Mathematical Literacy P2 Nov 2012 Memo Eng hlayiso.com

Downloaded from hlayiso.com NSC – Final Memorandum + NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2012 FINAL MEMORANDUM MARKS: 150 Symbol Explanation M Method M/A Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG Reading from a table/Reading from a graph SF Correct substitution in a formula O Opinion/Example P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off J Justification PLEASE NOTE: 1. If a candidate deletes a solution to a question without providing another solution, then the deleted solution must be marked. 2. If a candidate provides more than one solution to a question, then only the first solution must be marked and a line drawn through any other solutions to the question. This memorandum consists of 19 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/November 2012 Final Memorandum QUESTION 1 [26 MARKS] Ques Solution Explanation AS 1.1.1 South-westerly  A 2A correct direction 12.3.4 1A Southerly L3 (accept abreviations for compass directions) 1A Westerly (2) 2A correct national 1.1.2 N5 OR N17  A 12.3.4 road L3 N17 accepted due to unclear provincial boundaries (2) 12.3.4 1.1.3 One possible route: A 1A N1 L2 From Bloemfontein turn onto the N1 and travel south until Beaufort West. Then turn onto the N12 until George. A 1A N12 and Beaufort West A second possible route:  A OR From Bloemfontein turn onto the N1 and travel south until the 1A N1 intersection with the N9. Then follow the N9 until George. A 1A N9 A third possible route: OR A From Bloemfontein turn onto the N1 and travel south until the 1A N1 intersection with N10. Then follow the N10 in a south easterly direction until the N2. Then follow the N2 in a westerly direction until George. A 1A N10, N2 A fourth possible route: OR A From Bloemfontein turn onto the N1 and later turn onto the N6 to 1A (N1) N6 and East East London. London, Then follow the N2 in a westerly direction until George. A 1A N2 A fifth possible route: A OR From Bloemfontein turn north onto the N1, turn right unto N5, 1A N1; N5 and take a right unto N3 pass Pietermaritzburg to Durban. Then at Durban turn south unto the N2, pass East London, Port Elizabeth and continue until George. A 1A N3 Durban; N2 NOTE: Follow the learners route. But leaners cannot go back to Kimberley (No N8 route). (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 12.1.3 1.2.1 Total amount for accommodation = R1 050 × 6 A 1A rate × 6 L2 = R6 300 CA 1CA simplification OR (due to language interpretation) Total amount for accommodation = R1 050 × 7 A Correct answer only– full = R7 350 CA marks (2) Note: Equation must 12.2.3 1.2.2 have a variable L3 (a) Total cost (in rand) = (60 × 4 × number of breakfasts) M 1M adding + (90 × 4 × number of lunches) M 1M multiplying cost + (120 × 4 × number of suppers) M 1M multiplying by 4 or number of people OR OR M M 1M adding Total cost (in rand) = (60 × x + 90 × y + 120 × z) × 4 1M costs in terms of Where x = number of breakfasts meals y = number of lunches M 1M variables explained and z = number of suppers OR OR Total cost (in rand) = (number of days × n × 60) + M 1M adding (number of days × n × 90) + 1M costs in terms of M (number of days × n × 120) meals Where n = number of people M 1M variable explained OR OR Total cost (in rand) M = (Sat + Sun + Mon + Tues + Wed + Thurs + Fri) cost 1M adding = 120n + 270n + 180n + 210n + 270n + 150 n + 60n) 1M costs in