SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2015
GRADE 12
SUBJECT: MATHEMATICAL LITERACY
TEACHER NOTES
(Page 1 of 64)
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© Gauteng Department of Education
TABLE OF CONTENTS
SESSION
1
2
3
4
TOPIC
DATA HANDLING
FINANCE
Break-even analysis
Small business
FINANCE
Tariff systems
MEASUREMENT
Conversions
Use of formulae
PAGE
3 - 15
15 - 31
32 - 45
46 - 64
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© Gauteng Department of Education
SESSION 1
TOPIC: DATA HANDLING
Note to teacher:
This part of Mathematical Literacy is not very difficult. The learners can get a lot of
marks for this section. Make very sure that they know the terminology very well.
The most difficult part of these questions will be when TWO sets of data is given
and the questions jump from the one set to the other.
Make sure that they practise these questions before the exams.
LESSON OVERVIEW: 1. Introduction session
2. Typical exam questions
3. Discussion of solutions
5 minutes
55 minutes
30 minutes
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 14 Minutes
(from NSC Nov 2013 Paper 1)
Thandeka has a shop with a scrapbooking department and a toy department. She kept
a record of the ages of the customers who visited the two departments on a particular
day.
Scrapbooking is a hobby which
involves cutting and pasting
photos, pictures and other
decorative items in a book
Ages of customers who visited the scrapbooking department
35
60
46
57
54
34
60
54
56
46
47
67
65
54
45
Ages of customers who visited the toy department
5
15
25
7
36
21
70
20
17
6
15
65
9
15
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1.1 Arrange the ages of the customers who visited the toy department in
ascending order.
(2)
1.2 Calculate the range of the ages of the customers who visited the toy
department.
(2)
1.3 Determine the modus of the ages of customers who visited the
scrapbooking department.
(2)
1.4 Calculate the mean (average) age of the customers who visited the
scrapbooking department.
1.5
(3)
The upper quartile of the ages of customers who visited the toy
department is 25. List the ages of the customers who visited the
toy department that are greater than the upper quartile.
(2)
[11]
QUESTION 2: 12 Minutes
(from NSC Nov 2013 Paper 2)
The percentage distribution (per province) of persons aged 20 years and older
whose highest level of education was Grade 12 in 2011 is shown in the table
below.
TABLE 5: Percentage distribution (per province) of persons with
Grade 12 as highest level of education during 2011
Province
KZN EC
FS
WC NC
NW
GP
MP
LP
Percentage 30,9 19,8 26,8 28,2 22,7 25,2 34,4 29,0 22,4
KEY: KZN – KwaZulu-Natal
EC – Eastern Cape FS – Free State
WC – Western Cape
NC – Northern Cape
NW – North West GP – Gauteng MP –
Mpumalanga
LP - Limpopo
2.1 Explain why the sum of the percentages in Table 5 is not 100%
(2)
2.2 Determine the province that had the median percentage of persons with
grade 12 as the highest level of education in 2011.
(2)
2.3 For the data given above, the 25th percentile is 22,55% and the 75th
percentile is 29,95%. Identify the province(s) whose percentage
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distribution is less than the lower quartile.
(2)
2.4 Give a reason why each of the following types of graphs is NOT suitable
to represent the above data:
(a) a pie chart
(2)
(b) a histogram
(2)
[10]
QUESTION 3: 17 minutes
(from NSC Feb/Mar 2013 Paper 1)
Mrs Botha conducted a survey each day for a week to determine the
approximate number of minutes that her Grade 8 and Grade 12 learners
watched television.
She recorded the results (in minutes) of her survey as follows:
GRADE 8
30
90
120
45
95
120
60
95
150
60
120
150
60
120
180
GRADE 12
0
30
30
30 40
45
45
60
60
50
60
60
150
150
60
180
3.1 Determine the sample size of the survey.
(2)
3.2 How many learners did not watch any television during the week?
(1)
3.3 Calculate the range of the time spent by the Grade 8 learners watching
television.
(2)
3.4 Write down the modal time the Grade 8 learners spent watching television.
(2)
3.5 Determine the median time the Grade 8 learners spent watching
Television.
(3)
3.6 Calculate the average (mean) time the Grade 12 learners spent watching
television.
(3)
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3.7 One of the Grade 12 learners is randomly selected. Determine the
probability that this learner spent 45 minutes daily watching television. (2)
[15]
QUESTION 4: 18 minutes
(from NSC Feb/Mar 2013 Paper 2)
The Olympic Games were held in London in
the United Kingdom during July and
August 2012.
In April 2012, the school principals in
Jacksonville decided to hold a Mini Olympics
in anticipation of the Olympic Games.
As part of the Mini Olympics, a diving
competition was held between the schools.
Two divers, Bongani and Graham, were
given the following scores by the nine
diving judges after completing their dives:
Picture of diver
Bongani’s
scores
Graham’s
scores
4.1
9½
9
8
9
7½
9
9
8½
8½
9
6½
9
7½
8
8½
9
9
9½
Use the scores to determine the following:
(a) Bongani's median score
(b) The range of Bongani's scores
(3)
(2)
4.2 In diving competitions, there is a rule whereby a diver's highest and lowest
scores a not considered.
(a) Give ONE valid reason why this rule is applied.
(2)
(b) Determine, showing ALL calculations, which ONE of the two divers
attained the higher mean score if the competition rule is applied.
(8)
[15]
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SECTION B:
CONTENT NOTES
TEACHER: Make very sure that every learner understands the terminology VERY
WELL
POPULATION: The entire group about which data is collected.
SAMPLE: A collection of people that represent the population.
ORDER (ARRANGE): Sort the data set in:
ASCENDING ORDER: (smallest value to largest value)
DESCENDING ORDER: (largest value to smallest value)
RANGE: The largest data value MINUS the smallest data value.
MEAN (AVERAGE): Add all the data values together and divide the answer by the
NUMBER of data values.
Always use the formula:
MODE: The piece of data found MOST often.
There can be more than one mode.
MEDIAN: After the set is arranged in size order the MIDDLE-MOST value is the
median.
When the set has an UNEVEN number of data values, there is a data
value in the middle.
When the set has an EVEN number of data values, the median will be the
MEAN of the middle TWO data values.
QUARTILES: Quartiles divide the data into four equal parts.
The MEDIAN is also Q2, sometimes called the 2nd Quartile.
The LOWER QUARTILE Q1 is the middle data value when looking
from the start of the arranged data set to Q2
Q1 is also the 25th percentile, sometimes called the 1st Quartile
The UPPER QUARTILE Q3 is the middle value when looking from
the median to the last data value.
Q3 is also the 75th percentile, sometimes called the 3rd percentile.
INTERQUARTILE RANGE: The difference between Q3 and Q1 . (Q3 Q1)
This section represents the middle 50% of the data set.
PERCENTILES: Divides the data in 100 equal parts.
Q1 is the 25th percentile
Q2 is the 50th percentile
Q3 is the 75th percentile
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SECTION C: HOME WORK
QUESTION 1: 18 minutes (from NSC Nov 2012 Paper 2)
On 14 February 2012 there was a queue of customers waiting to eat at Danny's
Diner, a popular eating place in Matatiele.
The time (in minutes) that 16 of Danny's customers had to wait in the queue is given
below:
30
B
15
42
45
26
36
32
A
38
40
35
34
41
B
28
B is a value greater than 20.
1.1
The range of the waiting times was 37 minutes and the mean (average)
waiting time was 34 minutes.
(a) Calculate the missing value A, the longest waiting time.
(2)
(b) Hence, calculate the value of B.
(4)
(c) Hence, determine the median waiting time.
(3)
1.2 The lower quartile and the upper quartile of the waiting times are
27 minutes and 41,5 minutes respectively.
How many of the 16 customers had to wait in the queue for a shorter time
than the lower quartile?
(2)
1.3 Danny's previous records, for 16 customers on 7 February 2012,
showed that the median, range and the mean (average) of the waiting
times were 10 minutes, 5 minutes and 10 minutes respectively.
Compare the statistical measures relating to the waiting times on 7 and
14 February and then identify TWO possible reasons to explain the
difference in these waiting times.
(4)
[15]
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QUESTION 2:
5 minutes (from NSC Feb/Mar 2012 Paper 2)
They went fishing during their camping trip and recorded the mass (in grams) of the eleven
fish they caught, as shown below.
1 513
875
3 025
912
1 809
1 003
1 794
1 628
958
1 052
The following are statistical measures relating to the above data:
Lower quartile = 958 g
Upper quartile = 1 794 g
Median = 1 513 g
Mean = 1 462 g
1 513
2.1 Determine the number of fish caught having a mass between the lower
and upper quartiles.
(2)
2.2 Which ONE of the following measures of central tendency (median, mean
or mode) will change, if a twelfth fish with a mass of 1 462 g was also caught?
(2)
[4]
QUESTION 3: 14 minutes (from NSC Nov 2011 Paper 1)
The Swartberg High Fundraising Committee intends opening a school uniform
shop in January 2012. They foresee that the shop will mainly be supported by
the new Grade 8 learners.
They surveyed a sample of Grade 8 learners in June 2011 to determine their
shoe sizes.
SHOE SIZES OF BOYS:
5
5
5
6
6
6
6 6½ 6½ 6½
7
7
7 7½ 8
8½
9 10 11
SHOE SIZES OF GIRLS:
3½
4
4
4½
5 5
5
5
5
5½
5½ 6
6 6
6
6 6
6½
7 8
3.1
What is the modal shoe size of the boys?
(2)
3.2
Determine the median shoe size of the boys.
(2)
3.3
Determine the median shoe size of the girls.
(2)
3.4
Which shoe sizes are NOT worn by the boys?
(2)
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3.5.
Write, as a ratio, the number of boys not wearing half sizes to the number
of girls not wearing half sizes.
QUESTION 4:
(3)
[11]
35 minutes (from NSC Nov 2011 Paper 2)
Bathini High School and Vuka Secondary School entered some of their
learners in a science competition. The scores (in percentages) for the first
round of the competition are given below.
59
90
4.1
4.2
67
67
67
89
67
BATHINI HIGH SCHOOL
67 67 72 78 87 87
VUKA SECONDARY SCHOOL
50 78 54 67 95 90 98
4.3
57
99
49
78
If a learner is selected randomly from Vuka Secondary School, determine
the probability that the learner scored more than 90%. Write the answer in
simplified form.
(3)
The following table shows the median, mode, mean and range for
the two schools:
TABLE 2: Median, mode, mean and range
NAME OF SCHOOL
MEDIAN MODE
Bathini High
72%
67%
Vuka Secondary
P
67%
(a)
(b)
90
MEAN
76,4%
Q
RANGE
R
48
Determine the missing values P, Q and R.
Which school performed better in the competition? Explain your
answer.
(8)
(3)
The following table shows the percentiles of scores obtained in the
science competition for the two schools:
TABLE 3: Scores for the two schools
60th
NAME OF SCHOOL
25th
Percentile Percentile
Bathini High
67%
75,6%
Vuka Secondary
57%
78%
75th
Percentile
87%
90%
(a) List the scores of the learners from Vuka Secondary School who scored at the
75th percentile or more.
(3)
(b) How many learners from Vuka Secondary School obtained scores that were
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less than the 25th percentile of Bathini High School?
(2)
[19]
SECTION D: SOLUTIONS FOR SECTION A
QUESTION 1
1.1
5
17
6
20
7
21
9
25
15
36
15
65
15

