Mark Scheme (Results) January 2021 Pearson Edexcel International GCSE Mathematics A (4MA1) Paper 1H

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International GCSE Maths Apart from questions 5(b), 15, 17, 18, 19, 23 and 24 (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be t... aken to imply a correct method Q Working Answer Mark Notes 1 e.g. 0.7 × 20 160 oe (= 14 112) or 0.3 × 20 160 oe (= 6048) 4 M1 e.g. “14 112” ÷ (9 + 5 + 2) (= 882) or (20 160 − “6048”) ÷ (9 + 5 + 2) (= 882) M1 M2 for 9 2 9 5 2 − + + × “14 112” oe e.g. 9 × “882” – 2 × “882” M1 6174 A1 Total 4 marks2 (a) 70 < s ≤ 80 1 B1 (b) 10 × 45 + 16 × 55 + 19 × 65 + 23 × 75 + 12 × 85 or 450 + 880 + 1235 + 1725 + 1020 (= 5310) 4 M2 f × d for at least 4 products with correct mid-interval values and intention to add. If not M2 then award M1 for d used consistently for at least 4 products within interval (including end points) and intention to add or for at least 4 correct products with correct mid-interval values with no intention to add “5310” ÷ 80 M1 dep on at least M1 allow division by their  f provided addition or total under column seen 66.4 A1 accept 66.37 – 66.4 Total 5 marks3 e.g. 30 × 20 × 125 (= 75 000) or 85 × 40 × 125 (= 425 000) or (60 30 (85 30) 40) 125( 500 000)  + −   = oe 4 M1 for a method to find the volume of water already pumped out or the volume of water left or the total volume of the container “75 000” ÷ 1.5 (= 50 000) or “75 000” ÷ 90 (= 833.3... or 2500 3 ) or 17 "425000" "75000"( 5.66... or ) 3  = or 20 "500000" "75000"( 6.66... or ) 3  = M1 M2 for "425000" "75000" × 1.5 oe (= 8.5) or "500000" "75000" × 1.5 oe (= 10) [Show More]

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