# Mathematics Grade 11 KZN September Test 2024 Paper & Memo

> Grade 11 Mathematics question paper and memo from the KZN September Test 2024, 19 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-11-mathematics-2024-kzn-september-test-question-paper-and-memorandum/8e87ece0-80a2-41a4-9a81-aabe6300fefd
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2024
- Exam period: KZN September Test
- Curriculum: CAPS
- Pages: 19

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## About this document

Mathematics · Grade 11 · KZN September Test · 2024. Question paper and memorandum, 19 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

‘anmorephysics.com

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFIC

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a
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Sseseensene?®

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cy
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o
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feseesesseses

MARKS: 75

TIME: 1% hours

This question paper consists of 8 pages and
2 diagram sheets.

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### Page 2

madoaKenl oaded from Stanmor ephy sics.com Septomber 2024 Common Teat

NSC — Grade 11

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions,

10.

This question paper consists of 6 questions.
Answer ALL the questions.

Number the answers correctly according to the numbering system used in this
question paper.

Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
determining your answers.

Answers only will NOT necessarily be awarded full marks.

You may use an approved scientific calculator (non-programmable and
non-graphical), unless stated otherwise.

If necessary, round off answers correct to TWO decimal places, unless stated
otherwise.

Diagrams are NOT necessarily drawn to scale.

TWO DIAGRAM SHEETS for QUESTION 2.1, QUESTION 2.2, QUESTION 6.1,
and QUESTION 6.2 are attached at the end of this question paper, Detach the
DIAGRAM SHEETS and hand it in together with your ANSWER BOOK.

Write neatly and legibly.

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### Page 3

M tics
“Bownloaded from Stanmoxepliyaics.com Sepsonmiber 2024 Conamen Test

QUESTION 1

The table below shows the marks, as a percentage, scored by 10 learners in a Maths Olympiad.

MARKS OBTAINED
(AS A PERCENTAGE) 65 | 59 | 79 | 86 | 48 | 94 | 77 | 51 | 66 | 80
1.1 Calculate the mean mark of the data. (2)
1.2 Write down the standard deviation for the above data. (1)
1.3 Calculate the number of marks that lie within one standard deviation of the mean. (2)
1.4 Two learners were omitted from the original set of data. Their scores were x and 58.
When the scores of these two leamers are included with the original set of data, then
the new mean is now 67,5. Calculate the value of x. (2)
[7]
QUESTION 2
The ages of people who attended a musical concert was recorded. The information was
summarised in the frequency table below.
2.1. Complete the cumulative frequency column in the table provided on DIAGRAM
SHEET 1. (2)
2.2 Use the grid provided on DIAGRAM SHEET | to draw an ogive (cumulative
frequency graph) to represent the data. Q)
2.3. Write down the modal class of the data. (1)
2.4 Youth are regarded as people below the age of 35. What percentage of youth
attended the concert? (2)

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[8]

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### Page 4

MiQeuwaloaded from Stanmorephysics.com September 2024 Common Test

NSC - Grade 11

QUESTION 3

3.1

3.2

A and B are two events such that:

© P (mot A) = 0,55
* P®)=0,4
© P(A or B) = 0,67

Are events A and B independent events? Justify your answer with relevant
calculations. (5)

A survey was done among a group of 100 tourists to find out which city in South
Africa they loved during their visit. They chose from Johannesburg (J), Durban (D)
and Cape Town (C). The results of the survey are listed below:

17 said they loved all three cities

30 said they loved Durban and Cape Town

27 said they loved Johannesburg and Cape Town
52 said they loved Durban

9 said they loved only Johannesburg and Durban
12 said they loved Johannesburg only

9 said they loved Cape Town only

The above information is represented in the partially completed Venn Diagram below:

Ss

3.2.1 Write down the values of a, 5, e and f. . (4)

3.2.2 Calculate the probability that a tourist selected at random loved Cape Town or
both Johannesburg and Durban. (3)

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### Page 5

MahemQwnloaded from Stanmgrephysics.com

3.3

Nelisiwe had a small box of 80 Smarties sweets.
40% of the Smarties are green, = of them are blue and the remainder are red.

