# Mark Scheme (Results) November Pearson Edexcel GCSE in Mathematics Linear (1MA0) Higher (Non-Calculator) Paper 1H

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1 Mark Scheme (Results) November 2013 Pearson Edexcel GCSE in Mathematics Linear (1MA0) Higher (Non-Calculator) Paper 1H

3 NOTES ON MARKING PRINCIPLES 1 All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. 2 Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. 3 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. 4 Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. 5 Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. 6 Mark schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows: i) ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear Comprehension and meaning is clear by using correct notation and labeling conventions. ii) select and use a form and style of writing appropriate to purpose and to complex subject matter Reasoning, explanation or argument is correct and appropriately structured to convey mathematical reasoning. iii) organise information clearly and coherently, using specialist vocabulary when appropriate. The mathematical methods and processes used are coherently and clearly organised and the appropriate mathematical vocabulary used.

5 11 Linear equations Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded. 12 Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another. 13 Range of answers Unless otherwise stated, when an answer is given as a range (e.g ) then this is inclusive of the end points (e.g 3.5, 4.2) and includes all numbers within the range (e.g 4, 4.1) Guidance on the use of codes within this mark scheme M1 method mark A1 accuracy mark B1 Working mark C1 communication mark QWC quality of written communication oe or equivalent cao correct answer only ft follow through sc special case dep dependent (on a previous mark or conclusion) indep independent isw ignore subsequent working

6 M1 for 6 4 (= 1.5) or 4 6 (= ) or 4 6 oe or sight of any one of the correct answers A1 for three correct A1 for all correct 2 (a) Plot (90,17) 1 B1 cao (b) Positive 1 B1 Positive (c) In range 16 to 20 2 M1 for a single straight line segment with positive gradient that could be used as a line of best fit or a vertical line from 110 or a point plotted at (110, y) where y is in the range 16 to 20 A1 for an answer in the range 16 to 20 inclusive cm 3 4 M1 for M1 (dep) for A1 for 120 B1 (indep) for cm 3

7 4 (a) 4y + 5x M1 5x or 5 seen (b) 3x(3x 2y) 2 B2 for 3x(3x 2y) (B1 for x(9x 6y) or 3(3x 2 2xy) or 3x(ax by) where a and b are integers not equal to zero) (c) 4x B1 cao (d) x 2 2x 15 2 M1 for 4 terms correct with or without signs or 3 out of no more than 4 terms correct with correct signs 5 (a) B1 oe (b) M1 for oe 6 (a) Shape with vertices at ( 1, 3), (0, 6), (2, 6), (1, 3) 1 B1 for correct shape in correct position (b) Rotation centre (0,0) 90 anticlockwise 3 B1 rotation B1 (centre) (0,0) B1 90 anticlockwise or 270 clockwise Note: award no marks if more than one transformation is given

8 7 (i) 20, 40, 60 12, 24, 36, 48, 60 3 and 5 or any multiple of 3, 5 4 M1 attempts multiples of both 20 and 12 (at least 3 of each shown but condone errors if intention is clear) or identifies 60 or a multiple of 60 M1 (dep on M1) for a division by 20 or 12 or counts up multiples or identifies a common multiple (implied if one answer is correct or answers reversed) A1 cheese slices (packets) 3, burgers (boxes) 5 or any multiple of 3, 5 OR 20 = 4 5 = = 4 3 = M1 for expansion of either 20 or 12 into factors M1 for demonstration that both expansions include 4 (or 2 2) for cheese slices (packets) 3, burgers (boxes) 5 (ii) 60 B1 for 60 or ft from their correct answer in (i) or ft common multiple M1 3x 5 = 19 x M1 for a correct operation to collect the x terms or the number terms on one side of an equation of the form ax + b = cx + d A1 for x = 6 M1 for substituting their value of x in the three expressions and adding or substituting their value of x after adding the three expressions

9 9 (a) Criticisms 2 B1 Qu 1 Overlapping boxes, no units B1 Qu 2 e.g. no time frame, non-specific responses, no number quantities, open to interpretation, no option for those who do not exercise (b) Question given 2 B1 for a correct question with a time frame B1 for at least 3 correctly labelled non-overlapping response boxes (need not be exhaustive) or at least 3 response boxes that are exhaustive for all integer values of their time unit (could be overlapping) NB Units must be included in either question or response boxes to score full marks [Do not allow inequalities in response boxes]

10 *10 Not enough, needs M1 for splitting the shape (or showing recognition of the absent rectangle) and using a correct method to find the area of one shape M1 for a complete and correct method to find the total area M1 for a complete method to find 70% of 19 (= 13.3) or 70% of their total cost or 70% of their area A1 114(m 2 ) and ( )133 or 114(m 2 ) and ( )13.3(0) and 108(m 2 ) C1 (dep on M2) for a conclusion supported by their calculations OR M1 for a complete method for the number of tins required for one section of the area of the floor M1 for a complete method to find the number of tins for the whole floor M1 for a complete method to find 70% of their total number of tins and multiply by 19 A1 ( )133 C1 (dep on M2) for a conclusion supported by their calculations M1 200 (3+2) (= 40) or an equivalent ratio seen M1 (dep) 3 40 (= 120) or 2 40 (= 80) or 120: 80 or 80:120 M1 a complete method to find 70% of their total number of large letters e.g (=56) M1 multiplies their three totals by the correct unit price and adds, e.g. 60(p) ( ) ( ) A1 164

