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MATHEMATICS – SEC LEVEL MS10 MARKING SCHEME – PAPER I [CORE] MAY 2010 SESSION Question Number Answer 1 Year 1: 160 000 × 1.08 = €172 800 Year 2: 172 800 × 1.08 = €186 624 Mark M1 M1 M1 A1 Notes 8% of 160 000 Adding increase to 160 000 8% of year 1 ‘new’ price €186 624 Total: 4 marks 2 25km × 6 = 150km 150km ÷ 18km = × 1.15 = €9.583 = €10 litres M1 M1 M1 Multiplying by 6 Dividing by 18 Multiplying by 1.15 A1 €10 Total: 4 marks Page 1 of 7 MATHEMATICS – SEC LEVEL MS10 3(a) 0.001 = (b)(i) M1 A1 = Convert to fraction or standard form 10‒3 or ‒3 ‒3 = 10 M1 A1 Pattern: +85 8 nights: 450 + (3 × 85) = €705 For identifying pattern +85 €705 (ii) Night: 1 110 2 3 110+85 110+(2×85) nth term = 110 + 85(n ‒1) = 25 + 85n (iii) 1215 = 25 + 85n 85n = 1190 n = 14 nights 4 … 110+(3×85) M1 M1 A1 M1 M1 A1ft Writing at least 3 consecutive values in terms of multiples of 85 Ability to generalize pattern [up to 85n seen] 25 + 85n [or equivalent] Equating nth term in part (ii) with 1215 [Award as long as n is included in equation] Making n subject 14 nights [ft for incorrect part (ii). Award only if the final value of n is given as a positive whole number, without any rounding] Total: 10 marks 4(a) (b) Red : Blue : Yellow = 1 : 3 : 4 (×1.5) = 1.5 : 4.5 : 6 Tins: 2 + 5 + 6 = 13 tins Banana : apples and apples : oranges =2:3 =5:6 (×5) = 10 : 15 (×3) = 15 : 18 So 10 bananas ≡ 18 oranges (÷2) 5 bananas ≡ 9 oranges B1 B1 M1 A1 4.5 litres of blue paint 6 litres of yellow paint Rounding from litres to number of whole tins of paint [One case enough] 13 tins M1 M1 Multiply or divide ratios to obtain same number of apples in each ratio Obtain equivalence between weight of bananas and oranges A1 5 bananas Total: 7 marks Page 2 of 7 MATHEMATICS – SEC LEVEL 5(i) MS10 M1 For circle and chord drawn accurately [± 2mm] [Ignore labeling and arc not required for BC] (ii) M1 A1 Arcs drawn such that ABC is an equilateral triangle Labelling points B, C and D [as long as ABC is equilateral, of side 8cm and with two vertices A and B on circumference] (iii) M1 A1 Arcs drawn to obtain perpendicular bisector of AB Labelling point P [even without arcs as long as accurate] (iv) M1 A1 Arcs drawn to obtain bisector of angle BCD Labelling point Q [even without arcs as long as accurate] A1 Accept PQ in the range from 4.4cm to 4.8cm, both values inclusive (v) [Refer to Construction] PQ = 4.6cm (approx.) Total: 8 marks 6(a)(i) (ii) BPC = xo (alternate angles) PBC = 90o (interior angles) x + y + 90 = 180 (sum of angles in triangle PBC = 180o) x + y = 90 M1 M1 M1 M1 AQB = PBC (alternate angles or equivalent reason) ABP = BPC (alternate angles) Since ∆AQB and ∆CBP are equiangular then they are similar. M1 M1 M1 Including reason Including reason For proper conclusion [Or award the third M1 if 3rd set of equal angles is given, including reason] M1 B1 Dimensions of shape B increased by a factor of 3 Shape, position and labeling [labeling of either B or C enough] B1 Shape and position [labeling of either B or C enough] If using triangle AQB, second M1 is automatically awarded Award only if at least two reasons are given for previous statements 10 (b)(i) C 5 B (ii) A 0 5 10 15 20 Page 3 of 7 MATHEMATICS – SEC LEVEL MS10 Total: 10 marks 7(a)(i) 2x + 2y = 37 2y = 37 ‒ 2x M1 Relationship between x and y established A1 [or equivalent, provided y subject] (ii) (iii) A1 x = 2y + 5 [or equivalent] x = 2y + 5 x = 37 ‒ 2x +5 3x = 42 x = 14 M1 M1 Substituting to obtain an equation in one unknown Substituting value of x (or y) obtained to find remaining unknown A1 Values of x and y both correct y = (37 ‒ 2×14) ÷ 2 (b) = (37 ‒ 28) ÷ 2 = 4.5 B1 M1 M1 Up to A1 Substituting coordinates of point in line to obtain equation in terms of c. [Award even if value of gradient is incorrect] Gradient = ‒6= or 0.2 +c ‒6=1+c [or equivalent] c = ‒7 Total: 10 marks Page 4 of 7 MATHEMATICS – SEC LEVEL 8(i) MS10 [Possibility Space] B3 Deduct one mark for each two incorrect entries P(sum = 8) = B1 B1 5 in numerator 36 in denominator [or award B2 for B1 B1 Correct numerator Correct denominator (ii) (iii) P(difference is 1) = (iv) [or award B2 for B1 , or %] %] 0 or ‘impossible’ [or equivalent] P(sum = 8 and difference is 1) = 0 Total: 8 marks 9(a)(i) M1 Eliminating denominators M1 Squaring both sides [up to M1 A1 (ii) c = ± 0.57735 c = ± 0.58 enough] Collecting like terms and c2 subject ± 0.58 only M1 Page 5 of 7 MATHEMATICS – SEC LEVEL (b) for any value of c So p is always smaller than 1 (×30) MS10 M1 M1 M1 A1 15x ‒ 10(x ‒ 3) = 6(7) LCM and correct numerators [even with LCM not crossed out] Expanding brackets Collecting like terms 2.4 [or equivalent] 15x ‒ 10x + 30 = 42 5x = 12 x = 12 ÷ 5 = 2.4 Total: 9 marks 10(i) (ii) Tan14 = 20÷ AD AD = 20 ÷ tan14 = 80.2156 = 80.2m M1 Trigonometric ratio and substitution A1 80.2m Cos(28 + 14) = (AD + DC) ÷ AB Cos42 = (16 + 80.2156) ÷ AB AB = 96.2156 ÷ cos42 AB = 129.4708 M1 Trigonometric ratio and substitution M1 A1ft AB subject Accept values in range 129.4m to 129.5m [ft for incorrect AD in (i)] M1 Trigonometric ratio or Pythagoras Theorem, including substitution A1ft 86.6m [ft for incorrect AD and/or AB in (i) and/or (ii)] M1 M1 A1ft Subtracting 20m from BC, provided BC > 20m Trigonometric ratio and substitution 76.5o [ft for incorrect BC in (iii). Award only if first M1 is awarded] AB = 129.5m [Range: 129.4 ‒ 129.5] (iii) Sin42 = BC ÷ 129.4708 [or by Pythagoras: BC2 = 129.4702 ‒ 96.21562] BC = 86.6328 (iv) BC = 86.6m [Range: 86.5 ‒ 86.7] tanBEF = (86.6328 ‒ 20) ÷ 16 BEF = 76.497 = 76.5o Page 6 of 7 MATHEMATICS – SEC LEVEL MS10 Total: 10 marks Page 7 of 7