Mathematical Literacy P1 May June 2017 Memo Eng hlayiso.com

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATIONS MATHEMATICAL LITERACY P1 2017 MARKING GUIDELINES MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RD Reading from a table/graph/diagram SF Correct substitution in a formula O Opinion/Example/Definition/Explanation R Rounding off NPR No penalty rounding or omitting units AO Answer only, full marks These marking guidelines consist of 12 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 2 DBE/2017 SCE – Marking Guidelines Question 1 [30 Marks] Ques Solution Explanation T/L MA F 1.1.1 R8,70 × 40 = R348 A 1MA multiplying with 40 L1 1A box price AO (2) F 1.1.2 A profit is made when the selling price is more than the L1 cost price. O 2O explanation OR OR O A profit is the amount added to the cost price 2O explanation OR OR Making more money than the cost price. O 2O explanation OR OR O Positive difference between income and expenditure. 2O explanation OR OR O Income is more than cost or expenses. 2O explanation OR OR O Gained/extra money from the sale of a product 2O explanation (2) (Except a correct example as an explanation) F 1.1.3 Amount = 40% × R435,04 MA 1MA calculate 40% of L1 = R174,016 R435,04 ≈ R174,02 A 1A VAT amount AO NPR (2) F 1.1.4 Total cost RT M 1RT all correct values L1 (a) = R10,04 + R8,70 + R20,66 + R6,73 + R29,99 1M adding at least 3 correct amounts = R76,12 CA 1CA total AO (3) F 1.1.4 R22 770 1MA dividing correctly L1 Selling price = MA (b) 230 1A selling price = R99,00 A AO (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 3 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L F 1.2.1 South African Revenue Services A 2A full name L1 (2) F 1.2.2 R61 296 RT 2RT correct amount L1 (2) F 1.2.3 R542 096,76 1MA dividing correctly L1 MA 12 = R45 174,73 A 1A monthly salary AO (2) F 1.2.4 Tax bracket 4 RT 2RT correct tax bracket L1 OR 406 401 – 550 100 RT OR RT 96 264 + 36% of taxable income above 406 400 (2) A Maps 1.3.1 1 unit on the map is 200 units in reality 2A explanation L1 (2) OR The real one is 200 times bigger A OR The drawing is 200 times smaller A C Meas 1.3.2 Perimeter = 4 cm + 2 cm + 4,25 cm + 2,55 cm M 1C converting L1 = 12,8 cm CA 1M adding 4 sides 1CA perimeter AO (3) A A D 1.4.1 January 2015 1A correct month L1 1A correct year OR (2) 01/2015 A A D 1.4.2 The price of cake went down/decreased/ dropped/ 2A description L1 declined / less (2) D 1.4.3 100% A 2A correct index L1 No penalty if % omitted Penalise if the index is given as R100 (2) [30] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 4 DBE/2017 SCE – Marking Guidelines QUESTION 2 [35 MARKS] Topic Finance Ques Solution Explanation T/L RT L1 2.1.1 R8 060,27 + R600 = R8 660,27 CA 1RT reading both correct amounts 1CA balance AO (2) M L1 2.1.2 R4 050,98 – R4 034,77 = R16,21 CA 1M subtracting 1CA interest AO (2) A L1 2.1.3 Accept any account number from 143260000 to 2A possible number 143269999 OR (2) Writing only the FOUR missing digits L1 2.1.4 Mdiso Khaile A 2A correct person (2) L1 2.1.5 0 OR none A 2A correct number (2) P 2.1.6 0 OR 0% OR impossible A 2A correct probability L2 (2) M A L1 2.1.7 R1,50 × 4 + R0,40 × 6 + R1,20 + R5,00 × 2 = R19,60 1M adding values (a) 1A correct values (2) L2 2.1.7 R19,60 M 1M dividing by 114% (b) Amount without VAT = = R17,19 114% 1M subtracting VAT amount = R19,60 – R17,19 M = R2,41 A 1A VAT amount OR OR M 14% 1M dividing by 114% VAT amount = × R19,60 1M working with ratio 114% M = R2,41 A 1A VAT amount AO (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 5 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L L1 2.2.1 Service charges A 2A correct item (2) M L1 2.2.2 R4 253 219 thousand – R4 165 225 thousand 1M subtracting correct = R87 994 thousand A values from table 1A difference in thousands (2) RT A L1 2.2.3 R2 878 830 thousand = R2 878 830 000 1RT correct expected ≈ R2,9 billion CA income 1A expanding the amount 1CA income in billions AO (3) M MA L2 2.2.4 B = 4 253 219 – (794 866 + 2 694 542 + 34 044 + 211 526) 1M subtracting 1MA adding correct = 518 241 CA values 1CA value AO (3) L3 2.2.5 Total income MA 1MA adding correct = 716 603 + 2 227 636 + 51 027 + 519 604 + 312 290 values = 3 827 160 A 1A income Total expenditure = 886 355 +34 657 + 481 980 + 71 180 + 1 780 120 + 238 + 875 072 1A expenditure A = 4 129 602 A = R3 827 160 – R4 129 602 = – R302 442 CA 1CA amount or (R302 442) It is a DEFICIT CA 1CA deficit (5) L2 2.2.6 Percentage increase Difference in renumeration = × 100% Original budget renumeration 1RT reading correct RT values 43 033 000 − 42 350 000 SF 1SF substitution = × 100% 42 350 000 1CA % increase ≈ 1,613 % CA AO (3) [35] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 6 DBE/2017 SCE – Marking Guidelines QUESTION 3 [28 MARKS] Topic Measurement Ques Solution Explanation T/L L1 3.1.1 B A 2A correct letter OR OF 325 × 325 × 325 A 2A dimensions (2) RT L2 SF 3.1.2 Area = 1 200 mm × 325 mm 1RT correct dimensions = 120 cm × 32,5 cm C 1SF substitution 1C converting = 3 900 cm2 CA 1CA area OR OR RT 1RT correct dimensions Area = 1 200 mm × 325 mm SF 1SF substitution A = 390 000 mm2 1A area = 3 900 cm2 C 1C converting AO (4) L1 3.1.3 24 1MA dividing number Number of boxes on ground = = 12 MA 2 of boxes by 2 1M multiplying area of Total area needed = 12 × 1 056,25 cm2 M 1 box by number of = 12 675 cm2 CA boxes in one layer 1CA area OR OR MA Total area = 1 056,25 cm2 × 24 = 25 350 cm2 1MA multiplying area by 24 25 350 cm 2 Total needed = M 1M dividing total area 2 by 2 1CA area = 12 675 cm2 CA AO (3) L1 3.1.4 600 : 325 RT A 1RT correct two values 1A ratio correct order = 24 : 13 S 1S simplification AO (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 7 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L L3 3.1.5 Volume = 1 500 mm × 475 mm × 462,5 mm SF 1SF substitution (a) = 1,5 m × 0,475 m × 0,4625 m C 1C conversion = 0,32953125 m3 CA 1CA volume CA M 1CA subtracting Inside volume = 0,32953125 m3 – (0,32953125 m3 × 9,36%) 1M multiplying by = 0,298687125 m3 9,36% ≈ 0,299 m3 OR OR L3 3.1.5 Volume = 1 500mm × 475mm × 462,5mm SF 1SF substitution (a) = 1,5 m × 0,475 m × 0,4625 m C 1C conversion = 0,32953125 m3 CA 1CA volume 100% – 9,36% = 90,64% A 1A subtraction M 1M multiply with Inside volume = 0,32953125 m3 × 90,64% ≈ 0,299 m3 90,64% (5) L1 3.1.5 6m 3 MA 1MA dividing (b) Number of boxes = 3 0,299 m A 1A simplification ≈ 20,066 1R rounding down ≈ 20 R AO (3) L2 3.1.5 Volume needed = 148 × 0,299 1A total volume (c) = 44,252 A 44,252 m 3 M 1M dividing by 6 m3 Truck loads = 6 m3 = 7,375333... ≈8 R 1R rounding up OR OR 148 M Truck loads = 1M working with ratio 20 from Q3.1.5(b) A = 7,4 1A total volume ≈ 8 R 1R rounding up AO (3) A 1A radius L1 3.2.1 5 14 inches OR 5,25 inches A 1A inches (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 8 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L L2 3.2.2 Volume (in cm 3 ) h= 1 2 4 × π × (diameter in cm) 20 000 cm3 1SF correct substitution h= 1 SF × 3,142 × (10 12 × 2,54 cm)2 C (20 000 and 10 12 ) 4 20 000 cm 3 1C convert inch