UNIVERSITY OF MALTA SECONDARY EDUCATION CERTIFICATE SEC MATHEMATICS May 011 MARKING SCHEME MATRICULATION AND SECONDARY EDUCATION CERTIFICATE EXAMINATIONS BOARD
SEC Mathematics May 011 Extracts from General Marking Procedures 1. Abbreviations used in Marking Schemes: M denote method marks. Such marks are awarded for using a suitably correct method to answer the relevant question. A denote accuracy marks following from a correct method. A marks are only awarded if there is explicit or implied evidence that the candidate is using an appropriate method. B denote accuracy marks, given independent of method used. In awarding B marks, no evidence is sought as to the method employed to answer the question. Aft denotes follow through accuracy marks. In some questions, the use of a previous answer is necessary in order to work out the final answer. When an the accuracy mark is deemed as ft, candidate will get the mark for an inaccurate answer as long as the given answer follows from correct accurate working using previous incorrect answer. FNW: In some questions/ part questions, although the total mark is broken down into M and A marks, the total mark is designated as FNW (full marks no working). In such cases, candidates still get full marks even if they do not show the method used provided that candidates give a FULLY CORRECT ANSWER.. Trial & Error Methods: Unless the marking scheme specifies to the contrary, candidates obtaining correct answers using trial & error methods are NOT penalised. 3. Do not follow the marking scheme blindly. If a candidate uses a valid method that is not mentioned in the marking scheme, use your discretion to give credit to the candidate s work within the parameters of the marks allotted to the question. When allotting marks in this manner, specify this by writing OS (Outside Scheme) next to the marks awarded for that question. 4. Misreading of questions: Sometimes questions are misread, e.g. a number is copied incorrectly from the exam paper. If the question has been simplified, do not award marks. Otherwise, mark initially so as not to penalise the misreading. Add the component marks and subtract one mark from the total mark obtained. Indicate this by writing MR next to the mark awarded for that question.
MATHEMATICS SEC MAY 011 SESSION Mental Paper Marking Scheme ALL QUESTIONS CARRY ONE B-MARK *IGNORE ALL UNITS* 1 1 COMMENTS Accept 0. 131 3 13 Accept 13 4 0.76 (b) The diagonals of a rhombus are perpendicular. 6 0 7 10 8 9 3 10 1cm 11 8 1 13 13 ½ 14 Accept 0.6 or 6 10 Accept. or ½ Units can be left out Accept 4 8 or 6 Accept 0. or 4 1 1 16 % Accept 17 4 18 04:30 or half past four Accept 4.30 or 4:30 19 1.11 0 3 Accept 3.00 or 3 or 3.0 3
MAY 011 SESSION (Core Paper) 1 (i) 30,407 or 30407 (ii) 3:3 or 3.3 or 3 hr 3 min 8 3. (iii) 1 or or 1. 3 1 1 Do not accept 1. (iv) 96 or 960 10 (v) 9.9 1 B mark each = marks (a) (i) Butter : Icing.Sugar = 3: 7 Butter = 3 0 10 g = 7 gms for 3 0 10 Full marks even for no working for 7 (gms) (ii) Icing sugar = 14 x 70 = 10,00 gms = 10. kg For mult. by 14 Icing sugar : Butter-cream = 7 : 10 Hence butter-cream = 10 10. kg = 1 kg for 10 10. kg 7 7 Do not accept 1000 gms for 1(kg) 1 A mark 4x 3 x 1 3(4x 3) - (x 1) 1x 9 x 1 7x 1 x 3 (b) (i). 3 10 10 10 10 1 1x 9 x 1 up to 10 or equivalent 1 M mark for x 3. 1 1 B mark (ii) 4 x 3 x 1 x 3 = 7 7 x 3 10 x 10 3 10 1 for cross-multiplying 1 M mark for 10 1 A mark 3 (i) The perimeter = circles = 1. 0 = 17.0796m To 1 decimal place, answer = 17.1m for r for r = 1. 1B mark 4
