1H 1 Mark Scheme Specimen Paper Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 1H 3 2 1 50, 100, 150, 200, 250, 300, 350, 400 and 80, 160, 240, 320, 400 OR 2 × 5 × 5 and 2 × 2 × 2 × 2 × 5 2 × 2 × 2 × 2 × 5 × 5 or 400 b 245 = 5 × 7 × 7 140 = 2 × 2 × 5 × 7; 120 ÷ 1002 (=0.012) or 810 ÷ 120 (=6.75) 810 ÷ “0.012” or “6.75” × 1002 Working 5400 ÷ (5 + 3 + 4) (=450) “450” × 5 or “450” × 3 or “450” × 4 a Q A1 M1 M1 A1 2 3 35 16 40 M1 3 3 2250, 1350, 1800 67 500 Mark Answer Notes Total 3 marks Total 3 marks LCM found or 4 40 pm Total 5 marks or lists at least 3 factors of each number (other than 1 and the number) (1, 2, 4, 10, 14, 35, 70, 140) (1, 5, 7, 35, 49, 245) M1 M1 A1 M1 M1 A1 Apart from Questions 4d, 10 (where the mark scheme states otherwise), the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method. International GCSE Maths 1H 2 5 4 d c a b s 1 2 2s 2 t or t a 2 a 5 3 1 or 100% − 80% (=20%) or 8 8 1 3 23 5 8 40 4 3 5 1 23 " " or " " or 1 - " " 5 8 8 5 40 6x – 5 = 2(x + 1) or 6x – 5 = 2x + 2 6x – 2x = 2 + 5 2s = at2 or m + 9m – 5m − 45 2 1.75 17 40 3 M1 M1 A1 A1 2 2s a A1 M1 2 m2 + 4m − 45 t = (±) B2 M1 2 3y(2y + 5) 3 oe eg. Total 3 marks may see decimal or percentage equivalents M1 A1 may see decimal equivalents M1 7 dep on at least M1 scored 4 Total 9 marks for a correct first step B1 for 3(2y2 + 5y) or y(6y + 15) M1 for 3 terms correct or 4 terms correct ignoring signs or m2 + 4m + ….. or …. + 4m – 45 3 8 7 6 a b c b a 6630 ( 78) 85 55 000 000 1.382 × 109 − 1.327 × 109 oe or 42 × 7 (=294) or 8 × 50 (=400) 8 × 50 − 42 × 7 6630 ÷ 85 × 100 or “78” × 100 6630 = 85% oe or 0.03 × 180 000 (=5400) “5400” + 180 000 5.5 × 107 93 000 000 Singapore 106 7800 185 400 B1 B1 M1 A1 2 M1 M1 A1 M1 A1 1 1 3 3 3 M1 M1 A1 M1 Total 3 marks Total 6 marks M2 for 6630 ÷ 0.85 M2 for 1.03 × 180 000 Total 4 marks or for 5.5 × 10n n ≠ 7 dep dep 4 10 9 y = −1.5 −26y = 39 21x + 35y = 21 21x + 9y = 60 e.g. (h = ) or or 26x = 91 x = 3.5 9x + 15y = 9 10 14.5 35x + 15y = 100 14.52 102 (=10.5) or 10 (x = ) cos 1 (=46.3…) 14.5 1 e.g. 20 "10.5" or 2 1 20 14.5 sin("46.3...") 2 e.g. (h2 = ) 14.52 102 or cos x = x = 3.5, y = −1.5 105 4 4 A1 one variable correct dep on M1 M1 (dep on M1) for method to find second variable A1 both variables dep on at least M1 Total 4 marks M1 for method to eliminate one variable (condone one arithmetic error) Total 4 marks (dep on M1) method to find area M1 cao complete method to find height or angle M1 A1 start to find height or angle M1 5 median = 26 or LQ = 20 or UQ = 29 iqr = 9 and median = 26 (1 + 4)2 (= 25) “5”2 × 8 – 8 oe 11 12 192 median = 26; iqr = 9 and two comparisons 4 B1 M1 A1 B1 3 M1 M1 A1 complete method Total 3 marks ft comparison of IQR eg. English results were more spread out ft comparison of median eg. Maths results were higher NB. In order to award both marks, at least one of the comparisons must be in context Total 4 marks 6 7 1 8 4 1 3 8 8 e.g. 7 = “ ”× 4 + c or y – 7 = “ ”(x – 4) 3 3 11 c=− 3 8 11 y =” ”x – “ ” 3 3 Angle DBA = 43o or Angle DAB = 90o 180 – 90 – 43 (=47) 13 14 47 with reasons 5 8x – 3y = 11oe M1 M1 A1 B2 Total 4 marks dep on M2 M1 A1 for a method to find c M1 for full reasons Angles in the same segment are equal; Angle in a semi-circle is a right angle Angles in a triangle add up to 180o (B1 for a correct and relevant reason using a circle theorem) Total 5 marks 4 for a method to find gradient M1 7 16 15 b a 3 × (3 + 8.5) = 5 × PR or 3 × (3 + 8.5) = 5 × (5 + PQ) (3 × (3 + 8.5)) ÷ 5 − 5 2 k or k = 0.8 0.0064 k q 0.8 20 10 P 0.8 q 1.9 0.0016oe P 3 2 3 M1 M1 A1 A1 M1 A1 M1 M1 Total 3 marks for a complete method for PQ Total 5 marks oe with P as the subject implies first M1 8 18 17 b a -4 -3 -2 -1 0 1 4 63 or 62 t or 62 10 2 3 1 4 63 + 62 10 2 3 1 4 62 t = 63 + 62 10 2 3 -5 (x – 5)(x + 1) x2 + 2x – 6x – 5 > 0 or x2– 4x – 5 > 0 1 2 3 4 7 x < −1, x > 5 5 x 4 3 1 Total 4 marks for forming a correct equation M1 A1 for total volume M1 Total 4 marks for one expression for an appropriate volume ft from (a) M1 M1 M1 A1 B1 9 20 19 4 for simplified answer, may be factorised for completion of proof Total 4 marks proof M1 A1 8n + 8 (4n2 + 6n + 6n + 9) – (4n2 + 2n + 2n + 1) M1 Total 4 marks (2n + 3)2 – (2n + 1)2 = A1 for algebraic representation of two consecutive odd numbers for correct expansion of at least one bracket 4 M1 22 M1 for finding area of 50 – 55 bar M1 for method to find total area (condone two errors) M1 (dep on M2) for complete method eg. 2n + 1, 2n + 3 "44" 150 "300" e.g. 8.8 × 5 (=44) e.g. 3.4 × 10 (=34) + 7.2 × 20 (=144) + 8.8 × 5 (=44) + 4 × 15 (=60) + 1.8 × 10 (=18) (=300) 10 21 x 3 5 x = 3 32 4 or (x – 3)2 = 5 6 (6)2 4 1 4 x or 2 1 Correct graph 6 A1 M1 (x − 3)2 − 5 or x = 3 eg. B2 A1 M1 dy = 2x – 6 or (x − 3)2 − 32 + 4 or dx '3 5 ' '3 5 ' 2 e.g. for fully correct labelled graph (see end of mark scheme) (B1 for parabola intercepting y-axis at (0, 4)) Total 6 marks or exact equivalents (ft providing previous M1 scored) for (3, −5) for correct method to find roots 11 23 22 31 (31) 4 210 2 AB = 180 sin(180 55 "25.6") sin 55 CAB = 25.6(1…) sin CAB sin 55 95 180 95sin 55 CAB = sin 1 or 180 eg. diagram drawn showing relative positions of A, B and C can be implied by angle ABC = 55o e.g. (x – 21)(x – 10) = 0 or 2 e.g. x2 – 31x + 210 = 0 or 7x2 – 217x + 1470 = 0 7 x7 or x x 1 7 x7 7 2 x x 1 15 217 21 5 5 21 must be selected as final answer Total 5 marks A1 Total 5 marks 1802 952 2 180 95 cos(180 55 "25.6" M1 dep or for M1 dep M1 M1 interprets information A1 correct quadratic ready to solve method to solve quadratic equation for equation formed M1 M1 M1 for a correct expression for one probability M1 12 13 May 18 Mark Scheme (Results) Summer 2018 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 1H b “56” ÷ 40 0.5 × 19 + 1.5 × 12 + 2.5 × 5 + 3.5 × 2 + 4.5 × 2 (=56) or 9.5 + 18 + 12.5 + 7 + 9 (=56) 1.4 4 A1 for 1.4 or 1 2 5 M1 dep on at least M1 f provided Allow division by their addition or total under column seen M2 for at least 4 correct products added (need not be evaluated) If not M2 then award M1 for consistent use of value within interval (including end points) for at least 4 products which must be added OR correct mid-points used for at least 4 products and not added International GCSE Maths 4MA1 1H Apart from questions 3c, 11b and 20 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an incorrect method, should be taken to imply a correct method. Question Working Answer Mark Notes a 0<p≤1 1 B1 1 14 2 Question 2 M1 “1275” ÷ 5 × 2 or 3 × 170 " (girls = ) A1 M1 M1 dep on M2 5 15 "170 (=1275) or G : A = 2 : 6 oe 2 510 M1 award of this mark implies the first M1 2 (of children) 9 2 3 2 (of total) 9 5 15 2 3 3 2 2 or G : C : A = : : : 3: 2 9 5 5 5 3 (girls = ) A1 M1 “255” × 2 or “1275” – “765” or “1275” ÷ 5 × 2 Alternative scheme M1 dep on M2 Notes “765” ÷ 3 (=255) or "765" ÷ 3 × 5 (=1275) Mark 5 M1 M1 award of this mark implies the first M1 510 Answer 7 × “85” + 170 (=765) or 9 × “85” (=765) or “595” + 170 (=765) or 170 × “3.5” + 170 (=765) Working 170 ÷ 2 (=85) or 170 ÷ 2 × 7 (=595) or 7 ÷ 2 (=3.5) 15 4 19 oe 2 Yes with reason 10 oe 12 c 1, 3, 5, 7, 9, 10, 11 b (ii) 1, 2, 3, 4, 6, 12 4, − 6 (ii) a (i) (x – 4)(x + 6) 5x + 15 = 3x – 4 or x + 3 = 3x 4 5 5 e.g. 5x – 3x = −4 – 15 y14 16m12 Answer Working d (i) c a b Question 3 2 1 1 1 1 2 3 Mark 1 2 A1 M1 A1 M1 B1 B1 B1 B1 a with a < 12 or 12 for 10 oe or 0.83(3…) or 83(.3..)% 12 10 and 12 used with incorrect notation E.g. 10 : 12 for 12 – 2 (=10) or e.g. no numbers in both A and C or A and C do not intersect or A and C do not overlap or A and C are mutually exclusive cao cao or ft from any (x + p)(x + q) cao dep on at least M1 for (x + a)(x + b) where either ab = −24 or a + b = +2 e.g (x – 6)(x + 4) M1 A1 ft from ax + b = cx + d for correctly isolating terms in x on one side of equation and constant terms on the other side M1 Notes if not B2 then B1 for am12 or 16mb or 24m12 b ≠ 0, 12 a ≠ 1, 16 for removing bracket in a correct equation or dividing all terms by 5 in a correct equation B1 B2 16 6 b Question a 5 "106.34" (=10.3…) 2 π × “10.3…” or 2 × π × "10.3..." 9.72 3.52 or 9.72 + 3.52 (=106.34) 0.5 × 105 – 8 or 0.0005 or 5 × 10n or 5.0 × 10n Working 32.4 5 × 10-4 Answer 80 000 4 2 Mark 1 A1 M1 M1 M1 A1 M1 B1 3.5 (=10.3…) cos 70.2 for answer in range 32.3 – 32.41 dep on M2 eg MN= M1 for a complete method to find MN eg cos70.2= 3.5 MN 1 1 or 2000 2 103 M1 for the use of MN and a correct angle (70.1… or 70.2, 19.8…) in a correct trig statement SC : B1 for for 5 × 10-4 or 5.0 × 10-4 Notes 17 b Question a 7 E.g. 252 000 ÷ 1.05 160 000 – “6400” – “6144” − “5898.24” 4 × (160 000 – “6400”) (= 6144) 100 4 × (160 000 – “6400” – “6144”) (= 5898.24) 100 4 × 160 000 oe (=6400) 100 Working 240 000 Answer 141 558 3 3 Mark A1 M2 A1 M1 M1 If not M2 then award M1 for 160 000 ×0 .96 (=153 600) or 160 000 × 0.962 (=147 456) M2 for 160 000 × 0.963 or 160 000 × 0.964 (=135 895.44..)) NB: An answer of 239 400 scores M0 M0 A0 If not M2 then M1 for x × 1.05 = 252 000 or 252 000 ÷ 105 oe SC If no other marks gained, award B1 for 160 000 × 0.12 oe (=19 200) or 160 000 × 0.88 oe (=140 800) or an answer of 140 800 or an answer of in the range 179 978 – 179 978.24 accept (1 – 0.04) in place of 0.96 throughout for 141 557.76 - 141 558 for a complete method (condone 4 years rather than 3) Notes 18 b (ii) Question a (i) 8 Working 2 1 23 × 35 × 5 × 74 4, 2, 1 Mark 1 Answer 3 × 73 A1 M1 B1 B1 for r = 1 or for p = 4 and q = 2 or correct representation of C in terms of prime factors on a Venn diagram for 23 × 35 × 5 × 74 oe or 23 337 720 Notes for 3 × 7 oe or 1029 3 19 Question 9 12.8 or sin 72 12.8 or cos(90 72) "13.4..."2 12.82 12.8 or 12.8tan(90 – 72) tan 72 12.8 or cos(90 72) or 12.8tan(90 – 72) or 5 × (“13.4(58…)” – “4.15(89…)” ) + 5 × 12.8 or 5 × (“13.4…” + “4.15…” + 12.8) – 10 × “4.15…” 12.82 "4.15..."2 13.4(58…) or 13.5 or One of (hypotenuse = ) or 4.15(89…) or 4.16 or AND One of (shortest side = ) 13.4(58…) or 13.5 sin 72 (hypotenuse = ) 12.8 or 4.15(89…) or 4.16 or tan 72 E.g.(shortest side) = 12.8 Working E.g. tan 72 12.8 or tan(90 72) o or a 12.8 12.8 12.8 or cos(90 72) sin 72 h h Answer 110 Mark 5 for method to use found lengths to find perimeter for answer in range 110 - 111 M1 A1 for a complete method to find both missing sides of triangle NB Could use Pythagoras’s theorem with side found – must be a complete correct method for a complete method to find one side of the triangle M1 M1 Notes substitutes correctly into a trig ratio (including the Sine rule) M1 20 b Question a 10 80 Reading from graph from cf = 52 (=118) or Reading from graph from time = 120 (=55) Alternative scheme 0.65 × 80 (=52) "55" 100 oe (= 68(.75)) 80 or 0.35 × 80 (=28) or e.g. 80 "55" 100 oe (=31(.25)) Reading from graph from time = 120 (=55) or 80 – 55 (=25) Working Readings from graph at cf 20 and cf 60 eg. readings of 103 and 123 No with correct figures No with correct figures Answer 20.5 3 3 Mark 2 eg. No with 28 and 25 or No with 31.25% (accept value in range 30% – 31.25%) or No with 68.75% and 65% (accept value in range 68% – 70%) A1 A1 M1 eg. No with 118 (minutes) or No with 52 and 55 accept reading in range 55 – 56 accept a value in the range 30 – 31.25 or a value in the range 68 – 70 for this mark unless clearly from incorrect working M1 M1 accept reading in range 55 – 56 for answer in range 19 – 21 M1 A1 M1 Notes 21 Question a 11 Alternative scheme 2x3 – 10x2 – x2 + 5x + 6x2 – 30x – 3x + 15 eg. 2x3 + 5x2 – 3x – 10x2 – 25x + 15 or 2x3 – 4x2 – 30x –x2 + 2x + 15 or 2x3 – 11x2 + 5x + 6x2 – 33x + 15 Working 2x – x + 6x – 3 or 2x2 + 5x – 3 or x2 + 3x – 5x – 15 or x2 – 2x – 15 or 2x2 – 10x – x + 5 or 2x2 – 11x + 5 2 2x3 – 5x2 – 28x + 15 Answer 2x – 5x2 – 28x + 15 3 3 Mark 3 A1 M2 A1 M1 M1 for a complete expansion with 8 terms present, at least 4 of which must be correct (dep) ft for at least half of their terms correct in second expansion (the correct number of terms must be present) Notes for expansion of any 2 of the 3 brackets (at least 3 of 4 terms correct) 22 Question b 11 6 6 60 6 96 (x = ) 1 5 1 oe 3 5 ( x 1)2 1 ( 0) 3 Alternative scheme e.g 3((x + 1)2 – 1) – 5 (= 0) or NB: denominator must be 2 × 3 or 6 and there must be evidence for correct order of operations in the numerator Accept 9.79 – 9.8(0) in place of 6 96 or 6 2 Working 0.633, −2.63 Answer 0.633, −2.63 3 Mark 3 for correct method to isolate x dep on M1 for answer in range 0.63 to 0.633 , −2.63 to −2.633 Award M2A1 for correct answer with correct working that would gain at least M1 A1 for completing the square dep on M1 for answers in range 0.63 to 0.633 , −2.63 to −2.633 Award M2A1 for correct answer with correct working that would gain at least M1 6 62 4 3 5 23 condone one sign error in substitution; allow evaluation of individual terms e.g 36 in place of 62 M1 M1 A1 M2 Notes If not M2 then award M1 for 23 13 alternate segment theorem b (i) (ii) 54 Answer 3, 4 see graph at end of mark scheme angle at centre is twice angle at circumference 27 Working (ii) a (i) Question (a) 12 (b) 1 1 1 1 Mark 1 3 B1 B1 B1 B1 B1 B3 "54" 2 dep on B1 in (b)(i) accept alternative reason angle between tangent and radius is 90o If answer for (b)(i) is ft from (a)(i) then reason must be angle between tangent and radius is 90o ft from (a)(i) for dep on B1 in (a)(i) accept alternative reasons eg. angle at circumference is half the angle at the centre NB. May shade wanted or unwanted regions; lines may be solid or dashed cao If not B2 then award B1 for line x + y = 4 drawn If not B3 then award B2 for x + y = 4 drawn (with no additional lines drawn) and a region identified that satisfies at least 3 of the 5 given inequalities for correct region identified Notes 24 d c b Question a 14 or 3 19 x 5 oe 4 19 3 oe or f(4) or 3 19 3 5 4 4y = 3x – 5 or 4x = 3y – 5 Working x > 19 1.75 oe 4x 5 oe 3 Answer −6.5 oe 2 2 2 Mark 1 B2 A1 A1 M1 M1 B1 for (x) > 19 or an equivalent statement in words If not B2 then award B1 for (x) ≥ 19 for 1.75oe (and no other solution) Notes 25 b Question a 15 2 1(3x 5) or 2(3x 5)(3x 5) 2(3x 5)(3x 5) Alternative scheme 6 x 10 9 x 2 25 (9 x 2 25)(6 x 10) (9 x 2 25)(6 x 10) 6 x 