Nelisiwe picks a Smartie out of the box, takes note of the colour and then eats it.
She then picks out a second Smartie and notes the colour before eating it.

3.3.1 How many red Smarties were in the box at the beginning?

3.3.2 Represent the above situation by means of a tree diagram. Indicate the
probabilities associated with each branch and the possible outcomes.

3.3.3. Calculate the probability that Nelisiwe will pick two Smarties of the same
colour.

September 2024 Common Test

(2)

(3)

G)
[20]

### Page 6

Pownloaded from Stanmorephysics.com September 2024 Common Test
NSC — Grade 1]

QUESTION 4
ank was 5591 at the end of 2014. At the end of 2022, the

4.1. The population of fish in a t A
average annual rate at which the

population of fish grew to 6423. Calculate the
population of fish grew over these 8 years. GB)
4,2- ~ Convert the nominal interest rate of 9% p.a. compounded half-yearly to an effective -

interest rate.

business that makes built-in cupboards. Today, he bought a board-

4.3. MrKheswa owns a
chine is expected to last for 6 years.

cutting machine for R250 000. The ma
lepreciates

4.3.1 Calculate the scrap value of the machine in 6 years’ time if it d
annually at 12% p.a. on a reducing-balance method. (2)

increases at the rate of 11% p.a.

43.2 The cost of a new similar machine .
he cost price of a new board-cutting

compounded quarterly. Calculate t!
machine in 6 years’ time.
in the old machine and replace it with a new similar

4.3.3. Mr Kheswa will trade-
machine in 6 years’ time. Calculate how much more will Mr Kheswa have to
a)

pay if he bought the new machine for cash.

G)

Mr Kheswa opened an investment account that pays interest at the rate of

12,25% p.a., compounded monthly. How much should he invest in this
account today so that it will accumulate to the cash amount required to buy the

new machine in 6 years’ time?

43.4
(3)

Exactly seven years ago, Krish deposited R25 000 in a bank account. The bank paid
interest at the rate of 8% p.a., compounded monthly for the first 4 years and then
changed to 12% p.a., compounded quarterly for the next 3 years. Krish withdrew

R10 000 three years after making the initial deposit. This is illustrated in the time-line

below.

44

8% p.a. compounded monthly 12% p.a. compounded quarterly

IN A
|

| jl -R 10,000
, T
0 il ) 1

ol ee

2 K

6)

Calculate how much Krish has in the account today.
[20]

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### Page 7

Mporttoaded from Stanmorephysics. com September 2024 Common Test

QUESTION 5

The sketch below shows the graphs of f(x)=sin(x+45°) and g(x) =coshx+g for x €[-90°;180°]..

5.1 Write down the values of 6 and g. (2)
$2 Write down the period of g. (1)
53 Write down the equation of A, if h(x) = g(x)+1. (1)
5.4 Calculate the values of x for which f(x)—A(x) =0 if 0°< x <180°. (5)
5.5 Use the sketch to write down the value(s) of x for which ; S(x)x4g(x)==2. Q)
{1]

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### Page 8

° tanMorephySics.COM  serrerber 2024 Common Test
Monimloaded from S ore ply: “

QUESTION 6

6.1 Inthe diagram below, AABC is drawn with C at the origin.

Use the diagram on DIAGRAM SHEET 2 to Prove that c* =a? +5? —2abcosC. (4)

6.2 Inthe diagram below,

SABC is right-angled at B, ACLCD. AC=15V5 , CD=10
and BCA =26,57°,

Calculate the length of BD, (5)

19]
TOTAL MARKS: 75

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### Page 9

TEAR OFF

iMDewrHoaded from Stanmorephysics.com eee ea

NAME & SURNAME:

DIAGRAM SHEET 1

QUESTION 2.1

Age group (x) Number of people yell
18<x<22 2
22<x<26 6
26<x<30 12
30<x<34 24
34<x<38 36
38<x<42 24
42<x<46 16
46<x<50 10

QUESTION 2.2
OGIVE

Cumulative frequency

ese rh aSeeeestrevnesrexraesress

4 18 2 26 30 M 38 a2 46 50 54
Age groups

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### Page 10

meguagloaded from Stanmorephysics.cCOM september 2024 Common Test
NSC - Grade 11

NAME & SURNAME:

DIAGRAM SHEET 2

QUESTION 6.1

QUESTION 6.2

A

15/5 D

26,57°

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### Page 11

Downloaded from Stanmorephysics.com

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 75

This marking guideline consists of 9 pages.