for 17.1(m) only 1 A mark (ii) Area of lawn = circles + square Area of Circle = r and area of square = 1 M mark Area of lawn = 981.748 + 6 = 1606.748m To 4 significant figures, answer = 1607(m ). for rounded answer (b) Fencing cost 8 rolls @ 70 = 60 Turf cost 1607m @ = 40,17 (or 1606.748m @ = 40,169) Labour expenses cost 7,600 Total cost to the nearest Euro, is 48,33 (or 4839) or in between. For addition 1 M mark For correct answer 4 (i) 3 packets (or lbs) weigh 1.364 kg 1. 364 packets weigh 3 kg = 10.00 kg 1 M mark FMNW To the nearest kg, answer is 10 kg (ii) 1.364kg = 3 packets 7 kg = 3 7 1. 364 = 1.396 packets Number of packets = 16 FMNW 1B mark (i) 18% of 480 = 86.40 OR 118% of 480 = 66.40 Total cost = 480 + 86.40 = 66.40(N0 marks for 18% only) FMNW (ii) 10% of 480 = 48 OR 90% of 480 = 43 Showroom price = (480 48) = 43(NO marks for 10% only) FMNW VAT = 1% of 43 = 90.7 Final cost = (43 + 90.7) =.7 OR 11% of 43 =.7 If final.7 only is given and no working shown throughout give marks only. (iii) Savings = (66.40.7) = 43.68 Percentage savings = 43. 68 100% = 7.71% 66. 40 To 1 dec. place answer = 7.7(%) 6 (i) Constructing an equilateral triangle, ABC of side 6 cms. ARCS seen A marks (ii) Bisecting angle BAC. ARCS seen A marks (iii) Constructing circle, centre O, passing through A, B and D. A marks
(iv) Marking E or D correctly. Measuring the length of EB = 3.4 cm ± 0.cm Measuring the length of OC =.cm ± 0.cm C E D A B 7 (i) 4y = x + 4 1B mark S x + 4 R y = ½ x + 1 or equivalent 1B mark x x (ii) perimeter = x + (x + 4) = 0 1Mmark or x + (4y) = 0 P 4y Q 6x = 4 x + 4x + 8 = 0 6x + 8 = 0 x = 7 (cm), y = 4. (cm) (iii) SP = 7cm and SR = 18cm 1Aft mark Area = 7 x 18 = 16(cm ) 1Aft mark 6
Instrument Frequency No. of students guitar 8x 00 piano x 1 drums 4x 100 violin x 0 none x Total 0x 00 (i) Every two correct entries get 1 mark 3A marks (ii) Guitar = Violin = 00 00 0 00 0 0 of 360 144 ; Piano = 1 0 0 of 360 90 ; Drums = 100 0 0 of 360 7 00 0 0 of 360 36 ; None = 0 0 of360 18 00 00 For at least 1 correct guitar piano drums violin none For drawing a Pie chart with labels in degrees or names For all correct angles drawn A marks [For every two wrong angles subtract ] (iii) Probability = 00 or x 18 or 0x 360 = 1 or equivalent but not in terms of x 0 7
9 (a) (i) A: A marks (ii) B: A ft marks (iii) C: A ft marks C A B (b) Reflection in the y-axis or equivalent. 1B mark 10 (i) Marking both angles of elevation correctly on the given diagram. (ii) Taking 1.m not 13.m 1B mark T tan 36 = 1. GF B 6 1.m GF.tan 36 = 1. J G 36 F 1. GF = 0 tan36 = 17.0or 17.m or more accurate TX TX TX (iii) tan 6 = BX GF 17. 0 TX = 17.(0) tan 6 = 8.39(14) BG = TF TX BG = 13. 8.39(14) =.1(086)m 1A FTmark 8