70 
1 ascending order
1 all values
Answer only: full marks
(2)
1.2
Range = 70 – 5
= 65 

1 identifying
range concept
1 simplification
Answer only: full marks
# from
Question 2.2.1
(2)
1.3
54 
2 correct mode
(for the incorrect
data set, if answer
15 max 1 mark)
Including 60
and/or 46, max 1
mark
(2)
+
1.4
Mean =
67+65+54+45
1 sum of data
1 dividing by
number of data
entries
15 
780
=
15
= 52 
1 solution
Answer only: full marks
(3)
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1.5
36; 65; 70 
2 correct values
(one or two values
correct, 1 mark)
Including an incorrect
value max 1 mark
(2)
QUESTION 2
2.1
The percentages given represent the number of people with 2 acceptable
Grade 12 as a percentage of the number of people 20 years explanation
and older in each province and not nationally. 
OR
Data is per province
2.2
(2)

The ascending order is
19,8 ; 22,4 ; 22,7 ; 25,2 ; 26,8 ; 28,2 ; 29,0 ; 30,9 ; 34,4
∴ Free State has the median percentage

 1 arranging in
ascending order
1 province
OR
OR
The ascending order is

EC; LP; NC; NW; FS; WC; MP; KZN; GP
∴ Free State has the median percentage 
1 ascending order
1 province
Correct answer only:
full marks
(2)
2.3

Eastern Cape and Limpopo
1 EC
1 LP

(2)

2.4(a) The percentages do not add up to 100%
2 explanation
OR
The degrees to not add up to 3600

OR
There are too many sectors
(2)

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2.4(b)

The histogram cannot be used since the data is qualitative
2 explanation
OR

The data is not continuous
OR
Data is not given in class intervals
(2)

QUESTION 3
3.1
15 + 16 = 31 
2 solution
(2)
3.2
1 (one) 
3.3
Range = (180 – 30) minutes
= 150 minutes 
1 solution
(1)
1 concept of
range
3.4
120 minutes  
3.5
Median = 95 minutes
3.6
Mean

1
simplification
(2)
2
simplification
(2)

2 solution
(2)

= 0+30+30+30+40+45+45+50+60+60+60+60+60+150+150+180
16 
1 adding
1 dividing
by 16
=
= 65,63 minutes

1
simplification
(3)
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Probability (a learner watching TV for 45 minutes)
3.7
=2
1
6 A
1 numerator
1 denominator

(2)

OR 1

8
OR 12,5 %

QUESTION 4
M
4.1
(a)
7
8
88
The median score = 9
9
9
9
9
9 12
1 arranging in order

2 correct identification
Answer only full marks
(3)
4.1
(b)
Range =
9 – 7

= 2
1 subtraction
1 simplification
Answer only full marks
(2)
4.2
(a)

To eliminate scores of judges who are biased.
2 opinion
OR
OR
Eliminating the highest and lowest scores will have
the effect that the mean is calculated without
extreme values

2 opinion
OR
Any other valid, well-thought out opinion
(2)
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4.2
(b)
9+8+9+9+9+812 +812 
1 concept of mean
Bongani’s mean =
7
=
1 correct numerator
1 correct denominator
61 
7
= 8,714....
= 8,71 
1 simplification
1 correct numerator
=
Graham’s mean

1 concept of mean
=

1 simplification
= 8,5714...
= 8,57 
∴Bongani attained the higher mean score 
1 conclusion
(8)
SESSION 2
TOPIC:
FINANCES: BREAK-EVEN ANALYSIS
NOTE TO TEACHER
 When you run a small business, you must be able to calculate the
number of items you need to sell in order to make a profit.


