a1M1: Correct method for finding the equation of the line through \(A\) and \(C\) – a correct equation can imply this mark – condone one sign error only

a1A1: Correct equation (any correct form (allow unsimplified or simplified incorrectly) – condone any inequality sign or equals)

a2A1: CAO (any equivalent form provided coefficients are integers)

Notes

b1M1: Correct objective line drawn (gradient of \(-5\) - acceptable minimum length is from \((0,1)\) to \((0.2,0)\)) or testing both \(A\) and \(B\) in the correct objective function

b1A1: CAO (\(B\) or (5, 3))

b2A1: \(P = 28\) (allow if seen in working for \(B\))

Notes

c1M1: Put expression for \(A\) > expression for \(B\) (accept any inequality or equals) or considers gradient of line segment through \(A\) and \(B\) with \(-k\). Condone \(x\) for \(k\) for the M mark only

c1A1: \(k \lt 4\)

c2M1: Put expression for \(A\) > expression for \(C\) (accept any inequality or equals) or considers gradient of line segment through \(A\) and \(C\) with \(-k\). Condone \(x\) for \(k\) for the M mark only

c2A1: \(k \gt -\dfrac{2}{5}\)

Ignore any consideration of vertex \(D\)

Notes

a1M1: Any one row correct (but ignore b.v. column)

a1A1: All four rows correct (but ignore b.v. column)

a1B1: b.v. column correct

Notes

b1M1: Correct pivot located (2 in column \(y\)), attempt to divide row

b1A1: Pivot row correct including change of b.v

b2M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(x\), \(z\), \(r\) or value) correct following through their choice of pivot from column \(y\)

b2A1ft: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(x\), \(z\), \(r\) or value) correct following through their choice of pivot from column \(y\)

b3A1: CAO – no follow through – all values and row operations correctly stated – allow if row operations given in terms of old row 1 – ignore b.v. column for this mark

Notes

c1B1ft: Follow their profit equation from (b) dependent on scoring both M marks in (b)

c2B1: CAO (no follow through) for slack variables (\(r = 0\), \(s = 45\), \(t = 20\))

Notes

d1B1: CAO

Notes

a1B1: Expression correct (or \(75x + 100y\)) together with ‘maximise’ or ‘max’ but not ‘maximum’ – isw if coefficients are subsequently simpified but either \(75x + 100y\) or \(0.75x + y\) must be seen at some point for this mark to be awarded. The ‘max’ must appear beside or suitably close to one of the correct two expressions

a2B1: CAO (\(x + y \geqslant 400\))

a3B1: CAO (\(y \leqslant 350\))

a1M1: \(5y \blacksquare 3x\) where \(\blacksquare\) is any inequality or equals. Accept \(3y \geqslant 5x\) for this mark. M0 if coefficients are not integers

a1A1: CAO (\(5y \geqslant 3x\))

a4B1: CAO (\(11x + 14y \leqslant 7700\))

Notes

In (b), lines must be long enough to define the correct feasible region and pass through one small square of the points stated:

\(x + y = 400\) must pass within one small square of its intersection with the axes – (0, 400) and (400, 0)

\(11x + 14y = 7700\) must pass within one small square of its intersection with the axes – (0, 550) and (700, 0)

\(5y = 3x\) must pass within one small square of (0, 0) and if extended pass through (500, 300)

\(y = 350\) must pass within one small square of (0, 350) and if extended pass through (500, 350)

b1B1: Any two lines correctly drawn

b2B1: Any three lines correctly drawn

b3B1: All four lines correctly drawn

b4B1: Region, R, correctly labelled – dependent on scoring the first three marks in this part

Notes

c1M1: Drawing their objective line (based on their answer to (a)) or its reciprocal – if their line on the graph is shorter than the length equivalent to that of the line from (0, 37.5) to (50, 0) then M0. Line must be correct to within one small square if extended from axis to axis. Their line must have a negative gradient

c1A1: Drawing the correct objective line – same condition that the line must be correct to within one small square if extended from axis to axis

c2A1: The correct V labelled or clearly identified on their graph – note that this mark is dependent on scoring at least B1B1B1B0 in (b) and the two previous marks in this part