10 9 x 2 25 (9 x 2 25)(6 x 10) (9 x 2 25)(6 x 10) E.g. 1 1 (3x 5)(3x 5) 2(3x 5) 1 4 1 5 x 5 1 5 4x x y 4 E.g. or 2 or or 2 20 y 4 y 2 y 256 x ya ky a 1 k or oe , a = 2, b = 5 b or b with 2 of k = x x 4 ya or with 2 of m = 4, a = 2, b = 5 mxb 8 Working 7 3x 2(3x 5)(3x 5) 7 3x 2(3x 5)(3x 5) y 4 x5 2 Answer 3 3 Mark 2 accept equivalents eg. 7 3x 18 x 2 50 M1 for two correct fractions with a common denominator A1 M1 for two correct fractions with a common denominator if there is any expansion at this stage then it must be correct M1 indep for (3x + 5)(3x – 5) 2 1 2 y y y2 A1 4 2 -5 0.25 for 5 or 5 or 5 or 0.25y x x x 4x Notes M1 for a correct first step leading to a correct partially simplified expression 26 (7 3x)(3x 5) (9 x 2 25)(6 x 10) A1 accept equivalents M1 Numerator expanded and then factorised correctly 27 Question 16 " 27 3 " or 125 5 3 " 125 5 " 27 3 98 27 or 0.216 or 125 – 98 (=27) 125 125 3 3 1 " " or h − " " h oe 5 5 5 k= 3 1 2 1 98 1 2 r h (kr )2 kh = r h oe 3 3 125 3 Alternative scheme 3 3 1 " " or h − " " h oe 5 5 3 1 Working 2 h oe 5 2 h oe 5 Answer 4 Mark 4 A1 for 2 h oe (may not be simplified) 5 for the length scale factor M1 M1 sets up an equation using scale factor 5 for 2 h oe (may not be simplified) for the length scale factor may be seen as a ratio E.g. 3 : 5 Notes M1 A1 M1 M1 M1 28 b Question a 17 8 10 with e.g. “60” ÷ 2 (=30) and “210” ÷ 7 (=30) e.g. 63 3 60 211 1 210 with e.g. “58” ÷ 2 (=29) and “203” ÷ 7 (=29) OR 63 5 58 211 8 203 e.g. 5 8 10 3 " " or " " 8 4 11 1 BC 7 11 4 2 Working Proof Answer (13, 12) 2 Mark 3 A1 2 proof with justification eg. BE 29 (or 7 2 AE 30 ) with ABE is a straight line or 7 210 ÷ 60 = 3.5 and 7 ÷ 2 = 3.5 so ABE is a straight line may work with A and E, in which case may need to ft for method mark from (a) for coordinates (5 – 2 + 10, 8 – 7 + 11) assigned to C M1 A1 M1 Notes or coordinates (5 – 2, 8 – 7) (= (3, 1)) assigned to A (may be seen in vector form) or (13, y) or (x, 12) given as coordinates for C M1 29 b (ii) Question a (i) 18 Working 3 1 (−2, −0.5) oe e.g. 2, 90, 1 Mark 1 Answer (3, −1) B3 B1 B1 If not B2 then B1 for any 1 correct value or the graph of y = sinxo for 0 ≤ x ≤ 360 If not B3 then B2 for any 2 correct values NB. 2 values from 2, 90, 1 OR 2 values from −2, 270, 1 NB: accept a value of (90 + 360n) in place of 90 or (270 + 360n) in place of 270 where n is an integer (could be negative) for all 3 correct values e.g. 2, 90, 1 or −2, 270, 1 Notes 30 Question 19 9 2 11 11 " " " " or " " " " 20 20 20 20 2 1 2 3 3 11 1 3 3 2 11 or 1 4 5 4 5 20 4 5 4 5 20 1 2 2 3 3 9 or 4 5 20 4 5 20 1 3 3 3 2 6 or or 4 5 20 4 5 20 Working 121 oe 400 Answer A1 for M1 121 oe or 0.3025 or 30.25% 400 M1 for a complete method Mark Notes 4 M1 for any one correct probability 31 20 2 12 or y 2 x 12 or gradient = 2 x 3 3 3 3 3 x 43 2 2 × 37 = − 3 × 4 + c Alternative scheme 2y = −3x + c oe y 3 37 " " 4 c or 2 or 1 oe 2 " " 3 c = 43 2 (gradient of perpendicular line =) 3 oe y 3x + 2y = 86 3x + 2y = 86 5 5 A1 for correct equation (equation in any form) for 3x + 2y = 86 oe for a simplified equation with integer coefficients e.g. 3x = 86 – 2y A1 A1 A2 M1 M2 3 y 37 " "( x 4) 2 for 3x + 2y = 86 oe for a simplified equation with integer coefficients e.g. 3x = 86 – 2y 3 y 37 ( x 4) 2 M1 for M1 (dep on previous M1) and ft from their gradient M1 ft from their gradient M1 32 Question 12 -2 -1 -2 -1 0 1 2 3 4 5 6 7 8 9 10 y 1 2 3 4 5 6 7 8 9 10 x 33 49 Jan 19 Mark Scheme (Results) January 2019 Pearson Edexcel International GCSE In Mathematics A (4MA1) Higher Tier Paper 1H 2y 1 5 5 E.g. 5y – 2y = 1 or 3 y 1 or 3 y 1 0 or 3y 1 5 5 (c) 5 y 2 y 1 or y e 2 3e 5e 15 (b) 1 oe 3 e 2 2e 15 3 2 e.g. 0.3 , 0.3333… dep on at least M1 for 1 oe 3 for collecting terms in y in a correct equation M1 A1 for a correct first step for 3 correct terms or for 4 correct terms ignoring signs or e 2 2e k for non-zero k or ... 2e 15 M1 A1 M1 Apart from Questions 1(c), 5, 6(c), 20 and 21 (where the mark scheme states otherwise), the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method. Question Working Answer Mark Notes 2 p (2 3q ) (a) 2 B2 If not B2 then award B1 for 1 2(2 p 3 pq ) or p (4 6q ) or 2p(a two term expression) or x(2 + 3q) where x ≠ 2p 50 Triangle at (−2, 1), (−2, 3), (−1, 3) (c) Answer Rotation, 90° clockwise, centre ( 2, 3) Triangle at (−2, 2), (−2, 4), (−1, 4) Working (b) Question (a) 2 2 1 Mark 3 B2 B1 B1 B1 B1 If not B2 then award B1 for a triangle of the correct size and orientation or the wrong size but enlarged correctly from (-4, 2) with a sf other than 0.5 e.g. a triangle at (4, −2), (4, 6), (8, 6) for rotation 90° clockwise or −90o (or 270o anticlockwise) (centre) ( 2, 3) −2 Note: Do not accept ( ) for centre 3 Award no marks is more than one transformation explicitly stated (the sight of a vector is not a second transformation) eg. moved and then rotated; rotation and translation cao Notes 51 Question 3 150 "0.1" (P(yellow)=) "0.26" 0.06 or 0.1 2 Working 1 (0.15 0.26 0.33) or 1 0.74 (=0.26) 15 Answer 4 Mark 15 scores M3 A0 150 independent mark Award for 150 × p where 0 < p < 1 M1 NB: An answer of for a complete method to find P(yellow) M1 A1 Notes can be implied by two values where P(brown) + P(yellow) = 0.26 (may be seen in table) M1 52 (b) Question 4 (a) 1126.5 1.19 oe or 0.0976(475...) 1236.5 1126.5 "110" or 1126.5 1126.5 1236.5 1236.5 1 or 1.09 764... 1 or 100 100 or 1126.5 1126.5 Working 1236.5 1236.5 − 1126.5 or 110 or or 1.09(7647...) 1126.5 1236.5 100 or 109(.7647...) or 1126.5 1341 9.76 3 3 Answer Mark A1 M2 A1 M1 M1 for 1340 – 1342 19 1126.5 oe or 214(.035) 100 if not M2 then award M1 for for 9.76 - 9.765 for method that would result in 9.76… or 0.0976… Notes 53 Question 5 A1 B1 For full reasons: Alternate angles are equal and angles in a straight line add to 180º OR Allied angles (or co-interior) add to 180o and angles in a straight line add to 180º dep on previous M1 NB : It must be clear which angles are being found A1 for substituting their value for x into the expression NOT used to form the equation solved dep on previous M1 M1 M1 Notes for forming an appropriate equation E.g. AFC = 145 and FCD = 145 OR AFC = 145 and BCF = 35 OR x = 5 from the solution of two equations 5 Mark OR forms a second equation in x Shown correctly with reasons Answer OR E.g. 4 x 15 30 x 5 180 AND 30x – 5 = 20x + 45 E.g. 20 × “5” + 45 (=145) or 4 × “5” + 15 (=35) or 30 × “5” – 5 (=145) Working E.g. 4 x 15 30 x 5 180 OR 20x + 45 + 4x + 15 = 180 OR 4x + 15 + 20x + 45 = 180 OR 30x – 5 = 20x + 45 x=5 54 7 (c) (b) Eg 1.88 10 7 + 3.10 108 + 2.64 108 + 7.18 10 7 or 18 800 000 + 310 000 000 + 264 000 000 + 71 800 000 with at least 3 numbers correct x2 5x 6 x 1 (c) (a) (0, 6), (1, 2), (2, 0), (3, 0), (4, 2), (5, 6) Working (b) Question (a) 6 9.88 10 6 6.646 108 oe 71 800 000 1.6 and 4.4 Correct curve 1 2 1 3 2 Answer Mark (6),2,(0),(0),(2),6 1 B1 A1 M1 B1 for 6.646 108 oe eg 664 600 000 for a complete method or for digits 6646 dep on M2 ft from their graph in (b) if at least 1 mark scored in (b) for y x 1 drawn M1 A1 or for y x 1 for at least 5 points plotted correctly (ft their table) for a correct curve Notes For both entries correct M1 A1 M1 B1 55 Question 8 15.8 4 A1 for 15.8 – 15.83 dep on previous M1 M1 2 "5.41" 5 M1 Notes NB: 4.8 may be seen on the diagram ft the candidate’s value for height for this mark (award of this mark does not depend on award of previous mark) Mark M1 Answer ( x ) 2.5 2 " 4.8 2 " or ( x ) "29.29 " or 5.41(202…) Working 1 1 5 h 12 oe or 2.5 h 6 oe 2 2 h 4.8 or 56 (c) (b) Question (a) 9 15 and 45 indicated on the cumulative frequency axis and readings taken from speed axis Working 13 15 correct cf graph 2 2 Answer Mark 3, 19, 43, 53, 58, 60 1 A1 M1 A1 M1 B1 accept 13 – 15 ft from a cf graph ft from a cf graph for a correct method to find LQ and UQ and intention to subtract Eg for a correct reading from 45/45.75 and 15/15.25 from vertical axis to find LQ and UQ and an intention to subtract accept curve or line segments accept curve which is not joined to (0,0) ft from (a) if only one addition error for at least 4 points plotted correctly at end of interval or for all 6 points plotted consistently within each interval in the frequency table at the correct height (Eg. using values of 5, 15, 25 etc on x axis) Notes 57 10 Question 13 𝐶𝐷 13 8 +"4.73" or tan(BAD) = 0.97(93...) E.g. (BAD =) tan–1 ("0.979") or 44.4(024...) E.g. tan(BAD) = E.g. (CD =) 13tan20 or 4.7(316...) E.g. tan20 = Working Working with CD and then triangle ABD 24.4 Answer 5 Mark Award M1A1M1M1A0 for an answer in the range 44.3 – 44.41 for 24.3 - 24.41 for a correct method to find angle BAD M1 A1 for a correct statement or equation including angle BAD as the only variable E.g. 13 2 13 cos 20 2 for a correct method to find CD 2 13 2 E.g. CD = − 13 cos 20 2 for a correct statement or equation including CD as the only variable M1 M1 M1 Notes 58 10 Question A1 for a correct statement or equation including angle BAC as the only variable M1 Award M4A0 for an answer in the range 44.3 – 44.41 for ans in range 24.3 - 24.41 E.g. 132 (13 tan 20) 2 sin BAC sin110 or 8 "18.1" 82 = “13.8”2 + “18.1”2 – 2 × “13.8”ד18.1”×cos BAC E.g. E.g. (AC =) for a correct method to find AB for a correct statement or equation including AC as the only variable E.g. AC2 = 132 + (13tan20)2 for a correct method to find AC M1 M1 Notes E.g. (AB = ) "13.8"2 82 2 13.8 8 cos(110) (=18.1(9..) or 18.2 5 Mark M1 24.4 Answer 13 or 13.8(3…) cos 20 Working Alternative mark scheme – working with AC and then triangle ABC 13 E.g. cos20 = AC 59 Question 11 7 1 10 x 3 x 6 or 6x x 6x Working 10 x 3( x 2) 10 x 3( x 2) E.g. or 6x 6x 6x 7x 6 6x Answer 3 7x 6 as the final answer 6x SC: If no marks awarded then award B1 7x 6 for an answer of 6x for for a correct single fraction with brackets expanded M1 A1 Notes for two correct fractions with common denominator or a single correct fraction Mark M1 60 (b) Question (a) 12 1 3× x2 − 9 3 Working 3 2 x 2 9 oe 3 x 3 oe Mark Answer B3 A1 M1 SC: If no marks awarded and M1 awarded in (a) then award B1 for “quadratic” < 0 if not B2 then award B1 for x 2 9 0 or x 2 9 oe or for ( x 3)( x 3) or for ( x ) 3 (values maybe seen in incorrect inequalities) if not B3 then award B2 for x3 or x 3 or −3 ≤ x ≤ 3 may be seen as two separate inequalities or for 1x2 – 9 oe 1 for 3× x2 oe or −9 oe 3 Notes 61 (b) Question (a) 13 Working 2 3 Correct Venn diagram 3 oe 18 Mark Answer A1 M1 A1 M2 ft from Venn diagram a for where a is an integer "18" and 1 a "18" or "3" for where b is an integer and b b "3" ft from Venn diagram Accept omission of 0 for the award of full marks NB: For the award of the method marks do not accept a blank outside the circles as 0 Notes for at least 4 correct entries If not M2 then M1 for 2 or 3 correct entries 62 (b) Question (a) 14 Working 21.76 k 43 oe or k 0.34 T kr 3 73.44 T 0.34 r 3 oe Answer 1 3 Mark B1ft A1 M1 M1 for their value of k if T kr 3 Only award if T is the subject Award M2A1 if T kr 3 on answer line and k given as 0.34oe in working space. Condone use of proportional sign in place of equals sign Award M2 if k 0.34 stated unambiguously (m = 2.94) for correct substitution into a correct equation; implies first M1 Do not allow T r 3 Notes Allow r mT 3 63 Question 15 2 1 4 4 " h " 2 3 3 3 4 and (2 " h ") 2 h 3 A1 dep on award of first M1 ft for candidate’s expression for r or h for correct expressions for volume of hemisphere and volume of cylinder ; both in terms of either r or h M1 1 4 3 r 3 and (2 r ) 2 " r " 2 3 4 Eg OR for a correct expression for either r or h Notes for use of, for example, r and 2r in an equation condone omission of flat surface area A1 M1 3 4 r or r h 4 3 4 Mark h 4.5 oe Answer 4 r 2 r = 2 (2r )h oe Working Eg 2 64 (b) Question (a) 16 a 4b a 4b a 4b a 4b Eg ( a 4b )( a 4b ) a 2 4b 2 a 2 b a 2 b a 2 b Eg or a2 b a2 b Working a2 b a2 b or 2.5 oe a 2 4 a b 4b a 2 4b Answer 1 3 Mark B1 A1 M1 M1 a 2 4b a 2 4b a 2 b a 4 a b 4b for or 2 denominator dep on M1 for correctly simplified 2 denominator by a 4b or a 2 b Notes For multiplying the numerator and 65 Question 17 sin x sin110 5.3 "7.7" or 5.3 "7.7" or sin x sin110 cosx = 2 4.1 "7.7 " Eg sin x (= 0.764(83…) sin110 5.3 (= 0.644(2…)) or "7.7" 4.12 "7.7 "2 5.32 5.32 = 4.12 + “7.7”2 – 2×4.1×”7.7”×cosx oe Eg 59.7 641... or 7.7(3073) or AC 2 = 59.7… (AC = ) 16.81 28.09 14.8(641...) or 2 Working (AC² = ) 4.1 5.3 2 4.1 5.3 cos(110) 2 40.1 Answer 5 Mark for 40.1 – 40.11 for isolating sinx or cosx M1 A1 dep on first M1 for correct use of sine rule or cosine rule ft for their value of AC or AC2 M1 M1 M1 Notes for the correct use of Cosine rule to find AC NB: there must be evidence of correct order of operations for this mark to be awarded 66 (b) Question 18 (a) Working 3 ( 4, 5), ( 2, 0), (0, 3), (2, 4), (4, 3) (6, 0), (8, 5) Answer Parabola through 1 2 Mark B1 B2 ( 4, 5), (8, 5) If not B2 then B1 For u-shaped parabola with minimum (2, 4) or For u-shaped parabola through ( 2, 0), (6, 0) or For u-shaped parabola through ( 4, 5), ( 2, 0), (0, 3), (4, 3)(6, 0), (8, 5) Notes For a parabola with minimum (2, 4) through at least 5 of 67 (b) Question (a) 19 y 9 x 3 or x9 y3 y 9 ( x 3) 2 or x 9 ( y 3) 2 ( y 3) 2 9 ( x 3) 2 32 or ( x 3) 2 9 or ( y 3) 2 32 or Working 3 x 9 y 3 Answer 4 Mark 1 A1 M1 M1 M1 B1 and for 3 x 9 oe M3A0 for 3 y 9 for completing the square Notes Accept g 1 ( x ) 3 68 Question 20 2 Eg ( n 10)( n 3) 0 or 2 1 13 ( 13) 2 4 1 30 Eg 2n 2 26n 60 0 or n 2 13n 30 0 3n 27 n 60 n n 2 n4 n5 1 n n 1 3 Eg 3(n 2 9n 20) n(n 1) or n4 n5 or n n 1 Working 10 Answer 6 Mark for a correct quadratic equation with fractions removed for a correct quadratic equation equal to 0 dep on M2 ft for method to solve 3 term quadratic M1 M1 M1 M1 Award M0A0 for an answer of 10 with no working and no justificastion NB. Award M5A1 for an 8answer of 10 with justification 6 5 1 e.g. 10 9 3 for correct answer from correct working for the correct equation M1 A1 Notes n4 n5 or n n 1 69 21 Question A1 A1 Eg 2 × 5.5 + 23 (=34) and 8 × 5.5 + 2 (=46) OR 8 × 5.5 + 2 (=46) and 20 × 5.5 − 52 (=58) M1 x 5.5 4 Mark M1 shown Answer (8x + 2) – (2x + 23) = (20x – 52) – (8x + 2) oe or (2x + 23) – (8x + 2) = (8x + 2) – (20x – 52) oe Working Mark scheme 1 (see next page for alternative mark scheme) (8x + 2) – (2x + 23) (= 6x – 21) or (2x + 23) – (8x + 2) (= −6x + 21) or (20x – 52) – (8x + 2) (= 12x – 54) or (8x + 2) – (20x – 52) (= −12x + 54) for 12 from correct working for a correct equation for a correct expression for the common difference in terms of x brackets must be present or removed correctly Notes 70 21 Question OR 2x + 23) + 12 = (8x + 2) and (8x + 2) + 12 = (20x – 52) and gets x = 5.5 both times x 5.5 or x = 1.5 from (2x + 23) − 12 = (8x + 2) or x = 3.5 from (8x + 2) − 12 = (20x – 52) 2 × 5.5 + 23 (=34) and 8 × 5.5 + 2 (=46) and 20 × 5.5 − 52 (=58) Alternative method – starts by assuming d = 12 E.g. (2x + 23) + 12 = (8x + 2) or (8x + 2) + 12 = (20x – 52) or (2x + 23) − 12 = (8x + 2) or (8x + 2) − 12 = (20x – 52) or 3 (2x + 23) + (8x + 2) + (20x – 52) = 2(2 x 23) 2 12 2 Working shown Answer 4 Mark A1 A1 M2 OR solves both (2x + 23) + 12 = (8x + 2) and (8x + 2) + 12 = (20x – 52) and gets x = 5.5 both times for explicitly showing both common differences are 12 for a correct equation If not M2 then award M1 for a correct expression for the common difference in terms of x brackets must be present or removed correctly e.g (8x + 2) – (2x + 23) (= 6x – 21) or (20x – 52) – (8x + 2) (= 12x – 54) Notes 71 89 May 19 Mark Scheme (Results) Summer 2019 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 1H 14 9 × 3 10 e.g. 14 9 126 21 1 4 3 10 30 5 5 14 9 126 6 1 4 4 or 3 10 30 30 5 7 3 14 9 21 1 or 5 4 1 3 10 5 5 126 30 126 21 1 ¸ = = = 4 or 27 27 30 5 5 e.g. Shown A1 M1 126 30 ¸ 27 27 126 must be 30 1 5 21 5 NB: use of decimals scores no marks Total 3 marks to the multiplication to seen or correct cancelling prior multiplication e.g. from correct working – either sight of the result of the Dep on M2 for conclusion to 4 42 10 ÷ or 9 9 or for both fractions expressed as equivalent fractions with denominators that are a common multiple of 3 and 9 eg. International GCSE Maths Apart from questions 1, 11, 12b, 15 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an incorrect method, should be taken to imply a correct method. Question Working Answer Mark Notes 14 10 1 3 M1 Both fractions expressed as e.g. and improper fractions 3 9 90 2 (c) (b) (a) 10 9 m/sec or 0.2 km/min or oe 3 3 12 km/h or 1h 45m + 1h 30m or 1 + 0.75 + 1.5 or 3h 15m or 3.25h or 195m oe (24 × 2) ÷ “3.25” oe eg (48 ÷ 195) × 60 oe 15 25 m/sec or 0.25 km/min or 4 6 15 km/h or 14.8 line from (12:00, 24) to (12:45, 24) to (14:15, 0) ‘before’ with reason 3 2 1 M1 A1 M1 B2 B1 e.g. before as gradient is steeper or before as speed before is 15 km/h speed after is 12 km/h or before as she goes over 11(allow 11-12) km in ¾ hour but only goes 9 km in ¾ hour after oe NB: any figures used for the reason must be accurate if they haven’t used ‘gradient is steeper’ oe If not B2 then B1 for a line from (12:00, 24) to (12:45, 24) or for a line from (t, 24) to (t + 1.5, 0) or for a time of 1.5 hours (oe) seen ft from their graph for total time when cycling ft dep on M1 for full method awrt 14.8 Total 6 marks 91 4 3 (b) (a) (d) (a) (b) (c) x² + 9x − 2x − 18 2 1 1 2 correct graph 9, 3, (−1), −3, (−3), −1, (3) x² + 7x 18 4cp2(4c3 + 5p) e4 y16 2 2 A1 B2 B1 B1 M1 A1 M1 B2 If not B2 then award B1 for at least 2 correct values dep on B1 ft from (a) for at least 5 points plotted correctly for the correct graph (clear intention to go through all the points and which must be curved at the bottom) Total 4 marks if not B2 then award B1 for any correct factorisation with at least 2 factors outside the bracket eg 4cp(4c³p + 5p²) , cp²(16c³ + 20p), 2p(8pc4 + 10cp²) etc or the correct common factor and a 2 term expression with just one error Total 6 marks for 3 correct terms or 4 correct terms ignoring signs or x² + 7x + c or .... + 7x 18 92 6 5 336 92 “480” × 0.7 12 × 8 × 5 (= 480) 3 A1 M1 M1 or Total 3 marks Total 4 marks 92 , oe with 92 seen 200 dep on M2 and probabilities between 0 and 1 Dep on M1 A1 M1 eg (0.18 + 4 ד0.07”) × 200 or 0.46 × 200 or 36 + 42 + 14 oe M1 M1 4 x = (1 – 0.18 – 0.26) ÷ (2 + 2 + 3 + 1) (=0.07) 2x + 0.18 + 2x + 3x + 0.26 + x = 1 or 1 – (0.18 + 0.26) (= 0.56) 93 8 7 (a) (b) (c) 0.08 × 170 000 (=13600) or 0.92 × 170 000 (=156400) e.g. 0.92 × (0.92 × “156400”) 132377 3 A1 (dep)for a complete method or 132376.96 (SCB2 for 170 000 × 0.924 )(=121786.(810)) (SCB1 for 170 000 × 0.24 (=40 800) or 170 000 ×0.76 (=129 200) or 170 000 × 1.08 (= 183 600) or 170 000 × 1.08³ (= 214151) or an answer of 129 200 or an answer of 214 151 – 214151.1(0)) Total 3 marks M2 for 170 000 × 0.923 If not B2 then award B1 for 320000 or 3.2 × 105 oe or 5 × 10n oe where n ≠ 6 Total 4 marks M1 B1 B1 B2 oe eg 170 000 ÷ 12.5 1 1 2 M1 5 700 000 4 × 10-3 5 000 000 or 5 × 106 oe 94 10 9 46.3 (8 =) 2 × 2 × 2 or 2³ or 23+n 2 2n 3 3 5 m A1 M1 A1 M1 2 "8.48.." 0.5 × π × (= 9π or 28….) 2 M1 M1 M1 AC 6 (sin 90) sin 45 5 62 62 ( 72 = 6 2 =8.4(85…)or 8.5) or 6(sin 90) AC = 6 2 =8.4(85...)or 8.5) oe sin 45 (d 2 =) 62 + 62 (=72) or 0.5 × 6 × 6 (=18) Total 5 marks Total 2 marks For clearly writing 8 as a product of prime factors or as 2³ for 46.2 – 46.3 1 6 6 2 sin 45 or 2 1 6 2 3 2 oe 2 For area of triangle, or may use 95 11 17.5 – 5.5 5.5 or 6.5 or 12.5 or 17.5 12 3 A1 M1 M1 • • Accept 6.4 9 for 6.5 and 17.49 for 17.5 for UB – LB where 15 < UB ≤ 17.5 and 5.5 ≤ LB < 6 dep on M2 Total 3 marks 96 12 (b) (a) 4 m 3 7 2m 3 4 m 2m 7 3 oe 3 Alternative 4m + 6m = 21 – 9 or 10m = 12 or −21 + 9 = −6m – 4m or −10m = −12 4m + 9 = 21 6m 4m + 9 = 3(7 2m) 12 oe 10 (2x – 3)(x – 2) 4 4 2 awarded M1 for a correct equation with m terms isolated on one side ft their equation if first M1 M1 Division of each term on LHS by 3 (from 4m + 9 = 7 – 6m oe)] m = − 0.2oe with working shown (from 4m + 9 = 21 – 2m oe) or m = 2 with working shown [SC: B2 for an answer of M1 for a correct equation with m terms isolated on one side ft their equation if first M1 awarded A1 dep on at least M2 in a correct equation M1 for correct expansion of bracket or (3 – 2x)(2 – x) (B1 for (2x + a)(x + b) where ab = 6 or 2b + a = −7 eg (2x + 3)(x + 2), (2x – 5)(x – 1)), etc or for M1 for removing fraction B2 97 12 (c) contd 1 1 1 1 y4 or 4 y y 4 or y 4 y 10m = 3 × 4 oe 3 y 4 12 oe 10 2 A1 M1 A1 or b = 3 4 dep on at least M2 correct equation Total 8 marks M1 For removing fraction in a fully 98 13 42 2 8 7 7 6 " " " or 117 5 14 13 10 9 7 6 42 14 0.35(897...) or 13 9 117 39 8 7 56 7 6 42 oe or 14 13 182 10 9 90 (c) " 8 7 14 10 (b) (a) 28 oe 195 2 oe 5 3 7 3 7 , , , 10 10 10 10 6 8 , 14 14 3 2 2 A1 M1 M1 A1 M1 B1oe B1 6 3 8 4 , in correct 14 7 14 7 3 7 3 7 , , , in correct 10 10 10 10 28 oe, e.g. 0.14(3589…) 195 from accurate working Total 7 marks for ft from (b) 7 ( 0.54 to 2dp 13 6 0.67 to 2dp) 9 ft from (a) ft from (a) positions. for positions. Allow decimals of 2dp or better (0.43, 0.57) for 99 14 7, 8, 9, 10, 11 eg 2, 4, 6 (a) (b) 1 2 B1 B2 for any 3 of 2, 4, 6, 8, 10 Total 3 marks completely correct. (B1 for 4 or 5 correct and no more than 1 incorrect or for all terms seen correctly placed in a Venn diagram or for a correct description of the numbers in the set but not listed, eg 7 ≤ x < 12) 100 15 5.4 54 3 2 99 54 252 14 and 99 990 55 990 990 55 5.4 54 3 11 3 14 or and 99 990 55 55 55 100 x x 5.4545... 0.05454... 5.4 and 252 14 or 990 55 1000x – 10x = 254.545... – 2.545... = 252 and 25.2 14 or 99 55 eg 100x – x = 25.454... – 0.254...= 25.2 and 1000x = 254.54... 10x = 2.5454... 100x = 25.454… x = 0.25454… show 2 A1 M1 for completion to 14 55 with intention to subtract. For 2 recurring decimals that when subtracted give a whole number or terminating decimal eg 25.2 or 252 etc eg 100x = 25.454…and x = 0.25454... or 1000x = 254.54... and 10x = 2.5454.... with intention to subtract. (if recurring dots not shown then showing at least the digits 25454, ie 5sf) or 0.2 0.054 and eg x 0.05454..., 100 x 5.4545... 101 16 100th term is 3 × 100 + 4 (= 304) and 100 × (7 + “304”) ÷ 2 100th term is 7 + (100 – 1) × 3 (= 304) and 100 × (7 + “304”) ÷ 2 or 100 (2 7 (100 1) 3) or 2 a = 7 and d = 3 15 550 2 A1 M1 Total 2 marks for a method to find the sum brackets (100 – 1) must be used correctly Total 2 marks 102 18 17 17.82 + 26.32 – 2 × 17.8 × 26.3 × cos36 e.g. 1008.5... – 757…. or 251(.06…) 3 24 36 or 2³ : 3³ oe or or 3³ : 2³ oe or 36 24 8 27 or oe 27 8 3 (b) 24 36 or 2 : 3 oe or or 3 : 2 oe 36 24 eg (a) 8 V oe 27 15.8 (A =) 2160 3 2 2 M1 M1 A1 A1 A1 M1 M1 V 3.375 Total 4 marks "2160" 960 for correct order of operations for ans in range 15.8 – 15.9 Total 3 marks oe eg or ft from For correct SF for volume ft from linear scale factor in (a) for a correct scale factor 103 19 16 ÷ 10 (=1.6) 21 ÷ 5 (=4.2) 48 ÷ 15 (=3.2) 15 ÷ 20 (=0.75) correct histogram 3 B3 For a fully correct histogram [If not B3 then B2 for 3 correct frequency densities (can be implied by heights) or 3 correct bars drawn If not B2 then B1 for 2 correctly calculated frequency densities (can be implied by heights) or 2 correct bars drawn.] Total 3 marks 104 Students can use other methods to gain the correct answer 20 angle ABD = 71 or 5 angle ACD = 71 or using O as centre of circle, angle ADO = 90 – 71 (=19) angle ADB = 71 or angle ACB = 71 or angle BAD = 19 × 2 (=38) or reflex angle BOD = 2 × 142 (=284) angle BCD = 142 142 dep clearly labelled or stated Clearly labelled or stated, from no incorrect working for their method dep on A1 for fully correct reasons for each stage of working, repeated if used more than once. eg alternate segment theorem, base angles in an isosceles triangle are equal, angles in a triangle sum to 180o, angle between tangent and radius(diameter) is 90° congruent triangles (equal triangles) oe opposite angles of a cyclic quadrilateral sum to 180o angles in the same segment angle at the centre is 2 × angle at circumference oe equal chords subtend equal angles at the circumference If not B2 then award B1 dep on M1 for any one correct circle theorem reason associated with angle(s) found Total 5 marks M1 A1 B2 clearly labelled or stated M1 105 21 A1 A1ft (r = ) 6 or (h = ) 18 24 M1 M1 1 4 r 3 + r 2 3r = 792π oe 2 3 5 M1 h 3 1 4 r 3 oe or r 2 3r oe 2 3 h = 3r or r 3 2 2 4 correctly 3 evaluated dep on M3 Total 5 marks their “6” × 4 or "18" 1 4 h h h 792 2 3 3 3 or 3 h oe stated or 3 1 4 h h or or h 2 3 3 3 used correctly for h = 3r or r 106 22 translation of 3 1 2 2 (x – 3)2 + 1 (b)(i) (b)(ii) 2 correct graph (see end of mark scheme) [must go through (60, 2), (150, 0), (240, −2), (330, 0)] and not through (0, 0) (a) B1 B1 B2 B2 6 )² + n (where n ≠ 1) or for 2 must be column vector For ft from (b)(i) 3 1 x2 − ax − ax + a2 + b with 2a = 6 or a2 + b = 10) for translation Total 6 marks (x – m)² + 1 (where m ≠ 3) or for (B1 for (x – point on graph at (150, y) and a point at (330, y)) or a clear translation of 30 (ie a 0 if not B2 then award B1 for a graph of the correct shape going through 2 or 3 of the given points or for a clear stretch of SF2 (ie a maximum point on graph at (x1, 2) and a minimum point at (x2, −2)) 107 (v =) 3t2 – 6 × 2t + 5 (+ 0) 24 (a =) 3 × 2t – 12 6t – 12 = 3 10 2 7 19 , or (6, 13) 2 2 19 7 12 oe or 1.5 oe 10 2 8 3 2 m× = −1 oe or m = 2 3 2 “13” = “ ”× “6” + c or c = 17 3 2 oe or y "13" " "( x "6") 3 23 3y + 2x = 51 2.5 oe 5 4 A1 2 x 17 3 M1 2 “ must come 3 A1 M1 M1 M1 Total 4 marks for differentiating at least 2 terms correctly dep ft dep on at least M1 for equating their acceleration in terms of t to 3 from correct working] for 3y + 2x = 51 or 3y = −2x + 51 etc but must be integer coefficients Total 5 marks [NB: “13”, “6” and “ Or for y for use of m1m2 = −1 M1 M1 M1 108 Q19 0 1 2 3 4 5 Frequency density 10 20 30 40 50 Height (h metres) 109 q22 -3 -2 -1 0 1 2 3 y 60 120 180 240 300 360 x 110 127 Mark Scheme (Results) Jan 2020 January 2020 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 1H 2 or 28 105 or 2, 2, 3, 5, 7 oe 7 28, 56, 84, 112… and 105, 210, 315, 420… or 2, 2, 7 and 3, 5, 7 or 420 2 A1 cao or a fully correct Venn diagram Total 2 marks or 2, 2, 7 and 3, 5, 7 seen (may be in a factor tree and ignore 1) for starting to list at least four multiples of each number M1for any correct valid method e.g. Question Working Answer Mark Notes Apart from questions (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method 1 (a) 2 M1 for a correct method to find one coordinate or for one 5 13 4 1 or coordinate correct or for (−1.5, 9) 2 2 (9, −1.5) A1 oe (b) −3 1 B1 (c) No with 1 B1 No (oe) and e.g. line goes through (100, −298) or (101.3(3..), reason 304 −302) or , 302 or (3 × 100) – 302 = −2 not (+)2 3 Total 4 marks 128 3 E.g. 12 × 9 (=108) or (9 – 6) × x (= 3x) E.g. 129 – ‘108’ (= 21) or ‘108’ + ‘3x’ = 129 E.g. ‘21’ ÷ (9 – 6) or 129 '108' x= 96 7 4 A1 Accept 7 cm Total 4 marks M1 (dep on M1) for 129 used correctly with another area or for a correct equation (ft) with bracket(s) expanded M1 for a complete method M1 for one correct relevant area 129 4 (c) (a) (b) 2 1 40 40 ‘144’ ÷ 40 or [(12 × 2.5) + (16 × 3.5) + (9 × 4.5) + (2 × 5.5) + (1 × 6.5)] ÷ 40 30 + 56 + 40.5 + 11 + 6.5 (= 144) or (12 × 2.5) + (16 × 3.5) + (9 × 4.5) + (2 × 5.5) + (1 × 6.5) 3 40 3.6 3<w≤4 2 1 4 Total 7 marks 3 a where 0 < a < 40 or where b > 3 where a and b are b 40 integers A1 0.075 oe M1 for A1 oe Allow division by their Σf provided addition or total under column seen correct midpoints used for at least 4 products and not added M1 (dep on at least M1) or M1 for consistent use of value within interval (including end points) for at least 4 products which must be added If not M2 then award B1 M2 for at least 4 correct products added (need not be evaluated) or 130 6 5 (a) (b) 22 500 000 oe e.g. 22.5 × 10 or 2.25 × 10n n ≠ 7 (‘75’ – ‘48’) + (‘45’ – ‘15’) 6 ‘45’ ÷ 3 (= 15) or ‘45’ ÷ 3 × 2 (= 30) 16 9 ‘75’ × (= 48) or ‘75’ × (= 27) 25 25 E.g. 9 (‘45’ ÷ 3 × 2) + (‘75’ × ) oe or 25 ‘27’ + ‘30’ or 120 ÷ (3 + 5) (= 15) ‘15’ × 3 (= 45) or ‘15’ × 5 (= 75) 2.25 × 107 0.000 78 57 1 2 6 A1 B1 M1 A1 M1 for a complete method M1 M1 M1 M1 M2 for 3 120 (= 45) or 8 5 120 (= 75) oe 8 Total 3 marks Total 6 marks 131 7 A1 dep on at least