MAY 011 SESSION Paper A 1 (a) x > 3 x > 1. 1B mark 1. x 3 or in words correctly (b) Let x be no. of hrs worked. Either he will be paid 9x Euro or 17 + x Euro for both expressions correct 9x > 17 + x for inequality 4x > 17 x > 43.7 hrs Least number of hours is 44. Accept also any number between 43.7 and 44 (i) f(18) = 1 1 4 (18) = 1 A mark g( 3) = 3( 3) = 7 (ii) y = 1 1 4 x 1 1 4 x y x 4y or multiplying by LCM f 1 (x) = 4x. Do not accept final answer in terms of y FMNW (iii) g(x) + 4f 1 (x) = 4. 3x + 4(4x ) = 4 for substituting f - - 1 (x)and g(x) 3x + 16x 8 = 4 3x + 16x 1 = 0 (3x )(x + 6) = 0 or using formula 3x = or x = 6 or correct values in formula x = 3 or x = 6 (both correct) 1 3 (i) volume of jewel box = 1/3 area of base x height = 1 x 3 For substituting correct values = x (cm 3 ) 1 x x (ii) volume of lid = = 3 9 7 x (iii) x = 49 forming equation 7 130x = 49 for multiplying by correct LCM 7 9
x = 10.808 x = 10.139 finding root x = 10.14(cm) to d.pl. only 4 (i) A = P(1 + r / 100 ) n or for correct long method A = 0,000(1 + 4. / 100 ) for correct substitution in either method A = 467.93 or 468 (ii) Interest = 4,68 0,000 = 4,68 Accept more accurate answer (iii) Tax = 0% of 468 = 913.60 (or 913.9) New amount left is 4,68 913.60 1,000 = 11,64(.40) (iv) P = P(1 + / 100 ) n gives = 1.0 n If LHS is twice Principal or correct substitution in long method or formula Now 1.0 1 = 1.796 Or amount = 099(.3) which is less than double. 1B mark Hence 1 years are not enough Age 1 3 4 6 7 8 No. 40 140 9 80 60 6 3 Total 40 180 7 3 41 480 1 0 For finding totals (i) Drawing of cumulative frequency curve. Correct Scale and axes 4 correct values and points plotted All plotted points correct For general correct shape 600 00 Number 400 300 00 100 0 0 0 40 60 80 100 Age (ii) the median age is reading on 60, which is about 33 years, (31 3) (iii) the percentage of people under 18 years that attend this gym is 70 (60 80) This amounts to 70 100 % 0 = 13.%, (11.% 1.%) 10
(iv) the number of people under 60 years is around 40 (440 460) So over 60 years no. is 0 40 = 70 (60 80) (subtracting from 0) 6 (a) (i) (x 1) 1B mark 1B mark (x + 1) (ii) Show that the sum of any three consecutive numbers is a multiple of 3. For adding correct terms (x 1) + x +(x + 1) = 3x If sum is shown as multiple of 3 M1 mark A1 ft mark Do not accept proof with values instead of x. (iii) Show that the sum of any four consecutive numbers is always even. (x 1) + x + (x + 1) + (x + ) = 4x + = (x + 1) For adding 4 consecutive terms M1 mark For showing is a factor A1 mark (b) (i) p = k x 10 = k/( 01 ) 10 = 0k k=½ FMNW 1 when x =, p = ¼. p= x p = cy 10 = c 1 c = 0 so that p = 0y (ii) y= 1 p and x = 0 p xy = 1 p or p 0 1 0. 1 = 100 1 Ignore explanation (iii) When p = 1, x = 1 30 and y = 1 0 103 3 60x + y = 60( 301 ) + 103 = 10 or.3 7 (i) Pts of intersection are (.3,.7), ( 0.8, 3.7) or near enough ( 0.). 11 1B, 1B marks
