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GRADE 12 MATHEMATICAL LITERACY PAPER 2 MEMORANDUM - AMENDED SENIOR CERTIFICATE PAST PAPERS AND MEMOS MAY/JUNE 2017
Share via Whatsapp Join our WhatsApp Group Join our Telegram GroupMATHEMATICAL LITERACY PAPER 2
GRADE 12
SENIOR CERTIFICATE EXAMINATIONS
MAY/JUNE 2017
| Codes | Explanation |
| M | Method |
| MA | Method with Accuracy |
| CA | Consistent Accuracy |
| A | Accuracy |
| C | Conversion |
| D | Define |
| J | Justification/Reason/Explain |
| S | Simplification |
| RD | Reading from a table OR a graph OR a diagram OR a map OR a plan |
| F | Choosing the correct formula |
| SF | Substitution in a formula |
| O | Opinion |
| P | Penalty, e.g. for no units, incorrect rounding off, etc. |
| R | Rounding Off |
| NP | No penalty for rounding OR omitting units |
| MCA | Method with consistent accuracy |
These marking guidelines consist of 15 pages.
KEY TO TOPIC SYMBOL:
F = Finance; M = Measurement; MP = Maps, plans and other representations
DH = Data Handling; P = Probability.
| QUESTION 1 [39 Marks] | |||
| Ques | Solution | Explanation | T&L |
| 1.1.1 | Probability = 3 ✓ 15 ✓ = 0,2 ✓ | 1A numerator 1A denominator 1CA simplification AO (3) | P L2 |
| 1.1.2 | 6 members scores decreased. As a percentage = 6 × 100% ✓ 15 = 40% ✓ | 1A no. decreased 1MA percentage with denominator 15 1CA simplification AO (3) | D L2 |
| 1.1.3 (a) | Arranging scores in ascending or descending order: 27; 28; 30; 32; 34; 38; 41; 42; 43; 43; 44; 46; 53; 56; 62✓ Median is 42. ✓✓ | 1MA ordered data 2A median AO (3) | D L2 |
| 1.1.3(b) | 43✓✓ | 2A mode (2) | D L2 |
| 1.1.3 (c) | IQR = upper quartile – lower quartile = Q3 – Q1 = 46 – 32✓✓ = 14✓ | CA from 1.1.3(a) 1RT 46 1RT 32 1CA IQR value (3) | D L3 |
| 1.1.4 | The interquartile range of 1st tournament is smaller than that of the 2nd tournament (i.e. 14 compared to 50)✓✓ Range of scores is smaller (i.e. 35) in the 1st tournament compared to a range of 90 points scored in 2nd tournament. Majority improved their scores.✓✓ OR Highest score by a player in 1st tournament is 38 points less than a player in 2nd tournament. The interquartile range of 2nd tournament is higher than that of the 1st tournament (i.e. 50 points higher than 14 points).✓✓✓ The lowest score of tournament 2 is 17 less than the lowest score in tournament 1.✓ OR Players' performance in Tournament 1 were more consistent because the IQR is smaller and also the range is smaller.✓✓✓✓ | 2J comparison 2J comparison 2J comparison 2J comparison OR 2J comparison 2J comparison (4) | D L4 |
| 1.2.1 | Points : 3 × 1 = 3✓ 8 × 2 = 16 3 × 3 = 9 Point scored = 3 + 16 + 9 = 28✓✓ Player F✓ OR 3 × 1 + 8 × 2 + 3 × 3 = 28 points✓✓✓ Player F✓ | 1MA point in relation to position (multiply) 1M adding points 1A accumulated points 1CA player 1MA balls multiply by points 1M adding 1A total points 1CA player AO (4) | D L3 |
| 1.2.2 | 45 cm : 3,66 m✓ 0,45m : 3,66 m✓ 15 : 122✓ OR 45 cm : 3,66 m✓ 45 cm : 366 cm✓ 15 : 122✓ | 1MAwriting in correct ratio 1C convert cm to m 1CA simplification (no units) OR 1MAwriting in correct ratio 1C convert m to cm 1CA simplification (no units) (3) | M L2 |