M2 awarded Total 9 marks M1 ft (dep on 4 terms) for terms in x on one side of equation; number terms on the other 2(‘6x – 15’) = 9 – x oe or 9 x 3(2x – 5) = oe 2 2 12x + x = 9 + 30 oe or x 9 6 x 15 oe 2 2 M1 for any 3 correct terms or for 4 out of 4 correct terms ignoring signs for m2 – 3m … or for …– 3m – 40 A1 B2 If not B2 then award B1 for 5(y + 4y2) or y (5 + 20y) or 5y(a + 4y) where a is an integer and a ≠ 0 or 5y(1 + by) where b is an integer and b ≠ 0 B1 M1 for expansion of a correct bracket M1 for removal of fraction or separating fraction (RHS) in an equation 1 4 2 2 2 × 3(2x – 5) = 9 – x oe or 3 1 (c) (d) E.g. 6 x 15 or 12 x 30 oe (b) m2 – 8m + 5m – 40 m2 – 3m – 40 5y(1 + 4y) (a) 132 9 8 24.3 or PQ (PQ =) 24.3 or cos 63 24.3 (PQ =) or sin 27 24.3 sin 90 PQ sin 27 PQ 24.3 or sin 90 sin 27 oe sin 90 sin 27 PQ 24.3 24.3 sin 27 = or PQ cos 63 = 53.5 Trapezium with vertices at (6, 3) (8, 3) (8, 6) (4, 6) 3 2 A1 Accept 53.5 - 53.53 M1 for a correct rearrangement for PQ M1 for a correct trigonometric ratio Total 2 marks Total 3 marks PQ '47.6914 '2 24.32 oe and RQ 24.3 tan 63 47.6914.. M2 for B1 for shape of correct size and orientation or 3 or 4 points plotted correctly B2 If not B2 then award 133 11 10 6000 × 1.0152 (= 6181.35) or 6000 + (0.015 × 6000) + (0.015 × (6000 + ‘90’)) (= 6181.35) or (1.015)2 (= 1.030225) or 6311.16 1.05186 6000 6311.16 ÷ ‘6181.35’ (= 1.021) (×100) or 6311.16 '6181.35' (= 1.021) (×100) or '6181.35' ‘1.05186’ ÷ ‘1.030225’ (= 1.021) (×100) y≥ 2.1 1 x 2 oe 3 x + y ≤ 4 oe x ≥ −1 oe 3 3 Total 3 marks A1 awrt 2.1 Total 3 marks M1 (dep on M1) for a complete method to find the compound interest multiplier (×100) M1 for working out the total amount after two years or working out the compound interest multiplier after two years or working out the compound interest multiplier after three years Accept < for ≤ and > for ≥ throughout 1 x 2 oe) 3 (If no marks gained B1 for understanding of equation x + y = 4 e.g. y > 4 − x) (SC B3 for x ≤ −1, x + y ≥ 4 and y ≤ (B2 for two correct inequalities B1 for one correct inequality) B3 for all 3 correct inequalities 134 14 13 12 (0.5 ×) 9.3 × 14.7 × sin106 or (9.3 × cos 16) × 14.7 or (9.3 × sin 74) × 14.7 (GE2 =) 9.32 + 14.72 – 2 × 9.3 × 14.7 × cos106 377(.9….) or 378 or 86.49 + 216.09 + 75.3… or 302.58 + 75.3…. (b) 19.4 131 70 18 8000 (a) 56 or '400 ' (=20) 0.14 ‘20’ × ‘20’ × ‘20’ oe 56 oe or 56 ÷ 0.14 (= 400) w2 [40.5, 43] ‘42’ ÷ 0.6 oe (b) 0.14 E.g. 56 – 38 (a) 3 2 4 3 2 Total 5 marks A1 for 19.4 – 19.5 Total 5 marks M1 for applying the area of a triangle formula using correct values (to find half of the area of the parallelogram) or for a correct method to find the area of the parallelogram A1 awrt 131 M1 for the correct use of the cosine rule M1 (dep on M1)for the correct order of operations M1 (dep on M2) for a method to find the volume of the cube A1 Total 4 marks M1 for a method to find w M1 for using the given formula correctly M1 for subtracting readings from 60 and 20 oe A1 for answer in the range 17 – 19 B1 M1 for complete method to find the number of men A1 135 15 (a) [(−2x2 + x + 15)(x + 1) =] −2x3 − 2x2 + 15x + x2 + x + 15 Shown A1 for at least 4 terms correct out of a maximum of 8 terms or [(−x2 + 2x + 3)(2x + 5) =] −2x3 − 5x2 + 10x + 4x2 + 6x + 15 or M1 for multiplying out two brackets correctly at least 3 terms correct M1 for at least 3 terms correct out of a maximum of 6 terms 3 E.g. [(2x2 + 7x + 5)(3 – x) =] −2x3 − 7x2 − 5x + 6x2 + 21x + 15 or (3 – x) (2x + 5) = −2x2 + 6x − 5x + 15 (= −2x2 + x + 15) (x + 1)(3 – x) = −x2 + 3x – x + 3 (= −x2 + 2x + 3) or (2x + 5)(x + 1) = 2x2 + 2x + 5x + 5 (= 2x2 + 7x + 5) or 6x2 −2x3 + 6x − 2x2 + 15x − 5x2 +15 − 5x M2 for at least 4 terms correct out of a maximum of 8 terms 136 15 (b) 2 E.g. 2 2 1 1 6 x 16 0 oe 6 6 (x = ) 2 2 4 6 16 oe 26 (accept + in place of ±) or E.g. dV 16 2 x 6 x 2 oe x d '16 2 x 6 x 2 ' 0 oe dV 16 2 x 3 2 x 2 oe x d 1.47 5 A1 dep on M1 for answer in range 1.47 – 1.5 from correct working (Must reject −1.80 to −1.81 if calculated) Total 8 marks M1 (dep on M1) for a complete method to solve their 3-term quadratic equation (allow one sign error and some simplification – 2 4 384 ) allow as far as 12 M1 (dep on M1) for equating their differentiated expression to zero A1 for a correct differentiated expression 16 or −2x or (3 × −2x2) M1 for the correct differentiation of at least 2 correct terms from 137 16 2 58.45 19.5 (= 27.4366...) 3.55 58.35 or 58.45 or 19.5 or 20.5 or 3.55 or 3.65 27.44 3 58.4 < aUB ≤ 58.45 and 19.5 ≤ cLB < 20 and 3.55 ≤ dLB < 3.6 A1 from correct working allow 27.4 – 27.5 where 2 aUB cLB d LB M1 for correct substitution into 3.64 9 for 3.65 20.4 9 for 20.5 or Accept 58.449 for 58.45 or B1 for any correct bound Total 3 marks 138 17 (a) or 36 24 3 24 3 48 or 36 2 24 3 48 or 36 6 2 2 12 48 84 48 3 or 36 12 12 12 12 4 12 12 or 36 12 12 12 12 4 12 6 6 6 2 12 6 2 12 2 12 2 Shown 3 A1 for fully correct working leading to given expression M1 (dep on M1) or for showing or stating 12 = 2 3 oe for the use of (a + b)2 = a2 + 2ab + b2 or M1 for correct expansion of brackets showing four terms (need not be simplified) 139 17 (b) 2 2 3 8 1 2 8 1 3 t 30 a 8 or 24 9t 10 729a 1 3 1 8 2 a 9a t5 3 a 9 10 or 10 or 10 or 4 or t t t 3a E.g. t15 729a 24 3a 4 or 30 5 or 12 27 a t t E.g. t10 9a8 3 1 t10 a 8 or t10 a 8 9 9 t10 where k is an integer ≠ 0 ka 8 A1 Allow or cube rooting or inverting or squaring or ka 8 where k is an integer ≠ 0 t 10 M1 for two of cube rooting or inverting or squaring M1 for one of Total 6 marks 140 18 5 4 11 220 11 5 11 10 550 55 oe or oe 16 15 14 3360 168 16 15 14 3360 336 24 84 210 168 ' ' 3 ' ' ' ' 3 ' ' oe or 3360 3360 3360 3360 80 140 60 210 140 60 1 ' ' 3 ' ' 3 ' ' 3 ' '3 ' '6 ' ' oe or 3360 3360 3360 3360 3360 3360 220 550 60 1 ' ' 3 ' ' 3 ' ' oe 3360 3360 3360 or 5 4 4 80 1 5 4 7 140 1 oe or oe or 16 15 14 3360 42 16 15 14 3360 24 5 4 3 60 1 5 7 6 210 1 oe or oe or 16 15 14 3360 56 16 15 14 3360 16 5 7 4 140 1 oe 16 15 14 3360 24 or 4 3 2 24 1 7 6 5 210 1 oe or oe or 16 15 14 3360 140 16 15 14 3360 16 5 4 3 60 1 oe 16 15 14 3360 56 4 7 6 168 1 4 3 7 84 1 oe or oe 16 15 14 3360 20 16 15 14 3360 40 990 3360 4 A1 for 11 10 9 oe 16 15 14 M3 for 990 33 oe e.g. or 0.29(464...) 3360 112 Total 4 marks M1 for a complete method LLX or LXX (X = not L) or LLB or LLO or LBB or LOO or LOB or BOO or BBO M1 for finding the following in any order M1 for finding BBB or OOO or LLL 141 19 6 6 or tan AHF = or ' 142 ' ' 106 ' 52 92 (= 106 ) ' 106 ' or cos 1 or ' 142 ' ' 106 ' 6 1 90 sin 1 or 90 cos ' 142 ' ' 142 ' ' 106 ' or 90 tan 1 6 sin FAH = ' 106 ' 6 or cos FAH = or ' 142 ' ' 142 ' ' 106 ' tan FAH = 6 E.g. 6 6 1 sin 1 ' or tan ' 142 ' ' 106 ' 106 ' cos AHF = or ' 142 ' sin AHF = (FH = GE =) E.g. (AH =) 62 52 92 (= 142 ) or 30.2 4 A1 for 30.2 – 30.3 ' 142 '2 ' 106 '2 62 cos 1 oe or 2 ' 142 ' ' 106 ' sin 90 sin 1 6 oe ' 142 ' Allow M1 for a complete method ' 142 '2 ' 106 '2 62 cos AHF = oe or 2 ' 142 ' ' 106 ' sin 90 sin AHF = 6 oe ' 142 ' Allow Total 4 marks M1 for a correct method for finding angle AHF or finding angle FAH M1 for working out AH or FH or GE 142 20 x = 1 a oe or y = 4 – a 4 graph drawn in shape of a quadratic with a minimum in any quadrant x = 1 , y = 4 (1 – 1)2– a Correct graph 4 y-axis intercept marked as (0, 4 – a) oe Total 4 marks Note: The 0’s can be ignored (as shown in the diagram) quadratic shape with minimum in the fourth quadrant and marked as (1, – a) oe a x-axis intercepts marked as (1 , 0) oe on the positive 2 a x-axis and (1 , 0) oe on the negative x-axis 2 M1 for finding the turning point (may be seen marked on the graph as (1, –a)) M1 for finding one of the intercepts (or award for any one correct coordinate shown on graph) a a (0, 4 – a) or (1 , 0) or (1 , 0) 2 2 Note: The 0’s can be ignored (as shown in the diagram) A1 for a fully correct graph M1 for a quadratic with a minimum 143 21 x or 2 4 16 4 3 y or x 2 1 y (x + 2)2 = y + 1 or (y + 2)2 = x +1 2 x 4 x (3 y ) 0 or y 2 4 y (3 x) 0 or (fg(x) = ) (x + 3)2 – 2(x + 3) oe (fg(x) =) x2 + 4x + 3 (x + 2)2 – 4 + 3 or (x + 2)2 – 1 2 x 1 5 A1 oe Total 5 marks Allow same equations with x and y swapped or correctly substituting into the quadratic formula M1 ft (dep on M2) for a correct rearrangement for their completed the square quadratic Correctly setting up an equation or M1 for substituting g(x) into f(x) A1 Allow y2 + 4y + 3 M1 ft (dep on M1) for correctly completing the square on their 3 term quadratic 144 22 2 23 2 78 4 3 495 k 13 169 165 0 or 78 23 2 12 4 3 36 or 2 j 2 4 12 0 12 eg 3k2 – 78k + 495 = 0 oe or 5j² − 60j – 140 = 0 oe or 5k2 – 150k + 1045 = 0 oe or 3j2 – 12j – 36 = 0 oe or k 15 1 k 15 2 gradient HM: eg oe 2 or k = 2j + 15 or j j6 2 4 2 eg (j – 6)(j + 2)( = 0) eg (k – 15)(k – 11)( = 0) or or (j – 6)2 + (k – 15)2 = 80 oe or j 6 k 15 , oe 2 2 or 2 2 j 4 196 100 k 1 oe gradient of JK = −0.5 or m × 2 = −1 j 24 k 15 1 or 2k – j = 24 or j = 2k – 24 or k oe 6 j 2 2 j = −2, k = 11 6 M1 (dep on M3) for a complete method to solve their 3term quadratic equation (allow one sign error in the use of the quadratic formula) or a correct method to eliminate either j or k k 15 j 24 eg 2k – 24 = oe or = 2j + 15 oe 2 2 A1 Total 6 marks or A correct equation for the gradient of HM in terms of j and k or a correct equivalent equation or for equating length HJ with length HK M1 (dep on M3) writing a correct quadratic expression in the form ax2 + bx + c (= 0) (allow ax2 + bx = c) or for finding the midpoint of M M1 for finding equation of JK in terms of j and k M1 for finding the gradient of JK using m1 × m2 = −1 M1 for expressing the gradient of JK in terms of j and k or a correct equivalent equation 145 ALT 22 or 2 j 8 64 28 0 25 2 12 4 5 44 j = −2, k = 11 or 2 k 1 1 143 0 25 E.g. (5k – 119)(k – 11)( = 0) or 2 174 174 4 5 1309 E.g. (5j – 22)(j + 2)( = 0) or 12 E.g. 5k2 – 174k + 1309 = 0 or 3k2 + 6k − 429 = 0 oe E.g. 5j2 – 12j – 44 = 0 or 3j2 + 48j + 84 = 0 oe j 6 k 15 , oe 2 2 k 15 1 2 2 or k − 2j = 15 or k = 2j + 15 or j6 4 2 k 15 j oe 2 (j – 6)2 + (k – 15)2 = 80 oe 2 2 or j 4 196 100 k 1 oe 6 A1 Total 6 marks M1 (dep on M3) for a complete method to solve their 3-term quadratic equation (allow one sign error in the use of the quadratic formula) allow ax2 + bx = c M1 (dep on M3) writing the correct quadratic expression in form ax2 + bx + c (= 0) M1 for finding the length of JK in terms of j and k or for equating length HJ with length HK M1 for expressing the gradient of JK in terms of j and k or a correct equivalent equation M1 for finding the midpoint of M 146 2H 1 Mark Scheme Specimen paper International GCSE Mathematics A 4MA1/2H 2 1 (d) (b) (c) Q (a) Working Answer × 8.5² ÷ 22.5 (area of trapezium =) (20 + 25) ÷ 2 × h oe (=22.5h) × 8.5² (=226.98...) 5q ≥ 31 or 2q + 3q ≥ 31 2 10.1 q ≥ 6.2 −2, −1, 0, 1, 2 12e f 9a8 9 Mark Notes 4 2 2 2 2 A1 M1 M1 M1 A1 B2 B2 M1 B2 Total 4 marks A correct method to find h A correct method to find the area of the circle Use of correct formula for trapezium B1 for 9or a8 For 5q ≥ 31 or 2q + 3q ≥ 31 or 5q = 31 or q = 6.2 for q ≤ 6.2 or an answer of 6.2 following q ≥ 6.2 in working oe, (q > 6.2 is M1 only) B1 for 4 correct and none incorrect or all correct with one addition. Total 8 marks B1 for 2 correct parts Apart from questions 10, 11, 14, 17 and 21 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from an incorrect method, should be taken to imply a correct method. International GCSE Maths 2 4 3 78, 76, 74 (b) (c) Correct reason 4n + 3 49 0.22 (a) (91 + 0.3 × “350”) ÷ 4 [ (91 + “105”) ÷ 4] oe 91 ÷ 0.26 (=350) or (0.3 ÷ 0.26) × 91 (=105)) (b) “0.44” ÷ 2 1 – (0.26 + 0.3) (=0.44) (a) B2 B2 B1 2 1 A1 M1 M1 M1 A1 M1 2 3 3 Total 5 marks B1 for 4n + x where x is any integer B1 for one correct term The first sequence is only odd numbers and the second is only even numbers Total 6 marks A correct method to find total number of bricks or number of blue bricks A correct method to find number of layers 3 5 4 18000 oe or 720 100 = 778.75 4 × (18000+’720’+’748.80’) 100 = 748.80 4 × (18000+’720’) 100 Eg 1.043 18000 OR 2248 3 A1 M1 M1 (M1 for 18000 × 1.04 or 18720 or 18000 × 1.04² or 19468.8 or18000 × 1.044 or 21057.45) 18000 1.043 OR M2 for Accept 1 + 0.04 as equivalent to 1.04 throughout SC: If no other marks gained, award M1 for 18000 × 1.12 oe or 20160 OR or 2160 Answers in range 2247 – 2248 Total 3 marks oe or 720 for completing method 4 eg 18000 100 for 4 8 12 8 or sin x = or cos x = 12 208 208 (31.50 ÷ 7) × 8 oe (=36) 8 ‘36’× 1.2 oe (x = ) 360 – (90 + 90 + 52) 8 12 8 x = tan-1 or sin 1 or cos 1 12 208 208 tan x = 7 6 43.2(0) 128 Correct reasons 33.7 3 4 3 A1 M1 M1 B1 M1 A1 B1 A1 M1 M1 Total 3 marks Correct method to find the amount Behnaz has Correct method to find the amount Ahmed has The angle between a tangent and a radius is 90° oe Angles in a quadrilateral add up to 360° oe Total 4 marks Accept answers which round to 33.7 Total 3 marks A complete method to find angle x A correct trig ratio for angle x 5 9 15 and 45 or 15.25 and 45.75 indicated on cumulative frequency axis or stated A vertical line from 48 up to the cf graph (c) (d) (a) (b) Approx 6 Approx 19 4, 18, 35, 48, 55, 58, 60 Points correct Curve or line segments 2 M1 2 M1 A1 A1 M1 B1 B1 1 2 If M1 scored ft from CF graph. If M1 not scored, ft from correct curve and, if answer is correct (±½sq) award M1A1 Total 7 marks If M1 scored ft from CF graph. If M1 not scored, ft from correct curve and, if answer is correct (±½sq) award M1A1 + ½ sq ft from points if 4 or 5 correct or if points are plotted consistently within each interval at the correct heights Accept curve which is not joined to the origin Correct cumulative frequencies 6 10 135° − 108° (=27°) 180 – 2 × 27 Alternative 360 ÷ 8 (=45) 180 – 45 (=135) 360 ÷ 5 (=72)_180 – 72 (=108) 72° − 45° (=27°) 180 – 2 × 27 360 ÷ 8 (=45) 360 ÷ 5 (=72)_ 126 126 5 5 M1 M1 A1 M1 M1 M1 M1 A1 M1 M1 Method to find interior angle of octagon or pentagon Method to find interior angle of both octagon and pentagon Method to find CAB or CBA Fully correct method to find angle y dep on at least M2 Total 5 marks Method to find exterior angle of octagon or