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### Page 12

Mathenati syn I loa di le d from Sta nae replys GSin fa om September 2024 Common Test

QUESTION 1
11 || 65+59+ 79+86+48+94+77+51+66+80
10
_ 105 YA 705
10
=I 5 ¥ CA answer
Answer only 2/2
(2)
¥ A 14,38
(1)
¥CA interval
dard deviation of
¥ CA answer
(2)
1.4 705+X+58 _ 67 5 VCA (sum of 10 marks) +X+58 _ 67 5
12 12
xX=47 ¥CAx=47
(2)
[7]

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“aeGwnloaded from Stanperenhysiasneom

September 2024 Common Test

QUESTION 2
2.1
Age group (x) | Number of people | Cumulative frequency
18<x<22 2 2
22<x<26 6 8 YA 8
26<x<30 12 20
30<x<34 24 44
34<x<38 36 80
38<x<42 24 104 ae ae
42<x<46 16 120
46<x<50 10 130 (2)
2.2
Ogive
132
i
120
116
108
104
100
g || &
S| 7 V .
“|| 2 A grounding
e 3 at (18 ; 0)
S 60
3 Fd YA minimum of
B ‘a 4 points
O| 4 correct
2
28 ¥CA smooth
20 increasing
16 curve
8
4
0
14
(3)
2.3 | 34<x<38 YA 34<x<38
(1)
= v
2.4 | No. of youth (under 35 years) = 52 Accept 50 to 54 CA 52
52
Percentage = eee Accept 38,46% to 41,54%
v CA answer
=~ 40% (2)
[8]

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“aeBGwnloaded from Stanperenhysiasneom

QUESTION 3

September 2024 Common Test

3.1

P(A) =1—P(not A)
=1-0,55

= 0,45

.. Aand B are independent since P(A and B) = P(A)xP(B)

YA 0,45

¥CA substitution
¥CA 0,18

YCA P(A)xP(B)

¥CA conclusion

(5)
3.2.1 | a=10 YA a=10
b=13 YCA b=13
e=13 YCA e=13
fei VCA f=17
(4)
3.2.2 10+17+13+9 9 49 9
P|C Jand D) | =———_——— + — ¥CA — YCA —
[Cor (and D) 100 100 100 100
58
=73, oF 0,58 ¥CA answer
100
(3)
mae Number of green sweets = 80x 2 32
100
Number of blue sweets = x80 =12 AY 32and 12
Number of red sweets = 80— 44 = 36 YCA 36
(2)

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wateBSwnloaded from Stanmerenhysigs.com

September 2024 Common Test

3.3.2

Ww OUTCOMES
77-B BB
32
B » ¢& BG
2
80 BR
GB
G G GG

¥CA branch at first
pick with
probabilities

¥ CA branches at
second pick
with
probabilities

2"¢ CA mark applies
only if without
replacement is
considered

¥ CA outcomes

79
(3)
3.3.3 | P(same colour) = P(BB)+P(GG)+P(RR)
¥CA any two
-R, 11 422, 31 36. 35 ites
= xy SxS yx probabilities of
80 79° 80 79° 80 79 same colour
= 49 oF 038 ¥CA 3" probability of
395 same colour
¥_CA answer
(3)
[20]

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“aeGwnloaded from Stanperenbysiasneom

QUESTION 4

September 2024 Common Test

41

A=P(1+i)"