| 1.2.3 | Shaded Area = πr2(hoop) – πr2(ball) = 3,142 × (22,5cm)2 – 3,142 × (12,4cm)2✓✓✓✓ = 1 590,6375 cm2 – 483,11392 cm2✓✓ = 1 107.52 cm2✓ OR Area of circle (hoop) = π × (radius)2 = 3,142 × (22,5)2✓✓ = 1 590,6375cm2✓✓ Area occupied by the ball = π × (radius)2 = 3,142 × (12,4)2✓✓ = 483,11392 cm2✓ Shaded area = 1 590,6375 – 483,11392 cm2 = 1 107,52358 cm2✓ | 1A radius hoop 1A radius ball 1M subtracting 1SF correct values 1CA area in cm2 1CA area occupied by the ball 1CA simplification OR 1A radius 1SF correct values 1CA area 1A radius of a ball 1CA area occupied by the ball 1M difference 1CA simplification NPR (7) | M L3 |
| 1.3 | Proportional price money: Y group share R8,1 mil × 3 = R2,7 mil 9 Each member of Y group will receive = 2,7million ✓ 5 = R0,54 mil.✓ 0,54 × 1 000 000 = R540 000 ✓ The player was correct. ✓ OR Group Y receives 3 of the share ✓✓ 9 Each member receives 1 5 A player from Y = 3 × 8,1 million 45 = 0,54 million ✓ = R540 000✓ The statement is correct✓ | 1MA getting 9 1M multiply by ratio 1CA price money to share 1M divide by 5 1CA each member's share 1C to 1000's 1O conclusion based on calculation 2MA correct ratio 1A each member's share 1M multiply with ratio 1CA simplification 1C conversion 1O conclusion [max 4 marks if divided by 15 first to get 0,54 mil Max 5 marks if dividing by 3 instead of working with the ratio 93] (7) | F L4 |
| [39] | |||
QUESTION 2 (37)
| 2.1.1(a) | Amount × (106,18%) = R14,44 ✓ K = R14,44 ÷ 106,18 % or 1,0618 ✓ = R13,599 = R13,6 ✓ | 1RT correct values 1A dividing by 106,18% or dividing by 1,0618 1R value in rand (3) | F L2 |
| 2.1.1(b) | Q = R11,50 - R10,88 x 100% ✓ R10,88 = 5,7 ✓ OR – 0,81 + 12,2 + 7,82 + 2,28 + 6,18 +5,24 + 10,07 +11,34 + Q = 6,00 x 10 ✓ Q = 60 – 54,32 ✓ = 5,68 ✓ | 1RT correct values 1M subtracting values 1F percentage change 1CA simplification OR 1RT correct values 1M mean concept 1M subtracting values 1CA simplification NPR (4) | D L2 |
| 2.1.1 (c) | E =0,99 + 17,32 + 15,07 + 5,99 + 9,42 + 8,16 + 4,46 + 9,04 + 10,27 + 15,64 ✓ 10 ✓ = 96,36 10 = 9,64✓ | 1MA adding values 1MCA mean concept ÷10 1CA mean value (3) | D L2 |
| 2.1.2 | Apr. 2015 to Jan. 2016: both prices increased.✓✓ Jan. 2016 to Apr. 2016: The price of the 600 g loaf of white bread remained the same (is constant).✓ The price of the 700 g loaf of white bread increased✓ OR Per period per bread✓ 600 g: Apr 2015 – Jan 2016 : The price increased.✓ Jan 2016 – Apr 2016: The price remained the same.✓ 700 g: Apr 2015 – Jan 2016 : The price increased.✓ Jan 2016 – Apr 2016 The price increased.✓ | 2J both increased 1J 600 g constant 1J 700 g increased 600g: 1J increased 1J constant 700g: 1J increased 1J increased (4) | D L4 |
| 2.1.3 | He will have to adjust his spending to cater for the increased price. That is money that he was saving to use for other things will be used for wheat products.✓✓ OR Will experience financial difficulties (i.e. unable to afford bread any longer).✓✓ OR If he buys the wheat products it will cost him more and he will have less money to spend on other stuff✓✓ OR Can buy less and less✓✓ OR Any other valid reason✓✓ | 2J explanation OR 2J explanation OR 2J explanation OR 2J explanation OR 2J explanation (2) | F L4 |
| 2.2 | Increase in 2017 = 6,6% × R6,72 ✓ = R0,44 ✓ Increased price = R6,72 + R0,44 ✓ = R7,16 ✓ Increase in 2018 = R7,16 × 6% = R0,43✓ Increased price = R7,17 + R0,43✓ = R7,59 OR 2017: R6,72 × 1,066 = R 7,16✓✓✓ 2018: R7,16 × 1,06 = R7,59✓✓✓ OR R6,72 × 1,066 × 1,06 = R7,59✓✓✓✓✓✓ | 1MA multiplying correct values 1A increase amount 1M adding 1CA increased price 1CA increase % 1CA increased price OR 1MA multiplying correct values 1A increase amount 1M adding 1CA increased price 1CA increase % 1CA increased price (6) | F L3 |