pentagon Method to find exterior angle of both octagon and pentagon Method to find CAB or CBA Fully correct method to find angle y dep on at least M2 7 11 6 x 4 15 20 x 2 10 26x = 39 or 6x + 20x = 20 + 4 + 15 6x + 20x = 39 oe 6x – 4 – 15 + 20x = 2 × 10 oe 2(3x 2) 5(3 4 x) 2 or 10 10 2(3x 2) 5(3 4 x) 2 or 10 2(3x − 2) – 5(3 – 4x) = 2 × 10 Eg 1.5 4 Expanding brackets For correct rearrangement of a correct equation with terms in x isolated Award full marks for a correct answer if at least M1 scored Total 4 marks M1 M1 A1 for clear intention to multiply all terms by 10 or a multiple of 10 or to express LHS as a single fraction with a denominator of 10 or a multiple of 10 M1 8 13 12 (a) (b) 3 5 2 3 7 2 8 x = −1 or x = 5 3x² − 12x – 15 = 0 (3x + 3)(x – 5) (=0) 20 (−1, 8) (5, −100) 3x² − 12x − 15 A1 M1 M1 3 Total 3 marks For at least 3 correct entries into Venn diagram 30 – (3 + 3 + 5 + 7 + 2 + 2) B2 B1 for 2 correct terms M1ft M1 Correct factorisation or correct use of quadratic formula A1 One correct pair A1 Both correct pairs Total 6 marks 2 4 9 15 14 (b) (a) 4(7 5) 49 5 0.3 × 0.9 (=0.27) 0.7 + ‘0.27’ e.g. 0.97 7 1 5 11 11 show 3 2 3 A1 M1 M1 A1 M1 M1 A1 M1 The correct product for fail, pass A fully correct method to find the probability that Sophie passes 1st or 2nd time oe Total 3 marks Total 5 marks dep on correct working seen For selecting 10x = 3.2424.... and 1000x = 324.2424... oe 321 990 For multiplying the numerator and denominator by (7 + 5 ) For a correct single fraction with brackets expanded in denominator 10 17 16 Greatest number of spheres = 12.55³ ÷ 135 (=14.641899...) Largest volume of cube = 12.55³ 12.45, 12.55, 135 or 145 14 (270, −1) (ii) (b) (180, 0) (a)(i) 4 4 Correct intersections of 0°, 180° and 360° with x axis A1 A1 Dep on M1 Total 4 marks Units must be consistent 12.55³ M1 M1 For sight of 12.45, 12.55, 124.5, 125.5, 135 or 145 B1 Total 4 marks Correct shape curve M1 B1 B1 11 2 ½q − ¼p MN is parallel to BD BD 4 MN (b) 1 1 MB p or BM p 4 4 1 1 BN (p q)or NB (p q) 2 2 1 2 2q − p (a)(i) (a)(ii) (6, −4) 9 19 (7, −4) (3, −12) (a) (i) (ii) (iii) (b) 18 1 1 1 1 A1 A1 A1 B1 M1 B1 B1 B1 B1 Total 5 marks With suitable reasons With suitable reasons For correctly giving MB or BM or BN or NB Total 4 marks 12 20 '15'2 '7.5'2 15 5 ( 16.7705...) 2 CSA = 15 16.77.. 1 (2h) 2 h (=562.5𝜋) or 3 1 1 r 2 r (=562.5𝜋) 3 2 1 4h2 h 562.5 or 3 1 1 r 2 r 562.5 3 2 3 562.5 h 3 ( 7.5) or r = 3 3375 4 790 5 A correct expression for l M1 Total 5 marks A correct equation for h or r M1 (786.5 – 791.7) A correct equation for the volume of the cone with (2h)² expanded M1 A1 A correct expression for the volume of the cone NB: other letters may be used rather than r and h M1 13 2 x 3 7 (2 x 3)(2 x 5) 5 7 2 4 x 25 2 x 5 5 7(2 x 5) 5 14 x 35 or 2 4 x 25 4 x 2 25 a + 2d = 19 10 (2a + 9d) = 290 oe 2 Eg 10a + 45d = 290 10a + 20d = 190 Or 5(2(19 – 2d) + 9d) = 290, a = 11, d = 4 10th term = 11 + 9 × 4 or 290 – 4.5(2 × 11 + 8 × 4) 21 22 47 40 14 x 4 x 2 25 5 4 A1 Total 5 marks A correct method to find the 10th term. M1 M1 A formula for term 3 A formula for the sum of the first 10 terms A correct method to find a or d M1 M1 A1 Total 4 marks Correct subtraction shown ((4x²−25) can be factorised) Correct single fraction, unsimplified ((4x² −25) can be factorised) 40 14 x oe e.g. 2 x 5 2 x 5 M1 M1 For inverting and factorising M1 14 15 Mark Scheme (Results) Summer 2018 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 2H 1 eg ef (18ef eg 6e (3 f 2 2 3 M bd a c c 2 12e2 ) 2ef ) , eg 2 f (9e f 6e ) 2 (c) 3 5x39 4 oe ac M bd or –ac = −M – bd or (b) (a) 6e2 f (3 f 2 2e) 3 5 M bd c x8 a 2 2 2 A1 M1 A1 M1 A1 M1 M bd M bd , a c c c 43 or x 8.6 or 5 Total 6 marks [−∞, 8.6) Any correct partially factorised expression with at least 2 terms in the common factor or for the correct common factor and a 2 term expression inside the brackets with just one error Accept x [must have been seen with a = to award accuracy mark] Accept as equation or with the wrong inequality sign. Also award M1 for an answer of 8.6 or 8.6 with an = sign or the incorrect inequality sign. oe, eg a For a correct first stage Apart from Questions 4, 9, 15, 16, 21(a) 21(b) and 22, where the mark scheme states otherwise, the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method. Question Working Answer Mark Notes International GCSE Maths A June 2018 – Paper 2H Mark scheme 16 3 2 Question 3 "21600"19200 "2400" (100) 19200 20000 (=2400) (100) or 100 8 or “21600” ÷ 19200 (×100) oe 19200 "3450 20000 (=21600) or (20 000 – 19200) + 100 8 20000 (=1600) 20000 100 8 “1610” – “460” or " 1 (7 – 2) × “230” or 7 × “230” – 2 × “230” or 2 3450 3450( 460) or (=230) or 267 267 7 72 1 3450( 1610) or 267 267 3 Working 12.5 1150 Answer 4 3 Mark A1 M1 M1 M1oe A1 M1 M1 9 8 Total 4 marks or 112.5% 21600 or for 1.125 or oe Total 3 marks Award M2 for 200001.08 or Notes 17 4 Question 109 109 with RHS shown as 56 56 32 35 (1) 56 56 1 53 56 eg eg 3 32 35 1 56 56 88 35 2 1 56 56 Alternative method 3 56 3 109 or 2 56 56 eg (3) Alternative method Or eg 8 25 713 200 91 or 56 56 56 56 109 53 = 1 56 56 13 25 and 8 7 Working correctly shown correctly shown correctly shown Answer 3 3 3 Mark 109 53 and 1 but 56 56 53 109 allow showing that 1 = on 56 56 A1 dep on M2 Total 3 marks complete correct method M1 3 109 or 2 56 56 two correct fractions with a common denominator, at least one correct the subtraction eg dep on M2 with sight of the result of two improper fractions, with a common denominator, at least one correct correct subtraction of fractional parts RHS in working the subtraction eg dep on M2 with sight of the result of correct improper fractions or two improper fractions with a common denominator, at least one correct two correct fractions with a common denominator M1 A1 M1 M1 A1 M1 M1 Notes 18 7 6 5 Working (b) (a) 4 70.5 63 3 Total 7 marks or for 33 – “96” or 33 to “96” oe M1 A1 for 39 + 57(=96) or 33 + 60(=93) Correct calculation for mean of class B Expression for total of both classes together or total for class A Expression for total of class B Total 4 marks Allow 21 cm oe if units shown Correct scale factor (given as a fraction or ratio) or correct equation in r or a correct expression for r. Allow 2.6666... to 1 dp rounded or truncated Award even if part of a calculation including 1 or 2 circles awrt 5.63 Correct reason for 90° angle [If used alternate segment theorem] Total 3 marks For 90° and 18° correctly identified in the working or on the diagram or for 90 – 18 or for other fully correct method Notes M1 A1 M1 (28 32)72.6 2875 (=”2256” ÷ 32) 32 Highest in A = 39 + 57 (= 96) Highest in B = 33 + 60 (= 93) (39 + 57) – 33 M1 M1 A1 M1 (28 + 32) × 72.6 – 28 × 75 (=2256) (28 + 32) × 72.6 (=4356) or 28 × 75 (=2100) 0.21 2 8 0.6 1.6 r 0.6 (=0.375) or (= = 2.6 ) or 1.6 0.6 0.56 1.6 3 0.560.6 or (r ) or 0.56 ÷ 2.6 oe 1.6 A1 (b) 5.63 M1 A1 M1 2 × 𝜋 × 0.56 × 1.6 2 3 Mark (a) 72 Answer B1 OQT 90 and OQP 18 or 90 – 18 o Angle between tangent and radius(or diameter) is 90 degrees Question 19 8 Question ( x ) (= 12.6 ) or 0.61566... 12.6 12.6 or sin 38 = x x 12.6 12.6 or sin 38 cos 52 cos 52 Working 20.5 Answer 3 Mark A1 M1 M1 "16.12..." oe x Total 3 marks Allow fully rearranged sine rule 20.4 – 20.5 Or (x =) 12.62 "16.12..."2 or "16.12..." (x = ) sin 52 Allow correct first stage of sine rule Accept decimal correct to at least 3SF sin 52 first stage to find x eg x² = 12.6² + “16.12...”² or 12.6 (=16.12...) and a correct tan 38 Or use of tan to find horizontal side 12.6 × tan 52 or Notes 20 11 10 9 (b) (a) Question 5x + 5y = 75 + 7x – 5y = 3 or 23 27 or 1 1 or and16 24 4 2 16 9 4 V 1.53 (= 14.1(37)… or ) 2 3 109.6 D oe 3 4 3 1.5 1324 135 27 23 “6.5” + y = 15 or x + “8.5” = 15 or 7 × “6.5” – 5y = 3 or 7x – 5 × “8.5” = 3 7(15 – y) – 5y = 3 or 7x – 5(15 – x) = 3 oe eg 7x + 7y = 105 − 7x – 5y = 3 Working 7.75 19 4 x = 6.5, y = 8.5 Answer 3 2 2 3 Mark 7.75 – 7.78 Total 3 marks dep M1 A1 Correct expression for volume. Total 4 marks M1 19 or for k = −6 × 4 + 5 Accept 13 A1 24 4 for 13 Accept 2 Total 3 marks Correct method to eliminate x or y: coefficients of x or y the same and correct operation to eliminate selected variable (condone any one arithmetic error in multiplication) or writing x or y in terms of the other variable and correctly substituting dep Correct method to find second variable using their value from a correct method to find first variable or for repeating above method to find second variable dep on first M1 M1 A1 M1 A1oe M1 M1 Notes 21 12 or 42 + 144 + “138” + (50 + 96) + DEP = “540” (where P is on AB and FE extended) oe Eʹ = “720” – “138” – 42 −50 – 96 – 144 (= 720 – 470 = 250) and E = 360 – “250” or E = “138” + 42 + 50 + 96 + 144 + 360 – “720” (= 830 – 720) or “138” + 42 + 50 + 96 + 144 + (360 – E) = “720” eg “138” + 42 + 50 + 96 + 144 + Eʹ = “720” (2 × 6 – 4) × 90 (=720) EDC = 180 − 42 (=138) 110 5 A1 M1 M1 M1indep M1 Total 5 marks from no incorrect working A completely correct calculation for the correct angle E obtuse angle of the hexagon and E is the interior (reflex) angle or for an answer of 250 from correct working Method to find sum of interior angles of hexagon or the correct sums for the interior angles of shapes used (eg 540° & 180° if the line through FE to point on AB drawn or 720° and 180° if line drawn from E parallel to AB or 540° & 180° if line through FE extended and joined to line through CB extended) oe dep on previous M marks Equation for E or E where E is the May be marked on diagram. NB: splitting the shape incorrectly (FDC and DEA are not straight lines) gains no marks for angles calculated from false information. However angles calculated that follow the scheme, such as EDC = 138° or interior angles of hexagon = 720° can be awarded. Other ways of correctly splitting the shape can be awarded full marks, eg FE to a point on AB or adding a parallel line eg from E parallel to AB NB: some students show lots of lines but actually work with the angles correctly so please check carefully. Question Working Answer Mark Notes 22 13 (b) (a) Question 5 4 4 5 4 3 5 4 4 8 or oe 10 9 10 9 10 9 10 9 10 9 5 4 5 1 4 1 1 or 1− oe 10 9 10 9 10 9 10 Working 90 52 4 4 1 5 3 1 5 4 , , , , , , , ,0 9 9 9 9 9 9 9 9 Answer 3 2 Mark A1 M1 M1 B2oe 0 or the branch 9 Total 5 marks oe decimals 0.577… or 57.7...% rounded or truncated to 2 or more sf crossed out or left blank Award M1 for one correct product (ft tree diagram) A fully correct method (ft tree diagram) shown), 0 can be Award B1 for any 3 correct. Decimals must be correct (recurring Notes 23 15 14 −0.8 0.6 2.2 8.305 0.655 7.65 2 A1 A1 -0.8 or 0.6 or 2.2 M1 B2 B2 M1 4 2 2 Mark Plot y = −2x + 3 x³ − 2x² − 3x + 4 = −2x + 3 correct curve (c) -6, 4, 0, -2, 4 Answer (b) Working (a) Question A1 M1 Total 2 marks seen). Accept 0.6549 dep on correct method shown For either bound correct (used or Any one correct x value at intersection of graphs (or one or more points given as coordinates) ft dep on second M1 (Award even if curve in (a) is incorrect) Accept −0.9 to −0.7 SC B2 for all correct Accept 0.4 to 0.7 solutions from Accept 2.1 to 2.4 graph of (not coordinates) y x3 2 x 2 x 1 ft (±1 square) dep on second M1 must be 3 values Total 8 marks Sufficient to cross curve at least once. For correct smooth curve. If B2 not awarded, award B1 for at least 5 points plotted correctly ft from table dep on B1 or B2 in (a) (plots ±1 sq) Award B1 for 2, 3 or 4 correct. Notes 24 16 (b) (a) Question 5 " t 2 " 5x 2 8 eg 10 = k × 22 or R = kt² oe 40 = k × 4² Working or k = 2½ t x 0.8 5 R t2 2 Answer 2 3 Mark A1 M1 A1 M1 M1 2 25 16 x 1 proportional to x ] 2 2 Total 5 marks [allow other clear arguments that clearly shows t is inversely required. eg accept t ft dep on answer of the form R kt Simplification of constant is not ft dep on answer of the form R kt 2 Award for R kt if the value of k is shown clearly in (a) or (b). Equation consistent with R t Substitute values at any point on the graph or find the value of k. (Implies first M1.) Allow readings from graph for t ± 0.1 and R ± 1 Notes 25 17 (b) (a) Question 5 ,3 3 23 (4) (4)2 43(15) (3x + 5)(x – 3)( < 0) or 3x² − 4x – 15 < 0 (or = 0) Working 3 5 x3 3x 4 x 15 2 Answer 4 2 Mark A1oe M1 M1 M1 B2 Allow x 5 , x3 3 Total 6 marks ft from “their (a)” (=0) for 3 term quadratic, for correct factorisation or correct use of quadratic formula to find the two critical values, allow 1 sign error. [−(−4) could be 4 and (−4)² could be 4²](condone missing brackets) Both critical values correct Accept -1.66… rounded or truncated to 3SF. Inequality signs needed Award B1 for any 2 or 3 of the 4 terms differentiated correctly. ft from (a) ie “their (a)” = 0 (or < 0) Notes 26 19 18 Question 2810 "740" 10 (= 0.3676 …) sin DBE (=0.9428…) cos DBE = (= 0.3952…) or tan DBE M1 M1 Total 3 marks 21.5 – 21.6 (=0.9299...) Total 3 marks 2 "640" "740" 10sin 90 or "740" "640" "740" 102 Allow use of sine or cosine rule Complete method to find BE or BE2 or BD or BD2 oe 101.5 to 101.6 196 100 64 1 196 64 100 180 cos1 cos 280 224 cos DBE A1 A1 102 82 142 cos 1 oe ie cos-1 of the correct angle or 2 10 8 M1 a fully correct method to find the largest angle eg Correct substitution in cosine rule for any angle or for 44.4... or 34.047....(the other 2 angles to 1dp or better) M1 Notes "640" "740" 3 3 Mark sin DBE 21.6 101.5 Answer "640" 10 (BD = 8 10 = 25.298….) 