Maths Lit Practice exam memorandum November Paper 2 Gr11 1

Study & Master Support Pack 2 | Grade 11 CAPS Mathematical Literacy Exemplar examination memorandum: November (Paper 2) This revision pack for Mathematical Literacy Grade 11 provides support for learners revising the most important concepts and principles covered in the CAPS curriculum, by providing exam practice. This revision pack includes a memorandum of the sample November examination (Paper 2). You have permission to print or photocopy this document, and to distribute it electronically via email or WhatsApp. For more information on our Study & Master CAPS-approved textbooks and valuable resource material, visit www.cambridge.org Brighter Thinking | Better Learning

Grade TOPIC11 NOVEMBER XX: XXX EXAMINATION Time: 2,5 hours Paper 2 MEMORANDUM Marks: 100 Question 1 b) Area to be painted = 2 × 43,368 m2 ✓ = 86,74 m2 ✓ 1. a) R250 ✓(1) Number of 5 ℓ tins = 2,409 ✓ b) R125 per person ✓(1) They need three 5 ℓ tins because two 5 ℓ tins c) Five people ✓(1) will not be sufficient to give the walls two coats d) As more people travel in the car, ✓ the cost of paint, ✓ (4) per person decreases. ✓(2) c) i) Area of kitchen floor: 4 m × 3 m =12 m2 ✓ 2. a) Turn right down Bumba Lane, ✓ cross over Main Road Cost of tiling kitchen: 12 × R32,50 ✓ = R390 ✓(3) and continue until you get to Thato’s house. Turn left ii) Area of a customer area floor into Zwemba Drive, ✓ take the first left into Second = l × b Avenue and carry on until you get to Joe’s house. Pick = (5 × 4) m2 ✓ up Joe. Turn around and continue along Second Avenue. = 20 m2 ✓ Turn left into Main Road and take the first Right into ✓ Cost of tiling front of shop: 20 × R32,50 = R650 ✓ First Avenue. Stop at the Office Block. Total cost: R390 + R650 = R1 040 ✓ (4) Accept logical route, subtract one mark (−1) if the route is not the shortest one. (3) 3. Height of the decoration b) Scale measurements: = 2,04 m + 10 cm ✓ width: 21 mm = 0,021 m ✓ = 2,04 m + 0,1 m length: 35 mm = 0,035 m ✓ = 2,14 m ✓ Actual measurements: Width of the decoration width: 0,021 × 20 000 = 420 m ✓ = 0,8 + 0,1 m + 0,1 m length: 0,035 m × 20 000 = 700 m ✓ = 1,0 m ✓ 420 × 700 = 294 000 m2 (4) Length of decoration = 2,14 m + 1,0 m + 2,14 m 3. a) 17:15 – 07:50: 9 hours 25 minutes ✓ = 5,28 m ✓ (4) She would be charged the rate for ‘More than 9 hours’, which is R12,00. ✓(2) 4. a) Volume = πr 2h b) 08:45 – 08:01: 44 minutes = 3,14 (25)2 ✓(32) cm3 ✓ This is less than one hour, so she would not be charged = 62 800 cm3 anything. = 62 800 ml ✓ 17:30 – 12:15: 5 hours and 15 minutes ✓ = 62,8 ℓ of water ✓ (4) She would be charged the rate for 5 to 7 hours, b) Volume of water + volume of rice which is R8,00. ✓(2) = 625 ml + 250 ml ✓ c) Period parked: 08:45 to 13:25 = 4 h 40 min. ✓ = 875 ml ✓ She would be charged the rate for parking for 3 to 5 _3 ​​  4 ​​of 62,8 ℓ = 47,1 ℓ ✓ hours, which is R6,00 ✓ Number of cups of rice: _______ 47 100 ml A charge of R8,00 is incorrect. ✓ ​​  875 ml ​​= 53,828 ml ✓ She is being overcharged by R2,00. (3) The maximum number of cups she can cook: 53 ✓✓  [19] 54 cups would be too many. (6)  [34] Question 2 Question 3 1. i = ___ 9 ​​ 100 ​​= 0,09 ✓ 1. a) Median for 11A: _____ 58 + 58 ​​  2 ​​ ✓ = 58 ✓ A = P(1 + i)n Median for 11B: _____ 59 + 60 = R120 000 ✓(1 + 0,09)8 ✓ ​​  2 ​​ ✓ = 59,5 = 60 ✓ (4) Mean for 11A = _______________ = R239 107,52 ✓ (4) 18 + 25 + 31 + … + 92 b) ​​    30 ​​ ✓ 2. a) Area of walls = 2(l × b) + 2(l × b) ____ 1 730 = ​​  30 ​​ = 2(4 × 3) + 2(5 × 3) ✓ = 57,66 = 24 m2 + 30 m2 = 57,7 ✓ = 54 m2 ✓ Mean for 11B = _______________ 11 + 26 + 30 + … + 98 Surface area to be painted ​​    30 ​​ = 54 m2 – 9 m2 – (2,04 m × 0,8 m) ✓ = ​​ ____ 1 623 30 ​​ ✓ = 54 m2 – 9 m2 – 1,632 m2 ✓ = 54,1 ✓ (4) = 43,368 m2 ✓(5) 1

c) Range for 11A: 92 – 18 = 74 ✓ 5. a) R500 ✓✓ (accept between R480 and R520; Range for 11B: 98 – 11 = 87 ✓ R475 only 1 mark) (2) 11A has a smaller range. ✓(3) b) After using the cellphone for 200 minutes ✓✓ d) Number of learners in Grade 11A who obtained (190 minutes 1 mark) (2) 50 marks or fewer: __ 10 6. Rafique’s total sales: R3 000 + R4 000 + R5 000 ​​  30 ​​× 100% ✓ = 33,33% ✓ = R12 000 ✓ Number of learners in Grade 11B who obtained Melokuhle’s total sales: R2 000 + R3 000 + R4 000 50 marks or fewer: = R9 000 __ 13 Rafique’s sales were higher. ✓(2) ​​  30 ​​× 100% ✓ = 43,33% ✓ (4)  [18] e) 11A performed better than 11B. ✓ Their mean is higher and their range is smaller. ✓ Fewer learners in Grade 11A obtained 50 marks or fewer. ✓(3) 2. a) P = __ 3 ​​ 30 ​​ ✓ ✓ __ 1 = ​​ 10 ​​= 0,1 ✓(3) b) P = __ 30 ​​ 30 ​​ ✓ = 1 ✓(2) c) P = __ 0 ​​ 30 ​​ ✓ = 0 ✓(2) __ 2 3. P = ​​ 60 ​​ ✓✓ = 0,0333... ✓ = 3,33% ✓ (4)  [29] Question 4 1. The basic cost is R190 ✓ (constant) and the cost for each minute (t) is R1,50. ✓(2) 2. C ✓ = 2,50t ✓(2) 3. 72 min. and 12 s ≈ 73 min. ✓ Cost: R2,50 × 73 ✓ = R182,50 ✓(3) 4. Cost of cellphone calls 740 B 720 700 680 660 640 A 620 600 580 560 540 520 500 480 460 440 420 400 380 360 340 Cost (R) 320 300 280 260 240 220 200 180 160 140 120 100 80 60 40 20 0 150 200 250 0 300 50 100 (5) Time (min.) 2