Mathematical Literacy Grade 11 P2 2016 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2016 MATHEMATICAL LITERACY P2 MEMORANDUM MARKS: 100 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table/Reading from a graph/Read from map F Choosing the correct formula SF Substitution in a formula J Justification P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off/Reason O Opinion This memorandum consists of 8 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2016) QUESTION 1 [23] Ques. Solution Explanation Level 1.1.1 Electricity  2A One expense Water  excluding cost of Transport  ingredients Accept any relevant answers. (2) L2 1.1.2 Costs for 24 muffins Muffin mix 500 g for 12 muffins 1 kg of muffin mix = R12,00 500 = ½ kg = R6,00  For 24 muffins = R6,00 × 2 = R12  1M For muffin mix Eggs: 6 = R10,50 10,50 2= ×2 6 = R3,50  1M for eggs Oil ½ cup = 250 mℓ 750 mℓ = R12,50 12,50  × 250 750 = R4,17  1M 1M for oil Milk 3 cups = 750 mℓ 1 000 mℓ = R11,50 11,50  750 mℓ = × 750 1 000 = R8,63  1M 1M for milk Total amount = R12,00 + R3,50 + R4,17 + R8,63 = R28,30  1CA Total amount OR 12,00 Muffin mix = × 500 1 000 = R6,00  10,50 Eggs = 6 = R1,75  12,50 Oil = × 125 750 = R2,08 √ 11,50 Milk = × 375 1 000 = R4,31 √ Total = 6 + 1,75 + 2,08 + 4,31 = 14,14 × 2 = R28,28 √ (5) L3 Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICAL LITERACY P2 3 1.1.3 Cost R28,30 for 24 muffins CA from 1.1.2 28,30  For 1 200 = × 1 200 24 = R1 415  Selling price = R30 per dozen 1MA expenses for 1 200 = 1 200 / 12 1 200 muffins = 100 dozen  Money received for 120 dozen = 100 × 30 1CA number of = R3 000  dozen Profit = R3 000 – R1 415 1MA Income for = R1 585  1 200 muffins Statement is valid  1CA Difference 1O OR 28,28 = 14,14 x 100 = R1 414 (expenses)  2 Income = 100 × R30  = R3 000  Profit = R3 000 – R1 414 = R1 586  Statement is valid.  (5) L3 1.1.4 In 25 minutes she makes 24 muffins 25  240 muffins = × 240 24 = 250 minutes  1MA Time for She will bake 10 times. baking Time spent on cleaning pans: 10 × 10 minutes = 100 minutes  1CA time for cleaning Total time = 250 + 100 = 350 minutes Time in hours = 350 / 60 = 5,833333333 hours 1CA Convert to = 5 hours 50 minutes  hours and minutes Time taken to reach destination: Time = 50 / 100 1C driving time = 0,5 = 30 minutes  Total time = 5h50 + 0h30 1CA adding time = 6 hours 20 minutes  Time arrived = 09:00 + 6 hours 20 minutes 1CA time of arrival = 15:20  and statement She will arrive on time. (6) L4 Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2016) 1.2 Cleaner = R50 × 5 × 20 = 5 000  1M Salary for 175 cleaner Driver = ( × 50) 100 Driver per month = R87,50  1M calculating 75% = R87,50 × 5 × 20 more = 8 750  Assistant = 50 + 25 1CA Salary for = 75 driver per month Assistant per month = 75 × 5 × 20 = 7 500  1M helper per Total spent on workers = 5 000 + 8 750 + 7 500 month = R21 250  1CA total (5) L3 [23] QUESTION 2 [27] Ques. Solution Explanation Level 2.1.1  5 16 16 21 22 34 34 35 35 35 42 42 42 42 2M Calculate the 30 number who × 50 = 15  100 passes 13 learners were able to get more than 30% and only one leaner got less than 30% therefore the learners 2O performance was good.  (4) L4 2.1.2 CA from 2.1.1 5+16+16+21+22+34+34+35+35+35+42+42+42+42 1M Adding values  14  1A Divide by 14 = 421 / 14 = 30,07  1 CA mean (3) L3 2.1.3 Mode = 42  1A mode Median: 5 16 16 21 22 34 34 35 35 35 42 42 42 42 35+34  2MA finding the = 34,5  2 median  Difference = 42 – 34,5 = 7,5  1CA Difference (4) L4 2.1.4 50 1M Calculate value × 50 = 25  100 of 50% 5 1M probability as a   × 100 14 fraction = 35,71% √ 1CA percentage Accept 35,7% (3) L3 Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICAL LITERACY P2 5 2.2.1 Mathematical Literacy = 50  2A First answer Mathematics = 50  1A Second