Mathematical Literacy Grade 11 P2 2024 Memo (Eng)

NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2024 MATHEMATICAL LITERACY P2 MARKING GUIDELINES MARKS: 100 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT Reading from a table/graph/diagram SF Correct substitution in a formula O Opinion/Explanation/Reasoning P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off/Reason NPR No penalty for correct rounding minimum two decimal places AO Answer only MCA Method with consistent accuracy RCA Rounding with consistent accuracy This marking guidelines consist of 11 pages. Hosted by www.ecexams.co.za.

2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2024) MARKING GUIDELINES NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out (cancelled) an attempt to a question and NOT redone the solution, mark the crossed out (cancelled) version. • Consistent Accuracy (CA) applies in ALL aspects of the marking guidelines; however, it stops at the second calculation error. • If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalize for every extra incorrect item presented. Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P2 3 KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, plans and other representations; P = Probability QUESTION 1 [20 MARKS] ANSWER ONLY FULL MARKS Ques. Solution Explanation Level 1.1.1 Analogue clock ✓A 2A correct type clock M (2) L1 1.1.2 Ten minutes past ten o’clock in the morning. ✓A 2A correct time M L1 OR Ten past ten in the morning. ✓A OR Ten past ten before noon. ✓A (2) 1.1.3 Time taken = 10:27 1M subtracting time M -10:10 M 1A correct time L1 = 00:17 ⸫ 17 minutes A (2) 1.2.1 Perimeter is the distance around an object or shape. ✓A 2A definition M (2) L1 1.2.2 Perimeter = 230 cm + 200 cm + 95 cm + 88 cm + 135 cm + 1MA adding all the M 112 cm MA correct values L1 = 860 cm A 1A perimeter Accept if calculated as follows: Perimeter = (230 × 2) + (200 × 2) MA 1MA multiplying length = 860 cm A and width by 2 1A perimeter (2) 1.3.1 43 1C conversion M Side or length = 100 C 1A answer L1 = 0,43 m A (2) 1.3.2 Area of a square = side × side CA from 1.3.1 M = 0,43 × 0,43 SF 1SF substitution L1 = 0,1849 1MCA area in m2 ≈ 0,18 m2 MCA NPR (Accept 0,185 m2) (2) 1.4.1 5 casters (wheels) ✓A 2A number of casters MP (2) L1 1.4.2 Tool = Allen key ✓A 2A correct tool MP (2) L1 Copyright reserved Please turn over Hosted by www.ecexams.co.za.