Notes

d1M1: Must have scored at least B1B1B0B0 in (b) and candidates must have drawn an objective line (but note that it does not need to be correct but must have negative gradient). Must be solving one of the following three pairs of equations only: \(11x + 14y = 7700,\ y = 350\) or \(11x + 14y = 7700,\ 5y = 3x\) or \(5y = 3x,\ y = 350\). Must be a correct method to solve simultaneous equations and must arrive at \(x = \ldots\) and \(y = \ldots\) but allow slips/errors. This mark can also be awarded for the correct exact coordinates stated with no working provided B1B1B0B0 in (b) and an objective line drawn

d1A1: Correct exact coordinates of V either derived or stated (so no working required) as either \(\left(\dfrac{2800}{11},\ 350\right)\) or \(\left(254\dfrac{6}{11},\ 350\right)\). Note that this mark is dependent on B1B1B1B0 scored in (b) and a correct objective line

Notes

e1B1: CAO in context – so not in terms of \(x\) and \(y\) only – dependent on B1B1B1B0 in (b) and a correct objective line

e2B1: CAO (allow 540.5) – dependent on B1B1B1B0 in (b) and a correct objective line – condone lack of units if given in £ (or 54050p – if given in pence, however, units must be given)

Notes

a1M1: Two correct equations (e.g. \(-2x - 6y + z + r = 40\)) or inequalities

a1A1: CAO (not equations)

Notes

b1M1: Correct pivot located (2 in \(x\) column), attempt to divide row. If choosing negative pivot M0M0

b1A1: CAO pivot row correct including change of b.v. (\(s\) must be changed to \(x\))

b2M1: (ft) All values in one of the non-pivot rows correct or one of the non-zero/one columns correct (that is one of the \(y\), \(z\), \(s\) or value columns correct) following through their choice of positive pivot

b2A1: CAO on all values for first iteration – ignore row operations and b.v. column for this mark

Notes

c1B1: CAO (must be strict inequality)

Notes

d1B1: CAO pivot row correct including change of b.v.

d1M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(y\), \(s\), \(t\) or value) correct following through their choice of pivot from column \(z\)

d1A1: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(y\), \(s\), \(t\) or value) correct

d2A1: CAO – both iterations - all values correct and all eight row operations correctly stated – allow if row operations given in terms of old row 2 – ignore b.v. columns for this mark

Notes

e1M1: Setting any of the expressions in terms of \(k\) from the \(P\) row from their second iteration > or \(\geqslant\) 0 - dependent on all M marks in (b) and (d) and both correct pivot rows (but ignore b.v. label)

e1A1: CAO (oe) – need not be simplified but must not be strict inequality – if no working shown and fully correct award both marks, if no working shown and incorrect M0 A0

Notes

f1B1: CAO

f2B1ft: Follow through their values - dependent on all M marks earned in (b) and (d)

Notes

g1M1: Either 160 or 200 seen (but not as a term in an equation)

g1A1: CAO

Notes

1B1: Expression correct (or \(0.4x + 0.6y + 0.85z\)) together with ‘maximise’ or ‘max’ but not ‘maximum’ – isw if coefficients are subsequently simpified but either \(40x + 60y + 85z\) or \(0.4x + 0.6y + 0.85z\) must be seen at some point for this mark to be awarded

2B1: CAO

1M1: Correct method: \(\dfrac{7}{20}(x + y + z) \bullet x\) where \(\bullet\) is any inequality or =. The bracket must be present or implied by later working. An exact equivalent answer (with or without integer coefficients but with correct inequality sign) with no working can score M1. Accept equivalent fractions or decimals for 7/20 but not 35% (unless later converted to a correct fraction/decimal)

1A1: CAO – answer must have integer coefficients with like terms collected i.e. \(k(13x \geqslant 7y + 7z)\) for any positive integer \(k\) - the correct answer with no working can score M1 A1

2M1: Correct method: \(\dfrac{1}{5}(x + y + z) \bullet z\) where \(\bullet\) is any inequality or = . The bracket must be present or implied by later working. An exact equivalent answer (with or without integer coefficients but with correct inequality sign) with no working can score M1. Accept equivalent fractions or decimals for 1/5 but not 20% (unless later converted to a correct fraction/decimal)

