MATHEMATICAL LITERACY PAPER 1
GRADE 12
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
JUNE 2018
| 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 [20 marks] | Explanation | Marks | |
| 1.1.1 | Cost for deposit = 1,50+ 0,25 × 2 000 ✓=1,5+5 MA 100 = R6,50 ✓CA | 1 MA (0,25% of 2 000) 1CA | L1 (2) |
| 1.1.2 | Minimum = R2,00 ✓✓ | 2 RT | L1 (2) |
| 1.2.1 | A= 231,70 – (23,45 + 90 + 23 + 45) ✓MA = 50,25 ✓CA | 1 MA Subtracting values from 231,70 1 CA Value of A | L1 (2) |
| 1.2.2 | Cost of total data used = 231,7 ×149 ✓ MA 1 000 = R34,52 ✓CA OR Cost of total data used = 149 × 231,7 ✓MA 1 000 = R34,52 ✓CA | 1 MA Total data as a fraction of 1GB × 149 1 CA Cost NPR | L1 (2) |
| 1.3.1 | Seat 20B ✓✓ RT | 2 RT Seat number | L1 (2) |
| 1.3.2 | Time for the flight = 13:30 – 12:05 ✓MA = 1hr 25 min ✓A | 1 MA Subtracting the times 1 A 1 : 25 is incorrect | L1 (2) |
| 1.4.1 | Bar scale / Graphic scale✓A Linear scale ✓ A Word scale ✓A Fractional or ratio scale ✓A | 2 A for any two scales given | L1 (2) |
| 1.4.2 | For every one unit on a drawing or on a map there are 400 000 units in reality ✓✓ A | 2 A Explanation | L1 (2) |
| 1.5.1 | Others = 2 100 000 - (503 096 + 439 719 + 221 121 + 219 007) ✓MA = 717 057 ✓ CA | 1 MA Subtracting the total from 2,1 million 1 CA | L1 (2) |
| 1.5.2 | Health % of the total= 221 121 ×100 MA 2 100 000 = 10,53% ✓CA | 1 MA Fraction of correct values × 100% 1 CA NPR | L1 (2) |
| [20] | |||
| QUESTION 2 [27 marks] | Explanation | Marks | |
| 2.1 | Inflation Rate = 369,50 − 343 ×100 ✓SF 343 = 26,50343×100 ✓S = 7,73% ✓ CA | 1 SF 1 S 1 CA NPR | L2 (3) |
| 2.2.1 | Original Salary = 25 750 1,085 ✓✓MA = R23 732,72 ✓CA | 1 A Correct value 25 750 1 M divided by 1,085 1 CA | L2 (3) |
| 2.2.2 | Annual Salary = 25 750 ×12 ✓MA = R309 000 ✓A | 1 MA Multiplication of correct values 1 CA | L1 (2) |
| 2.3.1 | D = 2 500 50 ✓MA = 50 tickets ✓A | 1 MA Division of correct values 1 CA | L1 (2) |
| 2.3.2 | Cost of renting = R3 600 ✓✓ Reading from given information | 2 A | L1 (2) |
| 2.3.3 | 1 A 1st point 1 A Line drawn | L2 (2) | |
| 2.3.4 | From graph = Expense line above income = Difference R600 ✓ RG = Loss ✓O OR Difference = Income ‒ expense = (60 x 50) ‒ 3600 ✓M = -600 ✓S = Loss ✓O | 2 RG 1O 1 M Subtraction of expense from income correct value 1 S 1 O | L3 (3) |
| 2.3.5 | Break-even is the point where cost of renting equals income ✓✓ A (J) OR No loss and no profit | 2 A Explanation | L1 (2) |
| 2.4 | Rent with VAT = 3 600 Rent without VAT = 3 600 =?3 130,43 ✓ M 1.15 VAT = 3 600 ‒ 3 130,43 ✓ M = R469,57 ✓ CA OR VAT = 15 × MA×3600 ✓M 115 = R469,57 ✓CA | 1 MA Divide by 1,15 1 M Subtraction 1 CA 1 MA Fraction 1 M Multiplication 1 CA | L3 (3) |
| 2.5.1 | R1 = 0,46406 CYN ? = 1 250 ? = 1 250 0,46406 ✓MA = R2 693,62 ✓A | 1 MA Division using correct values 1A | L2 (2) |