INCOME is the money generated by selling the product.
COST is calculated by adding the operating costs and production costs
together.

Two graphs are drawn on the same grid, the point where these two lines
intersect is called the BREAK-EVEN POINT.

Learners must be able to read the profit or loss from the graph

Learners will be able to use this knowledge when they start up their own
small businesses.
LESSON OVERVIEW: 1. Introduction session
2. Typical exam questions
3. Discussion of solutions
5 minutes
55 minutes
30 minutes
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SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 24 minutes (from various sources)
John manufactures sandals using old tyres and tyre tubes. He decides to sell his
product at an informal Sunday Market.
To rent space at the market costs John R80 per day. The production costs of one
pair of sandals come to R70. John wants to sell the sandals at R90 per pair.
1.1 Complete the following tables for INCOME and COSTS for John’s sandal
business.
On ANNEXTURE A:
COSTS
Number of
Sandals (pairs)
Amount
INCOME
Number of
Sandals (pairs)
Amount
0
1
5
10
20
Do on
annexture
150
430
780
Do on
annexture
0
1
5
10
15
Do on
annexture
90
450
900
Do on
annexture
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(4)
1.2 Use the information on the tables to draw line graphs for COSTS and
INCOME on the grid on ANNEXTUE A. Label the lines.
(6)
1.3 What is the point oF intersection of the graphs called?
(1)
1.4 What is the importance of this point of intersection in this case?
(2)
1.5 How many pairs of sandals must John sell to start showing a profit?
(2)
1.6.1 Show on the grid, where the profit on 17 pairs of sandals can be determined.
(2)
1.6.2 Calculate, by using the graph, the profit on 17 pairs of sandals. Show
calculatioins.
ANNEXTURE A
1.1
COSTS
Number of
Sandals (pairs)
Amount
INCOME
Number of
Sandals (pairs)
Amount
0
0
1
5
10
150
430
780
1
5
10
90
450
900
(3)
20
15
1.2
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2000
JOHN'S STALL
RAND
1500
1000
500
0
0
5
10
15
20
PAIRS OF SANDALS
[20]
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QUESTION 2: 9 minutes (from NSC Nov 2012, paper 2)
2.1 Use the graph above to determine the exact number of items sold that will
give a loss of R1 400.
(3)
2.2 Mr Stanford stated that the company would break even if 40 items were
sold at R137,50 each.
Verify whether Mr Stanford's statement is correct or not. Show ALL the
necessary calculations.
(4)
[7]
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QUESTION 3: 33 minutes (from NSC Feb/Mar 2009 paper 2)
The Hospitality Studies department of Ses’fikile High School bakes brown bread in
order to raise funds for the shortfall incurred in their day-to-day expenses.
The school charges the Hospitality Studies department a fixed weekly cost of R400,00
for water and electricity
The cost of producing one loaf of brown bread, including labour and ingredients, is
R3,50. The brown bread is sold at R6,00 a loaf.
3.1
3.2
If one loaf of brown bread requires 450 g of flour, determine the maximum
number of loaves of brown bread that can be baked from a 12,5 kg bag of
flour.
(4)
The table below shows the weekly cost of making the bread.
TABLE 1: Weekly cost of making brown bread
Number of loaves
0
Total cost (in rand)
400
40
540
80
680
120
160
A
B
960
1
240
300
1
450
The formula used to calculate the total cost per week is:
Total cost per week = Fixed weekly cost + (number of loaves of bread ×
cost per loaf)
Use the given formula to determine the values of A and B in TABLE 1. (4)
3.3
The table below shows the weekly income from selling the bread.
TABLE 2: Weekly income received from selling bread
Number of loaves
Total income (in rand)
0
0
40
240
120
C
150
900
D
960
250
1
500
Determine the values of C and D in TABLE 2.
300
1
800
(4)
3.4 Use the values from TABLE 1 and TABLE 2 to draw TWO straight
line graphs on the same grid using ANNEXURE A, showing the
total COST per week of making bread and the INCOME per week
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3.5
3.5.1
from selling bread. Clearly label the graphs 'COSTS' and
‘INCOME'.
(8)
Use the tables or the graph drawn on ANNEXURE A to answer the
following questions.
3.5.2
How many loaves of bread must they sell to break even and
describe what is happening at the break-even point?
What income would they receive if 230 loaves were sold?
(3)
(2)
3.5.3
Estimate the number of loaves baked if the total cost is R840.
(2)
3.6
Determine, by calculation, whether Ses'fikile High School will make a
profit or a loss if they bake 300 loaves of bread during the week, but only
sell 250 of these loaves of bread.
(5)
[32]
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ANNEXURE A
QUESTION 3.4
INCOME AND COSTS
A
m
ou
nt
in
ra
n
d
Number of loaves of bread
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SECTION B:
CONTENT NOTES
Follow these steps when doing break-even point questions:
(1) Summarise the situation
(2) Formulate equations for cost and income
(3) Complete the table
(4) Draw the graphs using the table
(5) Analyse the graph
The graphs are always straight lines, BUT in the second homework question you will
find a line that is not completely straight.
Look out for questions which ask: ”How many products must be sold to start showing
a profit.” The answer is NOT the break-even point, but actually the first integer
AFTER the break-even point.
START-UP COSTS: Costs that need to be paid before the business starts to
operate, eg: stoves, fridges (small bakery); welding machine (burglar bar
business).
OPERATING COSTS: Costs that the business have to pay regularly while the
business is operating.
PRODUCTION COSTS: Costs that result directly from the production or supply or
service.
BREAK-EVEN POINT: Where income and costs are equal for a certain number of
products.
PROFIT: When the income is higher than the costs. (higher than the break-even
point)
LOSS: When the costs are higher than the income (the break-even point is not
reached yet).
Before the exams you must definitely practise these questions, as they often
carry many marks.
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SECTION C: HOME WORK
QUESTION 1: 29 minutes (from NSC Nov 2008, paper 2)
Mrs Maharaj makes duvet sets, which she sells at the local street market at R150,00
per set (including VAT).
• If she makes 50 or less duvet sets per month, her production costs are R100,00
per set.
• If she makes more than 50 duvet sets per month, her production costs are reduced
by 15% per set.
Mrs Maharaj has to pay R8 400 annually for the rental of her stall and she has
weekly transport costs of R75.
1.1
Mrs Maharaj prepares a monthly budget.
1.1.1 Show that her fixed cost for the month of February is R1 000,00.
(2)
1.1.2 How does her fixed cost for February compare to her average monthly
fixed costs? Show ALL calculations.
(5)
1.2
Calculate the production cost per duvet set if 90 sets are made per month.
(2)
1.3 The table below shows Mrs Maharaj's production cost for different quantities
of duvet sets made in February.
TABLE 3: Cost of duvet sets made in February
Number of
0
30
50
51
56
duvet sets
Total cost per
month (in
1000
4000 6000 5335 5760
rand)
60
70
D
6100
C
7800
The formula used to calculate the total cost is:
Total cost = fixed monthly cost + (number of duvet sets × cost per set)
Use the formula and the given information to determine the missing values C and D.
(5)
1.4 Mrs Maharaj draws two graphs to represent her income and expenses for
different quantities of duvet sets. The graph showing her INCOME for different
quantities of duvet sets has already been drawn on ANNEXURE A.
Use the values from TABLE 3 to draw a second graph on ANNEXURE A
showing the total EXPENSES for February of making different quantities of duvet
sets. Label the graph as 'EXPENSES'.
(7)
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1.5 Use the two graphs to answer the following questions:
1.5.1 How many duvet sets must Mrs Maharaj sell to break even?
(2)
1.5.2 What profit will she make if all 80 duvet sets are sold?
(3)
1.5.3 Suppose Mrs Maharaj makes 80 duvet sets, but only sells 70 of them.
Calculate her profit for February.
(3)
[29]
ANNEXTURE A
INCOME AND EXPENSES
12000
11000
10000
9000
Amount in Rand
8000
7000
6000
5000
4000
3000
2000
1000
0
0
10
20
30
40
50
Number of Duvet sets
60
70
80
25
© Gauteng Department of Education
QUESTION 2:
12 minutes (from NSC Nov 2013, paper 1)
Mrs Louw bought a knitting machine for R5 600,00 to make the jerseys. It will
cost her an average of R60,00 (including wool and electricity) to make one longsleeved jersey (irrespective of the jersey size). The school shop buys a longsleeved jersey for R95,00 and then sells it to the learners.
TABLE 4 below shows the relationship between the costs and income for making
and selling 200 long-sleeved jerseys.
TABLE 4: Costs and income for making and selling 200 long-sleeved jerseys
NUMBER OF JERSEYS MADE
0
Costs (in rand)
Income (in rand)
5 600
0
60
A
120
160
180
200
9 200 11 600 12 800 15 200 16 400 17 600
5 700
9 500 11 400 15 200
B
19 000
1.3
Determine the missing values A and B.
1.4
On ANNEXURE B, the line graph showing Mrs Louw's income from the
sale of 200 long-sleeved jerseys is drawn. Draw another line graph on the
same grid representing the costs of making 200 long-sleeved jerseys.
(4)
1.5.
(4)
Determine the minimum number of jerseys Mrs Louw should make and sell to
start showing a profit.
(2)
[10]
26
© Gauteng Department of Education
ANNEXTURE B
TABLE 4: Costs for making and selling 200 long-sleeved jerseys
Number of jerseys
made
Costs (in rand)
0
60
5 600
9 200
A
120
160
180
200
11
12
15
16
17
600
800
200
400
600
COSTS AND INCOME FOR MAKING 200 LONG-SLEEVED JERSEYS
A
m
ou
nt
in
ra
n
d
Number of long-sleeved jerseys
27
© Gauteng Department of Education
SECTION D: EXAM QUESTIONS SOLUTIONS
QUESTION 1 – SESSION 2
1.1
80
1480
0
1350
1.2
1
1
1
1 (4)
JOHN'S STALL
2000
RAND
1500
1000
500
0
0
5
10
PAIRS OF SANDALS
15
20
1.3
 INCOME starts in (0;0)
 INCOME goes to end of grid
 INCOME goes through (10;900)
COST goes through (10;780)
COST points connected
COST both lines labelled
Break-even point
1.4
It is where costs and number of sandalsare equal
2
1.5
5 and more
2
1.6.1
On the graph at 17 pairs:
(6)
1