terms of = 1 260 n M days Where n = number of people M 1M variable explained 270 × number of people/meals - (1 mark only) (3) REFER TO CANDIDATE’S 12.2.3 1.2.2 FORMULA L3 (b) Correct answer only– full Total cost (in rand) marks = (60 × 4 ×S S× 4 × 5) 5) + (90 × 4 × 4) + (120 1S correct substitution of number of people = 1 200 + 1 440 + 2 400 CA 1S correct substitution CA of number of meals = 5 040 1CA simplification OR 1CA total Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/November 2012 Final Memorandum Ques Solution Explanation AS OR Total cost (in rand) 1S correct subst. no. = (60 × x + 90 × y + 120 × z) × 4 S S of people = (60 × 5 + 90 × 4 + 120 × 5) × 4 1S correct subst. no. of meals = 1 260 × 4 CA 1CA simplification = 5 040 CA 1CA total OR (using equation from 1.2.2 (a) working with daily cost) 2S substitution of no. Total cost (in rand) = 1 260 × 4 S S of people = 5 040 CA CA 2CA total OR (calculating total daily costs) Cost of meals: Saturday = R120 × 4 = R480 2S correct subst. Sunday = (R60 + R90 + R120) × 4 = R1 080 daily cost Monday = (R60 + R120) × 4 = R720 S Tuesday = (R90 + R120) × 4 = R840 Wednesday = (R60 + R90 + R120) × 4 = R1 080 Thursday = (R60 + R90) × 4 = R600 S Friday = R60 × 4 = R240 Total cost (in rand) = 480 +1 080 +720 +840 + 1 080 + 600 + 240 CA 1CA simplification = 5 040 CA 1CA total OR (calculating total cost of types of meals) Total cost of breakfast = R60 × 5 × 4 = R1 200 S 2S correct subst. meal cost Total cost of lunches = R90 × 4 × 4 = R1 440 S Total cost of suppers = R120 × 5 × 4 = R2 400 Total cost (in rand) = 1 200 + 1 440 + 2 400 CA 1CA simplification = 5 040 CA 1CA total (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 12.1.3 1.2.3 Cost for nature walk = (R120 × 2) +(R100 ×2) M/A 1M/A expression L4 = R440 CA for cost 1CA simplification Cost for game park = R200 × 4 = R800 A 1A cost for game park Cost for boat cruise = (R200 × 2) + (R150 × 2) M/A 1M/A expression for = R700 CA cost 1CA simplification Total entertainment cost = R440 + R800 + R700 + R2 000 = R3 940 CA 1CA total cost Six day option: Total cost for the trip (accom. + meals + long dist. + local + ent) M/A 1M/A adding all =R6 300 + R5 040 + R1 602,86 + R513,60 + R3 940 = R17 396,46 CA costs 1CA total cost OR Seven day option: Total cost for the trip (accom. + meals + long dist. + local + ent) M/A 1M/A adding all =R7 350 + R5 040 + R1 602,86 + R513,60 + R3 940 costs = R18 446,46 CA 1CA total cost ∴ Mr Nel's estimate was CORRECT J 1J verification (9) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/November 2012 Final Memorandum QUESTION 2 [34 MARKS] Ques Solution Explanation AS 12.4.3 2.1.1(a) A – 15 = 37 M A = 37 + 15 M 1M concept of range L3 OR A = 52 A = 52 A 1A simplification Correct answer only– full marks (2) Refer to value of A 2.1.1(b) The mean for 16 customers is 34 minutes in 2.1.1(a) 12.4.3 L3 ∴ total waiting time = 16 × 34 = 544 M 1M total waiting time Total of known waiting times = 30 + 15 + 45 + 36 + 52 + 40 + 34 + 42 + 26 + 32 + 38 + 35 + 41 + 28 1M total of known = 494 M times Difference is 544 – 494 = 50 S 1S difference of the ∴ 2 customers have a total waiting time of 50 minutes totals 50 ∴B= = 25 CA 1CA value of B 2 OR OR Mean M 1M adding all the = 30 + 15 + 45 + 36 + 52 + 40 + 34 + B + B + 42 + 26 + 32 + 38 + 35 + 41 + 28 values 16 M 1M dividing by 16 = 34 494 + 2B = 34 16 2B = (34 × 16) – 494 S OR (34 × 16) − 494 S 1S simplification = 50 B= 2 ∴ B = 