Pearson Edexcel GCSE in mathematics november 2021, latest complete ,marking scheme. Download for HIGHER GRADES

Document Content and Description Below

Guidance on the use of abbreviations within this mark scheme M method mark awarded for a correct method or partial method P process mark awarded for a correct process as part of a problem solving qu... estion A accuracy mark (awarded after a correct method or process; if no method or process is seen then full marks for the question are implied but see individual mark schemes for more details) C communication mark awarded for a fully correct statement(s) with no contradiction or ambiguity B unconditional accuracy mark (no method needed) oe or equivalent cao correct answer only ft follow through (when appropriate as per mark scheme) sc special case dep dependent (on a previous mark) indep independent awrt answer which rounds to isw ignore subsequent workingPaper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 1 (a) x > ‒1 B1 cao (b) Diagram drawn C2 for a fully correct diagram, eg ‒5 ‒4 ‒3 ‒2 ‒1 0 1 2 3 4 5 (C1 for drawing a line from −3 to 4 or (indep) for an open circle at 4 or (indep) for a closed circle at −3 ) Condone arrow heads or line ending to denote the ‘end’ of the line 2 (a) 12 M1 for a correct factor tree for either 60 or 84 with no more than one arithmetic error or for listing factors of 60 or 84, at least 4 correct for either (with no more than 1 incorrect in either list), could be in factor pairs or for the prime factors of 60 (2, 2, 3, 5) or 84 (2, 2, 3, 7) Condone the use of 1 in any factor tree 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 A1 for 12 or 2×2×3 oe SC B1 for answer of 4 or 6, if M0 scored 2,2,3 is not enough, it must be a product (b) 120 M1 for a correct factor tree for either 24 or 40 with no more than one arithmetic error or for at least 3 multiples of both 24 and 40 (can include 24 and 40) or for the prime factors of either 24 (2, 2, 2, 3 ) or 40 (2, 2, 2, 5) or for a common multiple from their lists ( ≠ 120 ) Condone the use of 1 in any factor tree 24: 24, 48, 72, 96, 120, … 40: 40, 80, 120, … For the list not containing 120, accept the first 3 correct multiples or one error in the first 4 multiples A1 for 120 or 2×2×2×3×5 oePaper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 3 (a) 80 M1 for a complete method eg 20 15 × 60 or 20 × 4 or 20 ÷ ଵ ସ A1 cao (b) Travel graph M1 for method to find distance travelled in last 20 minutes, eg 75 × 20 60 (= 25) Can be implied by a distance of 25km drawn on the graph C2 for a fully correct travel graph (C1 for horizontal straight line from (10 15, 20) to (10 25, 20) or for a line of the correct length and gradient to indicate a speed of 75km/h eg a straight line from (10 25, 20) to (10 45, 45)) 4 (a) (10), 5, (2), 1, 2, (5), 10 B2 for all 4 values correct (B1 for 2 or 3 correct values) (b) Graph M1 ft (dep on B1) for plotting at least 5 of their points correctly A1 for a fully correct curve drawn Accept a freehand curve drawn that is not made of line segments (c) ‒0.65 to ‒0.8 and 2.65 to 2.8 M1 for y = 4 drawn or intersection with y = 4 or y = x2 ‒ 2x ‒ 2 drawn or 1 correct value ft a quadratic graph If answers stated as coordinates, award M1 for both coordinates and M0 for one coordinate A1 ft a quadratic graph or for answers in the range 2.65 to 2.8 and ‒0.65 to ‒0.8Paper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 5 41.6 P1 for start of process to find the length of the hypotenuse, eg (hyp2 =) 82 + 102 (= 164) Note lengths may be seen on the diagram P1 for complete process to find hypotenuse, eg 8 10 2 2  or 64 100  or 164 (= 12.8…) P1 (dep P2) for complete process to find the required perimeter, eg 8 + 8 + 10 + “12.8” + “12.8 – 10” or 16 + 4√41 8 + 8+ “12.8” + “12.8” oe is acceptable for this mark A1 for answer in the range 41 to 42 If an answer in the range 41 to 42 is given in the working space then incorrectly rounded, award full marks. 6 (a) 17.8 M1 for tan 56 = 12 x or (BC) =12 × tan 56 oe or alternative method to find BC For any alternative method candidates must arrive at an equation with BC as the only unknown A1 for an answer in the range 17.7 to 17.8 If an answer in the range 17.7 to 17.8 is given in the working space then incorrectly rounded, award full marks. (b) 33.6 M1 for cos x = 15 18 or cos x =0.83.. or x = cos-1 15 18 or alternative method to find x For any alternative method candidates must arrive at an equation with x as the only unknown A1 for an answer in the range 33.5 to 33.91 If an answer in the range 33.5 to 33.91 is given in the working space then incorrectly rounded, award full marks.Paper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 7 1.6 P1 for 1.8 × 80 (= 144) or 1.2 × 40 (= 48) or for 192 or for 80 : 40 = 2 : 1 P1 for (“144” + “48”) ÷ (80 + 40) or 192 ÷ 120 or for (1.8 × 2 + 1.2) ÷ 3 or 4.8 ÷ 3 A1 oe 8 Error in inequalites C1 for identifying incorrect inequalities Acceptable examples gives at least one correct inequality eg (10 < t ≤ 20) should be 0 < t ≤ 20 it should be t ≤ 20 (all) inequalities should start with 0 should start with 0 Not acceptable examples 10 < t ≤ 20 is wrong the numbers have been added wrongPaper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 9 (a) 138 M1 for upper quartile = 188 or lower quartile = 50 or an indication that they are trying UQ ‒ LQ Could be written on the grid A1 cao (b) Yes, with reason C1 Yes, with reason Acceptable examples Yes, because the median is at 2 hour (120 min) Yes, since 50% is at the 2 hour mark Yes, because the middle is at 2 hours Not acceptable examples No ….. The median is at the 2 hour mark Yes, because 50% is exactly half way between “188” and “50” (c) statement C1 Acceptable examples The median is lower on Tuesday (higher on Monday) The upper quartile is lower on Tuesday (higher on Monday) There may just have been one person waiting for 210 mins We don’t know how many people were waiting for each time Not acceptable examples The range is bigger for Tuesday (smaller for Monday) The IQR is smaller for Tuesday (bigger for Monday) M T Shortest time 20 20 Lower quartile 50 50 Median 120 100 Upper quartile 188 140 Longest time 200 210 Range 180 190 IQR 138 90Paper: 1MA1/2H Question Answer Mark Mark scheme Additional guidance 10 344 580.48 P1 for a start to the process to find the initial investment eg 344 605 ÷ 1.025 oe (= 336 200) or for 1.0253 (= 1.07689….) P1 for complete process to find original investment, eg 344 605 ÷ 1.0253 oe (= 319 078 to 320 265) P1 for [initial investment] × 1.022 × 1.035 oe [initial investment] must be clearly what they believe to be that and cannot be 344605 A1 for answer in the range 343 587 to 344 581 11 (a) (9, 7.5) M1 for x coordinate = PO (6) × 32 (=9) or y coordinate = OQ (3) × 52 (= 7.5) or PO (6) × 52 (=15) or OQ (3) × 32 (= 4.5) A1 cao (b) y = ̶2x + 3 P1 for process to find the gradient of the line, eg 3 ÷ 6 (=0.5) or y = mx + 3 Could use P and R or [Show More]

Last updated: 1 year ago

Preview 1 out of 26 pages