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GRADE 12 MATHEMATICAL LITERACY PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2016

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MATHEMATICAL LITERACY P1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
SEPTEMBER 2016

Symbol Explanation
M Method
A Accuracy
CA Consistent accuracy
RT/RG/RM Reading from a table/Reading from a graph/Read from map
RP Reading from the plan
SF Substitution in a formula
S Simplifications
P Penalty (no units, incorrect rounding off etc.)
O Opinion
J Justification
R Rounding
NPR No Penalty for Rounding


QUESTION 1

Quest. Solution Explanation Marks
1.1.1  South African Revenue Services ✓✓  2A  (2)
1.1.2 15 years 9 months ✓
= 15 x 12 + 9 ✓
= 189 months ✓		 1 A 15 years 9
months
1M Conversion to
months
1 CA		 (3)
1.1.3 Five hundred and six thousand ✓
Four hundred and seventy four rand ✓		 2A In words

 (2)
1.1.4  R122 138,71 – R104 227 ✓✓
= R17 911,71 ✓		 1A Correct values
1M Subtraction:
1CA 		 (3)
1.1.5 122 138,71 ✓: 506 474 ✓
1 : 4,15 ✓		 2M Ratio of Correct
Values
1A 		 (3)
1.1.6 R631,94 ✓✓ 2RT  (2)
1.1.7 R631, 94✓✓
    110%
= R574,49 ✓x 12 ✓
= R6 893,89 ✓
OR
R631,94 x 𝟏𝟎   ✓
                𝟏𝟏𝟎
= R57,45 ✓
R631,94 – R57,45 ✓
= R574,49 x 12 ✓
= R6 893,89 ✓
OR
𝑹631,94  x 100 ✓✓
    𝟏𝟏𝟎
= R574,49✓ x 12 ✓
= R6 893,89 ✓		 1M Correct Values
1M dividing by 110%
1CA
1M x12
1CA
(Accept 6893,88)
1M Multiplying by the
fraction
1S
1M subtraction
1M Multiply by 12
1CA
1M Multiply 100
1M Denominator
1CA for R574,49
1M multiply by 12
1CA		 (5)
1.2.1 $225 + $200 + $175 + $50 ✓✓
= $650		 2A Adding all the values

 (2)
1.2.2 $175 x 100 ✓
$650
= 26,9% ✓
= 27% ✓		 1M dividing by $650 and multiply by 100
1CA
1CA

 (3)
1.2.3 500 x $15,00 ✓
= $7 500 ✓		 1M identifying 500 and $15,00
1A		 (2)
1.2.4 7500 x 15,409095 ✓
= R115 568,2125 ✓
= R115 568,21 ✓		 1M multiplying by R15,409095
1S
1 CA (two decimal places)		 (3)
1.2.5  $200 ✓
 500
= $0,4 ✓		 1M for 200
1CA

 (2)
1.2.6 1 April 2016 – 30 April 2016 ✓✓ 2 RT (2)

[34]

QUESTION 2

Quest. Solution Explanation Marks
2.1.1 V =𝜋𝑟2h ✓
= 3,142 x (2,5 cm)2 x 12,5 cm ✓
= 245,47 cm3✓✓		 1A converting radius
1SF
1CA answer
1 unit
NPR		 (4)
2.1.2 No. of candles=5 000 𝑐𝑚3 245,47 𝑐𝑚3
                        245,47cm3
= 20,4 ✓
= 20 candles ✓		 1M for 5 000
1M
1CA answer		 (3)
2.1.3 Candle weight = density x volume
= 0,93g/cm-3 x 245,47 cm3 ✓✓
= 228,29 g ✓		 1SF
1M using 245,47 cm3
1CA answer
NPR		 (3)
2.1.4 TSA = 2 x (2,6 x 2,8) + 2 x 6,1(2,6 + 2,8) ✓
= 2(7,28) + 12,2 x (5,4) ✓
= 14,56 cm2 + 65,88 cm2 ✓
= 80,44 cm2 ✓ OR
𝑇𝑆𝐴=2𝑥(𝑙 x 𝑤)+2(𝑙 x ℎ)+2(𝑤 x ℎ)
= 2(2,8 x 2,6) + 2(2,8 x 6,1) + 2(2,6 x 6,1) ✓✓
= 2(7,28) + 2(17,08) + 2(15,86) ✓
= 14,56 + 34,16 + 31,72
= 80,44 cm2 ✓

1SF
1S
1S
1CA answer

2 SF
1S
1CA answer

 (4)
2.1.5 Diameter = 2,5 × 2 = 5 cm ✓
No. of candles along the length = 15=3✓
                                                       5
No. of candles along the width = 15=3 ✓
                                                     5
Total number of candles for the first layer
= 3 × 3 = 9 ✓		 1M diameter
1M length candles
1M width candles
1CA Check 2.1.1 for radius

 (4)
2.1.6 (a) 312℉ ✓✓ 2RD (2)
  (b) ℃ = (312°− 32°) ÷ 1,8 ✓
= 280° ÷1,8 ✓
= 155,6 ℃ ✓
Accept 155,56 ℃		 1SF
1S
1A penalise if ℉ is written in the answer

 (3)
2.2.1 2,3 m – 0,25 m ✓✓
= 2,05 m ✓		 1M Conversion to metre
1M subtraction
1 CA answer		 (3)
2.2.2 𝐴=𝑙 𝑥 𝑤
12 m2 = (2,3m +1,7m) x w ✓
12 m2 = 4 m x w ✓
 4m           4m
3 m = w		 1M adding 1,7
1M dividing by 4
1A

 (3)

[29]