25 CA = 25 CA 1CA value of B Correct answer only - full marks (4) 12.4.3 2.1.1 Waiting times are: M/A (Using A and B values L3 (c) 15; 25; 25; 26; 28; 30; 32; 34; 35; 36; 38; 40; 41; 42; 45; 52 calculated above) 1M/A arranging 16 terms 34 + 35 M in ascending order Median = 1M median concept (even 2 = 34,5 CA number of terms) 1CA simplification (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 2CA correct number 2.1.2 4 CA Note if B is greater than 27 answer can be 2 (2) 2.1.3 The mean, median and range for 7 February are less than 2O comparing the measures 12.4.4 those for 14 February. O Accept a comparison table L4 This means that his customers had to wait for a shorter time of correct values on 7 February than on 14 February. O Any two of the reasons below: 2J conclusion • It could be that more people came to eat at his eating place on 14 February, because of Valentine's Day. J • He had less staff on the 14th, J • He had the same number of staff but did not anticipate the increased number of customers. J • His equipment was faulty on the 14th – people had to wait longer to be served J • The electicity was off for a while J OR The mean, median and range for 14 February are more than those for 7 February. O This means that his customers had to wait for a longer time on 14 February than on 7 February. O Any two of the reasons below: • It could be that less people came to eat at his eating place on 7 February, because of Valentine's Day. J • He had more staff on the 7th, J • He had the same number of staff but did not anticipate the difference in number of customers.J • His equipment was working well on the 7th – people did not wait long to be served J • No electicity problems on the 7th J OR Any other valid, well thought out reason will be accepted (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 12.1.1 2.2.1 Percentage ordering chicken = 15% A 1A percentage (2) ordering chicken 12.4.4 If 20% of the total = 40 (2) 40 1M finding 1% L2 ∴ 1% of the total = = 2 M (2) 20 ∴ 15% of the total = 15 × 2 A 1A multiplying by 15 L3 = 30 CA 1CA simplification (2) OR OR M 20% : 40 = 15% : x A 1M using proportion 15% 1A percentage x= × 40 S ordering chicken 20% 1S expression for x 1CA simplification = 30 CA OR OR 1M finding total no. of 20% of total = 40 customers 40 Total = M 1A total number of 20% customers = 200 A 1A percentage A ordering chicken ∴ 15% of 200 = 30 CA 1CA simplification Correct answer only– full marks (4) M A 3 2.2.2 P(not lamb) = 1 – 25% = 75% OR 0,75 OR 1M subtracting from100 % 4 1A simplification OR Percentage not ordering lamb = 10 + 15 + 20 + 30 = 75 M 1M adding percentages A 3 P(not lamb) = 75% OR 0,75 OR 1A simplification 4 OR Number of people not ordering lamb M = 20 + 30 + 40 + 60 = 150 1M adding actual numbers 150 3 P(not lamb) = = OR 0,75 OR 75% A 1A simplification 200 4 Correct answer only - Full marks (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 2.3.1 Two of the following possible reasons: • To protect the base of the drum from burning. • To bring the fire closer to the grid. • To spread the coals evenly. (Perfect the braaing) • To use less coal. • To stabilise the drum. • To retain the heat of the burning coals. • The sand can be used to put out the fire. 2O reason Accept any two valid reasons. O O 2O reason (4) 12.3.1 2.3.2 Volume of the braai drum = 108 ℓ L4 = 108 × 1 000 000 mm 3 = 108 000 000 mm 3 C 1C volume in mm 3 572 mm Radius of the