NATIONAL
SENIOR CERTIFICATE
GRADE 12
JUNE 2022
MATHEMATICAL LITERACY P1
MARKING GUIDELINE
MARKS: 100
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuarcy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT/RG/RM | Reading from a table/ graph/ map |
| F | Choosing the correct formula |
| SF | Correct substitution in a formula |
| J | Justification |
| P | Penalty, e.g for no units, incorrect rounding off etc. |
| R | Rounding off/ Reason |
| AO | Answers only |
| NPR | No penalty for correct rounding off to minimum of two decimal places |
MARKING GUIDELINES
NOTE:
LET WEL:
| QUESTION 1 (20 Marks) | |||
| Que | Solution | Explanation/Marks AO: FULL MARKS | T/L |
| 1.1.1 | 18,25 = 1825 ✓M 100 10000 = 73 ✓A 400 | 1M fraction 1A answer in a reduced form (2) | F L1 * |
| 1.1.2 | % of price = 100 – 18,25% OR | 1M subtraction OR 1M % calculation (3) | F L1 * |
| 1.2.1 | Difference = R469 – (– R447) ✓CA = R916 million ✓RT | 1 RT for the two correct values 1 CA answer (2) | F L1 |
| 1.2.2 | Total = 265+277+326+390+447+458+486 – (469+300) = 1880 million ✓CA | 1M addition (+) and subtraction (–) of the values 1CA (2) | F L1 |
| 1.3.1 | Weekend wage rate = 3/2 × 25 ✓MA = R37,50 ✓A | 1MA multiplication 1A answer | F L1 * |
| 1.3.2 | ✓M Earnings = 6 × 25 + 37,50 × 4 ✓MA = R300 ✓CA | 1M multiplications 1MA addition 1CA answer (3) | F L1 * |
| 1.4.1 | Discrete ✓✓A | D L1 | |
| 1.4.2 | Game ✓✓RT | D L1 | |
| 1.4.3 | Total games = 4 + 6 + 5 + 4 + 1 + 2 + 2 = 24 games ✓M ✓CA | 1M adding the games 1CA answer (2) | D L1 |
| [20] | |||
| QUESTION 2 [18 MARKS] | |||
| Que | Solution | Explanation/Marks AO: FULL MARKS | T/L |
| 2.1.1 | Time 4 hours ✓✓RT | 2RT (2) | F L2 |
| 2.1.2 | From graph: 2 welders complete 1 frame in 4 hours 2 : 1 20 : ? frame in 4 hours Frames = 20×1 ✓✓M 2 = 10 frames ✓ OR n × t = 8 20 × t = 8 ✓ t = 8/20 = 0,4 hours to make 1 frame by 20 welders ✓S In four hours= 4/0,4 ✓M = 10 frames ✓A | 1M value from graph 1M numerator 1M denominator 1A answer OR 1SF substitution 1S simplification for 2,5 frames done in 1 hour by 20 welders 1M multiplication 1A answer (4) | F L3 |
| 2.2.1 | ✓M A = 28−25,81 × 100% ✓MA 25,81 = 8,485% = 8,5% ✓CA | 1M correct values for numerator and denominator M % calculation 1CA (3) (NPR) | F L2 |
| 2.2.2 | Cost: Up to 6 kℓ = R0 = R0 ✓M 6 – 25 kℓ = 19 k × R23,60 = R448,40 ✓M 25 – 30 kℓ = 5 kℓ × R32,20 = R161,00 ✓M ✓M TOTAL COST = R448,40+R161,00 = R606,40 ✓CA | 1M cost in block 1 1M cost in block 2 1M cost in block 3 1M addition all costs 1CA answer (5) | F L3 |
| 2.3.1 | Salary B = R3 192,05+15 761,80 ✓M = R18 953,85 ✓CA | 1M adding the two balances 1 CA answer (2) | F L2 |
| 2.3.2 | Bank fees for March = 42,37+17,47+100,88 ✓M = R160,72 ✓CA | 1M adding fees of March 1CA answer (2) | F L1 |
| [18] |
| QUESTION 3 [21 MARKS] | |||
| Que | Solution | Explanation/Marks AO: FULL MARKS | T/L |
| 3.1 | 2020 ✓ A Reason: Covid-19 pandemic ✓J | 1A year 1J reason (2) | D L1 |
| 3.2 | ✓M C = 25 285,1 – (2093,5+2092,8+2249,4+ 1988,8+1750,5 +1964,7+2067,1+2204,4+2308,0+2267,8+2493,4) = 1804,7 ✓M ✓CA | 1M subtracting from 25 285,1 1M addition of all other values 1CA answer (3) | D L2 * |
| 3.3 | descending order: ✓RT 2493,4; 2308,0; 2267,8; 2249,4; 2204,4; 2093,5; 2092,8 2067,1; 1988,8; 1964,7: 1804,7;1750,5 | 1RT all values including value from 3.2 1CA order with value from 3.2 (2) | D L2 * |
| 3.4 | ✓RT Range = 2 262,3 – 33,8 ✓M = 2 228,5 million ✓CA | 1RT highest and lowest values 1M concept of range 1CA answer (3) | D L2 |
