MATHEMATICAL LITERACY PAPER 2
GRADE 12
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
JUNE 2017
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 |
AO | Answer only |
NPR | No penalty for rounding |
QUESTION 1 [30]
Ques. | Solution | Explanation | Level |
1.1.1 | Because of the fixed monthly cost✓✓ OR No charge for copying✓✓ | 2A Explanation (2) | L2 F |
1.1.2 | You only pay the fixed cost and not for any copies made up to a certain point.✓✓ Accept any logical explanation | 2A Explanation (2) | L2 F |
1.1.3 | Company A: Cost for one copy 4 000 = 1000 + (3 500 – 1500) x Cost per copy✓ 4 000 – 1000 = 2 000 x cost per copy✓ 3 000 = 2 000 cost per copy Cost per copy = 3 000/2 000✓ = R1,50✓ OR Using any other points from the graph | 1M Subtracting 1 500 1M Subtracting 1000 1M Dividing by 500 1CA (4) | L3 F |
1.1.4 | Cost for Company B = 1 500 + (Number of copies – 500) x 1,25✓✓✓ OR Cost of Company B = 1 500 + 1,25 x number of copies above 500 | 1A Fixed Cost 1A no of copies more than 500 1A Cost per copy (3) | L4 F |
1.1.5 | The intersection points indicate the point where the number of copies and the cost for two or more companies ✓✓are the same | 1A Refer to number of copies the same 1A Refer to cost the same (2) | L3 F |
1.1.6 | 4 000 ; 4 750✓✓ | 1A Number of copies 1A Cost (2) Accept 4 700 on the vertical axis | L2 F |
1.1.7 | Company A✓✓ | 2A Correct choice (2) | L2 F |
1.1.8 | 4 001 copies✓✓ | 2A Correct minimum (2) | L3 F |
1.2.1 | Number of coloured pencils across = 79 mm / 7 mm✓ = 11,28571429✓ = 11 pencils✓ Number of pencils down = 18 cm / 17,5 cm✓ = 1,028571429✓ = 1 pencil✓ Total number of pencils in one container = 11 x 1 = 11 pencils✓ Number of pencils in 3 containers = 11 x 3✓ = 33 pencils✓ | 1M Dividing diameters 1 CA Simplification 1 R Number of pencils 1M Dividing heights 1 R Number of pencils 1CA Number of pencils 1M Multiply by 3 1CA (8) | L3 M |
1.2.2 | Probability of taking a red pencil from a container = 9 33 ✓✓ = 0,272727272 = 0,273✓✓ | CA from 1.2.1 1A Numerator 1A Denominator 1R to 3 decimal places (3) | L2 P |
[30]
QUESTION 2 [25]
Ques. | Solution | Explanation | Level |
2.1 2.1.1 | It means half (50%) of 12 year old boys are taller or shorter than other boys.✓✓ OR The boy has an average height✓✓ | 2A Explanation (2) | L4 D |
2.1.2 | His height-for-age puts him between the 5th and the 25th percentile, therefore he is below average height-for-age ratio.✓✓ | 2A Below average(2) | L3 D |
2.1.3 | The curves are the steepest between 5 and 7 years ✓✓ Accept between 4 and 8 years | 1A 5 years 1A 7 years (2) | L4 D |
2.1.4 | Height: 1,27 m = 127 cm✓ Height in inches = 127 cm✓ 2,54 cm = 50 inches✓ According to the graph with a height of 127 cm OR 1,27 m OR 50 inches a boy will be 13 years. OR At the age of 15 years a boy will be 53 inches Statement invalid✓ | 1C m to cm 1M Dividing by 2,54 1CA Inches 1A Explanation 1O Not valid (5) | L4 M |