1
1
28
© Gauteng Department of Education
Or
On the graph shown:


1.6.2
1530 – 1270 AS PER DRAWN GRAPH
1
1
1
=R260
QUESTION 2 – SESSION 2
2.1 For 30 items:

1 cost
1 income
Cost = R5 000 RG
Income = R3 600 RG
Loss = R5 000 – R3 600
= R1 400
30 items A
1 number of items
Correct answer only full marks
(3)
2.2 Cost of 40 items = R5 500
 OR 40
R50,00 + R3 500
1 cost
income
×
Income from 40 items = R137,50 × 40 
= R5 500 
Or Cost =
1 finding total income
1simplification

At 40 items, Cost = Income
Mr Stanford's statement is CORRECT.
1 verification
(4)
29
© Gauteng Department of Education
QUESTION 3 – SESSION 2
3.1
12,5kg
Number of loaves =

1 dividing

1 converting to
grams
450g
12500g
=
450g
= 27,78 
1 simplification
1 rounding down
27 loaves 
OR
OR
 
12,5kg
= 27,78 
Number of loaves =
0,450 kg
27 loaves 
3.2
1 dividing
1 converting to
kilograms
1 simplification
1 rounding down
(4)
Total cost = Fixed cost + (number of loaves × cost per loaf)
A = 400 + (120 ×R3,50) 
1 substitution
= R820
1 total cost
AND
1 240 = 400 + (B ×R3,50) 
840 = (B ×R 3,50)
240 = B 
1 substitution
1 number of loaves
(4)
3.3
Income = number of loaves ×
price of loaf
C = 120 × R6,00
1 substitution
= R720,00
1 income
AND
960 = D ×R6,00
1 substitution

D =
= 160 loaves
1 number of loaves
(4)
30
© Gauteng Department of Education
3.5.1
160 loaves must be sold 
1 reading from graph

At this point both the cost and the income are the same and
are equal to R960.

3.5.2
R1 380 
1 income is R960
1 cost is R960
(3)
2 reading from graph
(from graph)
(2)
3.5.3
125 loaves [Accept any whole number value between 120
and 130] 
2 reading from graph
( from graph)
(2)
3.5.4
Cost of making 300 loaves = R1 450
Income from selling 250 loaves = R1 500
Profit = R1 500 – R1 450
= R50 
3.6
The maximum number of batches per day = 6
The maximum number of loaves baked each day
= 6 × 20 loaves 
= 120 loaves 
So, the order for 110 loaves may be accepted. 
1 reading cost from
graph
1 reading income from
graph
1 profit
(from graph)
(3)
2 reading from the
time line
1 multiplying
1 maximum number
of loaves
1 conclusion
(5)
31
© Gauteng Department of Education
SESSION 3
TOPIC: FINANCES: TARIFF SYSTEMS
NOTE TO TEACHER:
 One often has to make choices between different tariff systems to live
economically.

By using graphs you can easily SEE which choice to make.

These graphs are similar to the INCOME and COST graphs of session
2, BUT now you are not determining profit or loss of a business, you are
now using the graphs to find the BEST OPTION for yourself..

You will draw two or MORE graphs on the grid, and where these lines
INTERSECT, (the break-even points for the 2 lines involved), the
situation depicted by the graphs has a “change”.