(ii) 4 y 1 x x x 4 x x x 3x 4 0 (Even if = 0 left out) 1A mark 3 9 3 x 4 or correct substitution in own equation x = 0.81 or.31 1A, s Must have 3 d.pl. but correct if two d.pl are correct (iii) ( 0.81, 3.70) and (.31,.70) 1A,s (iv) Area of triangle OAC = ½ base x height = ½ x1 x.70 = 1.3(1) Ignore rounding and units 8 (i) Angle OAB = x (OA = OB) w = 90 x; (PQ tangent, OA radius) for reason either OA = OB or tgt perp to radius y = 90 x; (180 90 x) for reason involving triangle ODA (ii) Δs ODA, ODB are congruent (RHS or other correct reasons) for reason of congruency for RHS or SAS proof Hence angle BOA = y But angle at circumference = ½ angle at centre Therefore z = y But y = w (both = 90 x) So w = z If alternate segment theorem is used without proof give 1
9 (i) angle ABE = 4 (90 4 ) A.6m B 1B mark Angle BEC = 90 ( 4 + 1) = 4 1B mark C Angle ECD = 180 ( 100 + 1) = 9 E 7.38m D (ii) BE = AB + AE =.6 +.6 =.33 For correct substitution in Pythagoras Th. BE = 7.4388 so that to 3 s.f, BE = 7.44m For using Sine rule to find CE 0 7. 38 sin100 giving CE = 0 sin9 For subst. and cross-mult. CE= 8.479m or 8.48(m) (iii) Area using ½ bc sina Area = ½ (7.4388)(8.479)(sin 4 0 ) for substitution = 1.8m or more accurate 10 H triangle HCB rt-angled at C Labelling x and x + 0. x HB + x + x + 0. = 6 HB =. x or equivalent C x + 0. B Pythagoras Theorem with correct substitutions (. x) = x + (x + 0.) 30. x + 4x = x + x + x + 0. For correct expansion of at least 1 bracket x 3x + 30 (= 0) (x 3)(x 10) = 0 or using formula x = 10 or 1. km (x has to be less than 6) x = 1. km x + 0. = km only Ignore extra answer with x = 10 13
MAY 011 SESSION Paper B 1 Write down: (i) a prime number between 0 and 8, (i) 3 1B mark (ii) a factor of 91, (ii) 7, 13,1 or 91 1B mark (iii) a multiple of. (iii), 0 or 7 or 100 1B mark (3 x ) = 486 so that 3 x = 43 FMNW x = 91 3 91 litres = gallons = 0 gallons 4. FMNW 4 63 80 13. 1 = 190.9 FMNW = 191 to 4 significant figures. In standard form, answer =.19(1)x10 3. 1Aft mark If only final answer in standard form given, without previous answers, give marks in all. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 31 0 1 8 6 30 3 6 4 37 (i) 0, 1,, 4,, 6, 6, 8, 30, 31, 3, 37 in order The median is 6 1B mark (ii) Total no. = 31 + 0 + 1 + + + 8 + 6 + 30 + 3 + 6 + + 4 + 37 = 3 The mean = 3 = 6.833 1 FMNW To the nearest integer mean = 7 6 PTR 100I I = P 100 TR 100 364 P 6 4. P = 13,00 1B mark 7 (i) of 80 = 3 are green 1B mark (ii) What fraction of the crayons are red or yellow? 1 4 of 80 = 0 are red and4 are yellow 44 are red or yellow For 0 red or total = 44 1B mark 14
Fraction = 44 11 80 0 (iii) Blue = 80 ( 3 + 0 + 4) = 4 1B mark 8 (i) For correct pattern 1B mark (ii) For every 3 correct entries 1B mark For all correct 1B mark Pattern 1 3 4 6 Tiles 1 4 9 16 36 (iii) n 1B mark 9 If f(x) = x and g(x) = x, (i) f(6) g( 6) = 6 ( 6) or 6. 6 or 36 + 1 = 48 FMNW (ii) f(x) + g(x) = x + x = x(x + ) 10 A = r rh A r = rh A r h = r or equivalent FMNW 11 (i) 1 3 4 6 A (A, 1) (A, ) (A, 3) (A, 4) (A, ) (A, 6) E (E, 1) (E, ) (E, 3) (E, 4) (E, ) (E, 6) I (I, 1) (I, ) (I, 3) (I, 4) (I, ) (I, 6) O (O, 1) (O, ) (O, 3) (O, 4) (O, ) (O, 6) U (U, 1) (U, ) (U, 3) (U, 4) (U, ) (U, 6) 10 correct entries Another 10 entries (ii) 1 1 30 or equivalent 1B mark (iii) For counting 9 possibilities 1B mark 9 3 or equivalent 30 10 1 (i) 3% on 40 = 17.0 or 13% on 40 1