185 =27.202 …) (BE = 2 BD2 82 242 (=64 + 576 = 640) oe (= 100 + 576 + 64 = 740) BE 2 102 242 82 cos A 102 82 142 14² = 10² + 8² − 2 × 10 × 8 × cosA or Working 27 20 Question 4 1 (= 8) or (=2) or 0.5 2 1 small square vertically = FD of 2 or 1cm vertically = FD of 10 oe eg (4 × 5 + 20 × 4 + 25 × 2 + 15 × 4) ÷ 5 or 4 + 40 × 0.4 + 50 × 0.2 + 30 × 0.4 or 4 + 16 + 10 + 12 oe height of last bar height of first bar eg 4 × 5 + 1 × 10 = 30 small squares for 6 babies or 30 ÷ 6 or 5 small squares represent 1 baby or Working 42 Answer 3 Mark A1 M1 M1 Total 3 marks Fully correct method, allow one error in products but must be the sum of 4 parts Start working with area being proportional to frequency or show the height of the first or last bar or show a correct scale on the frequency density scale, with no inconsistent values. eg could be awarded by seeing total of little squares ÷ 5 oe Notes 28 21 (c) oe 2 (b) Working ( x 3 2)2 (3 2)2 1 3 1 2( 3 1) 2 3 2 or or 3 1 3 1 31 2 95 and 45 (a) Question 2 ( x 3 2) 19 1 3 5 5 shown Answer 2 2 2 Mark A1 M1 A1 M1 A1 M1 a =3 2 or b = − 19 or ( x 3 3 1 3 1 Total 6 marks 2)2 18 1 or for 3 1 3 1 dep on M1 or seeing multiplication by 3 5 2 5 but we must see where these come from Rationalise denominator – award for or for 45 = 3 × 3 × 5 and 20 = 2 × 2 × 5 dep on M1 cao with sight of Notes 29 23 22 Question 0 M1 2 (2a (30 1)d ) A1 M1 (2a (36 1)d ) oe M1 30 2 36 5 M1 (2a (48 1)d ) 4 2 (2a (36 1)d ) oe 2 96a + 1392d = 0 oe eg 4a + 58d = 0, 2a + 29d = 0 or a = −14.5d etc 2 2 48 36 (2a (48 1)d ) or 48 21 or 16 16 41(80) A1 M1 (x – 4)(x + 20) (= 0) 16 576 A1 2 M1 x² + 16x – 80 (= 0) M1 5 × (5 + 4 + 7) = x × (2 × “8” + x) 6 Mark A1 4 Answer r = 8 or d = 16 7 × 4 = 2(2r – 2) or 7 × 4 = 2(d – 2) Working Incorrect working giving the radius as 16 cm gains M0A0 Notes Total 5 marks Indep Allow substitution of any ‘found’ values of a and d For a correct equation. For a correct expression for the first 48 terms or the first 36 terms Total 6 marks dep on first 2 method marks Correct factors or evidence of correct use of quadratic formula. Accept 5 × 16 = x(2r + x) Or a correct equation in r eg 5.5² − 1.5² = 4r − 4 30 47 Jan 19 Mark Scheme (Results) January 2019 Pearson Edexcel International GCSE In Mathematics A (4MA1) Higher Tier Paper 2H 3 2 Question 1 12 x 1 90 x 3 “6” × 2 (=12) or”6” × 13 (=78) or 3(12 + x) = 90 + x (“78” ÷ 2 ) – “12” or 2x = 54 or “78” × 3/2 –“78” – “12”oe 90 ÷ (2 + 13 ) (= 6) or Working 73 ÷ 200 (=0.365) or 73 × 100 (= 7300) or 1 cm = 2 m oe “0.365” × 100 or “7300” ÷ 200 73 ÷ 2 4 3 2 36.5 4n + 3 27 Mark Answer A1 M1 M1 M1 A1 B2oe M1 M1 200 M2 for dep on a correct method for “78” and “12” 2 13 90( 12) or 90( 78) 15 15 M2 for If not B2 then award B1 for answer of 4n + k (k ≠ 3) or n = 4n + 3 e.g. 7 + 4(n – 1) or 4n + (7 – 4) etc allow Tn = 4n + 3 or x = 4n + 3 etc 100 ÷ Allow their incorrectly 73 converted 73 m ÷ 200 oe Notes 48 5 Question 4 5 A1 M1 (67 – 2×”11”) × (123 – 2×”11”) (45 × 101) or 123 × 67 – 12 × “28” ×” 11” (8241 – 3696) 4545 M1 67 – 56 (=11) or 67 – 2×”28” (=11) or 123 – 4×”28” (=11) 4 M1 Fully correct Venn diagram dep on M2 for method to find length or width for method to find other dimension Notes B4 fully correct Venn diagram with labels A and B (If not B4 then B3 for 3 correct regions, B2 for 2 correct regions B1 for 1 correct region) e.g. “56” ÷ 2 (=28) 2 4 6 12 B Mark M1 1 3 Answer 8 10 123 – 67 (=56) or 2x = 123 – 67 or 2x + y = 67 or 4x + y = 123 oe (x = length of tile, y = width of tile) A 5 7 9 11 Working 49 6 Question (b) (a) 6 4 96 120 16 20 4 5 2 × 2 × 2 × 2 × 2 × 3 or 2 × 2 × 2 × 3 × 5 e.g. 2 96 120 2 48 60 2 24 30 3 12 15 4 5 Working 2 2 646 800 Mark 24 Answer A1 A1 M1 M1 or 2³ × 3 oe for 2m × 3n × 5p × 7q × 11r with at least two of m = 4, n = 1, p = 2, q = 2, r = 1 (or omission of one with others fully correct) NB: e.g.24 could be 2 ×23 or prime numbers may be seen in a Venn diagram – if so must be correctly placed or 24 × 3 × 52 × 72 × 11 oe Notes for one number written as product of prime factors number may be at the end of factor trees or on ‘ladder’ diagrams or Use of table method (allow 1 error), 2 examples shown but could have 2, 3, 4, 6, 12, 24 along the side or at least 2 factors for each (excluding 1, 96, 120) 50 (b) Question (a) 7 687 700 ÷ 0.92 (=747 500) or 687 700 ÷ 1.15 (=598 000) or 1.15 × 0.92 (=1.058) 687 7000 ÷ (0.92 × 1.15) 8500 × 0.023 (=195.5) or 8500 × 1.023 (=8695.5) ((8500 + “195.5”) × 1.023) × 1.023 Working Mark 3 3 Answer 9100 650 000 complete method for 9100 – 9100.1 (answer for 600(.1) gains M2A0) a correct first step Dep on M1 for completely correct method M1 A1 M1 M1 A1 M1 Notes M2 for 8500 × 1.0233 (M1 for 8500 × 1.023n) 51 (b) Question (a) 8 3.5 V 3.5 (V ) 0.65 630 1000 oe 60 60 630 × 1000 (=630 000) 60 × 60 (=3600) eg 630 ÷ 60 (=10.5) 630 000 ÷ 60 (=10 500) 1000 ÷ 60 (=16.66...) 1000 ÷ (60 × 60) (=0.277...) 1 ÷ (60 × 60) (= 0.000277...) 0.65 Working 175 5.38 Answer 3 3 Mark 3500 0.65 (M2 for 630 ÷ 3.6) Fully correct method M1 A1 for converting 630 km to m or 1 hour to seconds or for correct operation(s) using at least 2 of the numbers 630, 1000, 60, 60 been converted eg V for answer in range 5.38 – 5.385 SCB1 for a “correct” equation involving V with digits 65 and 35 where units have M1 A1 M1 M1 Notes 52 10 Question 9 4 1 or c = 1 2( 0) y = “1.5”x + c or y = mx + 1 or eg y – 4 = m(x – 2) 3 ÷ 2 (=1.5) or eg y = 2x – 9 and 4x + 5(2x – 9) = 4 4x + 5y = 4 10x – 5y = 45 With the operation of adding Working e.g. 4x + 5y = 4 4x – 2y = 18 with the operation of subtraction y = 1.5x + 1 oe x = 3.5 oe, y = −2 Answer 3 3 Mark A1 M1 A1 M1 M1 M1 oe eg y 4 3 ( x 2) 2 (dep) for substitution of found variable into one equation or correct method to eliminate second variable Dep on M1 for correct method to find gradient – may see this on grid. For c = 1, could be (L =) mx + 1 oe or for 1.5x + c for use of y = mx + c with either m or c or for (L =) 1.5x + 1 Notes for correct method to eliminate one variable – multiplying one or both equations so the coefficient of x or y is the same in both with the intention to add or subtract to eliminate one variable(condone one arithmetic error) or isolating x or y in one equation and substituting into the other equation 53 Question 11 On the whole students have higher marks in Science The spread of results is greater for Science Results are more consistent for Maths Comparisons in context: eg The median is greater for Science/less for Maths The IQR (or range) is higher for Science/less for Maths The median is 2.5 marks higher for Science The IQR (or range) is 7 marks more for Science Working Basic comparisons from information: eg one for IQR and one for median Answer Two comparisons Mark 2 B2 NB; any numbers used must be correct for the award of the mark (B1 for 1 or 2 basic statements or for 1 statement in context) Notes For 2 comparisons in context or 1 basic comparison and 1 comparison in context 54 (d) (c) Question (a) 12 (b) 6a r 5r m2 × 5r = 6a + r 5rm2 – r = 6a m2 2(e2 – 9) or (2e – 6)(e + 3) or (e – 3)(2e + 6) Working r 6a 5m 2 1 2(e – 3)(e + 3) Answer 1 27x6y15 4 2 Mark 1 2 M1 M1 A1 M1 A1 M1 B1 B2 6 a oe 1 5m 2 6a 6a in working if 2 5m 1 5m 2 1 alone is given as answer r NB: to award A1 we must see or for r If not B2 then B1 for any two correct terms in a product Notes 55 13 Question "264 8 x " "39 x " or “273” – “264” "264 8 x " =7 oe eg“264 + 8x” = “(39 + x)” × 7 "39 x " (4 + 13 + 16 + 6 ) × 7(=273) oe or 4 × 5 + 13 × 6 + 16 × 7 + 8x + 6 × 9 (20 + 78 + 112 + 8x + 54) or 264 + 8x (4 + 13 + 16 + 6 + x) × 7 (=7(39 + x) = 273+ 7x) or Working 9 Answer 4 Mark for use of mean M1 M1 A1 M1 Notes at least 3 products correct with intention to add 56 (b) Question (a) 14 0.35 × 0.35 or 0.35 × 0.65 or 0.65 × 0.35 or 0.65 × 0.65 0.35 × 0.35 + 0.35 × 0.65 + 0.65 × 0.35 or 1 – 0.65 × 0.65 Working 0.5775 Answer 0.65 0.35, 0.65 0.35, 0.65 3 Mark 2 A1 M1 M1 B2oe oe e.g. 231 , 0.58 or 58% or better 400 ft from (a) Notes for all correct If not B2 then award B1 for 0.65 in any of the 3 possible positions NB all values may be given as fractions ft from (a) 57 (b) Question 15 (a) 1 2 1 (8 x 2 12 x 6 x 9) 133 2 or (4x − 25)(2x + 11) (=0) 6 36 8800 6 36 8800 6 8836 or or 28 16 16 or 8x² − 12x + 6x – 9 = 266 eg. or 0.5(4x + 3)(2x – 3) oe e.g. ( x 5 3 x 2) (2 x 3) Working 6.25 oe shown Answer 3 3 Mark for completion to given equation dep on M2 A1 dep on M1 and 6.25 oe alone given as final answer ft from an incorrect 3 term quadratic equation (if student gains M1 and shows both answers the 2nd M1 can be awarded) 28 Condone one sign error in substitution; allow evaluation of individual terms e.g. 36 in place of (−6)² [allow −62 or 62 in place of (−6)2, throughout allow + rather than ± ] or (4x ± 25)(2x ± 11) (=0) 6 ( 6) 2 4 8 275 M2 If not M2 then award M1 for A1 M1 for correct equation with brackets expanded Notes M1 correct algebraic expression for area 58 18 17 16 Question (a) (b) (c) 2 960 4 405 3 (=1.3...)or 3 (=0.75) 405 3 960 4 4.55 25 7.5 7.5 or 8.5 or 4.65 or 4.55 25 or 15 g(−1.5) = 1 ÷ (1 – 2 × −1.5) (=0.25) or 1 fg(x) = 4 3 oe 1 2x 3 4 928 or 928 oe 4 3 2 e.g. 3 Working 4 2 3.25 oe 0.26 oe 3 1 1 Mark 522 −11 0.5 oe Answer A1 A1 M1 M1 M1 LB1 with 4.55 ≤ LB1 UB LB 2 < 4.6 and 20 < UB ≤25 and 7.5 ≤ LB2 < 8 for 0.26 from correct working for g(−1.5) must be the correct calculation alone. for a complete method M1 A1 B1 B1 M1 Notes for a correct linear scale factor M1 59 Question 19 2.5 × 2 + 4 × 3 + 3.4 × 5 + 2.2 × 5 + (1 ×) 15 or 5 + 12 + 17 + 11 + 15 (=60) or e.g. 100 + 240 + 340 + 220 + 300 (=1200) Working At least 2 of: 2.5 × 2 (=5) or 4 × 3 (=12) or 3.4 × 5 (=17) or 2.2 × 5 (=11) or (1 ×) 15 or (1 ×) 10 (=10) or e.g. at least 2 of 100, 240, 340, 220, 300 or 200 1 oe 6 Answer 3 Mark A1 M1 M1 1 or 16.6 % or 0.16 or 1 in 6 6 (percentage or decimal rounded or truncated to 3 or more sig figs) for for method to find total number of people (allow one error) or total number of squares/blocks for method used (allow one error) Notes for working with area of at least 2 bars could be using freq density × mins or use of counting squares or blocks 60 Question 20 Alternative method angle OCB = 90 − x angle BOC = 180 – 2(90 – x) (=2x) angle AOB = 2x and angle CDA = 2x Alternative method angle CDB = x or angle CAB = x angle ACB = x angle ACQ = 2x and angle CDA = 2x Working angle CDB = x or angle CAB = x angle CBA = 180 – 2x angle CDA = 180 – (180 – 2x) = 2x proof with reasons 5 5 5 proof with reasons proof with reasons Mark Answer B1 A1 M1 M1 M1 B1 A1 M1 M1 M1 B1 A1 M1 M1 M1 dep for any one appropriate circle theorem reason for complete proof with full reasons angle between tangent and radius is 90o oe, angles in a triangle sum to 180o, isosceles triangle, angle at centre is twice angle at circumference oe dep on M1 for any one appropriate circle theorem reason for complete proof with full reasons alternate segment theorem, isosceles triangle dep on M1 for any one appropriate circle theorem reason for complete proof with full reasons alternate segment theorem, angles in a triangle sum to 180o, isosceles triangle, opposite angles of a cyclic quadrilateral sum to 180o Notes 61 21 Q20 contd 6 6 oe x ( 33) or (gradient = ) 4 4 6 2 m× = −1 or (gradient of M =) oe 4 3 2 k 6 " " 3 4 5 y eg angle ABC = 180 – 2x Angle CAB = angle ACB = [180 – (180 – 2x)] ÷ 2 = x BCQ = CAB = x 12 4 M1 dep A1 M1 M1 M1 B1 A1 M1 M1 or complete method to find equation of line (3y = −2x + 28) and then substitution of x = −4 Dep on M1 for any one appropriate circle theorem reason For complete proof with reasons e.g. opposite angles of cyclic quadrilateral sum to 180° angles in triangle sum to 180° isosceles triangle alternate segment theorem Alternative method where students assume CDA = 2x and must work to show that BCQ = x 62 Question 22 3 5 3 ( 1.62...) 5 ] r 3 2 × sin 1 oe 5 AVB 3 AVB e.g. sin oe eg sin 2 5 2 l ( 0.9772...) and l r 3 oe or l 5 [r 3 5 3 r 2 : r 2 rl 3 : 8 or r 2 : rl 3 : 5 or r 2 3 and rl 5 8 r 2 3( r 2 rl ) or 5 r 2 3 rl or Working r2 3 or 2 r rl 8 73.7 Answer 6 Mark A1 M1 M1 M1 M1 M1 awrt 3 sin 1 ( ) 36.86.... 5 Notes 63 Question 23 or oe 1 2 3 4 BC 4 3 12 5 1 3 4 BC oe or 7 12 5 1 AB 4 1 3 DC 3 4 12 Working e.g. AB AD DB 2 1 3 7 41 cao Answer 5 Mark A1 M1 M1 A1 M1 No isw Notes for a correct vector equation for AB 64 4 A 7 11 5 9 3 1 12 2 6 4 B 10 8 65 6b A 7 5 3 7 2 C 3 11 5 2 B 66 82 May 19 Mark Scheme (Results) Summer 2019 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 2H 2 Two pairs of intersecting arcs with equal radius centre D and E ‘5400’ ÷ ‘48’ 112.5 4 M1 for consistent use of value within interval (including end points) for at least 4 products which must be added 760 + 1260 + 1725 + 1250 + 405 (= 5400) M1 for 2 pairs of arcs that intersect within guidelines or correct perpendicular bisector without arcs. Allow division by their Σf provided addition or total under column seen A1 oe accept 112 or 113 from complete working Accept 112.5 with no working Do not accept 112 or 113 with no working Total 4 marks M1 dep on at least M1 correct midpoints used for at least 4 products and not added or If not M2 then award or Question Working Answer Mark Notes Apart from questions 9, 16, 20, 23 (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method 1 95 × 8 + 105 × 12 + 115 × 15 + M2 for at least 4 correct products added (need not be 125 × 10 + 135 × 3 (= 5400) evaluated) or 83 3 (b) (c) (a) e.g. Examples There are no members that are in both A and B No members in common (in A and B) No numbers the same (in A and B) B has even numbers. A has odd numbers except 2 which is not in B Nothing in A is in B oe No overlap A and B don’t share any numbers 1 and 9 1, 2, 8, 9 Correct statement Correct bisector with arcs 1 2 1 2 Total 2 marks B1 B2 for fully correct (B1 for 3 or 4 correct with no more than one addition or a fully correct Venn diagram) B1 for a statement which indicates correct meanings for intersection and empty set A1 84 (b) '732 ' '480 ' 100 or '480 ' e.g. 1 + 0.2 (= 1.2) or 100(%) + 20(%) (= 120(%)) or '732 ' 100 100 or 152.5 – 100 or '480 ' '732 ' 1 100 or 0.525 × 100 '480 ' ‘252’ ÷ ‘480’ × 100 or e.g. 4 × 120 (= 480) e.g. 120 ÷ 2 × 5 (= 300) or 120 × 0.4 × 7 (= 336) or (120 – ‘60’ – ‘48’) × 8 (= 96) or 120 × 0.1 × 8 (= 96) e.g. (120 ÷ 2 × 5) + (120 × 0.4 × 7) + ((120 – ‘60’ – ‘48’) × 8) (= 732) or (120 ÷ 2 × 5) + (120 × 0.4 × 7) + (120 × 0.1 × 8) (= 732) or ‘300’ + ‘336’ + ‘96’ (= 732) 5 (a) π × 72 × 20 (= 3078.76...) or 980π 4 52.5 3080 5 2 Total 2 marks A1 accept 53 M1 M1 for a complete method to find the percentage profit M1 for a complete method to find the total income M1 M1 for a method to find the income for one of the selling prices M1 for complete method to find volume A1 for answer in range 3077.2 − 3080 Total 4 marks 85 e.g. 15 ÷ 1.2 or 15 ÷ 120 × 100 or 15 × 100 ÷ 120 15 (= 0.125) oe 120 12.5(0) 3 A1 accept (£)12.5, (£)12.50p, 1250p if the £ sign is crossed out Total 8 marks M1 dep 86 5 ALT (b) (a) e.g. 15 ÷ 1.2 or 15 ÷ 120 × 100 or 15 × 100 ÷ 120 15 (= 0.125) oe 120 e.g. 1 + 0.2 (= 1.2) or 100(%) + 20(%) (= 120(%)) or 12.5(0) 52.5 3 5 M1 for a method to find the profit of one of the books e.g. 120 ÷ 2 × 1 (= 60) or 120 × 0.4 × 3 (= 144) or (120 – ‘60’ – ‘48’) × 4 (= 48) or 120 × 0.1 × 4 (= 48) e.g. (120 ÷ 2 × 1) + (120 × 0.4 × 3) + ((120 – ‘60’ – ‘48’) × 4) (= 252) or (120 ÷ 2 × 1) + (120 × 0.4 × 3) + (120 × 0.1 × 4) (= 252) or ‘60’ + ‘144’ + ‘48’ (= 252) ‘252’ ÷ ‘480’ × 100 oe A1 accept (£)12.5, (£)12.50p, 1250p if the £ sign is crossed out Total 8 marks M1 dep A1 accept 53 M1 M1 for a complete method to find the percentage profit M1 for a complete method to find the total profit M1 4 × 120 (= 480) 87 7 6 19.5 ÷ 2.5 or 19.5 × 0.4 oe or (b) 19.5 a e.g. 30 × 26.8 (= 804) or 13 × 25 (= 325) or (26.8 – 25) × 30 or 1.8 × 30 e.g. (30 × 26.8 − 13 × 25) ÷ (30 – 13) (= 28.1764...) or (‘804’ – ‘325’) ÷ (30 – 13) (= 28.1764...) or (‘804’ – ‘325’ ÷ 17) (= 28.1764...) or ((26.8 – 25) × 30) ÷ 17 + 25 (= 28.1764...) or ‘1.8’ × 30 ÷ 17 + 25 (= 28.1764...) 