Using this method one can determine which investment is the best, OR
which cell phone contract is the best etc

Before the exams these questions must definitely be practised, as they
can sometimes carry a lot of marks.
LESSON OVERVIEW:
1. Introduction session
2. Typical exam questions
3. Discussion of solutions
5 minutes
55 minutes
30 minutes
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1: 34 minutes (from NSC NOV 2013 paper 2)
32
© Gauteng Department of Education
1.1 Interpret the horizontal part of the line graph for Option A
1.2 Payment Option B starts at point P.
(a) Explain why point P is represented by an open circle on the graph.
(b) Describe in detail the cost of driving lessons if option B is used.
(2)
1.3 The graphs intersect at points Q and R. Interpret the graphs at point Q.
(2)
(3)
(2)
1.4 Zaheera budgeted R1 200 for her driving lessons.
Explain which option would be better for:
(a) Zaheera
(2)
(b) Toni
(2)
1.5 In an attempt to further reduce the total cost of her driving lessons,
Zaheera asks a friend to teach her some basic driving skills. After a
series of free lessons with her friend, she realises that she only
requires 6 hours of lessons from a driving school.
Identify the option she should now choose. Explain your answer.
1.6 Calculate the difference in cost for a learner using OPTION A and
another learner using OPTION B if they both require 30 hours of lessons.
(3)
(5)
[28]
33
© Gauteng Department of Education
QUESTION 2: 22 minutes (from NSC Feb/Mar 2013, paper 1)
Nandi is considering changing her hairstyle and visits a local hair salon to
determine the cost of styling her hair. She has a choice between hair extensions or
hair relaxing.
The pictures below compare relaxed hair and hair extensions.
Original hair
Original hair
Hair extensions
Relaxed hair
The cost of the two choices is shown below.
COST OF HAIR RELAXING
COST OF HAIR EXTENSIONS
R140,00 per treatment, including
moisturising gel and one hair wash
R500,00, including one hair wash
Weekly hair wash at R40,00,
including moisturising gel
Weekly hair wash at R40,00
Treatment must be repeated every
four weeks or monthly.
Extensions last for 6 months or 24
weeks.
2.1
Calculate the cost of hair relaxing for the first four weeks.
Use the formula:
Cost for the first four weeks (in rand) = 40 + (3 × cost of a hair wash)
(2)
2.2
Calculate the cost of hair extensions for the first four weeks.
Use the formula:
Cost for the first four weeks (in rand) = 500 + (3 × cost of a hair wash)
(2)
34
© Gauteng Department of Education
2.3
Nandi wants to convince her father that in the long run, the cost of hair
extensions will be cheaper than the cost of hair relaxing.
The accumulated cost for each choice over a 37-week period is given in
the table below.
TABLE 3: Comparison of accumulated costs after the first week of
each month
Time period (in
weeks)
1
5
B
Accumulated
cost of hair
relaxing (in
rand)
140
A
920
1 440 1 700 1 960 2 480
Accumulated
cost of hair
extensions (in
rand)
500
660
980
1 300 1 920 2 080 2 400
a)
b)
c)
2.4
21
25
29
37
Calculate the missing values A and B.
(4)
Which hairstyle will be cheaper over the first 21 weeks? (2)
Calculate how much more Nandi will pay over a 37-week
period for relaxing her hair compared to wearing hair
extensions.
(2)
The graph showing the cost of hair relaxing over a period of 9 months is
given on ANNEXURE A. Draw a labelled line graph of the cost of hair
extensions over a period of 37 weeks on ANNEXURE A.
(6)
[20]
35
© Gauteng Department of Education
ANNEXURE A
QUESTION 2.3(d)
COMPARISON OF ACCUMULATED COSTS
Co
st
in
ra
n
d
Number of weeks
36
© Gauteng Department of Education
SECTION B:
CONTENT NOTES
The steps for doing these questions are the same as for break-even point questions
(1)
(2)
(3)
(4)
(5)
Summarise the situation
Formulate the equations
Complete the table
Draw graphs using the table
Analyse the situation by using the graphs
The points of intersection divide the area of the grid into REGIONS – on the grid
these are bounded by the points of intersection.
In these REGIONS different advantages and disadvantages for the graphs apply.
To describe the regions you need to revise inequality notation and terminology:
less than
(always used for the first region)
more than
(always used for the last region)
(always used for the other regions)
Graphs in this section of the work do not always have to by straight lines, it can also
be a curve.
You need to know how to write down the coordinates of the points of intersection.
Eg. (the x-value; the y-value).
So when interpretation is asked, you first give the interval (eg: less than 25 days),
and then state the best option (eg Contact A is the cheapest).
37
© Gauteng Department of Education
SECTION C: HOME WORK
QUESTION 1: 20minutes
(from NSC Feb/Mar 2012, paper 1)
AA High School is considering renting a photocopier. They approach two companies
(Company A and Company B) and obtain the following quotations:
Company A:
Rental of R800,00 per month, which includes 3 000 free copies per month.
Thereafter a charge of 5 cents per copy applies.
Company B:
Rental of R600,00 per month, which includes 2 500 free copies per month.
Thereafter a charge of 10 cents per copy applies.
TABLE 3: Monthly cost (in rand) of renting a photocopier
Number of
0
2 000 2 500 3 000 4 000 6 000
copies made
7 000
8 000
Company A
800
800
800
800
Q
950
1 000
1 050
Company B
600
600
P
650
750
950
1 050
1 150
1.1
Determine the missing values P and Q.
(4)
1.2
Write down a formula that can be used to calculate the total cost per
month of renting a photocopier from Company B.
(3)
1.3
The line graph illustrating the total rental cost for Company B has been
drawn on ANNEXURE B. On the same system of axes, draw a line graph to
illustrate the total rental cost for Company A.
(4)
1.4
Determine the number of photocopies made if the total rental cost for both
companies is the same.
(2)
1.5
AA High School makes an average of 7 000 photocopies per month.
Calculate how much the school will save by choosing the cheaper rental
option and identify the company that charges the lower total rental cost.
(3)
[16]
38
© Gauteng Department of Education
1.3 ANNEXTURE B
1 200
Company B
1 100
1 000
900
800
Co
st
in
ra
nd
700
600
500
400
300
200
100
0
0
QUESTION 2:
1 000
2 000
3 000
4 000
5 000
6 000
7 000
17minutes (from NSC Nov 2011, paper 2)
Timothy is a newly qualified marketing graduate. He has been offered two
positions, one as a medical sales representative for Meds SA and the other
as a tobacco sales representative for ABC Cigs.
The formula for calculating the monthly salary for a medical sales
representative is:
Salary = R3 000 + R500
×
number of days worked.
As a tobacco sales representative, he will earn a salary of R750 per day for
each day worked in a month. He will only receive a salary if he works for one
or more days in a month.
39
© Gauteng Department of Education
8 000
2.1
Write down a formula that can be used to calculate the monthly salary of
a tobacco sales representative.
(2)
2.2
Draw TWO line graphs on the same grid on ANNEXURE A to represent
the monthly salaries for both the positions of medical and tobacco sales
representatives. Clearly label each graph.
(8)
2.3
Use the graphs drawn on ANNEXURE A, or otherwise, to answer the
following.
(a) After how many working days will the two salaries be the same? (2)
(b) Suppose Timothy worked at Meds SA for 18 days. How many days
would he have to work at ABC Cigs to earn the same salary?
(2)
[14]
2.2
ANNEXTURE A
SALARIES FOR THE TWO POSITIONS
16
000
14
000
12
A 000
m
ou 10
nt 000
ea
8 000
rn
ed
in 6 000
ra
nd
4 000
2 000
0
0
5
1
Number0of days
worked
1
5
2
0
40
© Gauteng Department of Education
SECTION D: EXAM QUESTIONS SOLUTIONS
QUESTION 1
1.1
No change in the cost after 15 hours. 
2 correct description
OR
Constant cost from 15 hours onwards. 
OR
For 15 hours or more of driving lessons there is a fixed rate of R1
500. 
1.2
(a)
No payment for zero lessons. 
(2)
2 correct description
OR
Payment will only be made once the driving lessons start. 
(2)
1.2
(b)

• A learner driver pays a basic amount of R600 for
the first two hours 
• Then R50 per hour for every additional hour. 