2A1: CAO – answer must have integer coefficients with like terms collected i.e. \(k(x + y \geqslant 4z)\) for any positive integer \(k\) - the correct answer with no working can score M1 A1

3M1: Correct complete method: \(\dfrac{x}{400} + \dfrac{y}{300} + \dfrac{z}{200} \bullet 1\) (oe) where \(\bullet\) is any inequality or =. An exact equivalent answer (with or without integer coefficients but with correct inequality sign) with no working can score M1

3A1: CAO – answer must have integer coefficients with like terms collected i.e. \(k(3x + 4y + 6z \leqslant 1200)\) for any positive integer \(k\) - the correct answer with no working can score M1 A1

Notes

ai1B1: CAO - allow in any equivalent form e.g. \(P - x - 2y - 5z = 0\) but not say \(P = x + 2y + 5z = 0\)

aii1M1: Two inequalities (or equations with slack variables) correct

aii1A1: CAO

Notes

b1M1: Correct pivot located (3 in the \(z\) column), attempt to divide row. If choosing negative pivot then M0M0

b1A1: CAO pivot row correct including change of b.v. (so \(r\) must be changed to \(z\))

b2M1: (ft) All values in one of the non-pivot rows correct or one of the non zero/one columns (\(x\), \(y\), \(r\) or value) correct following through their choice of pivot

b2A1: CAO on all values for the first iteration – ignore row ops and b.v. column for this mark

b3M1: Their correct pivot located following their first iteration, attempt to divide row. If choosing negative pivot M0M0 - however, allow recovery for the third and fourth M marks only if positive pivot chosen for the second iteration after a negative pivot chosen for the first iteration

b3A1ft: Their pivot row correct including change of b.v. following their first iteration

b4M1: (ft) All values in one of the non-pivot rows correct or one of the non zero/one columns (\(x\), \(r\), \(s\) or value) correct following through their choice of pivot

b4A1: CAO for all values and row operations for both iterations - including all eight row operations stated correctly (ignore b.v. column for this mark)

If pivoting on any other positive value for the first iteration then candidates can score in (b) and (c):

(b) M0A0M1A0 M1A1M0A0  (c) M1A0 (so max. of 4/10)

Notes

c1M1: Their correct values stated for at least \(P\), \(x\), \(y\), \(z\) from their ‘optimal’ iteration so there must be no negatives in the profit row. Two M marks in (b) must have been awarded – the numerical value of \(P\) must be explicitly stated and not as part of an equation

c1A1: CAO for all seven values explicitly stated

Notes

a1B1: Any two correct (accept strict inequalities) – accept equivalent inequalities

a2B1: CAO (accept equivalent inequalities)

Notes

b1B1: CAO for both integer coordinates – accept \(x = 4,\ y = 2\), etc.

b1M1: Using simultaneous equations to find the non-integer vertex – must get to \(x = \ldots\) and \(y = \ldots\) Must be a correct method to solve simultaneous equations but allow slips/errors. If no working present then this mark can be awarded for an awrt (11.1, 5.56) or (11.1, 5.55)

b1A1: CAO – must be exact (condone correct recurring decimal notation). If correct answer seen with no working then award M1 A1 in this part. ISW if correct exact answer seen which is then given in non-exact form

Notes

c1M1: Testing all three of their vertices in the correct objective function

c1A1: Correct three values for \(P\) (accept awrt 38.9 for 350/9) and correct optimal vertex either stated or clearly indicated (allow awrt (11.1, 5.56) or (11.1, 5.55))

Notes

d1M1: WAY 1 – point testing: Their attempt at \(\left(\dfrac{100}{9},\ \dfrac{50}{9}\right)\) evaluated in \(Q\) compared to (0, 10) evaluated in \(Q\) - allow any inequality sign or equals

d1A1: \(\lambda \lt 5\) (CAO so allow equals used throughout and then the correct inequality at the end but A0 if incorrect inequality seen in working or if non-exact values used in working)

d2M1: Their attempt at \(\left(\dfrac{100}{9},\ \dfrac{50}{9}\right)\) evaluated in \(Q\) compared to (4, 2) evaluated in \(Q\) – allow any inequality sign or equals