6423 = 5591(1+i)"

vA correct substitution
into correct formula

ti = 9/9423 YA Li = e238
5591. 5591
i=0,01749...
r=1,75%p.a. ¥ CA answer
(3)
4.2
Ling -{1+te =) VA formula
asic =l4 aT VA correct substitution
+ leg =| 4+ into correct formula
= ooo o, a) »} x100
=9,20% p.a. ¥CA answer
(3)
4.3.1 | 4 =P(1-i)" a
5 vA correct substitution
= 250 000(1--0,12) into correct formula
= R116 101,02 ¥CA
CA only if iandn
are correct (2)
4.3.2 | A=P(1+i)"
cont YA f=
=R250 ooo( +924)" 4 oo
4 ¥CA correct substitution into
= R479 406, 53 correct formula
¥CA_ answer
(3)
4.3.3 | Top-up amount = R479 406, 53 — R116 101, 02
= R363 305, 51 CA answer
(1)

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### Page 17

vateBSwnloaded from Stanmerenhysigsncom

September 2024 Common Test

4.3.4 A=P(1+i)"
YA i= — and n=72
0.1225 \" ¥CA correct substitution
So peop P(t 12 } into correct formula
P = R174 857,32 ¥ CA answer
(3)
44
8% p.a. compounded 12% p.a. compounded
f —R10 000 4 i
I | | ! ! | |
I T T T T T 1
0 ai 2 3 4 5 6 (yéars)
A=P(1+i)"
0,08 \*"* 32
=25 000( + coal VA 0 soa na
12
A; = R31 755,93— R10 000 YCA R31 755,93—R10 000
= R21 755, 93
12
A, = R21 755, 99(1+ oe)
12
= R23 561, 66
3x4
A, =R23 561,66(1+922)
= R33 593, 29
OR

| E

3a2 12 2x4 0,08 \*"”
A=R|| 25 ovo{a+ 92) —10 000 [+92] x49?) Ay 25 o0o{ 1+ 9.
12 12 4 12
iA
=R33593, 29 avai 000
rm (12922)
12
3x4
vA (+3?)
4
¥CA R33593, 29
(5)
[20]
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“aeGwnloaded from Stanperenhysiasneom

September 2024 Common Test

QUESTION 5
5:1. be2 VA b=2
gett YA q=-1
(2)
5.2 | Period = 180° vA Answer
(1)
5.3 h(x) = cos 2x-1+1
= cos 2x ¥CA cos 2x
(1)
54 f(x) =h(x)
sin(x+45°) =cos 2x
sin(x+45°) =sin(90°-2x) YA. co-ratio
X+45° =90°— 2x +k.360° or x + 45° =180° — (90° — 2x) + k.360° ¥ CA both equations
3x = 45° + k.360° or — x = 45° + k.360° Ven Denccene
X=15°+k.120 or x=-45°-k.360°;k eZ VA keZ
JX =15° or x=135° (5)
OR
OR
Y Aco-ratio
_ ¥ CA both equations
ap Sa eee Bee” ¥CA general solution
-3x 0° — 45° + k.360° YCA both values of x
ikeZ YA keZ
(5)
5.5 1
= f (x)x4g(x) =-2
2 (x)490%) NOTE:
=| Correct answer only:
F£O)xg09) =—1 award full marks. VA (X)x g(x) =—1
— VA X= 45°
(2)
[14]

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### Page 19

“aeBGwnloaded from Stanperenhysiasneom

QUESTION 6

September 2024 Common Test

6.1

B(acosC ; asinC)

C(0; 0) A(b; 0)

AB’ =(% Xe) +(Va -Ya)
? = (b—acosC)’ +(0—asinC)’
=b* —2abcosC +a’ cos*C +a’ sin? C
* =b? +a? (cos*C +sin’ C)-2abcosC
a

c? =a? +b? —2abcosC

VA B(acosC ; asinC) and
A(b ; 0)

“A substitution into distance
formula

vA Expansion

YA cos?C+sin?C=1

(4)
6.2 In AABC:
a BC
cos ACB = AC YA Correct ratio and
BC substitution
cos 26,57° = ——
155
BC =15V5 x cos 26,57°
BC = 29, 998.../30 units. YA BC=29,998.../30 units
In ABCD:
BD? = BC’ + CD’ - 2(BC)(CD)cos BCD
2 2 3
= (30) +(10)° -2(30)(10)cos116.57 VA 116.57°
=1268,374509 ¥CA correct substitution into
BD = 35, 61units correct formula
¥ CA Answer
(5)
[9]

TOTAL MARK: 75

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