4.2 2.5 or 0.4 or 0.7 or 1.4(2857……) 15 6 4.2 6 or or or oe 6 15 6 4.2 (a) 28.2 7.8 10.5 3 2 2 Total 4 marks A1 accept 28.15 – 28.2 (accept without working) (Accept 28 from complete working) Total 3 marks M1 for a complete method to find the mean mark for the girls M1 for finding the total marks for the boys or the total test marks A1 oe A1 oe M1 If using DF ft their answer from part (a) M1 for a correct scale factor, accept ratio notation eg 6 : 15 88 x 1000 oe 60 60 (x) × 1000 or (x) ÷ 60 or (x) ÷ 60 ÷ 60 or (x) × 1000 ÷ 60 oe Using elimination then substitution 9 e.g. e.g. x + 2y = −0.5 3x + 6y = −1.5 + 6x – 2y = 32 − 3x – y = 16 (7x = 31.5) (7y = −17.5) e.g. e.g. ‘4.5’ + 2y = −0.5 x + 2 × ‘−2.5’ = or −0.5 3 × ‘4.5’ – y = 16 or 3x – ‘−2.5’ = 16 8 x = 4.5 y = −2.5 5 x 18 3 3 x 1 x or 3.6 3.6 5 oe 18 Total 3 marks A1 (dep on first M1) for both solutions Total 3 marks for repeating above method to find second variable M1 (dep) for substituting their value found of one variable into one of the equations or M1 for a correct method to eliminate x or y: coefficients of x or y the same and correct operation to eliminate selected variable (condone any one arithmetic error) A1 accept 0.27x or 0.27 x or M1 (dep) for a complete correct method M1 for at least one of × 1000 or ÷ 60 or 89 16 ' 2.5' x 3 x = −0.5 – 2‘–2.5’ or e.g. Using substitution 9 3(−0.5 – 2y) – y = 16 (7y = −17.5) or 16 y 2 y 0.5 3 (7y = −17.5) or 0.5 '4.5' 2 y 3' 4.5' 16 y (7x = 31.5) e.g. x 2 3x 16 0.5 (7x = 31.5) or 0.5 x 3x 16 2 x = 4.5 y = −2.5 3 A1 (dep on first M1) for both solutions Total 3 marks M1 (dep) for substituting their value found of one variable into one of the equations M1 for correctly writing x or y in terms of the other variable and correctly substituting 90 10 0≤x≤2 1≤y≤3 (b) x ≥ 0, x ≤ 2, y ≥ 1, y ≤ 3 or y = 5x – 3 oe (a) 2 2 Accept <, ≤, > and ≥ throughout (SC B2 y > 3, y < 1, x < 0, x > 2) Total 4 marks (Treat double−ended inequalities as two separate inequalities) (B1 for 2 or 3 out of 4 inequalities correct) B2 fully correct oe y = 5x or y = 5x + a or y = bx – 3 or (L=) 5x – 3 If not B2 then B1 for B2 fully correct equation eg y = 5x + −3 or y − −3 = 5(x – 0) 91 11 (8.3 × 103) × 50 (= 415 000 or 4.15 × 105 ) or (4.2 × 104) ÷ 50 (= 840 or 8.4 × 10² ) or (4.2 × 104) ÷ (8.3 × 103) (= 5(.060...)) 1.15 × 0.92 (= 1.058) oe or 105.8 (b) (c) e.g. 200 1.15 0.92 200 number or variable n 1.15 0.92 where n is a n ±(7.7 × 104 – 9.5 × 103) or ±(7.7 × 104 – 0.95 × 104) or ±(77 000 – 9 500) or ±67 500 oe (a) 5.8 No supported by correct comparabl e figures in the same form 6.75 × 104 2 2 2 A1 NB. −5.8 (M1A0) decrease of 5.8% (M1A0) condone x × 1.15 × 0.92 oe NO and 5(.060...) M1 NO and 840 and 8 300 or NO and 8.4 × 10² A1 for NO and 415 000 and 42 000 or NO and 4.15 × 105 A1 allow −6.75 × 104 allow ±6.8 × 104 M1 for a relevant calculation M1 for clearly subtracting the correct values 92 12 16.7 (= 24.48686...) sin 43 16.7 (=17.90855...) tan 43 16.7 (= 24.48686...) sin 43 CD ED CD 16.7 + 21.2 × 2 + ‘24.5’ + ‘17.9’ (= 101.495...) and or 16.7 (=17.90855...) tan 43 ED 101 4 A1 accept 101 – 102 Total 4 marks NB. Sine rule must be in the correct form to give the answer M1 (dep on M2) complete method with no extra sides ED '24.48....'2 16.72 (=17.90855...) CD 16.72 '17.90....'2 (= 24.48686...) use of Pythagoras theorem NB. Sine rule may be used M1 for a correct method to find both CD and ED or (E is the point on line AD from where a vertical line is drawn downwards from point C) M1 for a correct method to find length CD or ED Total 6 marks 93 13 (c) (d) (b) (a) e.g. readings of 11−13 and 22−24 indicated on horizontal axis or 23 − 12 For correct use 20 and 60 (20.25 and 60.75) indicated (horizontal line or mark) on the cumulative frequency axis and their readings taken from time taken axis 9 – 13 Correct cf graph 17 − 19 7, 17, 29, 48, 66, 80 2 1 2 1 A1 accept 9 – 13 ft from a cumulative frequency graph dep on M1 in (b) Total 6 marks A1 accept curve or line segments accept curve that is not joined to (0,0) B1 ft from a cumulative frequency graph dep on M1 in (b) M1 for a complete method to ft from a cumulative frequency graph dep on M1 in (b) for all 6 points plotted consistently within each interval in the frequency table at the correct height for at least 4 points plotted correctly at end of interval or M1 ft from (a) if only one addition error B1 cao 94 15 14 x 2 y2 3 y2 4 A1 accept x 2 y 2 oe y2 3 Total 4 marks M1 for isolating terms in x and factorising the correct expression of the equation Total 3 marks y2 + 2 = x(3 – y2) oe M1 squaring both sides to get a correct equation A1 1 6 1 6 5 or (= ) 3 9 3 9 12 or M1 for multiplying by the denominator and expanding the bracket 3x 2 x 1 3 52 122 6 1 6 1 5 or (= ) or 9 3 9 3 12 If not M2 then M1 for 2 5 122 or xy2 + y2 = 3x – 2 oe y2 13 2 52 12 or M2 for 95 16 2 1 2 1 = 4 2 4 4 8 oe 4 2 4 8 2 8 or 2 1 4 2 44 8 or 2 1 4 2 4 16 8 or 2 1 e,g, 4 8 2 1 8 + 6√2 3 A1 (dep on M2) or for stating a = 8 and b = 6 4 2 4 8 2 8 4 2 4 4 8 or or 2 1 2 1 4 2 4 16 8 2 1 e.g. Using −√2 − 1 Total 3 marks condone missing brackets M1 (dep) for expansion of numerator with at least 3 terms correct oe M1 for rationalising the denominator by multiplying numerator and denominator by √2 + 1 (or −√2 − 1) 96 17 (b) (a) 3 67.5h ' 20 ' oe h2 20h = k × h3 oe y = kx3 or ky = x3 20x3 h2 1.5h y 2 3 20x3 oe h2 A1 accept 3 3h h or 2 2 Total 5 marks (a) or in part (b) M1 ft, dep on at least M1 in part (a), complete method to find x 20 oe is seen in part h2 20h = k × h3 oe M2 for Award 3 marks if answer is y = kx3 and k A1 for y M1 (NB. Not for y = x3) Constant of proportionality must be a symbol such as k M1 substitution of x and y into a correct formula 97 18 ‘−10’[(x – ‘2.4’)2 – ‘2.4’2] oe ‘−10’ × − ‘2.4’2 or ‘−10’ × − ‘5.76’ (x = 2.4) 48 × ‘2.4’ – 10 × ‘2.4’2 57.6 5 A1 accept 58 from correct working Total 5 marks finding max value of A from completing the square or for isolating x and substituting into A completing the square M1 ft if previous M1 awarded or M1 for differentiating a correct expression for A (allow 1 error) and equating to zero M1 for a complete expression for A (6 sides) with brackets expanded x2 + x2 + 48x − 12x2 (= 48x − 10x2) ‘48 – 20x’ = 0 M1 for finding an expression for the area of one face x2 oe or x(12 − 3x) oe 98 19 BC sin 95 '30.56 ' sin 180 95 47 sin 95 BC 250 + 0.5 × ‘30.56’ × ‘22.43’ × sin(180 – 95 – 47) (= 461.03....) or 250 + 0.5 × ‘30.56’ × ‘18.88’ × sin(47) (= 461.03....) (= 18.8(8524...)) (= 22.4(3407...)) or '30.56 ' sin 47 sin 95 '30.56 ' oe sin 95 AB sin 180 95 47 or sin 47 461 6 (AC=) 30.5(5579...) or 30.6 A1 accept 461 - 462 22.39 – 22.47 for AB 18.8 – 18.92 for BC M1 for a complete method to find total area sight of values in the ranges Total 6 marks M1 (dep on previous M marks) for a correct method to find a missing length or M1 dep on M1 for correct substitution into sine rule If this mark is awarded then ft on the remaining M marks A1 AB '30.56 ' oe M1 for using the area formula correctly 250 = 0.5 × 26 × AC × sin(39) oe 99 20 y = 4.5 and y = 3 x = 4.5 and x = 6 2×2 e.g. 15 2 15 2 2 x 27 4 4 e.g. 2 2 21 21 2 x 54 4 4 2×2 −(−15) ± √(−15)2 − 4×2×27 y = −(−21) ± √(−21) − 4×2×54 2 allow 2y2 – 15y = −27 oe e.g. (2y – 9)(y – 3) (= 0) allow 2x2 – 21x = −54 oe e.g. (2x – 9)(x – 6) (= 0) x = (9 – y)2 – 3y(9 – y) + 2y2 (= 0) e.g. 6y2 – 45y + 81 (= 0) or 2y2 – 15y + 27 (= 0) x2 – 3x(9 – x) + 2(9 – x)2 (= 0) e.g. 6x2 – 63x + 162 (= 0) or 2x2 – 21x + 54 (= 0) (4.5, 4.5) and (6, 3) 5 Total 5 marks A1 (dep on M1) both x-values or both y-values A1 (dep on M1) oe Must be paired correctly 21 441 432 ) 4 allow ax2 + bx = c M1 (dep on M1) for a complete method to solve their 3-term quadratic equation (allow one sign error and some simplification – allow as far as M1 substitution of linear equation into quadratic A1 (dep on M1) writing the correct quadratic expression in form ax2 + bx + c (= 0) 100 20 Alt (9 – y – y)( 9 – y – 2y) (= 0) (9 – 2y)(9 – 3y) (= 0) oe y = 4.5 and y = 3 (x – (9 – x))(x – 2(9 – x)) (= 0) (2x – 9)(3x – 18) (= 0) oe x = 4.5 and x = 6 (x – y)(x – 2y) (= 0) (4.5, 4.5) and (6, 3) 5 A1 (dep on M1) oe Must be paired correctly Total 5 marks A1 (dep on M1) both x-values or both y-values M1 (dep on M1) A1 (dep M1) substitution of L into their factorised C M1 for a method to factorise C 101 21 (= √29) or 3x (= 33.854...) or ' 20 x ' (CAF =) tan 1 3 (= 33.854...) or ' 20 ' ' 20 ' (CAF =) cos 1 ' 29 ' (= 33.854...) or 3 (CAF =) sin 1 (= 33.854...) ' 29 ' (CAF =) tan 1 e.g. (AF =) √(√20) + (3) 2 33.9° 3 A1 answers in the range 33.85 – 33.9 (CAF =) tan 1 1.5 (= 33.854...) ' 5' AB : BC : CF = 2 : 1 : 1.5 x can represent any number Total 3 marks M1 for a complete method to find angle CAF using length AC or for a complete method to find angle CAF using length AF with or without x or AC 2 22 12 5 e.g. AB : BC : CF = 2 : 1 : 1.5 2 2 (AF =) √(4)2 + (2) + (3) (= √29) or 2 x can represent any number M1 for a method to find an expression for length AC or length AF with or without x or (AC =) √(4)2 + (2)2 (= √20) or (AC =) √(4x)2 + (2x)2 (= √20x) or e.g. 102 22 x(2x + 5)(3x − 1) or (2x + 5)(3x2 − x) and (2x + 5)(2x – 5) oe x(2x + 5)(3x − 1) or (2x + 5)(3x2 − x) or (2x + 5)(2x – 5) oe x 3x 1 2x 5 3 A1 accept 3x 2 x oe Do not ISW 2x 5 Total 3 marks denominator into 2 brackets where one of the factors must be (2x + 5) and numerator into 2 or 3 factors where one of the factors must be (2x + 5) M1 for a correct factorisation of the denominator into 2 brackets where one of the factors must be (2x + 5) or numerator into 2 or 3 factors where one of the factors must be (2x + 5) M1 for a correct factorisation of the 103 23 3 2 t 3 t 4 or t t 1 t t 1 3 2 + t t 1 t 3 t 4 23 = t t 1 35 3 t 3 t 3 3 or t t 1 t t 1 3 t 3 + t t 1 t 3 3 12 = or t t 1 35 3 t 3 12 2 oe t t 1 35 2×2 2 −(−37)±√(−37) − 4×2×105 37 2 37 2 e.g. 2 t 105 4 4 e.g. t = e.g. (2t – 7)(t – 15) = 0 allow 2t2 – 37t = −105 e.g. 2t2 – 37t + 105 (= 0) or RR and GG method RG and GR method 12 5 A1 (dep on A1) cao 7 ) 2 Total 5 marks Can be implied by answers of 15 (and 37 1369 840 ) or 4 A1 (dep on M2) writing the correct quadratic expression in form ax2 + bx + c (= 0) allow ax2 + bx = c M1 (dep on A1) for a complete method to solve the 3-term quadratic equation (allow one sign error and some simplification – allow as far as M1 dep on M1 for a correct equation M1 for one correct product 104 23 Alt 3 2 or x3 x2 x x 1 x3 x2 3 2 + x3 x2 x x 1 23 = x 3 x 2 35 3 x or x3 x2 x 3 x3 x2 3 x + x3 x2 x 3 12 = or x 3 x 2 35 3 x 12 2 oe x 3 x 2 35 2×2 2 −(−25)±√(−25) − 4×2×12 2 2 25 25 e.g. 2 x 12 4 4 e.g. x = e.g. (2x – 1)(x – 12) = 0 allow 2x2 – 25x = −12 e.g. 2x2 – 25x + 12 (= 0) or RR and GG method RG and GR method 12 5 ) A1 (dep on A1) cao 1 2 Total 5 marks can be implied by answers of 12 (and 25 625 96 ) or 4 A1 (dep on M2) writing the correct quadratic expression in form ax2 + bx + c (= 0) allow ax2 + bx = c M1 (dep on A1) for a complete method to solve the 3-term quadratic equation (allow one sign error and some simplification – allow as far as M1 dep on M1 for a correct equation M1 for one correct product 105 (a) (b) 2 2 2 x 5 y 41 oe 2 2 y 2 x 5 41 oe 9 2 y 2 x 5 52 or 2 y 2 x 5 5 9 or e.g. 9 y 2 x 2 10 x 2 y 2 x 10 x 9 or 2 Note: Allow candidates to swap x and y when finding the inverse 24 5 x 41 2 13 4 1 A1 oe M1 M1 dep Total 5 marks B1 M1 for a correct equation for a first step in order to complete the square 106 (b) (a) x 5 x 41 y oe 2 5 x 41 2 4 A1 oe M1 Total 5 marks M1 dep B1 20 400 8(9 y ) or 4 20 400 8(9 y ) x 4 1 M1 for a correct first step 13 2 x2 - 20 x + (9 - y) = 0 Note: Allow candidates to swap x and y when finding the inverse 24 Alt 107 (b) (a) 2 x 5 2 y 41 oe 2 2 x 5 41 y (= 0) 9 2 2 x 5 52 y (= 0) or 2 2 x 5 52 9 y (= 0) or 2 5 x 41 2 4 A1 oe M1 Total 5 marks M1 dep B1 e.g. 1 M1 for a correct first step 13 2 x2 20 x (9 y) 0 Note: Allow candidates to swap x and y when finding the inverse 24 Alt 108 120 Mark Scheme (Results) Jan 2020 January 2020 Pearson Edexcel International GCSE In Mathematics A (4MA1) Paper 2H 2 1 (b) (a) Q Working Answer 32.4 × 1003 eg 78 × 74 = 712 or 78 ÷ 73 = 75 or 75 × 74 or 74 ÷ 73 = 7 or 78 × 7 or 7’12’ ÷ 73 = 7’12’−3 32 400 000 79 x 7 Mark Notes 2 2 1 M1 A1 A1 M1 B1 for 32.4 × 1003 oe for 32 400 000 accept 3.24 × 107 Total 2 marks Total 3 marks for one correct step – must be written as a power of 7 for 79 Apart from Questions 3, 7b, 12, 17, 20, 22 the correct answer, unless clearly obtained by an incorrect method, should be taken to imply a correct method International GCSE Maths 121 3 eg 14 5 19 3 70 57 70a 57a or or or 3 5 15 15 15a 15a 10 12 22 4 3 7 oe 15 15 15 70 57 127 7 22 7 8 or 7 8 15 15 15 15 15 15 7 127 or if shows 8 at the beginning then show that the 15 15 127 addition comes to 15 14 19 10 12 10a 12a ( ) or (4) ()(3) or (4) ()(3) 3 5 15 15 15a 15a Shown 3 for correct fractions with a common denominator of 15 or a multiple of 15 dep on M2 for a correct answer from fully correct working or shows that 