# Mathematical Literacy Grade 11 September Test 2018 QP & Memo

> Grade 11 Mathematical Literacy question paper and memo from the September Test 2018, 20 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/math-lit-sept-qp-memo-2018/576f7eb3-08cc-41af-ba2e-9a216224d800
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2018
- Exam period: September Test
- Curriculum: CAPS
- Pages: 20

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## About this document

Mathematical Literacy · Grade 11 · September Test · 2018. Question paper and memorandum, 20 pages. Read online or download the PDF.

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Mathematics Common Test September 2018

NSC - GRADE 11
INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions:
1. This question paper consists of 6 questions.
2. Answer ALL the questions.

Clearly show ALL calculations, diagrams, graphs, et cetera, which you have used in
determining the answers.

we

4, Answers only will not necessarily be awarded full marks.

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

6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise.
7. Diagrams are NOT necessarily drawn to scale.

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

9. Write neatly and legibly.

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Mathematics Common Test September 2018

NSC - GRADE 11

QUESTION 1

Ld For two events A and B, it is given that P(A) = 0,4 and P(B) = 0,3.
Calculate P(A or B) if:

1.1.1 A and B are mutually exclusive. (2)
1.1.2) A and B are independent. (3)

1.2 A survey was carried out among 100 learners about which movies they have
watched recently. The results are given below:

° 43 watched Braven (B)

¢ 41 watched Hereditary (H)

° 50 watched The Kissing Booth (K)

° 6watched all 3 movies

° 7 watched Braven and The Kissing Booth but not Hereditary
e 18 watched Hereditary and The Kissing Booth

* 15 watched only Hereditary

The above information is represented in the Venn diagram below.

(SLN

e

1.2.1 Write down the values of a, b, c and d in the Venn diagram above. (4)

1.2.2. Calculate the probability that a learner selected at random from this group
has not watched Braven or Hereditary or The Kissing Booth. (3)

1.2.3 Calculate the probability that a learner selected at random from this group

has watched at least 2 of these movies. (3)
{15}

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Mathematics Common Test September 2018

NSC - GRADE 11

QUESTION 2

2.1 Jane bought a laptop for R9 600. Calculate the book value of the laptop after
3 years, if it depreciates at 20% p.a. on the reducing balance method.

(2)
2.2 Sandra invests R20 000 in a savings account that pays interest at the rate of 12%
p.a., compounded monthly. How much will she have accumulated in this account
after 4 years? (3)
2.3. Thuli has two sons: Bongani and Sbu. They share the same birthday.
On the day that Bongani had his 12" birthday and Sbu his 15" birthday, she
opened an account for each of them, and invested some money in each account.
The total amount that she invested in the two accounts was R150 000.
Both accounts earn interest at 9,6% p.a., compounded quarterly.
She divided the R150 000 between the two accounts in such a way that each of
the boys will receive the same amount of money on his 25" birthday.
How much money does she invest for Bongani? (5)
[10]

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Mathematics Common Test September 2018
NSC - Ate 1]

QUESTION 3

ABCD is a cyclic quadrilateral and CB is extended to E.
DA = 155mm and DC = 136mm. ABE =81° and CE= 193mm. AD | GE.

193 mm

3.1 Give a reason why D=sl°. (1)
3.2 Calculate the length of AC. (3)
3.3 Calculate the size of DAC. (3)
3.4 Calculate the area of AACE (4)
[11]

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Mathematics Common Test September 2018

NSC - GRADE 11

QUESTION 4

The organiser of the regional school soccer league decided to have a trophy made to
present to the winning team of the season. He had it made in the shape of a cylinder
with a sphere on top of it. The radius of the cylinder is 5 cm and its height is 20 cm. The
total volume of metal used to make this trophy is 2000 cm’.

\

= 1 eth
3
Vearh
3
= Jr
SOCCER 3
TROPHY
Calculate the radius of the sphere. [5]

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Mathematics

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Common Test September 2018
NSC - GRADE I

Give reasons for your statements and calculations in questions 5, 6 and 7.

QUESTION 5

In the figure, O is the centre of the circle and A, B, D, E, and F are points on
the circumference such that AB//FOC. DB,=22° and C= 78°.