4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2024) 1.4.3 M8 × 15 mm A 1A M8 × 15 mm MP M5 × 18 mm A 1A M5 × 18 mm L1 (Accept any order) (2) [20] Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P2 5 QUESTION 2 [20 MARKS] Ques. Solution Explanation Level 2.1.1 N2 ✓A 2A correct national road MP (2) L1 2.1.2 Southeast OR SE ✓A 2A correct direction MP (2) L1 2.1.3 Map distance = 7 km × 100 000 C 1C conversion MP = 700 000 cm 1M dividing by scale L2 700 000 ⸫ 56 000 M 1CA map distance = 12,5 cm CA 1R rounding ≈ 13 cm R OR OR 56 000 Map distance = 100 000 = 0,56 m C 1C conversion 7 ⸫ 0,56 M 1M dividing by scale = 12,5 cm CA 1CA map distance ≈ 13 cm R 1R rounding (4) 2.2.1 Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 1SF substitution MP 𝑇𝑖𝑚𝑒 Accept if calculated 1M changing subject of L3 100 km/h = 𝑇𝑖𝑚𝑒 11,2 𝑘𝑚 SF the time as follows: formula ⸫ Time 11,2 𝑘𝑚 = 100 𝑘𝑚/ℎ M 1C converting hours to = 0,112 × 60 minutes = 0,112 hours × 60 C = 6 minutes, 43,2 sec 1CA time = 6,72 minutes CA (full marks) NPR (Accept 7 minutes) (4) M 1M multiplying by 2 MP 2.2.2 Total Distance = (11,2 km × 2) + 156 km M 1M adding 156 km L2 = 178,4 km A 1A total distance (3) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2024) 2.2.3 178,4 CA from 2.2.2 F Number of litres of petrol = 100 × 5,7 M 1M dividing by 100 L4 = 10,1688 litres A and multiplying ⸫ Cost = R24,45 × 10,1688 M with 5,7 = R248,63 MCA 1A number of litres ⸫ Her statement is valid O 1M multiply with R24,45 OR 1MCA petrol cost 1O opinion OR 5,7 Number of litres of petrol = 100 × 178,4 M 1M dividing by 100 and multiplying = 10,1688 litres A with 178,4 ⸫ Cost = R24,45 × 10,1688 M 1A no. of litres = R248,63 MCA 1M multiply with ⸫ Her statement is valid O R24,45 1MCA petrol cost 1O opinion (5) [20] Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P2 7 QUESTION 3 [34 MARKS] Ques. Solution Explanation Level 3.1.1 Volume is the amount of space inside the fish tank. ✓A 2A definition M (2) L1 3.1.2 Volume of a cylindrical prism = π × radius2 × height 1SF substitution M 38,8 ft3 = 3,142 × 2,12 × height SF 1M divide correct values L2 38,8 ft3 = 13,85622 × height 1CA height 38,8 ⸫ Height = 13,85622 M = 2,8 foot CA (3) 3.1.3 ℃ = (℉ - 320) × 9 5 1SF substitution M 5 1A temperature L4 = (72 ℉ - 320) × 9 SF 1O opinion = 22,222… ≈ 22,2 ℃ A ⸫ Hein’s claim is valid O (3) 3.1.4 Volume of stones = 87% – 75% M 1M subtracting M = 12% A percentages L3 12 1A correct % ⸫ 100 × 38,8 M 1M multiply with 38,8 = 4,656 ft3 CA 1CA volume of stones ≈ 4,7 ft3 R 1R rounding OR OR 87 Volume of water and stones in tank = 100 × 38,8 M 1M calculating 87% = 33,756 ft3 75 Volume of water before stones added = 100 × 38,8 M 1M calculating 75% = 29,1 ft3 ⸫ Volume of stones = 33,756 ft3 – 29,1 ft3 M = 4,656 ft3 CA 1M subtracting volumes ≈ 4,7 ft3 R 1CA volume of stones 1R rounding (5) 3.1.5 1 1A correct fraction P Probability = 7 A × 100 % M 1M multiply with 100% L2 = 14,2857… 1CA probability as % ≈ 14,29% CA NPR (Accept 14,3% OR 14,286%) (3) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

8 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2024) 3.2.1 Total SA = [ 2 (l × w) + 2 (l × h) + 2 (w × h)] 1C conversion M C 1SF substitution L2 = [ 2 (240 × 112) + 2 (240 × 70) + 2 (112 × 70)]SF 1S simplification = 53 760 + 33 600 + 15 680 S 1CA total SA = 103 040 mm2 CA (4) 3.2.2 2 500 Number of complete pallets = 500 MA 1MA dividing correct M values L1 = 5 pallets A 1A number of pallets (2) 3.2.3 9 187 1M dividing weight by 5 M Weight of one pallet = 5 M 1C conversion L2 = 1 837,5 kg 1 837,5 1CA weight of one pallet ⸫ = 1 000 C = 1,8375 ton CA OR OR Weight of one pallet = 500 × 3,675 M 1M multiplying 500 with = 1 837,5 kg 3,675 1 837,5 ⸫ = 1 000 C 1C conversion 1CA weight of one pallet = 1,8375 ton CA (3) 3.2.4 Volume of a rectangular prism = Length × Width × Height M SF = 240 mm × 112 mm × 70 mm 1SF substitution L4 = 1 881 600 mm3 ⸫ 1 881 600 × 2 500 M 1M multiply volume with = 4 704 000 000 mm3 2 500 bricks 4 704 000 000 ⸫ 1 000 000 000 C 1C convert volume to m3 = 4,704 m3 CA 1CA volume of all bricks ≈ 4,7 m3 1O opinion ⸫ Hein’s claim is correct O OR OR Volume of a rectangular prism = Length × Width × Height C SF 1C conversion = 0,24 m × 0,112 m × 0,070 m 1SF substitution = 0,0018816 m3 ⸫ 0,0018816 × 2 500 M 1M multiply volume with = 4,704 m3CA 2 500 bricks ≈ 4,7 m3 1CA volume of all bricks 1O opinion ⸫ Hein’s claim is correct. O (5) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P2 9 3.2.5 Total cost = R2,60 × 2 500 M 1M multiply cost with F = R6 500 + R650 (delivery) M number of bricks L2 = R7 150 CA 1M adding delivery costs ≈ R7 200 R 1CA total cost 1R rounding (4) [34] Copyright reserved Please turn over Hosted by www.ecexams.co.za.