11 12 y = 3x B1 for axes scaled and labelled x drawn (Table of values) y M1 for at least 2 correct attempts to find points by substituting values of x M1 ft for plotting at least 2 of their points (any points from their table must be correctly plotted) A1 for correct line between x = 2 and x = 2 (No table of values) M1 for at least 2 correct points with no more than 2 incorrect points M1 for at least 2 correct points (and no incorrect points) plotted OR line segment of y = 3x + 2 drawn A1 for correct line between x = 2 and x = 2 (Use of y = mx + c) M1 for line drawn with gradient of 3 OR line drawn with y intercept at 2 M1 for line drawn with gradient of 3 AND with y intercept at 2 A1 for correct line between x = 2 and x = 2 SC B2 (indep of B1) for correct line segment between x = 0 and x = 2 (ignore any additional incorrect line segment(s))

12 = = = M (= 350) or (= 363) M1 (dep) finding the difference in their totals e.g OR 10 (35 33) = = (a) 1 5 (b) 1 9 OR 1 B1 oe 1 B1 cao M1 10 (35 33) (=20) or 11 (35 33) (=22) M1 (dep) or (c) M oe or

13 15 6x + 8y = 10 6x 9y =27 y = 1 3x 4 = 5 3x = 9 x = 3 x = 3, y = 1 4 M1 for a correct process to eliminate either variable (condone one arithmetic error) for either x or y M1 (dep on M1) for correct substitution of found value into one of the equations or appropriate method after starting again (condone one arithmetic error) OR 9x + 12y = 15 8x 12y = 36 x = y = 5 4y = 4 y = 1 OR M1 for full method to rearrange and substitute to eliminate either variable (condone one arithmetic error) for either x or y M1 (dep on M1) for correct substitution of found value into one of the equations or appropriate method after starting again (condone one arithmetic error) Trial and improvement scores 0 marks unless both x and y are correct = = = (3,6,7) to ( 2,2,5) ( 5, 4, 2) ( 2 5, 2 4, 5 2) M (= 6) oe, can be implied by M ( 7, 2, 3) 2 M1 for midpoint plus change or complete method for 2 out of 3 coordinates, can be implied by 2 correct values

14 18 (a) 68 1 B1 cao *(b) = = = Yes as 28 > 20 or 35% > 25% or 53 < 60 3 M1 for reading a value from graph at time = 60 (=28, accept 27 to 28) M1 for (= 35) or (= 20) C1 (dep on M2) for correct decision based on their figures OR M1 for (= 20) M1 for reading a value from graph at cf = 20 (=53, accept 52 to 54) C1 (dep on M2) for correct decision based on their figures (c) 28, 53, 68, 76, 96 Box plot plotted 3 B1 for ends of whiskers at 28 and 96 with a box B1 ft for median at 68 inside a box B1 for ends of box at 53 (accept 52 to 54) and M1 for (= 0.3) or (= 0.6) M1for A1 for 0.82 oe OR M1 for (= 0.3) or (= 0.6) M1 ( ) + ( ) + ( ) A1 for 0.82 oe

15 20 (a) 4 3 M1 for correct expansion to 32x 8 or multiplying both sides by 3x or dividing both sides by 4 M1 for a compete and correct method to isolate the x terms and the number terms (condone one arithmetic error in multiplying out the bracket) (b) 2( y 6) ( y+ 3) ( y+ 3)( y 6) y 15 ( y+ 3)( y 6) 3 M1 for common denominator of ( y+ 3)( y 6) 2( y 6) y+ 3 M1 for oe ( y+ 3)( y 6) ( y+ 3)( y 6) 2( y 6) ( y+ 3) or oe ( y+ 3)( y 6) y 15 y 15 A1 for or 2 ( y+ 3)( y 6) y 3y M1 y = kx 2 oe or 36 = k 3 2 A1 k = 4 M1 (dep on M1) (y =) 4 5 2

16 * y y y M1 ADC = 2 A y C2 (dep on M1) for both reasons Angle at centre is twice the angle at the circumference Opposite angles in cyclic quadrilateral add to 180 (C1 (dep on M1) for one appropriate circle theorem reason) OR M1 reflex AOC = 360 y 360 y A1 oe 2 C2 (dep on M1) for both reasons Angles around a point add up to 360 Angle at centre is twice the angle at the circumference (C1 (dep on M1) for one appropriate circle theorem reason) 23 Triangle with vertices at ( 1, 4), ( 1, 5), ( 3, 4.5) 2 M1 for correct shape and size and the correct orientation in the wrong position or two vertices correct

17 24 (a) AB = a + b 1 ON = OA AB 3 a b 3 M1 for correct vector equation involvingon, eg. ON = OA + AN, may be written, partially or fully, in terms of a and b, e.g. ( ON = ) ON = a + 2 ( a + b) 3 = 1 3 a b OR ON = OB BA ON = b + 1 ( b + a) 3 = 1 3 a+ 2 3 b a AB M1 for showing answer requires AN = A1 1 3 a b oe 2 AB or BN = BA (b) OD =OA +AC +CD = a + b + b = a + 2b OD = 3( 3 1 a+ 3 2 b ) OD = 3 ON Proof 3 M1 for a correct vector statement for OD or ND in terms of a and b, e.g. OD = a + b + b oe or 2 ND = ( b + a) + b + b oe 3 A1 for correct and fully simplified vectors for ON (may be seen in (a)) and for OD (= a + 2b) or ND (= 2 3 a b) C1 (dep on A1) for statement that OD or ND is a multiple of ON (+ common point)

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