d2A1: \(\lambda \gt -4\) (CAO - see d1A1). Do not award this mark if candidates give both correct answers and then give an answer of \(0 \lt \lambda \lt 4\) or if any additional answers seen

d1M1: WAY 2 – objective line: \(-\dfrac{2}{\lambda}\) compared to either \(-\dfrac{2}{5}\) or \(\dfrac{1}{2}\) or \(-2\) (oe e.g. \(\dfrac{2}{\lambda}\) with \(\dfrac{2}{5}\), \(\dfrac{\lambda}{2}\) with \(\dfrac{5}{2}\), etc.) so correctly comparing the gradient of the new objective line with any one of the three line segments that define \(R\) – accept any inequality or equals

d1A1: \(\lambda \lt 5\) or \(\lambda \gt -4\) (CAO so allow equals used throughout and then the correct inequality at the end but A0 if incorrect inequality seen in working)

d2M1: \(-\dfrac{2}{\lambda}\) compared to both \(-\dfrac{2}{5}\) and \(\dfrac{1}{2}\) (oe) – so correctly comparing the gradient of the new objective with the correct two line segments that give the correct optimal vertex - accept any inequality or equals

d2A1: \(\lambda \lt 5\) and \(\lambda \gt -4\) only (CAO - see d1A1) – note that this mark is dependent on all correct three vertices that define the feasible region and must come from correct comparisons with \(-\dfrac{2}{5}\) and \(\dfrac{1}{2}\). Do not award this mark if candidates give the correct answer and then give an answer of \(0 \lt \lambda \lt 4\) or if any additional answers seen

Note that the correct answers in Way 2 must come from \(-\dfrac{2}{\lambda} \lt -\dfrac{2}{5}\) and \(-\dfrac{2}{\lambda} \gt \dfrac{1}{2}\)

Notes

a1B1: CAO – accept any exact equivalent inequality (isw if simplified incorrectly)

Notes

In (b):

\(7x + 8y = 112\) must pass within one small square of its intersection with the axes – (0, 14) and (16, 0)

\(20x + 65y = 520\) must pass within one small square of its intersection with the axes – (0, 8) and (26, 0)

\(-x + 24y = 24\) must be sufficiently long to define the feasible region and pass within one small square of (0, 1) and (24, 2) if extended

\(x = 2\) must pass within one small square of (2, 0) and (2, 8)

b1B1: Any two lines correctly drawn

b2B1: Any three lines correctly drawn

b3B1: All four lines correctly drawn

b4DB1: Region, R, correctly labelled – not just implied by shading – dependent on scoring the first three marks in this part

Notes

c1B1: CAO - \(k(2x + 3y)\) where \(k \in \mathbb{R}\) – condone equal to \(P\) or equal to a constant

Notes

d1M1: Drawing either the correct objective line or their objective line (based on their answer to (c)) or the reciprocal of the correct objective line or the reciprocal of their objective line – if their line is shorter than the length equivalent to that of the line from (0, 1) to (1.5, 0) then M0. Line must be correct to within one small square if extended from axis to axis

d1A1: Drawing the correct objective line – same condition that the line must be correct to within one small square if extended from axis to axis

d2A1: Correct V labelled clearly on their graph – please note that this mark is dependent on scoring at least B1B1B1B0 in (b) and the two previous marks in this part – by clearly labelled the vertex should either be labelled ‘V’ or circled or clearly distinguishable from the other three (but A0 if not clear e.g. other vertices circled too)

Notes

e1M1: Simultaneous equations being used to find V. Must have scored at least B1B1B0B0 in (b) and candidates must have labelled one of their vertices as V (oe – see above). Must be solving either the pair of simultaneous equations \(20x + 65y = 520\) and \(7x + 8y = 112\) or \(7x + 8y = 112\) and \(-x + 24y = 24\). Must be a correct method to solve simultaneous equations and must arrive at \(x = \ldots\) and \(y = \ldots\) but allow slips/errors. This mark can also be awarded for the correct exact coordinates stated with no working provided B1B1B0B0 in (b) and a vertex labelled as V

e1A1: Correct exact coordinates of the correct V derived with working (not just stated) as either \(\left(\dfrac{624}{59},\ \dfrac{280}{59}\right)\) or \(\left(10\dfrac{34}{59},\ 4\dfrac{44}{59}\right)\) or stated just in terms of \(x\) and \(y\). Note that this mark is dependent on B1B1B1B0 scored in (b) and all three marks in (d). ISW if correct exact values seen followed by decimal approximations