| 2.3.1 | V = 690 mm × 445 mm × 180 mm ✓✓✓ = 55 269 000 mm3 | 1SF correct values 2CA volume P if unit is wrong (3) | M L2 |
| 2.3.2 | Number of crates lengthwise or 2000 | 1C conversion 1M dividing 1CA number length wise 1CA number 1M finding the total number 1CA number of crates 1J conclusion (7) | M L3 |
| 2.3.3 | Number of loaves = 80 × 8 = 640✓ Cost price per bread = R5350 640 = R8,36 Number of loaves to break even = FC SP - CP = R1 1720,70 ✓ R11,50 - R8,36 = 548 ✓ | 1A total number of loaves 1M dividing 1CA cost price 1SF substitution (at least 2 correct values) 1CA number of whole loaves (5) | F L3 |
| [37] |
QUESTION 3 (38 marks)
| Ques | Solution | Explanation | T&L |
| 3.1.1 | Total population = 22 574 500 ✓ 41,4% ✓ = 54 925 790,75 ✓ ≈ 54 925 800 people ✓ | 1RT correct values 1M dividing by % 1CA population 1R number of people (4) | D L3 |
| 3.1.2 (a) | P(White female) = 2 325 100 55 908 900 ✓ = 0,042 OR 4,2% OR 1 ✓ 24 | 1MA numerator and denominator 1CA simplification AO (2) | P L3 |
| 3.1.2 (b) | Total males = RSA population – Female population = 55 908 900 – 28 529 100 ✓ = 27 379 800 ✓ P(male) =27 379 800 ✓= 0,489721672 ≈ 0,49 OR 48,97% 55 908 900 OR P(female) =28 529 100 = 0,51027 ✓... ≈ 0,51 or 51,03% 5 908 900 P(male) = 1 – 0,51027 ✓.. or 1 – 0,51 or 100% – 51,03% = 0,489721672 or 0,49 or 49,97% ✓ | 1MA difference 1CA males total 1CA probability OR 1A P(female) 1M subtracting from 1 1CA P(male) (3) | P L3 |
| 3.1.3 | 2016 = 684 100 × 100% ✓✓ 28 529 100 = 2,3979024 ≈ 2,4% ✓ 2015 = or 2014 = 673 900 × 100% 664 900 × 100% ✓ 28 078 700 27 635 900 = 2,4% = 2,4% ✓ OR 2014: 100% – (80,2% + 8,9% + 8,5%)= 2,4% ✓✓✓ 2015: 100% – (80,4% + 8,9% + 8,3%)= 2,4% ✓✓ 2016: 100% – 80,6% – 8,9% – 8,1% = 2,4% | 1MA numerator and denominator 1M multiply by 100% 1CA percentage 1MA numerator and denominator 1CA percentage OR 1MA subtracting from 100% 1M adding other values 1CA percentage 1MA another year 1CA another year (5) | D L4 |
| 3.2.1 | Total distance of a space and a post = 100 mm + 40 mm✓ = 140 mm or 0,1 m + 0,04 m = 0,14 m Distance between posts that must have a space and a post = 3 460 mm – 100 mm✓ = 3 360mm or 3460m - 0,14m = 3,360 m Number of small posts = 3360 140 = 24✓ or 3,360 0,140 = 24m✓ | 1A correct distance 1M subtracting 1M dividing by 140 1CA number of small post [Accept 26 full marks] (4) | M L2 |
| 3.2.2 | Direct sunlight coming into the rooms through the windows for much longer.✓✓ OR Sun spend most of the time on the north side of the house.✓✓ OR It is the side on which the sun shines most of the time during the day. ✓✓ | 2J sun and time OR 2J direction and time OR 2J sunshine (2) | MP L4 |
| 3.2.3 | Open outward because they have short width✓✓ OR Designed to store things, as such they will obstruct inward opening of the doors✓✓. OR Storage space will be lost if doors open inwards✓✓ OR Other rooms open inward because it is the entrance to the room.✓✓ | 2O wideness OR 2O purpose OR 2O space OR 1O way of opening 1O purpose (2) | MP L4 |
| 3.2.4 | Carpeted floor = Area of a Passage + Dining + Living rooms DR area = 3,3274 × 3,6576 = 12,17029824 m2✓ LR area = 4,5720 × 4,2672✓ = 19,5096384 m2✓ Area of passage = 15% of (12,17 + 19,51) m2 = 15 % of 31,68 m2✓ = 4,751990496 m2✓ Total area = 12,17 m2 + 19,51 m2 + 4,75 m2✓ = 36,43 m2✓ ≈ 37 m2✓ | 1SF finding area 1CA area of DR 1CA area of LR 1M finding 15% 1CA area of passage 1M adding 3 or 4 values 1CA total area 1R rounding [Max 6 marks if total area is calculated] (8) | M L3 |