Total insurance = 40 + 17.0 = 607.(0) FMNW (ii) Method 1:New insurance = 11% of 40 = 04 or 1% of 40 + 40 1M mark No-claim bonus = % of 04 = 16 or % of 40 or previous answer Amount to be paid = (04 16) = 378 Method : 7% of 40 = 337.0 11% of 337.0 or 1% of 337.0 + 337.0 = 378 M1, M1, A1 13 (a) (i) Time = [(3x4) + 3] + 49 60 = 7 49 60 = 7.8167 hrs To 3 places of decimal time = 7.817 hrs (ii) Speed = distance = 380000 time 7. 8167 km/hr Answer = 01 km/hr FMNW 1Aft mark (b) Time = 380000 447 = 8.8(33) hrs = 8 hrs 3 mins 14 G B x A z D y C E F EITHER Triangle BCE is isosceles (or radii) Hence angle x =. Angle y = + = 0 (exterior angle of triangle or angle at centre) Angle ABE = 90 + = 11 For using 90 In Δ ABE, angle z = 180 (11 + ) or z = 180 (90 + 0) For using either angle sum of triangle or angle between radius and tangent angle z = 40 16
1 (i) BC = AC AB (Pythagoras Theorem) = 13 1 = 169 144 = BC = (cm) 1B mark Area of Δ ABC = ½ (x1) (including correct substitution) = 30cm (ii) Rotational symmetry of 90 or Rotational symmetry of order 4 (iii) the area of the tile = 4 x 4 = 76cm FMNW (iv) Horizontal line Vertical line Subtract 1 mark if any extra lines of symmetry are added 16 For scale drawing, for using the correct given scale for placing letters correctly for correct diagram (i) BS =. ± 0.1cm representing 0m ± 10m or just for.cm or 4.3cm BD = 4.3 ± 0.1cm representing 430m ±10m. or either 0m or 430m FMNW. (ii) Bearing of B from Y is 63 ± For measuring just inner angle between 4 o and 10 o For giving final answer FMNW 17
17 (i) the gradient of the line l 1 = 6 0 3 0. 1B mark l 1 (ii) the equation of the given line l 1 is y = 3 x + 6. (gradient or intercept) Equation exact (follow through from (i)) 1Aft mark (iii) the equation of a line l is y = 3 x 3. follow through from (i) 1Aft mark (iv) Drawing: For each correct intercept on axes A marks If the line is drawn correctly from wrong equation in (iii), then subtract 18 (i) Let a and b be the number of hours assigned to machines A and B respectively. a + b = 4...(i) (ii) 0a + 40b = 1000...(ii) (iii) Eqtn (i) 40 : 40a + 40b = 960 Subtracting: 10a = 40 or reduce to 1 equation a = 4 hrs or b = 0 hrs (iv) Cost is (30x4) + (0x0) = 10 + 400 = 0 FMNW (v) Machine A takes 1000 = 0 hrs. 0 Cost = (0x30) = 600 Extra cost = (600 0) = 80 For both steps 80 100 Percentage increase = % = 1.384% 0 To 1 decimal place answer = 1.4(%) 18
19 9 (i) Vol = 4 100 = 1017.373 cm3 For correct volume of cylinder. For 9% of volume Answer = 1017cm 3 or more accurate For correct answer 1017. 37 (ii) No. of lamps = 1 = 8.14 but correct answer = 8 lamps 1Aft mark (iii) Let x be number of lamps 4x + 1x = 1017(.37) For 149 or 149x 149 x = 1017(.37) For vol/149 x = 6.83 1Aft mark giving number of lamps = 6 Accept trial and error method 0 (i) To show that the triangles, BCD and ABE are similar: In Δs BCD, ABE A B angle BDC = angle BEA = 30 (given) x x no reason needed here angle BCD = angle ABE (alternate s) 30 or C D Remaining angles CBD = BAE Reason must be given Hence Δs are equiangular, so similar or 30 Δ BCD and ΔABE are similar in that order Or the remaining angles are equal. E (ii) To calculate the length of CD: BE CE = BC BC = 10 6 = 4cm BC CD BD = (= ) AB BE AE so that CD = 4 10 = 8cm FMNW 19