QUESTION 3

Quest. Solution Explanation Marks
3.1.1 North east ✓✓ 2A (2)
3.1.2 1 : 75 ✓✓ 2RP (2)
3.1.3 𝐿𝑒𝑛𝑔𝑡ℎ=9 𝑐𝑚 𝑥 75100✓
                   100
= 6,75 m✓
Width = 1,3 x 75 ÷100 = 0,975 m = 1 m ✓		 1M
1A answer for length
1A answer for width		 (3)
3.1.4  3.14 2M Drawing

 (2)
3.1.5 15 ✓✓ 2RP (2)
3.1.6 7 ✓
16 ✓		 1M numerator
1M denominator		 (2)
3.2.1 4 ✓✓ 2RD (2)
3.2.2 From the reception go straight along the orchard and turn right, ✓ then go down pass the playground and turn leftand go straight you will get 11b. ✓ 3RD (3)
3.2.3 7 ✓✓ 2RD (2)
3.2.4 9 ✓✓ 2RD (2)
3.2.5 Table tennis OR Pool table ✓✓ 2RD any facility (2)

[24]

QUESTION 4

Quest. Solution Explanation Marks
4.1.1

Av weight = 92+94+96+98+102+108+110+112+115+116+117+120×2+125 ✓
                                                               14

= 1 525✓
      14

= 108,93 kg ✓ (Accept 108,929)

 1M
1S
1CA		 (3)
4.1.2 1 569 ✓✓ 2A (2)
4.1.3 186 cm ✓✓ 2A (2)
4.1.4
Interval   Tally   Frequency 
0 – 30   a  6✓
31 – 60   b  10 ✓
61 – 90   c  7 ✓
91 – 120   ///  3 ✓
121 – 150  //  2 ✓

1 mark (both tally and frequency) x 5 =5

 1 x 5 = 5 (5)
4.1.5 Probability is the chance or likelihood of an event happening. ✓✓ 2M Definition (2)
4.1.6 4  ×100 ✓
28
= 14,3% ✓		 1 M Fraction multiply by 100
1CA

 (2)
4.1.7

4.17b

✓ 1 Mark per correctly plotted bar joined to an existing one

    
4.1.8 (a) 1 , 2 , 2 , 4 , 10 , 19 , 20 , 38 ✓
= 14 ✓
    2
= 7 ✓		 1M Correct values
1M
1CA

 (3)
  (b) Range = 64 ✓– 1 ✓
= 63 ✓		 1Correct values
1M Subtracting
1A		 (3)
  (c) Line graph ✓✓ 2A (2)
4.2.1 White ✓✓ 2A (2)
4.2.2 A = 41 000 938 +4 586 838 +4 615 401+1 286 930+280 454 ✓
= 51 770 561 ✓		 1M Adding
1A		 (2)
4.2.3 Whites and Coloureds ✓✓ 2A (2)
4.2.4 Indian / Asian ✓✓ 2A (2)
4.2.5

B + B+ 79,2 + 2,5 + 0,5 = 100%✓
= 100% – 82,2
2B = 17,8✓
B = 8,9%✓
         OR                                                                 OR
4 586 838 ✓ x 100%✓                         4 615 5401 ✓ x 100% ✓
51 770 561                                           51 770 561
= 8,859                                                    = 8,915 

= 8,9% ✓                                                 = 8,9%✓

 1M Adding to make 100
1S value of 2B
1A
1M fraction with correct Values
1M multiply by 100
1A

 (3)

[40]

QUESTION 5

Quest Solution Explanation Marks
5.1.1 Panado Medical Centre ✓✓ 2RT (2)
5.1.2 R24,46 ✓✓ 2RT (2)
5.1.3 R89,80 – 24,46✓
= R65,34 ✓		 1M Subtraction
1A		 (2)
5.1.4 R38,91 ✓✓ 2RT (2)
5.1.5 R24,46 +R309,70 + R108,49 +R38,91 +R13,10 + R5,02 ✓
= R499,68 ✓		 1M Adding
1A		 (2)
5.1.6 Pain located in other parts of the lower abdomen. ✓✓ 2 RT (2)
5.1.7 R13,10 x 14% ✓
= R1,83 + R13,10 ✓
= R14,93 ✓
OR
R13,10 x 114% ✓✓
= R14,93 ✓		 1M
1Adding
1A		 (3)
5.1.8 60 days ✓
= 2 months ✓		 1M
1CA ( give a mark if answer is 30 days only)		 (2)
5.2.1 Morning + Evening
(10 𝑚ℓ+15 𝑚ℓ+10 𝑚ℓ +10 𝑚ℓ+15 𝑚ℓ+10 𝑚ℓ) 
= 70 𝑚ℓ✓
OR
(10 𝑚ℓ x 4) + (15 𝑚ℓ x 2) ✓
= 40 𝑚ℓ + 30
= 70 𝑚ℓ ✓		 1 M
1CA		 (2)
5.2.2 10𝑚ℓ+10𝑚ℓ ✓ =20 𝑚ℓ✓ 
OR
10 𝑚ℓ x 2 ✓
= 20 𝑚ℓ ✓
OR
100 𝑚ℓ – (20 𝑚ℓ x 4) ✓
= 100 𝑚ℓ – 80 𝑚ℓ
= 20 𝑚ℓ ✓
OR
100 𝑚ℓ – (10 𝑚ℓ x 8) ✓
= 100 𝑚ℓ – 80 𝑚ℓ
= 20 𝑚ℓ ✓		 1M
1A		 (2)
5.3.  60  ✓ =  3   ✓
100         5		 2A (2)

[23]

TOTAL: 150

Last modified on Friday, 11 June 2021 08:52
Published in Grade 12 Exam Papers and Memos 2016 September

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