answer (3) L2 2.2.2 Mathematics = 83  3A Correct exam number (3) L2 2.2.3   1A for left 60 – 54 = 6  1A for right 1CA difference (3) L2 2.2.4 Ratio = 1 : 30 Number of learners = (9 × 6) + (10 × 10) + (6 × 10)  1M Total number = 54 + 100 + 60  = 214  1CA Total 214 / 30 = 7,1 1CA Dividing by 30  8 invigilators will be needed  1R (4) L4 [27] QUESTION 3 [21] Ques. Solution Explanation Level 3.1.1 1A Subjects Learner A Learner B APS score APS score 1A For each IsiXhosa 6 5 2 correct English FAL 5 5 for both learners Mathematics 4 5 Life Orientation 3 3 1A Accounting 4 3 Economics 3 5 1A LO Business Studies 6 5 Total: 31 31 1CA Adding all Both learners have the same APS score. O values 1O Conclusion (6) L4 Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2016) 3.1.2 Percentages of 2 Grade 12 learners 80 70 60 50 Percentage 40 CA from Learner A 3.1.1 30 Learner B 20 1A for each two 10 compound bars 0 Subjects (5) L2 3.1.3 Equal chances  2O (2) L2 3.2.1 Rent = R1 650 × 11 1M = R18 150  Calculating Food = 1 500 × 10 rent = R15 000  1M Calculating Total amount = R18 150 + R15 000  food = 33 150  1M Addition 1CA Total (4) L2 3.2.2  2O Share room with someone OR to save on costs OR no privacy. Explanation (2) L4 [19] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2016) MATHEMATICAL LITERACY P2 7 QUESTION 4 [31] Ques. Solution Explanation Level 4.1 Triangular window: 1C covert to m Base = 150 cm = 1,5 m Height = 100 cm = 1 m   Area of triangular window = ½ base × height 1SF = ½ × 1,5 × 1  = 0,75 m2 1CA answer Total = 0,75 × 4 = 3 m2  1A radius Circular window: 1SF Radius = ½ × 1,5 = 0,75 m  1CA area of four  Area of circular window = 𝜋 × r2 circular window = 3,142 × 0,75 × 0,75  = 1,767375 m2 1CA area of four Total = 1,767375 × 4 square windows = 7,069 m2  1CA adding the areas Square window: Width = 1,5 m Length = 1,5 m 1CA total of  Area of square window = S2 areas = 1,5 × 1,5 = 2,25 m2 Total = 2,25 × 4 1M area of 1 = 9 m2  classroom wall Total area of windows for all four walls = 3 + 7,069 + 9  = 19,069 m2  Area of wall of 1 classroom = l × b 1M for ×4 area = 12 × 5 of 4 classrooms = 60 m2  Total area for four classrooms = 60 × 4  = 240 m2  1CA answer Area of wall to be painted = total area of walls – total area of 1CA subtraction windows = 240 – 19,069  = 220,9305 m2 1CA Wall area = 221 m2  to be painted (14) L3 Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2016) 4.2 Litres required for the walls: CA from 4.1.1 8 m2 = 1 ℓ 1MA 221 m2 = 1 /1 × 221 / 8 = 27,625 litres  1CA If 5 litre tins are bought: 27,625 / 5 = 5,525 = 6 tins of paint  (5 ℓ) 1M × by 105 Cost = 6 × 105  = R630 1M If 20 litres tins are bought: 27,625 / 20 = 1,38 = 2  1CA Cost = 2 × 405 = R810  1O Their statement is valid.  (6) L4 4.3.1 5 ℓ tins: Radius = 9 cm Diameter = 18 cm 1M number of tins Box A: over length Number of tins over the length of the box: 130/18 = 7  Number of tins over the height of the box: 1M number of tins 25/24 = 1,04 =1  over width Number of tins over the width of the box: 104/18 = 5,77 = 5  Total tins for Box A = 7 × 1 × 5 1M number of boxes = 35 tins  over height Box B: l = 65 cm 1CA total number of b = 49 cm tins in Box A h = 52 cm Number of tins on the width of the box: 1CA total number of 65/18 = 3,6 = 3 boxes in Box B Number tins along the length of the box: 49/24 = 2,04 = 2 1O No. of tins on height = 52/18 = 2,88 = 2 Total number of tins in box B = 3 × 2 × 2 = 12 tins  Box A can carry 28 tins which is more than 2 times 12 so their argument is valid.  (6) L3 4.3.2 Tins in Box B = 12  1CA from 4.3.1 Tins in 20 boxes = 12 × 20 1CA number of tins = 240  1M × by 240  1A Price of paint Amount lost = R105 × 240  1CA Total amount = R25 200  (5) L3 [31] TOTAL: 100 Copyright reserved Please turn over