10 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2024) QUESTION 4 [26 MARKS] Ques. Solution Explanation Level 4.1.1 Graphic scale A OR Bar scale A OR Linear scale A 1A bar scale MP Ratio scale A OR Numeric scale A 1A number scale L1 (2) 4.1.2 Scale = 6,7 cm A 1A measuring scale MP 6,7 cm = 12 miles in cm L3 6,7 𝑐𝑚 12 𝑚𝑖𝑙𝑒𝑠 = 6,7 M 1M divide by 6,7 6,7 1CA scale 1 cm : 1,791044776 miles CA 1M multiply 10,5 ⸫ Actual distance = 10,5 cm × 1,7910… M cm with the scale = 18,80597015 miles 1CA actual distance ≈ 18,81 miles CA NPR (Accept 18,806 miles OR 19 miles) (5) 4.1.3 Time = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 MP 𝑆𝑝𝑒𝑒𝑑 If calculated L4 = 121,4 𝑚𝑖𝑙𝑒𝑠 as follows, 1SF substitution SF 85 𝑚𝑖𝑙𝑒𝑠/ℎ DO NOT 1C converting = 1,4282… hours penalize: ⸫ 0,4282… × 60 C hours to minutes = 25,694… minutes 1CA travel time Travel time ⸫ Time = 1 hour 25 minutes 42 seconds CA = 1h26 min Arrival time Arrival time = 13:30:00 (departure) = 13:30 01:25:42 (travelling) M 1M adding time 01:26 1CA arrival time 00:25:00 (stop at petrol station) 00:25 = 15:20:42 CA 1O opinion = 15:21 (6) ⸫ His claim is invalid. O Probability = 8 A 4.1.4 2 1A numerator P A 1 1A denominator L2 (Accept simplified form = 4) A (2) A 4.2.1 Radius 9,5 = 2 = 4,75 cm M 1M finding radius M 1SF substitution L2 Area of a circle = π × radius2 1CA area of white = 3,142 × 4,752 SF cloth = 70,891375 ≈ 70,89 cm2 CA (3) 4.2.2 Length of wooden frame = 22 inches × 2,54 C 1C converting M = 55,88 cm A length L3 Area of rectangle = length × width 1A answer 2 682 = 55,88 × width SF 2 682 1SF substitution ⸫ Width = M 55,88 1M dividing area = 47,9957 … by length ≈ 48 cm CA (Aanvaar 47,996 cm) 1CA width (5) Copyright reserved Please turn over Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICAL LITERACY P2 11 4.2.3 48 CA from 4.2.2 M Width of one rectangular bar = 6 MCA 1MCA dividing width by 6 L4 = 8 cm CA 1CA width of rectangular bar ⸫ Lwandile’s claim is valid.O 1O opinion OR OR 1MCA dividing width by 6 47,996 1CA width of rectangular bar Width of one rectangular bar = 6 MCA 1O opinion = 7,9993… cm (3) ≈ 8 cm CA ⸫ Lwandile’s claim is valid. O [26] TOTAL: 100 Copyright reserved Please turn over Hosted by www.ecexams.co.za.