4(b) Odd B1 4(c) Either B1 5(a) Correct straight (if not drawn with a ruler then intention to be straight) line graph from (0, –1) to (4, 7) with 1mm (½ square) tolerance Allow a dashed line B3 B2 correct line but not from (0, –1) to (4, 7) for at least a continuous x distance of 2. B2 all integer points (any others must also be correct) between 0 and 4 plotted but line not drawn B2 correct but more than ½ square from tolerance Only one of these may be awarded. B1 straight line graph through (0, –1) of any length even if crooked later but not x = 0 or y = –1 B1 Single straight line graph with gradient 2 of any length B1 two correct points calculated (eg in table) or plotted Any line that is not straight is B0 although the B1 for two points calculated or plotted may still be gained 5(b) 1.5 B1 Correct (eg from algebra) or ft their graph if y = 2 drawn to the graph and then a vertical line to x-axis
Q Answer Mark Comments 6(a) Histogram or frequency polygon with mid-points of bars and vertices of polygon at (5, 36), (15, 34), (25, 18) and (35, 12) B2 B1 one error Ignore lines before (5, 36) and after (35, 12) if polygon drawn 6(b) 6  (18 + 12) NB table can be seen if necessary. M1 30 oe 600 100 180 A1 SC1 30% stated as answer SC1 for 420 as answer 7 2  4 + 3  3 + 5  1 or 8 + 9 + 5 M1 22 has to come from correct working (30 – their 22) ÷ 4 M1Dep Their 22 + 4a = 30 2 A1 First M must be scored 7 Alt 1 Guess a value for a and correctly works out ∑xf M1 Guesses a second value nearer to the correct answer and correctly works out ∑xf M1Dep 2 A1 First M must be scored
Q Answer Mark Comments 8(a) 3(x – 5) B1 8(b) 5y + 20t – 10 B2 B1 for 2 correct terms. Penalise any incorrect further working. Eg 5y + 20t – 10 = 25yt – 10 is B1 5y + 20t – 1 = 25yt – 1 is B0 (error in expansion and incorrect further work) 5y + 20t – 10 = 5(y + 4t – 2) given as answer is B1 as shows a misunderstanding of expanding brackets. 8(c) 3w + 6 = 2w – 1 M1 w + 2 = w – 3w – 2w = –1 – 6 M1 This mark is for rearranging their expansion correctly to get w terms one side and number terms on the other. w – w = – – 2 (oe) –7 A1ft ft on one error
Q Answer Mark Comments 9 (550 – 250) ÷ 3 M1 J + W = 250 or J + 4W = 550 100 A1 3W = 300 or W = 100 250 – their 100 M1Dep 100 + J = 250 or 400 + J = 550 150 A1 9 Alt 1 – (= ) M1 Their = 300 or = 100 A1 250 – their 100 M1Dep 150 A1 9 Alt 2 550 marked by top division and 250 marked by bottom division on same diagram M1 300 indicated as difference on diagram or 350 and 450 written by intermediate divisions A1 100 marked between any two divisions is M1, A1 150 marked at bottom M1Dep
150 stated as answer A1 9 Alt 3 Guesses a value for weight of jug, subtracts from 250, multiplies answer by 4 and adds to their value M1 Correct calculations A1 Guesses a second value for weight of jug nearer to 150 and correctly calculates all values M1Dep 150 A1 Q Answer Mark Comments 10 y ≤ 5 or 5 ≥ y B1 Any order. Penalise first use of > or < only. Penalise first use of >= or => or =< or <= only. Accept 2 < y ≤ 5 or 2 ≤ y ≤ 5 x ≥ 2 or 2 ≤ x B1 Accept 2 ≤ x < 5 or 2 ≤ x ≤ 5 y ≥ x or x ≤ y B1 oe Sc1 y = 5 and x = 2 and y = x or Sc1 y ≥ 5 and x ≤ 2 and y ≤ x 11 3 + 4 + 5 (= 12) B1 3 and 4 must be used
48 ÷ their 12 (= 4) M1 ‘Their 12’ means their addition of 3 + 4 + 5 or their total if they think that a pentagon does not have 5 sides 20 A1ft ft on B0 Accept 12 : 16 : 20 12 2a + 2c = 5a – 5b or 2a + 2c = 5(a – b) 2c = 5a – 2a – 5b or 2c = 5(a – b) – 2a M2 M1 if one expansion, sign or rearrangement error on any line –2c = is OK if rest correct c 5b 2 or equivalent expression A1ft ft on one error Must have c = on answer line If question simplified by an incorrect expansion 2a + c to give c = … (see exemplar below) then they must simplify their answer Do not award if incorrect further work 12 Alt a + c = 2.5(a – b) c = 2.5(a – b) – a M2 M1 if one expansion, sign or rearrangement error on any line c = 2.5(a – b) – a or equivalent expression A1 ft ft on one error Must have c = on answer line Do not award if incorrect further work Q Answer Mark Comments 13(a) 20 B1 13(b) 9 B1 13(c) 11 and 3 seen M1 Could be written on diagram
n (2n) 2 B1 This mark is for combining terms inside bracket correctly n2 B1 1n2 is OK 18 (x2 + 2x – 3) – (x2 + x – 3) M1 Or attempt to ‘balance’ equations y = x A1 – 2.3 and 1.3 A1 ft ft if M awarded and their line drawn Q Answer Mark Comments 19(a) 3  3 = 9 and √3  √3 = 3 or 3  3 = 3 B1 √3  √3 is essential seen or implied Accept (31.5)2 = 33 = 27 oe 19(b) (3√3)2 – (3√2)2 (= 27 – 18) or (3√2)2 + AD2 = (3√3)2 M1 Invisible brackets must be recovered for M1 √9 or 3 A1 3√2 + √2 (= 4√2) M1Dep Dep on M1, not on A1 as well 0.5  their base  their 3 M1Dep 6√2 A1
19(b) Alt 1 (3√3)2 – (3√2)2 (= 27 – 18) M1 Invisible brackets must be recovered for M1 √9 or 3 A1 Area ABD = 0.5 their 3  3√2 and area ADC = 0.5  their 3  √2 M1Dep Dep on M1, not on A1 as well Area ABD = 0.5 their 3  3√2 + area ADC = 0.5  their 3  √2 M1Dep Sum of two correct areas for their AD 6√2 A1 19(b) Alt 2 (3√3)2 – (3√2)2 (= 27 – 18) M1 Invisible brackets must be recovered for M1 √9 or 3 A1 3√2 + √2 (= 4√2) M1Dep Dep on M1, not on A1 as well Sin ABD = their 3 ÷ 3√3 and Area = 1 0.5 √ √2 M1Dep 6√2 A1