1 R600
1 time period
1 rate in rand
(3)

1.3
At point Q, both Options cost the same at the same time.
1 same cost
1 same time
OR
OR


There were 10 hours of driving that cost R1 000 for both
Options.
1 time
1 cost
Accept " breakeven
point " ONLY 1 mark
(2)
41
© Gauteng Department of Education
1.4
(a)


With Option B Zaheera will get 14 hours of driving lessons.
1 correct option
1 justification
OR


Zaheera must choose Option B to get 2 more hours of driving
lessons than in Option A.
1.4
(b)

Toni would benefit more from Option A. She still gets
R1 200 but in a shorter time than Option B 
(2)
1 correct option
1 justification
OR


Option A, she will have 2 hours to train someone else.
(2)

1.5

Option A is cheaper for Zaheera.
1 correct option
2 justification
OR


She must choose Option A she will pay R600 for the driving
lessons.
(3)
42
© Gauteng Department of Education
1.6
Option A:

1 cost option A
Cost for 30 hours = R1 500
Option B:


Cost for 30 hours = R600 + (R50 per hour × 28 hours)
= R600 + R1 400
= R2 000 
∴ Difference in cost = R2 000 – R1 500
= R500 
1 difference in
cost
OR
1 cost option A
OR
Option A:
1 basic rate
1 rate multiplied by hours
1 cost

Cost for 30 hours = R1 500
Option B:
Cost for 30 hours


= R600 + (R100 per two hours × 14 two hour periods)
= R600 + R1 400
= R2 000 
1 basic rate
1 rate multiplied by
period
1 cost
1 difference in cost
OR
∴ Difference in cost = R2 000 – R1 500
= R500 
1 rate
1 extra cost
OR
Option B:
For 22 hours it costs R1 600

It is increasing with R100 every 2 hours
∴ Extra cost = 4 × R100 = R400

Cost for 30 hours = R1 600 + R400
= R2 000

Option A:
1 cost
1 cost option A
1 difference in cost

Cost for 30 hours = R1 500
∴ Difference in cost = R2 000 – R1 500
= R500 
43
© Gauteng Department of Education
QUESTION 2 – SESSION 3
2.1
Cost for the first four weeks (in rand) = 140 + (3 × 40) 
= 260
2.2
1 simplification
(2)

Cost for the first four weeks (in rand) = 500 + (3 × 40) 
A = R140 + R260

= R400

920 = 400 + B × 40
520 = B × 40

A = R400 
OR
500 + 40 × (B – 1) = 980 
OR
B = 13 
(2)
1 substitution
1 value of A
OR
2 reading from
graph
1 substitution
1 value of B
40 (B – 1) = 480
B-1
B = 13
B=13
1 substitution
1 simplification
= 620 
2.3(a)
1 substitution
=12

OR
OR
2 reading B from
graph
OR
1 list of values
OR
1 value of B
(4)
140; 400; 660; 920; 1 180; 1 440; 1 700  So, B
=1+3×4
= 13 
2.3(b) Hair extensions 
2 conclusion
(2)
3.3(c) R2 480 – R2 400 
1 correct values
1 simplification
= R80 
44
© Gauteng Department of Education
(2)
3.3(d)
COMPARISON OF ACCUMULATED COSTS
1 (1 ; 500)
1 (25 ; 1 920)
1 (29 ; 2 080)
1 (37 ; 2 480)
1 joining the
points
1 labelling the
graph
(6)
45
© Gauteng Department of Education
SESSION 4
TOPIC: MEASUREMENT: CONVERSIONS USE OF FORMULAE
TEACHER NOTES:
 In Mathematical Literacy most of the calculations are based on direct
proportion (when you buy MORE apples, you will pay MORE money)
 Conversion questions can sometimes be very short BUT remember
when
the question is worth more than ONE mark, learners are required to
show STEPS.
 Although some of the memoranda attached to your teacher notes state
learners can have all the marks for the ANSWER ONLY, this changed
with CAPS and there will no longer be all the marks given for “answer
only”
 For any conversions in which the metric system is used, learners must
know the relationships from memory. Make sure the learners know
these well.
 When doing conversion between different systems the conversion
factors or tables will be given.
 Conversions are a basic skill and is used in finances, data,
measurement and map work.






When using formulae there are some definite rules.
A formula has a left hand side, an equal sign and a right hand side.
The complete formula must be written down, and NOT only the right
hand side of the equation.
The substitution that has to be done, MUST be shown, the best is to
always write them in brackets.
Using CAPS a lot of formulae such as area and volume will not be
given any more and you will have to learn them by heart.
Learners need to get a lot of practice doing conversions and using
formulae.
LESSON OVERVIEW:
1. Introduction session
2. Typical exam questions
3. Discussion of solutions
5 minutes
55 minutes
30 minutes
46
© Gauteng Department of Education
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
36 minutes (from NSC 2013 papers)
1.1 10 mℓ of sugar weighs 8 g. Calculate the weight of 245 mℓ of sugar.
(2)
1.2 A few weeks after the swimming competition, an Australian tourist who had
been a spectator at the competition deposited 1 500 Australian dollars (AUD$)
into the club's bank account as a donation. The bank converted this amount
to rand as R14 595,00. Calculate the exchange rate, in rand per AUD$, used
by the bank.
(2)
1.3 The scale used on the layout plan is 1 : 58.
Calculate the actual length of the table on the layout plan if its scaled
length is 2,26 cm.
(2)
1.4 Approximately 2,5 kg of oranges are used to make 1 ℓ of juice. The juice is
poured into 5 ℓ plastic bottles. Determine the number of 5 ℓ bottles of juice that
can be made from 400 kg of oranges.
(3)
1.5
Convert 450 metres to kilometres.
(2)
1.6 Write 5,34 million as an ordinary number.
(1)
1.7 Calculate the price per egg if half a dozen eggs cost R7,92.
1.8 Convert 18 gallons to litres where 1 gallon = 4,546 litres.
(2)
(2)
1.9 Determine the cost of 15,76 litres of fuel if fuel costs R9,92 per litre.
1.10 Convert 1 260 seconds to hours.
(2)
(2)
1.11 Determine the price per gram (rounded off to the nearest cent) if
200 g of peanuts cost R9,96.
(2)
1.12 Convert
cup to millilitres. 1cup=250ml
1.13 Convert 5 ounces to grams. 16 ounces
(2)
480g
(2)
1.14 Thabo bought goods in Ghana to the value of 1 345 cedi.
1 cedi = 4.41 ZAR. Calculate the value, in rand, of the goods Thabo
had bought.
(2)
1.15 Determine the total length, in miles, of the South African coastline if the
coastline of the Eastern Cape is approximately 500 miles long. The East
Coast measures 800km and the South African coastline is 2798km. (3)
1.16 Annie measured the length of the coastline of South Africa on her map
47
© Gauteng Department of Education
and found it to be 223 mm long. The coastline is 2798km in real life.
Determine the scale of the map in the form 1 : …
Round off the answer to the nearest hundred thousand.
(4)
[35]
QUESTION 2:
30 minutes (from NSC 2013, papers)
2.1
The time (in seconds) taken by a moving object to cover a distance of
50 m is given by:
Time (in seconds) =
Where:
s = average speed in metres per second
d = distance in metres
Calculate the time taken if the object is moving at an average speed of
metres per second.
(2)
2.2
The total value of the demo toys that children can play with is currently
R15 000 and the depreciation rate is 17,5% per annum. Thandeka
uses the straight-line depreciation method to determine the value of the
demo toys.
Calculate the depreciated value of the demo toys at the end of 4 years.
Use the formula A = P(1 – i × n)
where A = the depreciated value
i = the annual depreciation rate
P = the present value
n = the number of years
(3)
2.3 The inside measurements of the walls of the pool are as follows:
Length = 50 m, breadth = 25 m and height = 1,5 m
The inside walls and the floor of the pool need to be repainted. Determine the
total area of the pool that will be repainted.
Use the following formula:
Area to be repainted = ℓ × b + 2h(ℓ + b)
where ℓ = length
b = breadth
h = height
(3)
2.4 Calculate the height of the water in the pool if the volume of water in the
pool is 1 500 m3.
Use the following formula:
48
© Gauteng Department of Education
Height of a rectangular prism =
2.5
(2)
The temperature of the water in the pool needs to be maintained at
22 oC. The temperature gauge used shows the temperature in
degrees Fahrenheit (oF).
Convert (rounded off to the nearest degree) 22 oC to degrees Fahrenheit.
Use the following formula:
Temperature (in oF) = 32 + 1,8 × (Temperature in oC)
Determine the following bearing in mind that the average diameter of an
orange is 90mm.
Surface area ( in mm2 ) of an orange
2.6.1
2.6.2
2.7
(3)
Volume ( in mm3 ) of an orange.
(2)
The following formulae may be used:
Surface area of a sphere = 4 × × r2
Volume of a sphere = × × r3
where = 3,14 and r = radius
(3)
The cylindrical section of a basket has a height of 25 cm and a diameter
of 30 cm. The space in the cylindrical basket not occupied by the oranges
is 113 040 mm3.
Franz states that a basket can hold at most 44 oranges.
Verify, by showing ALL the necessary calculations, whether Franz's
statement is correct.
The following formula may be used:
Volume of a cylinder = × r2 × h
The average diameter of an orange is 90mm
where = 3,14, r = radius and h = height
(7)
[28]
SECTION B: CONTENT NOTES
1. CONVERSIONS
Try this method and you are guaranteed to always have conversions correct
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EXAMPLE
Peter mixes cement. The mixture needs 13,1kg sand and 5,4kg cement. If he has
8kg of cement, how much sand must he put into the mixture?
METHOD
WHAT TO DO
1. Write down the key to 13,1kg sand 5,4 kg cement
your conversion
2. Write down the other
number BELOW the
same name as in the
1st line
3. Take the last number
you wrote down 8,
multiply it with the
number above-andacross from it 13,1
AND now divide by
the number next to
13,1 which is 5,4
4. What you find now is
that you have an
answer BUT it is on
the WRONG side of
your
conversion.
Therefor just write an
= below the arrow
and write the answer
down again (the
marker must be able
to look at the last
number to SEE that
your answer is
correct
13,1kg sand
5,4 kg cement
8kg cement
13,1kg sand
5,4 kg cement
8kg cement
13,1kg sand
19,41kg sand
5,4 kg cement
8kg cement
19,41 kg sand
Remember the new CAPS syllabus does not supply formulae with questions as in
the past. Key to conversions of Imperial systems will be given, BUT al the keys that
has metric connotations must be known by memory. Also often used formula such
as area and volume must be learned.
2. USING FORMULAE
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