| 3.5 | ✓M Mean income for 2018 = 24846,4 = 2070,53 million ✓A Mean income for 2020 = 9818,5 = 818,21 million ✓A Double mean income for 2020 = 818,21×2 = 1636,42 Million Mean income for 2018 (2 070,53) is greater than double mean income for 2020 (1636,42) Statement Valid ✓J | 1M concept of mean 1A mean for 2018 1A mean for 2020 1M comparing values of mean 2018 and double mean income for 2020 1J valid statement. NPR | D L4 * |
| 3.6 | From 2018 December income dropped right through up to July 2019; then increased from August 2019 to December 2019. It remained high up to March 2020. ✓J Then it dropped drastically in from April 2020 and remained low in 2020. ✓J | 1J justification for the period Dec 2018 to July 2019 1J justification for the period August 2019 to 2020 (2) | D L4 |
| 3.7 | May and June | 1A first months 1A second months. CA from 3.2 (2) | D L2 |
| [20] |
| QUESTION 4 [20 MARKS] | |||
| Que | Solution | Explanation/Marks AO: FULL MARKS | T/L |
| 4.1.1 | Values of dependent variable at break-even point Income = R300 ✓RT Expenses = R300 ✓RT | 1RT value for income 1RT value for expenses | F L2 |
| 4.1.2 | Total sales in a week = 37 packets ✓RT From Graph: Income = R555 ✓RT Expenses = R385 ✓RT Profit = R555 – R385 = R170 ✓CA OR Total sales = 37 ✓RT Income=37 × 15 = R555 ✓SF Expenses = 200+37× 5 = R385 ✓SF Profit = R555 – R385 = R170 ✓CA | 1RT adding sales from table 1RT reading income from graph 1RT expenses from graph 1CA answer for profit OR 1RT total sales 1SF for income 1SF for expenses 1CA answer for profit (4) | F L2 |
| 4.2.1 | Year 2009 ✓✓RT | 2RT for the year (2) | F L2 |
| 4.2.2 | Fees in 2015 = 1,093 × R12 500 = R13 662,50 ✓M Cost of fridge in 2015 = 1,04×R12 500 = R13 000 ✓M Difference = R13 662,50 – R12 500 = R662,50 ✓CA | 1M value from multiplication with education inflation rate. 1M value from multiplication with general inflation rate 1CA answer (3) | F L4 |
| 4.2.3 | The graph shows education has constantly outstripped general inflation. ✓✓J | F L4 | |
| 4.3.1 | Arrangement of currencies: £; €; $; P; R; ¥ ✓RT ✓A | 1RT all currencies | F L3 |
| 4.3.2 | 1¥ = R0,1383 3974,85 = R? Cost of 1 in Rands = 3974,85× 0,1383 ✓M = R549,72 ✓A Cost of 500 DVD players = 500 × 549,72 = R274 860,88 ✓CA | 1M converting the Japanese yens to Rands 1A cost of one DVD 1CA answer for cost of 500 DVDs (3) | F L2 |
| [18] |
| QUESTION 5 [25 MARKS] | |||
| Que | Solution | Explanation/Marks AO: FULL MARKS | T/L |
| 5.1.1 | Tax bracket = 4 ✓✓ RT | 2RT bracket (2) | F L1 |
| 5.1.2 | R128 650 ✓✓ RT | 2RT value of threshold (2) | F L2 |
| 5.1.3 | Monthly income = R35 455 | 1MA multiplication by 12 and annual income 1A the annual pension 1CA taxable income 1M use of correct tax bracket 1CA tax payable before rebates 1RT Total value of rebates 1M subtracting rebates and tax after rebates (7) | F L4 |
| 5.2.1 | ✓RT 2,27%; 5,04%; 5,05%; 5,90%; 6,68%; 7,24%; 13,38%; 16,15%; 38,28%. ✓M Median value = 6,68% giving EC ✓CA | 1RT all values from graph 1M arranging in order descending or ascending 1CA median value: EC | D L2 |
| 5.2.2 | Q1 = 5,04+5,05 ✓M 2 = 5,045% ✓A Q3 = 13,38 + 16,15 2 = 14,765% ✓A IQR = Q3–Q1 = 14,765% – 5,045% ✓M = 9,72% ✓CA | 1M concept of getting Quartile 1 1A for Q1 1A for Q3 1M method of subtracting Q3-Q1 1CA answer (5) | D L3 |
| 5.2.3 | Probability is the chance that an event is likely to happen. ✓✓A | 2A explanation (2) | P L1 |
| 5.2.4 | Probability for GP = 0,3828 ✓CA Probability for EC = 0,0668 ✓CA Probability for a car to be in GP OR EC = 0,3828 + 0,0668 = 0,4496 ✓A | 1CA converting 5 to decimal for QP 1CA converting to decimal for EC 1A answer (3) | |
| [24] | |||
| TOTAL: | 100 |