Calculate the size of each of the following angles, giving reasons:

5.1 0, (2)
52 B, (2)
53. £ (2)
5.4 B, 3)

[9]

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Mathematics Common Test September 2018

NSC - GRADE 11
QUESTION 6

6.1 In the figure below AB is a tangent at A to the circle with centre O. D and E are
points on the circle

Prove the theorem that states that DAB= E.

(5)
6.2 Inthe figure PT is a tangent to the circle at T and TR is a chord. PR cuts the

circle at Q. S is a point on TR so that PTSQ is a cyclic quadrilateral. PS, TQ and

SQ are joined.

Let S, =x

p
Q
of S :
T

6.2.1 Prove that PS isa tangent to circle QSR at S. (5)

6.2.2 Prove that PT = PS. (6)
[16]

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Mathematics Common Test September 2018

NSC - GRADE 11

QUESTION 7

O is the centre of the circle. BOA is a diameter and cuts chord TC at N such that
NA 4

BONA LTC. —-=—. TC =24 units.
BN 9
Let NA = 4x.
B
oO
Cc
T NO | N
A
7.1 Write down, giving a reason, the length of TN. (2)
i Write down, in terms of x, the lengths of:
7.2.1 BA (1)
7.2.2 ON 2)
7.3. Calculate the length of the radius of the circle. (4)
19]
TOTAL: 75

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TFA R-OFE SHEET.

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Mathematics

DIAGRAM SHEETS: HAND IN WITH YOUR ANSWER BOOK

NAME:

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10

NSC - GRADE 11

QUESTION 5

QUESTION 6.1

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GRADE:

Common Test September 2018

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Mathematics 11 Common Test September 2018
NSC - GRADE 11

pz

QUESTION 7

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education

; Department:
Education
PROVINCE OF KWAZULU-NATAL

NATIONAL
SENIOR CERTIFICATE

=—s

. MATHEMATICS :
C COMMON TEST :
: SEPTEMBER 2018 ‘
t MARKING GUIDELINE i
MARKS: = 75
This marking guideline consists of 7 pages.
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Mathematics 2 Common Test September 2018
GRADE 11
Marking guideline
QUESTION 1
1.1.1 | P(A or B) = P(A) + PB)

= 0,4 + 0,3 ¥ subst. into correct formula
=0,7 ~ answer
(2)
1.1.2 | P(A or B) = P(A) + P(B) — P(A). P(B) Y correct formula
= 0,4 + 0,3 — (0,4)(0,3) ¥ subst. into correct formula
= 0,58 Y answer
(3)
1.2.1 |a=12 Ya=12
b=8 vb=8
c= 22 ¥co=22
vd =25
d=25
(4)
1.2.2. | Number that did not watch any of these movies
= 100-(22+8+6+7+25+ 12+ 15) 100 — (22+8+ 6+7+25+12+15)
=§ v5
5
P(did not watch any of these movies) = in 0,05 or 5% | “answer
(3)
1.2.3 6 12 v¥ number that watched at least 2 of
P(watched at least 2 of these movies) Se ese vomies waren BOESES

ao or 0,33 or 33%
100

¥ dividing by 100
¥ answer (3)

15]

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3

Common Test September 2018

Mathematics
GRADE 11
Marking guideline
QUESTION 2
2.1 | 4=P-i)"
Book value = 9 600(1—0,2)’ ¥ substitution into
correct formula
=R4915,20 ~ answer
(2)
2.2 | 4=P(i+i)"
0,12 viz 8
A=20 ooo( 1-42)" 12
12 Y substitution into
7 correct formula
= R32 244,52 TS answer
(3)
2.3. | Let the amount that Thuli invested for Bongani be x
-. the amount that Thuli invested for Sbu is150 000-—x
axl3 4x10 0,096 ax13
x [1 ad ~ (150000-){1+ 22 va{is® A a")
4 4
0,096)”
0,096 0,096" 0,096)” ¥ =(180000-9[ 1+ 4 )
x( owe =150000{1 - —x|1l+—
4 4
0,096)" 0,096)" 0,096)"
o[(1-222) (122%) =150 000] 1+ ri Y simplification
150 al 0, ome)
4
a= 40
0, 096)’ 0,096
1+ FT + ao
Y answer
x=R64399,02 (5)
OR OR

4x13 4x10
x (1228) = (150 no0-)[1+ a 7)

3,43239883x = (150 000—.x)(2,582249878)
3,43239883x = 387 337,4817 —2,582249878x
6,014648708x = 387 337,4817