Always always write down the complete formula.
Even when a formula is given, the whole formula must be rewritten in the test
or exam.
When doing the substitution, do use brackets.
Formulae are usually for the calculation of something with a UNIT. Do write
down the units after the answer. It does not always count a mark, but
sometimes it does, and you must not throw valuable marks away
SECTION C: HOME WORK
QUESTION 1: 26 minutes (from NSC 2012 papers)
1.1 Convert 23,005 litres to millilitres. (2)
1.2 Determine the total price of 2,5 kilograms of meat costing R63,99 per
kilogram. (2)
1.3 Convert R3 850 to euros (€) if the exchange rate is €1 = R10,2584. (2)
1.4 Calculate the petrol consumption (in litres per 100 km) if Mr De Haan
covered a distance of 325 km using 12,5liters of petrol. (2)
1.5 The distance measured on the map from Mr De Haan's house to the
entrance of the Bayview Hospital is 8,9 cm.
Calculate the actual distance (in km) if 1 cm on the map represents 0,3 km.
1.6 Convert the 6 inches height of the box in centimeters if 1 inch = 2,5 cm.
(2)
(2)
1.7 The length of the Jetstream in the picture is 9,9 cm, while its actual length is
19,25 m.
Determine the scale (rounded off to the nearest 10) of the picture in the
form 1: …
(4)
1.8 A nautical mile is a unit of measurement based on the circumference of the
earth.
1 nautical mile = 1,1507 miles
= 6 076 feet
= 1,852 kilometres
Calculate the maximum operating altitude (to the nearest nautical mile)
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of the Jetstream, that has a maximum operating altitude of 25 000 feet.
(3)
QUESTION 2:
2.1
30 minutes (from NSC 2012, papers)
A spin wheel is divided in 24 equal parts The wheel has a radius of 60 cm.
(a) Calculate the circumference of the wheel.
Use the formula:
Circumference of a circle = 2 × × radius, using π = 3,14
(2)
(b) Calculate the area of ONE of the sectors of the wheel.
Use the formula:
where π = 3,14 and n = number of sectors
2.2
(3)
Kedibone has a cheque account with Iziko Bank. The bank charges a service fee
up to a maximum of R31,50 (VAT included) on all transaction amounts.
TABLE 2 below shows five different transactions on Kedibone's cheque account.
NO. DESCRIPTION OF
TRANSACTION
1
2
3
4
5
TRANSACTION
AMOUNT
(IN R)
4 250,00
344,50
Debit order for car repayment
Debit order for cell phone
contract
Personal loan repayment
924,00
Vehicle and household insurance B
Cheque payment
403,46
SERVICE FEE
(IN R)
31,50
A
14,59
11,85
8,34
2.2.1 Calculate the missing value A, using the following formula:
Service fee (in rand) = 3,50 + 1,20% of the transaction amount
(3)
(3)
2.2.2
Calculate the missing value B, using the following formula:
Amount (in rand) = service fee − 3,50
1,20%
(3)
(3)
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2.3
Jabu Ndou requires a cylindrical water tank to collect rainwater from his roof.
This water will be used for irrigating his garden.
Outer height
Cylindrical rainwater tank
2.3.1 Jabu wants to know how much rainwater the tank can hold. The inner radius of
the tank is 0,998 m and the inner height of the tank is 2,498 m.
(a) Calculate the total volume, rounded off to THREE decimal places, of the water tank.
Use the formula:
Volume of a cylinder = × (radius)2 × height,
and using = 3,14
(3)
(b)
2.3.2
(3)
Determine the height, rounded off to THREE decimal places, of the water in
the tank when it is 80% full.
(2)
The outside walls and roof of the rainwater tank need to be painted. The
outer radius of the tank is 1 m and the outer height of the tank is 2,5 m.
Calculate the surface area of the tank that will be painted using the
formula:
Surface area of the tank = × radius × (2× height + radius),
and using
(5)
(5)
2.3.3 Suppose the tank filled up at an average rate of 5 mm per minute.
Calculate how long it took (in hours) for the water in the tank to reach a height
of 1 200 mm, if the tank was initially empty.
Use the formula:
Time (in hours) =
(3)
[24]
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SECTION D: EXAM QUESTIONS SOLUTIONS
QUESTION 1
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1.1
1 mℓ of sugar weighs 0,8 g 
245 mℓ of sugar weighs (0,8 × 245) g
= 196 g 
OR
1 multiplication with
correct values
1 mass of sugar