| 3.2.5 | Labour Cost: R1 600 + 37 × R70✓ = R1 600 + R2 590 = R4 190 Number of boxes = 37 ÷ 2,15 ✓ = 17,209 ≈ 18 Cost for boxes flooring: 18 × R299,90 = R5 398,20 ✓ Number of underlay rolls: 37 ÷ 10 = 3,7 ≈ 4 Underlayer: 4 × R56,90 = R227,60 ✓ Total cost = R4 190 + R5 398,20+ R227,60 ✓ = R9 815,80 ✓ The budget is sufficient. ✓ | Area CA from 3.2.4 above 1MA finding labour 1CA labour cost 1M dividing by 2,15 1CA cost of boxes 1CA underlayer cost 1MCA adding all 3 different cost types 1CA total cost 1O conclusion (8) | F L4 |
| [38] |
QUESTION 4 [36 marks]
| Ques | Solution | Explanation | T&L |
| 4.1.1 | Tax bracket 3, 4 and 5 ✓✓✓ [Accept Tax bracket 1] OR $37 001 – $87 000 ✓ $87 001 ̶ $180 000. ✓ $180 001 and over.✓ [Accept $0 – $1 200] | 1RT bracket3 1RT bracket 4 1RT bracket 5 OR 1RT tax bracket 1RT tax bracket 1RT tax bracket (3) | F L2 |
| 4.1.2 | Pay extra tax (2% on taxable income)✓✓ OR The levy is an extra (additional, more) tax on their income.✓✓ OR Higher income earners are subjected to an extra tax in addition to usual income tax paid. ✓✓ | 2O reason OR 2O reason OR 2O reason (2) | F L4 |
| 4.1.3 | Tax due 2016: = $54 547 + 45% × ( $289 303,26 ̶ $180 000)✓✓ = $54 547 + 45% × $109 303,26 =$54 547 + $49 186,47 =$103 733,47✓ Medical levy = $289 303,26 × 2% = $5 786,07✓ Total due = $103 733,47 + $5 786,07 = $109 519,54✓ Tax due 2017: = $54 232 + 45% × ($311 001 ̶ $180 000)✓✓ = $54 232 + 45% × $131 001 = $54 232 + $ 58 950,45 = $113 182,45✓ Medical levy = 2% × $311 001 = $6 220,02 Total for 2017: $113 182,45 + $6 220,02 ✓ = $119 402,47 Tax due difference: $119 402,47 – $109 519,54 ✓ = $9 882,93.✓ The statement is VALID. ✓ | 1RT tax bracket 1 SF correct substitution 1CA tax due 1MA levy value 1CA total due 1RT tax bracket 1SF correct values 1CA tax due 1CA total 1M finding difference 1CA simplification 1O conclusion (12) | F L3/4 |
| 4.2.1 | Mary Rose restaurant; Denmark hotel; Civic Centre✓✓✓ | 3A venues Accept hotel (3) | MP L2 |
| 4.2.2 | Because it runs over the river.✓✓ OR Portions of the river not visible from above where the highway crosses or passes over the river.✓✓ | 2O reason OR 2O reason (2) | MP L4 |
| 4.2.3 | North west OR NW OR West of North✓✓ | 2RT direction (2) | MP L2 |
| 4.2.4 | Turn right walk along Walker Str✓ Turn right into Strickland Str✓ Pass South Coast Highway And turn left into Mount Shadforth Rd✓ Restaurant will be on his right OR Turn SW into Walker Street and proceed.✓ At the corner turn NW and continue.✓ Cross South Coast Highway Turn W into Mount Shadforth Rd.✓ The restaurant is on the northern side of the road. | 1A route and turn 1A route and turn 1A turn and road OR 1A route and turn 1A route and turn 1A turn and road (3) | MP L3 |
| 4.2.5 | Measured distance between = 23 mm✓✓ Scale 23 mm is 100 m ✓ How long it will take him = Time = Distance ✓ Speed = 100m 1,1m/s = 90,91 seconds ✓ In minutes 90,909 ÷ 60 = 1,52 minutes. ✓✓ No. He can walk in less than 2 minutes at that speed. ✓ OR 2 min = 120 sec Distance = 1,1 m/s x 120 s = 132 m Measured distance = 23 mm ✓✓ Scale 23 mm = 100 m ✓ He will have passed the Indigo Cuisine ✓ [Accept measurements 23 mm to 25 mm] | 2MA measuring 1C using scale 1F formula 1A dividing by speed 1CA calculating time 1C divide by 60 1CA minutes 1O conclusion OR 1C multiply by 60 1A time in seconds 1A multiply with speed 1F formula 1CA distance 2MA measurement 1C using scale 1O conclusion (9) | MP L4 |
| [36] |
TOTAL: 150