2.1.5 | Boys with Down Syndrome develop differently than normal boys.✓✓ Accept any other relevant answer | 2A Explanation (2) | L4 D |
2.2 2.2.1 | Range = Highest value – Lowest value 11 = A – 8✓ A = 19✓ | 1M Concept of range 1CA Value of A (2) | L2 D |
2.2.2 | Mean = 8 + 9x4 + 10x2 + 11x2 + 12x3 + 13x3 + 14x3 + 15x3 + 16x8 + 17x4 + 18x2 + 19 ✓ 36 ✓ = 499 36 = 13,86 ✓ | CA from 2.2.1 1M Adding all 36 values 1A Dividing by 36 1CA Simplification NPR (3) | L3 D |
2.2.3 | B = 11+ 12 ✓ 2 ✓ = 11,5 ✓ C = 14+ 15 2 = 14,5 ✓ D = 16 ✓ | 1A Identifying the correct values 1CA Value of B 1M Concept of median 1CA Value of C 1CA value of D (5) | L2 D |
2.2.4 | P(girl not 16 years and younger) = 7 36 ✓ | 1A Number older than 16 1A No. of girls (2) | L2 P |
[25]
QUESTION 3 [26]
Ques. | Solution | Explanation | Level |
3.1 3.1.1 | Per annum if rounded = R3 651 x 11 = R40 161 ✓ Actual amount per annum = R40 166 ✓ Rounding to a whole number gives a difference of R5,00 ✓✓ | 1MA 3 651 x 11 1RT Correct value 2O Comparison(4) | L4 F |
3.1.2 | Grade 11 Learner = R104 670 – ( R104 670 x 0,05)✓ = R104 670 – R5 233,50 ✓ = R 99 436,50✓ Grade 3 Learner = R5 807 x 11✓ = R63 877 4 = R15 969,25 ✓ Grade 7 Learner = R68 373 ✓ 11 = R6215,73 x 3 ✓ = R18 647,18 ✓ Total cost for the first term = R99 436,50 + R15 969,25 + R 18 647,18 ✓ = R134 052,93 ✓ | 1M Calculate 5% 1M Subtract 5% 1CA Cost 1M Multiply by 11 1M Dividing by 4 1CA Cost 1M Dividing by 11 1M Multiply by 3 1CA Cost 1M Adding 1CA Total Cost (11) | L3 F |
3.1.3 | Smaller classes✓✓ OR Individual and Special attention✓✓ OR More extra-mural activities✓✓ Accept any relevant answer | 2O Reason (2) | L4 F |
3.2 3.2.1 | Total Parts = (6 tubes x 4 each) + 1 + 4 + 4 + 8 + 4 ✓ = 45 parts✓ | 1A Total tubes 1A Other parts 1CA Total parts(3) | L2 M&P |
3.2.2 | Fix the tubes together✓✓ Connect the ends by the plastic parts supplied✓✓ Throw over the cover and tie at tube No.6✓✓ | 2A Tubes first 2A Other parts second 2A Cover (6) | L4 M&P |
3.2.3 | Small parts is choking hazard✓✓ | 2R Reason (2) | L4 M&P |
[28]
QUESTION 4 [17]
4.1 4.1.1 | Original capacity =95000 1,2559 ✓✓ = 75 642,9652 ✓ = 75 643 ✓ | 1M Dividing 1A Correct % 1CA Answer 1R Rounding (4) | L3 M |
4.1.2 | North stand ✓ South stand ✓ A better view over the pitch ✓✓ | 1A North Stand 1A South Stand 2R Reason (4) | L2& L4 M&P |
4.1.3 | For Non Manchester United fans ✓✓ OR For wheel chairs ✓✓ | 2O Reason (2) | L4 M&P |
4.1.4 | For surface water to run off easily ✓✓ Accept any relevant answer | 2O | L4 M&P |
4.1.5 | If 100 yards = 91,44 metres Than 1 yard = 0,9144 metres ✓ Length = 115 x 0,9144✓ = 105, 156✓ ≈ 105 m Width = 74 x 0,9144 = 67,6656✓ ≈ 68 m Area = length x width = 105 m x 68 m✓ = 7 140 m2 | 1A Simplify 1C Convert to metres 1CA Length 1MA Convert to metres 1M Multiply approximated length and width (5) | L3 M |
[17]
TOTAL : 100