braai drum = = 286 mmA 1A value of radius 2 M 1M using 12 cylinder Volume of the braai drum = 12 × π × (radius) 2 × (height) SF 108 000 000 mm 3 = 12 × 3,14 × (286 mm) 2 × (height) 1SF substitution into formula 2 × 108 000 000 mm 3 M Height = 1M Finding expression for 3,14 × (286 mm) 2 height = 840,99 mm CA (840,56... mm using π ) 1CA for height only ≈ 841 mm But length of grid = 1% more than height of drum 1% of 840,99 mm = 8,4099 M 1M calculation percentage M CA 1M increasing by 1% ∴ Length of grid = 840,99 mm + 8,4099 = 849,41 mm 1CA length of grid OR OR M M 1M increasing by 1% ∴ Length of grid = 101% of 840,99 mm = 849,40 mmCA 1M calculation percentage 1CA length of grid No penalty if answer is rounded to 850 mm (9) [34] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/November 2012 Final Memorandum QUESTION 3 [26 MARKS] Ques Solution Explanation AS 12.2.1 3 500 A 3.1.1 Number of R2,00 tickets per seller = 1A using 3 500 L3 number of sellers A 1A dividing by number of sellers OR OR 7 000 A 1A using 7 000 ÷ 2 Number of R2,00 ticket per seller = 2 × number of sellers A 1A dividing by number of sellers OR 7 000 3 500 Number of R2,00 tickets per seller = = 2n n where n = number of sellers (2) 12.1.1 3.1.2 Indirect/Inverse proportion A 1A correct type of L2 (a) proportion two answers zero marks (1) 12.2.1 3.1.2 1A finding the number of L2 3 500A tickets (b) P= OR P : 70 = 50 : 250 A 250 A 1M dividing by 250 CA CA 70 = 14 = 50 × = 14 1CA correct value of P 250 M 3 500 Q= = 28 CA 1CA correct value of Q 125 Correct answer only - Full marks (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/November 2012 Final Memorandum 12.2.2 3.1.2 L2 (c) R2 Tickets A A CA A 1A correct plotting of point (20;175) 1A correct plotting of point (140;25) 1A one other point plotted correctly 1CA joining the plotted points by a "smooth" curve (section from 20 ticket sellers to 100 ticket sellers) (4) 12.1.2 3.2.1 Fewer tickets have to be sold.  J 2J reason for decision (1) OR 12.2.3  J To reduce the number of sellers. (1) OR L4 To raise the money faster (in a shorter time)  J OR To raise more money/to buy more computers  J (2) 12.1.2 3.2.2 Fewer people can afford (too expensive) to buy the R5,00 2J disadvantage (1) tickets. 12.2.3 OR (1) Some of the sellers might not be able to sell all their tickets L4 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 12.2.1 R 7 000,00 M 1M dividing by R5 (3) 3.2.3 Number of tickets to be sold = R5 12.2.2 = 1 400 A 1A number of tickets to be (5) sold L3 (4) 1 400 L4 (4) Number of tickets per person = CA 1CA formula number of sellers OR Showing values in a table/co-ordinates - 3 marks The possible points learners can use: (other point values can be used) 10 20 35 50 100 140 140 70 40 28 14 10 R2 Tickets A A CA A A R5 Tickets 4CA any 4 points plotted correctly 1CA joining the plotted points by a smooth curve (8) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/November 2012 Final Memorandum Ques Solution Explanation AS 12.1.1 3.2.4 At R2 per ticket 50 tickets must be sold RG 1RG reading from graph (1) At R5 per ticket 20 tickets must be sold RG 1RG reading from graph 12.2.3 Difference = 50 – 20 (2) = 30 tickets CA 1 CA difference in number of L3 tickets OR OR 3 500 Number of R2,00 tickets per person = 70 = 50 M 1M calculating the number of R2,00 tickets 1 400 M Number of R5,00 tickets per person = 1M calculating the number of 70 R5,00 tickets = 20 Difference = 50 – 20 tickets = 30 tickets CA 1CA difference in number of tickets Answer only – Full marks Accept values from 29 to 32. (refer to candidate's graph) (3) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/November 2012 Final Memorandum QUESTION 4 [27 MARKS] Ques Solution Explanation AS 12.4.4 4.1.1 Avro A 1A correct aircraft J L4 It is the only one that can take MORE than 37 passengers 2J justification (himself plus 37 others) (3) 12.3.2 4.1.2 Scale is 9,9 cm to 19,25 m M C 1M scale concept (1) or 9,9 cm to 1 925 cm OR 0,099 m : 19,25 m 1C converting to the same unit 12.3.3 1CA dividing to bring to a unit (3) CA 1925 OR 19,25 ratio L3 Scale = 1 : CA 1: 9,9 0,099 1CA rounding off = 1 : 194,44 = 1 : 190 CA Reversed ratio maximum 2 marks No conversion maximum 2 marks Correct answer only- full marks (4) 12.3.2 4.1.3 Maximum Operating Altitude = 25 000 feet RT 1RT reading from the L3 25 000 M table = nautical miles 1M dividing by 6076 ft 6 076 = 4,1145… nautical miles ≈ 4 nautical miles CA 1CA nearest nautical mile (3) 12.2.1 4.1.4 Distance = average cruising speed × time L3 (2) 510 km = average cruising speed × 39 minutes SF 1SF substitution L4 (2) 510 km Average cruising speed = 39 minutes 510 km = 1C converting to hours 0,65 h C = 784,62 km/h CA 1CA average speed Ms Bobe was travelling in the SUKHOIJ 1J identification of Aircraft OR C OR )km = 325 kmSF 39 Distance (Jetstream) = (500 × 1SF substitution 60 1C converting to hours 39 Distance (Sukhoi) = (800 × )km = 520 kmCA 1CA distance travel 60 39 Distance (Avro) = (780 × )km = 507 km J 60 1J identification of Ms Bobe was travelling in the SUKHOI Aircraft Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 15 DBE/November 2012 Final Memorandum Ques Solution AS Ques OR 4.1.4 Comparing time cont distance Time = speed 1SF substitution 510 SF CA C Time (Jetstream) = h = 1,02 hours = 61,2 minutes 1CA time taken 500 510 1C converting to minutes Time (Sukhoi) = h = 0,6375 hours = 38,25 minutes 800 510 Time (Avro) = h = 0,6538... hours = 39,23 minutes 780 Ms Bobe was travelling in the SUKHOI J 1J identification of Aircraft (4) fuel capacity (in kg) 12.3.2 4.1.5 Fuel capacity (in litres) = L2 (2) 820 g L3 (1) 9 362 kg SF = 1SF substitution 820 g 9 362 000 g C 1C converting to grams = 820 g = 11 417,07317 ≈ 11 417 CA 1CA nearest litre OR fuel capacity (in kg) Fuel capacity (in litres) = 820 g 9 362 kg SF = 1SF substitution 820 g 9 362 kg C 1C converting to kilograms = 0,820 kg = 11 417,07317 ≈ 11 417 CA 1CA nearest litre No conversion - maximum 2 marks (3) 12.4.4 4.2.1 Johannesburg to Polokwane: SA 8809 A 2A correct flight number L3 Polokwane to Johannesburg: SA 8816 A 1A correct flight number (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 16 DBE/November 2012 Final Memorandum Ques Solution AS 12.4.2 4.2.2(a) L3 A CA A A 1A drawing the horizontal line at 4 1A plotting (Saturday; 2 ) 1A plotting (Sunday; 3) 1CA joining the plotted points (4) 12.4.4 4.2.2 (b) Saturday A 1A correct day L4 Not many people travel on Saturday, as most business meetings are scheduled during the week. O 2O own opinion based on candidates graph OR If people go away for the weekend on holiday, they travel there on a Friday and travel back on Sunday. O OR Possible religious reason O OR Any other valid reason O (3) [27] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 17 DBE/November 2012 Final Memorandum QUESTION 5 [37 MARKS] Ques Solution Explanation AS 12.2.2 5.1.1 For 30 items: L3 Cost = R5 000 RG 1RG cost Income = R3 600 RG 1RG income Loss = R5 000 – R3 600 = R1 400 ∴ 30 items A 1A number of items Correct answer only - full marks (3) 12.2.2 5.1.2 Cost of 40 items = R5 500 RG OR 40 × R50,00 + R3 500 1RG/A cost Or L4 Cost = income Income from 40 items = R137,50 × 40 M = R5 500 A 1M finding total income At 40 items, Cost = Income 1Asimplification ∴ Mr Stanford's statement is CORRECT. CA 1CA verification (4) 12.1.1 5.2.1 N is the total sales. 