| 2.5.2 | Total parts 2 + 3 = 5 ✓MA Value of each part = 26945 = R538,80 ✓S Wife has one part more; she got R538,80 ✓CA more OR Total parts = 5 ✓MA Wife got = 3 × 2 694=?1 616,40 5 Husband got = 2 × 2 694=?1 077,60 5 Wife got 1 616, 40 - 1 077,60 ✓MA = R538,80 ✓CA more OR Total parts 2 + 3 = 5 ✓MA Value of each part = 1250 = CYN 250 ✓S 5 Wife has one part more; she got CYN 250 ✓CA more | 1 MA value of 5 1 S 1 CA 1 MA value of 5 1 Subtraction 1 CA (Allow if calculated in Chinese Yuan) 1 MA value of 5 1 S 1 CA | L2 (3) |
| [27] | |||
| QUESTION 3 [16 marks] | Explanation | Marks | |
| 3.1 | 1,8 m ✓✓ RD OR Radius = 3,6 ✓ M 2 = 1,8 m ✓A | 2 RD 1 M Divide by 2 1 A | L1 (2) |
| 3.2 | C=?? C= 3,142 x 1,8 m ✓ = 5,66 m ✓ OR C=3,142 x (3,6 m) ÷ 2 ✓ =5,66 m ✓ | 1 SF 1 CA from 3.1 1 SF 1 CA NPR | L2 (2) |
| 3.3 | A + A + A + 0,4 m +1,1 m + 1,75 m = 5,8 M ✓ 3A + 3,25 m = 5,8 m 3A = 5,8 m – 3,25 m ✓S 3A = 2,55 m✓M/A 3 3 A = 0,85 m ✓CA | 1 M Addition 1 S 1 M/A Division 1 CA | L1 (4) |
| 3.4 | TA= ?r2+ (length x breadth) x 2 = 3,142 x 1,82 + (5,8 m x 4,9 m) x2 ✓✓ = 10,18008 m2 + 28,42 m2 x 2 ✓ = 10,18008 m2 + 56,84 m2 = 67,02 m2 ✓ = 67 m2 ✓ | CA from 3.1 2 SF 1 S 1 CA 1 Rounding off | L2 (5) |
| 3.5 | 5,8 m + 1,8 m = 7,6 m ✓ x 2 ✓ = 15,2 m ✓ | 1 M Adding (Value 7,6 m) 1 M (Multiplication by 2) 1 CA | L2 (3) |
| [16] | |||
| QUESTION 4 [11 marks] | Explanation | Marks | |
| 4.1 | 16 ✓✓ RD | 2 RD | L1 (2) |
| 4.2 | (a) AA = 48 ✓✓A | 2 A Number | L1 (2) |
| (b) BB = 65 ✓✓A | 2 A Number | L1 (2) | |
| 4.3 | Q 5 ✓✓ RD | 2 RD Award 1 mark for 5Q | L1 (2) |
| 4.4 | Turn it over once in a clockwise direction in such a way that the left side is now the top side, the bottom side is now the left side and the top side is now the right-hand side. ✓✓✓ | 3 A Explanation | L2 (3) |
| [11] | |||
| QUESTION 5 [26 marks] | Explanation | Marks | |
| 5.1.1 | 18 ✓✓ RM | 2 A RM Total number of data values | L1 (2) |
| 5.1.2 | Willowmore ✓ RM Port Elizabeth ✓ RM East London ✓ RM Aliwal North ✓ RM | 4 A RM (1 Mark for each town or city) | L1 (4) |
| 5.1.3 | 14; 14; 15; 15; 15; 16; 16; 17; 17; 17; 17 ✓✓ RM | 2 A RM Arranged in ascending order | L1 (2) |
| 5.1.4 | 16 ✓✓ CA | 2 CA From 5.1.3 Median | L2 (2) |
| 5.1.5 | ?1 =15 ✓ A and ?3 = 17 ✓A Interquartile range = 17 ‒ 15 ✓ 1 M = 2 ✓ CA | 2 MA for ?1 and ?3 1 M subtraction 1 CA | L2 (4) |
| 5.1.6 | Mean = 16+18+21+22+27+28+28 = 160 ✓✓ MA 7 7 = 22,86 = 23 ✓A | 2 MA For addition and division 1 CA NPR | L2 (3) |
| 5.1.7 | 28 ✓✓ RM | 2 RM | L2 (2) |
| 5.1.8 | Difference = 28 ‒ 17✓ MA = 11 ✓CA | 1 MA Subtraction 1 CA | L1 (2) |
| 5.2.1 | Probability is the chances of an event to happen ✓✓ | 2 A Explanation | L1 (2) |
| 5.2.2 | P (City or town with temperature less than 170°?) = 8✓ = 4 18✓✓ 9 | 1 A for 8 the numerator 1 CA From 5.1.1 1 A NPR | L2 (3) |
| [26] | |||
TOTAL: 100