to cm = 558,717431cm 2 ≈ 35,8 cm CA 1CA height NPR AO (3) [28] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 9 DBE/2017 SCE – Marking Guidelines QUESTION 4 [23 MARKS] Topic Maps, Plans and other Ques Solution Explanation T/L L2 4.1.1 North West or NW A 2A direction (2) O L1 4.1.2 It indicates the BORDER between South Africa and 2O explanation Botswana Accept: border /fence/ boundary (2) A L1 4.1.3 Travel from Johannesburg to Zeerust via Koster, then 1A Koster or N14 A A 1A Zeerust or N4 and then from Zeerust to Abjaterskop Gate 1A Abjaterskop Gate or R49 OR A A A Take the N14, N4, then turn on to the R49 (3) M RT L1 4.1.4 Distance = 221,2 km – (62,4 km + 88,1 km) 1M subtracting 1RT correct distances = 70,7 km CA 1CA distance AO (3) A M CA L2 4.1.5 Via Koster: 70 km + 71,9 km + 35,2 km = 177,1 km 1A correct distances 1M adding 1CA shortest route distance CA from 4.1.3 (3) L1 4.2.1 Left-hand side A 2A correct side (2) MA RT 1RT 31 cottages L2 4.2.2 3 × 31 = 93 CA 1MA multiply 3 1CA number of guests AO (3) L2 4.2.3 Walk towards reception and pass between reception 1A passing reception and cottage number 17. A 1A passing ablusion Continue pass the ablusion block A 1A crossing road Cross the road to the swimming pool A OR OR Turn right into the road passing the petrol station, 1A passing petrol station, reception reception and shop A and shop Turn left into the road A 1A turn left into road Continue straight, the swimming pool is on your right- 1A swimming pool on your right hand side A hand side (3) A 1A numerator P 4.2.4 2 1A denominator L2 P (not a night drive) = or 66,67% or 0,67 3 A (2) [23] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 10 DBE/2017 SCE – Marking Guidelines QUESTION 5 [34 MARKS] Topic Data Ques Solution Explanation T/L L1 5.1.1 Free State A 2A correct province (2) L1 5.1.2 1RT all correct values RT M 66 007 + 24 475 + 74 823 + 96 057 + 57 108 +34 936 + 1M adding (min 8 prov.) 8 972 + 26 194 + 36 451 = 425 023 CA 1CA total teachers AO (3) L2 5.1.3 1RT correct values RT 1MA % calculation 6 156 × 100% MA 25 720 1CA % schools CA AO ≈ 23,93% NPR (3) L2 5.1.4 Total number of learners 1RT correct values LSR = 1SF substitution Total number of schools 2 129 526 RT SF = 1CA ratio 2 649 AO ≈ 803,898 ≈ 804 CA NPR (3) L1 5.1.5 30,1 A 2A mode (a) (2) A L1 5.1.5 31,5 30,1 30,1 30,0 29,8 29,4 28,9 28,5 27,2 A 1A all the values (b) 1A correct order (2) L2 5.1.5 2A median (c) 29,8 A CA from Q5.1.5 (b) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 11 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L 5.1.6 Teacher - School Ratio in the public schools and independent schools, by province 30 28,2 A 25 A 22 A 20 Teacher-School Ratio 18,5 A 17,5 A 16,3 15,6 15,7 A 15 14 11,5 10 5 0 Gauteng Limpopo KwaZulu-Natal Eastern Cape North West Northern Cape Free State Mpumalanga Western Cape Provinces L2 6 × 1A for each correct bar (6) Note: If the candidate redrew the grid: • Correct scale used – maximum 6 marks • Unclear scale used – maximum 3 marks Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P1 12 DBE/2017 SCE – Marking Guidelines Ques Solution Explanation T/L A L1 5.2.1 0,1 = 10% CA 1A identifying the correct value 1CA writing it as a percentage (2) L1 5.2.2 (a) R N OR N R A 2A outcome at (a) (b) D L OR L D A 2A outcome at (b) (4) P RT 5.2.3 5 1 1RT correct probability L2 0,05 = = CA 1CA simplified fraction 100 20 AO (2) RT L1 5.2.4 1 562 × 0,8 = 1 249,6 CA 1RT correct values ≈ 1 249 or 1250 R 1CA simplification 1R rounding AO (3) [34] Copyright reserved