1M concept L2 (4) 16 % of N = 800 M 1M finding an L3 (3) 100 M expression for N N = 800 × 16 = 5 000 A 1A total sales OR OR 16% of the sales = 800 800 1M finding unit value 1% of the sales = M 16 800 ∴ 100 % of the sales = × 100 M 1M finding 100% 16 ∴ N = 5 000 A 1A total sales OR OR 21 % of total sales = 1 050 M 1M concept 100 Total sales = 1 050 × M 1M finding an 21 expression for N ∴ N = 5 000 A 1A total sales K= 750 × 100 M 1M concept 5 000 = 15 CA 1CA simplification Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 18 DBE/November 2012 Final Memorandum Ques Solution Explanation AS L = 17% of total sales 17 1M finding 17 % L= × 5 000 M 100 1CA simplification = 850 CA OR OR 16% of the total is 800 800 1% of the total is 16 ∴ 17% of the total is 800 × 17 M 1M finding unit value 16 ∴ L = 850 CA 1CA simplification Please note Correct answer only full If L is found first: marks M CA The values need not be a N = 350 + 750 + 1 050 + 850 + 800 + 900 + 200 + 100 calculated in the same = 5 000 CA order as on the memo (7) 12.1.1 5.2.2 Vivesh's % (value of M) L4 M 900 000 M 1M expression for % = × 100% OR 900 ×100% 5 000 000 5 000 = 18% CA = 18% CA 1CA simplification OR 100% – (7 + 15+ 21 + 17 + 4 + 2 + 16)% M = 18% CA M 1M calculating Vivesh's bonus = 18% of R300 000 percentage 1CA simplification = R54 000 CA ∴ The objection is NOT VALID. CA 1CA conclusion (5) 12.1.1 5.2.3 R50 000 A 2A correct basic bonus L3 (a) (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 19 DBE/November 2012 NSC – Memorandum Ques Solution Explanation AS 12.1.1 5.2.3 Total bonus amount =6,5 % × R5 500 000 L4 (b) = R357 500 A 1A total bonus Sales up to and including 10% : 3 persons Sales of more than 10% up to and including 20% : 4 persons Sales of more than 20% : 1 person Bonus amount remaining M M = R357 500 – (3 × R10 000 + 4 × R50 000 + R100 000) 1 M finding the total basic = R357 500 – R330 000 bonus = R27 500 CA 1M finding the difference 1CA simplification R 27 500 1M dividing by 8 Amount each will receive = M 8 1CA simplification = R3 437,50 CA Mabel's total bonus = R100 000 + R3 437,50 1CA Mabel's bonus (must include R100 000) = R103 437,50 CA O ∴ Mabel's bonus is NOT MORE THAN than R104 000. 1O verification (8) 12.4.6 5.3.1 Vivesh's sales in 2012 was more than double his sales in 2011. L4 Vivesh was the top salesperson in 2012. O O 2O interpretation OR There is an increase in percentage sales from 12% to 28% OR Any other numerical comparison (2) 12.4.6 5.3.2 He read Mabel's and Henry's combined sales of 2011 and 2012 2O errors L4 as the sales for 2012. O J Henry's sales for 2012 were only 25%, Mabel's sales were 21% 1J Henry & Mabel and the person with the highest sales was Vivesh with 28% J 1J mention Vivesh as highest (4) 12.4.6 5.3.3 Any TWO of the following: L2 • Different type of Bar graphs O 1O bar graphs • Line graphs O • Pie charts 1O line graphs OR 1O pie charts (2) [37] TOTAL: 150 Copyright reserved