Notes

f1M1: Testing any two of (11, 4) or (9, 5) or (10, 5) or (10, 4) or (11, 5) in a correct objective function or the correct pair of inequalities. Note candidates may reject a point after testing in only one correct inequality which is acceptable – this mark is not dependent on any previous mark

f1A1: CSO (all previous 12 marks must have been awarded) – must have tested (11, 4) in the correct objective function or correct pair of inequalities – accept \(x = 11\) and \(y = 4\) or stated as a pair of coordinates

f1B1: CAO – this mark is not dependent on any previous mark and condone lack of units

Notes

a1B1: e.g. identifies \(x\), refers to basic variable (oe)

Notes

b1M1: Correct pivot located (2 in column \(z\)), attempt to divide row

b1A1: Pivot row correct including change of b.v. (so the \(r\) must be replaced with a \(z\))

b2M1: All values in one of the non-pivot rows correct or one of the non zero-and-one columns (\(y\), \(r\), \(s\) or value) correct following through their choice of pivot from column \(z\)

b2A1ft: Row operations used correctly at least twice, i.e. two of the non zero-and-one columns (\(y\), \(r\), \(s\) or value) correct following through their choice of pivot from column \(z\)

b3A1: CAO – no follow through – all values and row operations correctly stated – allow if row operations given in terms of old row 1 – ignore b.v. column for this mark

Pivoting on the 3 in the \(z\) column (can score a maximum of B1 M0A0M1A1A0 B1 B0B0 – so 4/9)

Notes

c1B1ft: Dependent on the second M mark earned in (b) – must be an equation containing \(P\) (please note that \(P = 13y + 2r - 5s + 27\) is incorrect)

Notes

d1B1: Must have gained both M marks in (b) and must refer to increasing \(y\), \(r\) or \(s\). Do not accept ‘negatives in profit row’ with no further explanation given

d2B1: CAO – dependent on the correct profit equation in (c). Specifically identifies \(s\) as the next variable that could be increased and states ‘not optimal’ (oe)

Notes

a1B1: \(x + y = 8\) correctly drawn. Must pass within one small square of (0, 8), (4, 4) and (8, 0)

a2B1: \(3y = 9 + 2x\) correctly drawn. Must pass within one small square (0, 3), (6, 7) and sufficiently long enough to define the feasible region

a3B1: \(4y = x\) correctly drawn. Must pass within one small square of the origin and (8, 2)

a4B1: \(x = 8\) correctly drawn. Must be sufficiently long enough to define the feasible region. This must be shown as a dashed line or distinctive from the other three lines (see note above)

Notes

b1B1: Region, R, correctly labelled – all lines must have been drawn correctly but condone \(x = 8\) not distinct from the other three lines (so must have scored either B1B1B1B1 or B1B1B1B0 in (a))

Notes

c1B1: Drawing a correct objective line – if their line is shorter than the length equivalent to that of the line from (0, 1) to (1.5, 0) then B0. Line must be correct to within one small square if extended from axis to axis

c1M1: Candidates must have drawn either the correct objective line or its reciprocal. If they have drawn the correct objective line they must be solving \(x + y = 8\) and \(4y = x\). If they have drawn the reciprocal objective line line they must be solving \(x + y = 8\) and \(3y = 9 + 2x\). Must get to either \(x = \ldots\) or \(y = \ldots\) (condone one error in the solving of the simultaneous equations). The correct exact answer \(\left(\dfrac{32}{5},\ \dfrac{8}{5}\right)\), or for the reciprocal (3, 5), can imply this mark

c1dA1: CAO – the correct exact coordinate \(\left(\dfrac{32}{5},\ \dfrac{8}{5}\right)\) or \((6.4,\ 1.6)\) or \(\left(6\dfrac{2}{5},\ 1\dfrac{3}{5}\right)\) - this mark is dependent on the correct objective line seen (so must have scored the B mark). If B1 awarded then the correct answer with no working scores M1A1