OR
245 ×
= 196 g 
1 multiplying
OR
by
1 mass of sugar
10 : 8 = 245 : x
=g
x
OR
= 196 g 
1 proportion
1 mass of sugar
(2)
1.2
R14595,00
Exchange rate =

AUD$1500,00

= R9,73/AUD$ OR R9,73 per AUD$
1 division with
correct values
1 simplification
If R1 = 0,102 AUD$, max 1
mark
Answer only: full marks
(2)
1.3
1 multiplying correct values
Actual length = 2,26 cm × 58 

= 131,08 cm OR 1,31 m OR 1310,8 mm 1 actual length
If 1,31 or 1310,8 (without
unit) max 1 mark
(2)
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1.4
Amount of juice (in litres)

400 kg
OR
=
2,5 kg

2,5 kg makes 1 
1 dividing by 2,5
400kg 
400 kg makes
1 simplification
2,5kg/
OR
= 160
= 160  
Number of 5 bottles
160
=
5
= 32

Number of 5 bottles
160
=
5
= 32 
1 using proportion
OR
1 simplification
1 : 2,5 = x : 400
2,5x = 400
x=
x = 160

1 simplification

OR
160
Number of 5 bottles =
5
= 32
1 mass of fruit

1 dividing by 12,5
OR
5  juice is made from 5 × 2,5 kg = 12,5 kg fruit
400kg
∴ Number of 5 bottles =

12,5kg
= 32


1 simplification
OR
1 using proportion
1 dividing by 2,5
1 simplification
OR

400 kg
= 80 kg /
5

80 kg/
Number of 5 bottles =
= 32

2,5 kg/
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1.5
450 m = (450 ÷ 1 000) km

1 answer
= 0,45 km
1.6
(1)
5,34 million = 5,34 × 1 000 000
1 solution
(1)
= 5 340 000 
1.7
1 dividing by 6
R7,92

Price per egg =
6
= R1,32
1 simplification

Answer only full marks
(2)
1.8
18 gallons = 18 × 4,546 litres 
1 multiplying by
= 81,83 litres 
1.9
conversion factor
1 solution
(2)
1 correct formula
1 substitution

Percentage decrease =
×100% 
=
8,66935.. %
≈ 8,67 % 
1 solution
(3)
1.10
1 260 seconds
=
= hours
hours
1 dividing by 3 600

OR 0,35 hours

Answer only – FULL
MARKS
1 simplification
(2)
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1.11
1 dividing by 200 g
Price per gram = R9,96 
200
= R0,0498
≈ R0,05 OR 5c 
Answer only – FULL
MARKS
1 simplification
(2)
1.12
cup = × 250 mℓ

1 multiplying
= 187,5 mℓ 
Answer only – FULL
MARKS
1 simplification
(2)
1.13
1 converting
1 ounce = 480 g = 30 g

16
∴5 ounces = 5 × 30 g
= 150 g 
1.14
1 simplification
1 345 cedi = 1 345 × R4,41000 
= R5 931,45 
(2)
1 multiplying by
correct rate
1 simplification
Answer only – FULL
MARKS
(2)
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1.15
2 798km:
800 km = 500 miles 
1 equating distances
= miles 
1 correct conversion
= 1 748,75 miles 
1 simplification OR
OR
l = the length of South African coastline
2798 =
l
1 concept
 =800 km = 500 Miles
1 manipulation

1 simplification

OR
l = 1 748,75 miles 
OR
1 concept
1 conversion
1 simplification
800 km = 500 miles
Answer only full marks
So 1 km
= miles 
∴ 2 798 km = × 2 798 miles

= 1 748,75 miles 
1.16
223 mm on the map represents 2 798 km 
223 mm on the map
represents 2 798 000 000
1 mm on the map
represents
= 12 547 085,2 mm 
Scale is 1: 12 500 000 
(
3
)
(3)
mm 
1 correct conversion values
1 conversion
1 simplification
1 rounding
(4)
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QUESTION 2

2.1

Time (in seconds)
OR 6
1 substitution
1 solution
=
OR 6,25
2.2
A = P( 1 – i × n)

= R15 000(1 – 0,175 × 4)
17,5
R15 000 1−
×4 
100
= R4 500 

OR
= R4 500 
(2)
1 substituting any
2 values correctly
1 substituting the
3rd value correctly
1 value
Answer only: full marks
(3)
2.3
Area to be repainted = × b + 2h(+ b)  
= [50 × 25 + 2 × 1,5(50 + 25)] m
= [1 250 + 3(75)] m2
= 1 475 m2 
1 substitution and b
1 substitution h
1 area
Incorrect use of
BODMAS, no CA
Answer only: full marks
(3)
2.4
1 substitution
1500m3
Height of a rectangular prism

=
50cm×25cm
1 height
Answer only: full marks
= 1,2 m

(2)
2.5
Temperature in ° F = 32 + 1,8 × (Temperature in ° C)
= 32 + 1,8 × (22) 
= 71,6 
1 substitution
1 simplification
1 rounding
(3)
≈ 72 
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2.6.1
Radius (in mm) =
= 45
1 value of radius

Surface area (in mm ) = 4 × 3,14 × 452
2
= 25 434


1 substitution
1 simplification
Accept
25
446,90 using π
Using diameter max
2 marks
NPR
Correct answer
only:
full marks
(3)
from 1.2.1
2.6.2
Volume (in mm3) = × 3,14 × 45 3 
= 381 510

1 substitution
1 simplification
Accept 381 703,51
using
NPR
π
Correct answer
only:
full marks
(2)
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2.7
Radius of basket =
1 radius of basket
= 15 cm 
1 substitution

Volume of basket = 3,14 × (15 cm)2 × 25 cm
= 3,14 × (150 mm)2 × 250 mm 
= 17 662 500 mm3 
1 converting to mm
1 volume of basket
Accept 17 671 458,68
using π
1 subtracting space

17662500mm −113040mm
3
3
The number of oranges =
3
381510mm

= 46
1 dividing by volume
of an orange
from 1.2.2
1 conclusion
∴ Franz’s statement is not correct

OR
OR
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OR
1 value of radius
Radius of basket =

= 15 cm

1 substitution 1 volume of
basket
Accept 17 671,46 using π
2
Volume of basket = 3,14 × (15 cm) × 25 cm
= 17 662,5 cm3 
1 dividing by volume of an
orange
17662,5cm −113040mm
The number of oranges =
381,51cm3 
1 subtracting space
381510mm
1 converting to cm
17662,5cm3 −113,040cm3 
=
3
1 conclusion

381,51cm
= 46
OR
(46 > 44)
Franz’s statement is not correct
1 radius of basket

1 substitution
OR
1 converting to mm
1 volume of
basket

Radius of basket =
= 15 cm
Volume of basket = 3,14 × (15 cm)2 × 25 cm
= 3,14 × (150 mm)2
= 17 662 500 mm3 

×
250
mm

Space in the basket for oranges (in mm3)
= 17 662 500 – 113 040 = 17 549 460

Space occupied by oranges (in mm3)
2
= 381 510 mm × 44 = 16 786 440 mm

2
1 subtracting space
1 calculating the space occupied
by the oranges
1 conclusion
Correct conclusion only: 1 mark
(7)
( there is space for more oranges)
Franz’s statement is not correct

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