Download [PDF] October 2018 CAIE P2 Mark Schemes 1112 Mathematics Cambridge Lower Secondary Checkpoint

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This document consists of 9 printed pages. IB18 10_1112_02/2RP © UCLES 2018 [Turn over Cambridge International Examinations Cambridge Secondary 1 Checkpoint MATHEMATICS 1112/02 Paper 2 October 2018 MARK SCHEME Maximum Mark: 50 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Markers were instructed to award marks. It does not indicate the details of the discussions that took place at an Markers’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the End of Series Report. Cambridge will not enter into discussions about these mark schemes. Mark scheme annotations and abbreviations M1 method mark A1 accuracy mark B1 independent mark FT follow through after error dep dependent oe or equivalent cao correct answer only isw ignore subsequent working soi seen or implied

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 2 of 9 Question Answer Marks Further Information 1 32.01 1 2 ($)74 2 56 103.60 × 40 or 1.85 or 0.71(...) or 7 5 M1 3 2 (°C) 2 For either 18 or 20 seen or For a mark on the graph at 17:30 B1 4(a) 2 1 4(b) 10 1 5 Millilitres or ml Kilograms or kg Kilometres or km 2 Allow any unambiguous indication of the correct answer. 2 correct B1 6 11 and 13 1 In either order. 7 20 1 Ignore any units given.

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 3 of 9 Question Answer Marks Further Information 8(a) 31 or 62 or 0.33(…) or 33(.3…)% 1 8(b) 12045 or an equivalent fraction e.g. 8 3, 249 or 0.375 or 37.5% 1 9 Point E marked on the grid at (2, 1) 1 Allow any unambiguous indication of the correct answer. 10 Ticks Women and correct figure for comparison, e.g. • (31 out of 80) = 38(.75%) • (41% of 80) = 32(.8) 2 Note other correct methods are acceptable. or 0.38(..) or 39(%) or 0.39 or 33 correct method for comparison, e.g. 8031 or 31 ? 80 or 0.41 × 80 oe M1 Note other correct methods are acceptable or 38(.75%) or 0.38(…) or 39(%) or 0.39 implied by 32.8 or 33

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 4 of 9 Question Answer Marks Further Information 11(a) 2(3 x + 5) + 2( x + 2) isw 1 Allow any equivalents e.g. 8 x + 14 , 6 x + 10 + 2 x + 4 , 2(4 x + 7) 11(b) 23 3 Correctly solving their linear equation from (a) M2 Implied by x = 6 Forming a correct equation e.g. 2(3 x + 5) + 2( x + 2) = 62 oe or 3 × their 6 + 5 M1 Allow their expression in (a) = 62 6 x + 10 + 2 x + 4 = 62 8 x + 14 = 62, 2(4 x + 7) = 62 12 Q C 1 13 24 (%) 2 or 600 144 or 0.24 or 600 744 or 1.24 or 124(%) M1

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 5 of 9 Question Answer Marks Further Information 14     2 Allow any unambiguous indication of the correct answer. 2 correct answers. B1 15 140(.088…) (pounds) 2 Award 2 marks for answers rounding to 140 correct to the nearest whole number 45.4 63.6 × 100 oe or 1.4…. or 0.454 or 0.71… or 2.2… seen M1 16 Saturday and a correct reason relating to the mean or average , e.g. It has the largest mean/ average value. 1 Do not accept Saturday has the smallest range, by itself.

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 6 of 9 Question Answer Marks Further Information 17 Algebraic method seen leading to ( x =) 7 ( y =) 3 3 Do not accept trial and improvement as a method. An algebraic method leading to either x = 7 or y = 3 M2 An attempt at eliminating either x or y e.g. • correct substitution and evaluation from incorrect first value implied by two values satisfying one of the original equations. • making the coefficients of x or y equal followed by an appropriate consistent subtraction or addition across all 3 terms • re-arranging one of the equations to make one variable the subject and then substituting their arrangement into the other equation, e.g. x + 2(24 – 3 x) = 13 or 3(13 – 2 y) + y = 24 M1 With no more than one arithmetic error. Can be implied by 5 y = 15 or 5 x = 35 With no errors.

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 7 of 9 Question Answer Marks Further Information 18 (8, 10) 3 (8, k) or ( k, 10) for D or for finding coordinates of C, i.e. (6, 6) B2 (6, k) or ( k, 6) for C or for sight of 2 2 1x x+ or with correct numbers or or M1 e.g. , Do not allow 2 2 1x x− 19(a) 3n + 11 2 Mark the final answer for 2 marks. Allow equivalent unsimplified, e.g. 14 + 3n – 3, 14 + 3( n – 1) Do not accept: n = 3 n + 11 (±) 3 n ± k seen or correct expression seen then spoilt B1 Allow for just 3n 19(b) 3 1, 52, 73 2 In correct order. Accept equivalent fractions or decimal equivalents for 2 marks or for B1: = 0.33(33 …), = 0.4, = 0.42(857 …) or 0.43 Any one term correct. B1 Regardless of order

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 8 of 9 Question Answer Marks Further Information 20 81 3+ x 2 Correct fractions with a common denominator e.g. 81 8 2+ +x x M1 M1 implied by correct unsimplified answer e.g. M1 for 321) 4( 32 8+ +x x , 32 4 1+ 2x 21 A complete trial and improvement method leading to the answer x = 8.6 Must include all three marking points below. 3 Ignore the final column in the table when marking. Any correct trial of a number between 8 and 9 A correct trial of x where 8.6 < x ≤ 8.65 8.6 in answer space. M1 M1 B1 For both M1 marks to be awarded, one appropriate trial to at least 1 decimal place and one appropriate trial to at least 2 decimal places must be seen, e.g. trial at 8.6 and trial at 8.65 22 72 (cm) and 53 (°) 1 Both correct for the mark.

1112/02 Checkpoint Mathematics - Mark Scheme October 2018 Page 9 of 9 Question Answer Marks Further Information 23 Shirt A and gives correct supporting ratios, e.g. 1.8(57…) : 1 and 1.5 : 1 2 Allow fractions : 1 and : 1 Allow rounded e.g. 1.9:1 and 1.5:1 Any of • 1.8(57…) or 7 13 • 1.5 or 2 3 B1 Allow rounded e.g. 1.9 24 8 (hours) 1 25 –2 1 26 1500 (kg) 2 150 000 (m 2) seen or for a correct method e.g. 15 × 10 000 × oe or for follow through of incorrect area conversion multiplied by 0.01 correctly. M1 e.g. 15 × 10 k × oe