Notes

d1B1: CAO or 17.6 or \(17\dfrac{3}{5}\)

Notes

e1B1: CAO vertex (7, 7) (accept \(x = 7,\ y = 7\))

e2B1: CAO value (35)

Notes

f1M1: \((k =)\ \dfrac{5}{3}\) or \(\dfrac{3}{5}\) or 1.6 or 0.6 or \(1\dfrac{2}{3}\)

f1A1: CAO \((k =)\ \dfrac{5}{3}\) (oe)

Notes

a1M1: Correct pivot located (4 in column \(x\)), attempt to divide row

a1A1: Pivot row correct including change of b.v.

a2M1: All values in one of the non-pivot rows correct or one of the non zero and one columns (\(y\), \(z\), \(s\) or value) correct following through their choice of pivot from column \(x\)

a2A1ft: Row operations used correctly at least twice, i.e. two of the non zero and one columns (\(y\), \(z\), \(s\) or value) correct following through their choice of pivot from column \(x\)

a3A1: CAO – no follow through – all values and row operations correctly stated – allow if row operations given in terms of old row 2 – ignore b.v. column for this mark

Pivoting on the 1 in the \(x\)-column

Pivoting on the 2 in the \(x\)-column

Notes

b1B1ft: Follow their profit equation from (a) dependent on scoring both M marks in (a)

b2B1: CAO (no follow through) for slack variables (\(r = 5\), \(s = 0\), \(t = 3\))

Notes

1B1: CAO – expression correct and ‘minimise’.

2B1: CAO

1M1: Correct method – must see \(\frac{1}{4}(x + y) \blacksquare x\) where \(\blacksquare\) is any inequality or =. The bracket must be present or implied by later working.

1A1: CAO – simplified – answer must have integer coefficients.

2M1: Correct method – one of \(2x \blacksquare y\) or \(x \blacksquare 2y\) where \(\blacksquare\) is any inequality or =.

2A1: CAO – answer must have integer coefficient.

Notes

In (a) lines must pass through one small square of the points stated:

\(7x + 8y = 840\) passes through (0, 105), (40, 70), (80, 35), (120, 0)

\(4y = 5x\) passes through (0, 0), (40, 50), (80, 100)

\(5y = 3x\) passes through (0, 0), (50, 30), (100, 60)

a1B1: One line other than \(x = 25\) or \(y = 25\) correctly drawn.

a2B1: Two lines other than \(x = 25\) or \(y = 25\) correctly drawn.

a3B1: All five lines correctly drawn.

a4B1: Region, R, correctly labelled – not just implied by shading – must have scored all three previous marks in this part.

Notes

b1M1: Drawing the correct objective line or its reciprocal. Line must be correct to within one small square if extended from axis to axis.

b1A1: Correct objective line.

b2A1: V labelled clearly on their graph. This mark is dependent on the correct five line segments that define the boundary of the feasible region.

Notes

cM1: Simultaneous equation being used to find their V (but not from \(x = 25\) or \(y = 25\)). Must get to \(x = \ldots\) and \(y = \ldots\)

cA1: Correct coordinates of V stated exactly as \(\left(\dfrac{840}{17},\ \dfrac{1050}{17}\right)\) or \(\left(49\dfrac{7}{17},\ 61\dfrac{13}{17}\right)\). If the correct coordinates are stated exactly with no working then this scores M1A0.

Notes

d1M1: Testing the correct inequalities for at least three of (49, 61), (49, 62), (50, 61), (50, 62).

d1A1: CAO (50, 61).

Notes

a1M1: Correct pivot located, attempt to divide row. If choosing negative pivot no marks.

a1A1: Pivot row correct including change of b.v.

a1B1: Row operations CAO – allow if given in terms of old row 2.

a2M1: (ft) Correct row operations used at least once, column \(x\), \(z\), \(s\) or value correct.

a2A1: CAO on numbers (ignore row operations and b.v.).

a2B1: Correct pivot located and b.v. changed. If choosing negative pivot 2B0 3M0.

a3B1: Row operations CAO.

a3M1: (ft) Correct row operations used at least once, column \(x\), \(r\), \(s\) or value correct.

a3A1: CAO on numbers (ignore row operations and b.v.).