127 and fully correct working RHS = 15 127 shows LHS = 15 M1 A1 Total 3 marks for correct improper fractions or fractional part of numbers written correctly over a common denominator M1 122 5 4 5x = ‘120’ or ‘120’ ÷ 5 or 180 – 30 – 10 – 20 (=120) 24 Fully correct angle bisector with all relevant arcs shown e.g. 30 + 4x + 10 + x + 20 = 180 or 5x + 60 = 180 oe 30 + 4x + 10 + x + 20 (= 5x + 60) or 180 – 30 (=150) 4 2 A1 M1 M1 M1 B2 Fully correct angle bisector with all arcs shown. B1 for all arcs and no angle bisector drawn or for a correct angle bisector within guidelines but not arcs or insufficient arcs Total 2 marks Allow 5x + 60 = n M2 for where n ≠ 180 or for 5x + 30 = 150 subtracting 30 from 180 oe for setting up the equation or for subtracting all numerical values of angles from 180 for correctly simplifying to ax = b or for dividing ‘120’ by 5 for 24 Total 4 marks 123 6 ‘0.45’ ÷ 3 (= 0.15) Or ‘0.45’ × 180 (= 81) Or 180 − 99 (= 81) ‘0.15’ × 180 Or ‘81’ ÷ 3 1 – (0.24 + 0.31) (= 0.45) Or (0.24 + 0.31) × 180 (= 99) 27 4 A1 M1 M1 M1 or for an answer of Total 4 marks 27 180 or for a correct equation for missing values eg x + 0.24 + 2x + 0.31 = 1 oe (can be implied by 2 probabilities that total 0.45 in table if not contradicted in working space) (or 0.15 correctly placed in table as long as not contradicted) 124 7 (b) (x − 8)(x + 5) (x ± 8)(x ± 5) 2 3 9 160 2 3 169 3 13 or 2 2 or (3) (3) 4 1 (40) 2 1 (a) 2x ˃ 4 – 7 or x + 3.5 > 2 8, −5 x ˃ −1.5 3 2 A1 M1 3 169 3 13 or 2 2 dep on at least M1 for correct values Total 5 marks M1 For a correct first step allow 2x = 4 – 7 or x + 3.5 = 2 or an answer of x = −1.5 or x < −1.5 or −1.5 A1 for x ˃ −1.5 oe M1 or (x + a)(x + b) where ab = −40 or a + b = −3 OR correct substitution into quadratic formula (condone one sign error in a, b or c and missing brackets) (if + rather than ± shown then award M1 only unless recovered with answers) 125 45 47 100( 9) and 100 ( 8.6) 500 545 (a) 545 – 500 ( = 45) or 592 – 545 ( = 47) 45 47 100 ( 9) or 100 ( 8.6) 500 545 0.15 × “1120” or “1120” – 952 oe (b) 952 ÷ 85 × 100 oe (=1120) Alternative mark scheme for 8(a) 545 545 100( 109) or ( 1.09) and 500 500 592 592 100 ( 108.6) or 1.086 545 545 8 168 No, 109(%) and 108.6(%) No, 9(%) and 8.6(%) 3 4 4 oe eg no and 1.09 and 1.086 M1 for a method to find price before discount M1 for a correct method to find discount A1 A1 952 15 85 Total 7 marks M2 for (if not M3 then award M2 for one of these expressions) M3 for both correct expressions which should lead to 109 or 1.09 and 108.6 or1.086 (allow 108 or 108.7 from correct working for 108.6 or 1.08 or 1.087 from correct working for 1.086 throughout) M1 for both correct expressions or having found “9%” finds 109% of 545: 1.09 × 545(=594.05) or 9% of 545 (49.05) or having found “8.6%” finds 108.6% of 500: 1.086 × 500(=543) or 8.6% of 500 (43) A1 for no oe, 9% and 8.6% seen or no oe and 9% and 594.05 or 8.6% and 543 or No, 49.05 > 45 or No 594.05 > 592 oe M1 may be seen as part of a calculation M1 for one correct expression (allow 8 or 8.7 from a correct expression for 8.6 throughout) 126 50 60 60 5 oe eg 50 ÷ 1000 18 1000 ÷ 60 ÷ 60 (= 0.27777…. or 5 ) 18 50 × 60 (= 3000) or 50 ÷ 1000 (= 0.05 or 10 or 50 × 60 × 60 (= 180 000) or or 60 60 ( 3.6) 1000 or 19.3 × 150 9 1 ) 20 180 2895 3 2 A1 M1 M1 M1 A1 Total 2 marks for 180 (SCB1 for both conversion factors correct but applying them wrongly 50 1000 ) eg 60 60 Total 3 marks (dep) for a complete method 1000 ÷ 60 ÷ 60 60 60 ( 3.6) 1000 or or for 50 with at least one of ÷ 1000 or × 60 for 19.3 × 150 for 2895 127 2 ( 4 12.566...) ‘12.566…’+ 15 + 17 2 '8' (AC =) 17 15 ( 64 8) 2 (AC2 =) 172 − 152 A1 M1 M1 “12.566” + 15 + 17 5 Total 5 marks dep on M2 for '8' or 12.5663… or 4π 2 for “12.566” + 15 + 17 and no additional values for awrt 44.6 Total 5 marks for a correct method to find one of the angles additional values for awrt 44.6 oe or 4π A1 2 '8' for ‘12.566’ + 15 + 17 and no 12.5663… dep on M2 for M1 M1 M1 M1 M1 M1 44.6 44.6 5 15 × tan (‘28.0724’)(= 8) or 15 ÷ tan (‘61.9275’)(= 8) '8' (= 4π = 12.566…) 2 Alternative mark scheme for 11 15 15 cos 1 (= 28.0724) or sin 1 (= 61.9275) 17 17 11 128 12 Cost per litre then conversion 2500000 5.952.. 4.2 105 770 000 (0.895..) 8.6 105 ‘5.952’ ÷ 6.57(=0.9059..) or ‘0.895’ × 6.57(= 5.882..) 8.6 105 8.6 105 ( 1.1168) or ( 0.1699..) 770 000 '5058900 ' Conversion then litres per amount of money 2500 000 (380517.5..) or 770 000 × 6.57(= 505 8900) 6.57 4.2 105 4.2 105 ( 0.168) or ( 1.103..) 2500 000 '380517.5' 4.2 105 ( 0.168) l/k 2500 000 A: l/$ to l/k ‘1.1168’ ÷ 6.57 (= 0.1699..)or D: l/k to l/$ ‘0.168’ × 6.57 (= 1.103..) Litres per amount of money and then conversion 8.6 105 ( 1.1168) l/$ 770 000 Arctic Oil and relevant figures Arctic Oil and relevant figures Number of litres per Krone for A l/$ to l/k for A or l/k to l/$ for D M1 M1 Litres per Krone or litres per $ for D Litres per Krone or litres per $ for A M1 M1 Price per litre in $ for A M1 M1 Conversion of Krone to $ or $ to Krone Price per litre in Krone for D M1 A1 for Arctic Oil with 1.1168… and 1.10376… or 0.168 and 0.1699.. Changing Krone to $ or $ to Krone M1 A1 for Arctic Oil with 1.1168… and 1.10376… or 0.168 and 0.1699.. Number of litres per $ for D M1 129 4.2 105 21 ( 0.488..) 5 8.6 10 43 ‘0.488..’ × 770 000 $ (=376046.511..)$ ‘376046.511..’× 6.57 =2470625.58..K or 2500 000÷6.57 = 380517..$ Arctic Oil and relevant figures For Arctic Oil with 5.952 and 5.882 or 0.895 and 0.9059 For Arctic Oil with 5.952 and 5.882 or 0.895 and 0.9059 M1 Multiplier for same amount of D as A or same amount of A as D A1 M1 Cost per litre in $ or cost per litre in Krone for A M1 Cost per litre in Krone or cost per litre in $ for D M1 Changing Krone to $ or $ to Krone A1 M1 Cost of equal amount of D as A or A as D M1 Converts so can compare costs – either K to $ or original A to K or $ to K or original D to $ Arctic Oil and A1 Arctic Oil and 779154.. or with relevant figures 2470625..(figures may be rounded) Or Arctic Oil with 5119047… and 5058900 or with 376046.. and 380517 Students may compare other equal amounts – please use the scheme that best fits their method and award marks appropriately. Total 4 marks Comparing equal amounts 8.6 105 43 ( 2.047..) 5 4.2 10 21 ‘2.047..’×2500 000 K (=5119047.619..)K ‘5119047.619’÷6.57 = 779154.88…$ or 770 000×6.57=5058900 K Conversion then cost per litre 2500 000 (380517.5..) or 770 000 × 6.57(= 505 8900) 6.57 2500000 '380517.5..' ( 5.952) or (0.9059..) 5 4.2 10 4.2 105 770 000 '5058900 ' ( 0.895) or ( 5.882..) 5 8.6 10 8.6 105 Arctic Oil and relevant figures 130 13 Angles in the same segment are equal, angle in a semicircle is 90° (or angle at centre is double angle at circumference oe) angles in a triangle add up to 180°/angles in a triangle isosceles triangle alternate angles vertically opposite angles (or vertically opposite) angles at a point opposite angles in a cyclic quadrilateral angle between tangent and radius (diameter) alternate segment theorem angles subtended by the same arc(or chord) at the circumference (or on the circle) Angle CAD = 28° or angle ACB = 32° or angle ACD = 90° or angle ABD = 90° 30° 4 A1 B2 M1 Total 4 marks For a correct answer of 30 Dep on M1 for all correct reasons for their method used (if not B2 then award B1(dep on M1) for a correct circle theorem reason) 131 15 14 (b) (a) 7 6 oe only 20 19 21 190 C, B, E 21 190 Correct probabilities on the tree diagram 3 2 2 B3 (B2 (B1 A1 M1 B1 B1 Total 3 marks Total 4 marks 21 oe or 0.11… (at least 2 190 for all 3 correct for 2 correct) for 1 correct) dp) for 13 7 and on the first branch 20 20 (0.65 and 0.35) 12 7 13 6 , , and on the for 19 19 19 19 second branch (accept 2 dp or better 0.6315..., 0.3684..., 0.6842..., 0.3157...) ft from (a) as long as probabilities less than 1 for 132 17 16 4 y2 1 y2 1 e.g. n2 – n2 + 2n – 1 or n2 + 2n + 1 − n2 e.g. n2 – (n – 1)2 or (n + 1)2 – n2 e.g. 2n – 1 is always odd 3 A1 M1 M1 A1 SCB1 for eg (2n)² − (2n – 1)² or (2n + 1)² − (2n)² oe Total 3 marks dep on M2 for eg 2n – 1 or 2n + 1 or – (2n + 1) oe and a suitable conclusion Correct expansion of brackets and correct signs or a correct result for setting up a correct algebraic expression (any letter can be used) for expanding the bracket and rearranging for x so that the terms in x are on one side of the correct equation 4 y2 1 4 y 2 1 for x 2 or x y 1 1 y2 (need to see x = somewhere) Total 4 marks M1 y 2 x x 4 y 2 1 or 4 y 2 1 x y 2 x or x for removing the fraction M1 x( y 2 1) 4 y 2 1 or 4 y 2 1 x(1 y 2 ) for squaring M1 4 x 1 x4 2 y ( x 4) x 1 or y 2 x 4 y 2 x 1 y2 133 18 (b) or all correct values in bars oe not added (1 × 7) + (2.5 × 6) + (5 × 4.8) = 7 + 15 + 24 (= 46) or no. of sml squares (48 × 5) + (6 × 25) + (7 × 10) = 240 + 150 + 70 (=460) 46 120 no. of sml squares = (10 × 7) + (5 × 34) + (9 × 10) + (6 × 25) + (15 × 48) = 70 + 170 + 90 + 150 + 720 (= 1200) (a) (0.7 × 10) + (3.4 × 5) + (1 × 9) + (2.5 × 6) + (4.8 × 15) = 7 + 17 + 9 + 15 + 72 (= 120) 9 15 46 120 2 3 M1 for a correct method to work out the area between 17 minutes and 35 minutes eg using frequency density or number of small squares oe A1 46 for oe (allow 2 dp or better 0.3833... or 38% or 120 better) M1 n q where n ˂ 15 or where q < 720 or for 15 720 r where r < 72 or 72 9 432 where m > 9 or where p > 432 m p 43.2 where t > 43.2 t A1 9 oe 15 Total 5 marks M1 for a correct method to work out the total area eg total frequency or number of small squares or other correct method (allow one error in method) [count use of 25 for 24 as one error] or all correct values in bars oe not added 134 19 (b) (a) 2 1 1 2 3 or m or − or −0.6 2 3 3 2 5 5 mp 3 5 4 (6) c oe eg 4 = −10 + c (c = 14) 3 5 y 4 ( x 6) 3 m 4x + k (oe) Eg 5x – 3y + 42 = 0 y= M1 4 numerical value k ≠ 7 Could be written in another form e.g. 3y + 12x = 20 for using 𝑚 for y = 4x or y = 4x + k where k is any M1ft dep on previous M1 for substituting (−6, 4) into linear equation formula 5 4 x c to find value of c or 3 5 y x 14 or y 1.66...x 14 3 A1 for correct simplified equation where all values are integers 10x – 6y + 84 = 0 or 3y = 5x + 42 oe Total 5 marks M1ft for using m1 m2 1 B1 1 135 21 20 (a)(i) (iii) (b) 18( 7 1) 7 1 eg x² + 3x + 10 oe or 3 9 eg ( x ) 2 4 6 oe 2 4 eg y – 6 = x² + 3x + 4 eg (x – 4)² + 3(x – 4) + 4 oe or 3 9 eg ( x 4) 2 4 oe or 2 4 3 7 3 eg 18 7 1 7 1 7 1 y = (x – 4)² + 3(x – 4) + 10 or 3 9 y ( x 4) 2 4 6 2 4 (0, 6) (2, 6) 3 7 3 2 2 3 A1 B1 B1 M1 A1 M1 M1 Total 4 marks 5 31 or oe eg y = ( x )2 2 4 y = x² − 5x + 14 oe for applying one of the transformations to the equation Total 3 marks 18 7 1 7 1 7 1 Dep on M1 for a correct numerator and multiplying out the denominator to 7 – 1 or 6 Dep on M2 Allow 3( 7 -1) for 136 x 2 2 1 1 (4 1 12) or 2 1 2 2x² + 2x – 24 (=0) or x² + x −12 (=0) or 2x² + 2x =24 or x² + x =12 (x + 4)(x − 3) (= 0) or x2 ( x 2)2 2( x 2) 24 y = 5 and y = −2 (−4, −2) and (3, 5) y 3 ( 3) 2 (4 1 10) or 2 1 2 2 3 3 y 10 0 2 2 2y² − 6y – 20 (=0) or y² − 3y – 10 (=0) 2y² − 6y = 20 or y² − 3y = 10 ( y 5)( y 2) 0 or 1 1 x 12 0 2 2 x = −4 and x = 3 (−4, −2) and (3, 5) Alternative mark scheme for 22 ( y 2)2 y 2 2 y 24 22 (−4, −2) and (3, 5) (−4, −2) and (3, 5) 5 5 A1 A1 M1ft A1 M1 A1 A1 M1ft A1 M1 which expanded give 2 out of 3 terms correct for both y values dep on M1 for both solutions dep on M1 Total 5 marks 3 9 40 ) or if factorising, allow brackets 2 for substituting linear equation into the quadratic equation for a correct equation in the form ay2 + by + c = 0 or ay2 + by = −c dep on M1 for solving their quadratic equation using any correct method (allow one sign error and some simplification – allow as far as for substituting linear equation into the quadratic equation for a correct equation in the form ax2 + bx + c = 0 or ax2 + bx = −c dep on M1 for solving their quadratic equation using any correct method (allow one sign error and some simplification – allow as far as 1 1 48 ) or if factorising, allow brackets 2 which expanded give 2 out of 3 terms correct) for both x values dep on M1 for both solutions dep on M1 137 23 uuuur 3 3 1 1 1 PM = a b 4a (2b 4a) a b 2 4 2 4 2 uuuur 1 AM 4a (2b 4a) ( 2a b) 2 uuuur 1 AM 2b (4a 2b) ( 2a b) 2 uuur 1 MA (2b 4a) 2b ( 2a b) 2 uuur 1 MA (4a 2b) 4a ( 2a b) 2 3 3 1 1 ( AP : PM ) a b : a b oe 2 4 2 4 3 3 ( AP : AM ) a b : 2a b ( 3 : 4) oe 2 4 1 1 ( AM : PM ) 2a b : a b (= 4 : 1) oe 2 4 3 3 1 1 AP 3PM oe eg a b 3( a b) oe 2 4 2 4 4 AM AP oe 3 AM 4PM oe 3:1 3 Total 3 marks For use of a correct ratio or fraction linking AP and PM or AP and AM or AM and PM (in either order) vectors must be in form pa + qb M1 A1 uuuur uuuur uuur for finding PM or AM or MA M1 138 24 3x 1 x 2 x 3 3 2 x e.g. A1 2 or isw for incorrect denominator expansion Total 4 marks 2 3x 1 oe 12 x 9 x 4 x 3 3x 1 or x 12 x 9 4 x 2 x 2 x 3 1 3x 3x 1 or x 2 x 3 3 2 x Writing 1st fraction as a fraction over a common denominator (can be 2 separate fractions) Complete factorisation of numerator or denominator of 2nd fraction may be partially simplified M1 M1 4 3x 1 2 x 5 2x 3 2 x 5 2 x 3 x 3 2 x 3 2 x 8 x 12 6 x 15 oe 2 x 5 2 x 3 M1 or x 3 2 x 3 2 x or 3x 1 2 x 5 2 x 5 2 x 3 4 2 x 3 3 2 x 5 139 25 50 2 50 50 1 k oe 2 33125 25 100 49k oe 1325 = 100 + 49k oe 1225 = 49k oe 33125 n = 50 25 3 A1 B1 M1 Total 3 marks (k may be written as d) For correct equation, using formula with a = 50 and n = 50 substituted (for this mark, allow n = 49) 140 26 1600 or "7.817..." "26.193..."( 643.315...) x "49.1196..." 2 "26.193..." 360 or x "643.315..." "26.193..."2 360 or 2 × π × “7.817…” (= 49.1196…) l "7.817..."2 252 ( 686.1154... 26.193...) 1 25 3 192 r ( 61.1(154..) 7.8176...) eg r 1 1600 r 2 25 oe 3 107° 6 M1 Dep on M2 correct method to find slant height of cone (radius of sector) M1 for using 𝐶 2𝜋𝑟 oe using figures from correct method or for using A rl using figures from correct method M1 x for using arc length = 2 r 360 or for using area of sector = x r2 360 A1 for 107° - 108° Total 6 marks M1 dep for correct rearrangement of volume formula for r M1 for substituting into volume formula for cone correctly and equating to 1600 141
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