_ 387337,4817
6,014648798 v
answer
x = 64339,02 (5)
NOTE: Answer only: full marks
[10]

4x13
v(t o2*)
4

4x10
v¥ =(150000—x) 14 2.098
4

Y simplification

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Mathematics 4 Common Test September 2018
GRADE 11
Marking guideline
QUESTION 3
3.1 Exterior angle ofa cyclic quadrilateral = opposite Y correct reason
interior angle. qd)
3.2 AC? = AD? +DC? -2(AD)(DC)cosD Y applying cosine rule
=155? +136° ~2(155)(136)(cos81’) ¥ substitution
= 35925, 72...
AC =189,54 mm v answer
(3)
3.3 sinDAC _ sinD Y applying sine rule
DC AC
sin DAC _ sin81°
= v oa
B6 189,54 substitution
DAC =45,13° Y answer
(3)
3.4
ACB= DAC = 45,13° [alt. 2’s; AD || BC] ¥ ACB= DAC
Area of AACE = (AC)(CE)sin ACB ¥ applying area rule
i :
= 3 (189,54) (193) sin 45,13" Y substitution
= 12962,73 mn? Y answer
(4)
[14]
QUESTION 4

Total V = V of cylinder + V of sphere
2000 = arh + Sa

2000 = x(5)’(20) + sm

= 5007 + 4a
3

4 ,_ 2000—500z

a

r -(3}136.5..
4

r* =102,4648....
r=4,68 cm

V arh + Sa
Y equating to 2000

Y substitution

making +? the subject of the
formula

Y answer

65]

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Mathematics

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5

GRADE 11
Marking guideline

Common Test September 2018

QUESTION 5
5.1 O, = 100° —_[ext. angle of A] YSYR
(2)
5.2 B, = 50° [Z at centre = 2 x Z at circumference] YSYR (2)
5.3 E = 130° [opp. angles of cyclic quad] vSYR (2)
5.4 ABC-C, [corresp. 2°s; AB |[FC] YSYR
= 78
B, = 78°— 50°
* 28° ¥ answer
3)
OR —
f = 78°-50° [ext. angle of A]
= 28° “Ss
B, = 28° [alt. 2°s; AB ||FC] Vanswer YR
(3)
1]

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Mathematics 6 Common Test September 2018
GRADE 11
Marking guideline
QUESTION 6
6.1 Construction : Join OA and OD ¥ for construction
Proof:
Let A, =x
wA, = 90° —x (tangent 1 to radius] ¥S/R
AO =OD [radii]
D, =A, =x [Z’s opposite = sides] v S/R
AOD = 180° —2x (sum of 4’s of A] YS/R
E = [AOD [4 at centre =2 x Z at YSYR
circumference]
= 90°-x 6)
E = A, OR
OR ;
Construction: Draw diameter AC, and join Cand D. | “ for construction
ae
oo. B
A, +A, =90° [tangent 1 to radius] VSIR
ADC=90° [Z in semicircle]
c+A, =90° [sum of 2’s of A] VSIR
nOnk,
But C=B [2Z’s in the same segment] YSYR 6)
5

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Common Test September 2018

T

Mathematics
GRADE 11
Marking guideline
Aue
6.2.1 t=8,= 2% [angles in the same segment] ¥ SYR
T,=R =x [tan-chord-theorem] VY SYR
§,=R
PS is a tangent to circle QST [converse: tan- YR
chord-theorem]
2 (5)
622 | Let hy
P, = i, =y {angles in the same segment] VY SYR
PTR = T+, =xty
§ = B+R =yptx [ext. angle of A] VSYR
§= PTR vS
« PT = PS [sides opp. equal angles] YR ©
[16]
QUESTION 7
71 TN = 12 units [line from centre of circle 1 to chord] YSYR
(2)
71 BA = 4x+9x = 13x v answer
(1)
7.2.2 132 :
OA = 7 [radius = * diameter] “length of radius
ow = oa = wa, = Rg
2 ¥ answer
(2)
7.3 TN’ +ON? =OT? [Pythagoras]
sx) (13x) “SR
(8
2 2 Y substitution
2 2
1444 25x _ 169x
4
36x° =144
v=4
xeD
13(2
radius = ie BG) ) =13 Yvalue of x
2 P) v answer (4)
[9]
TOTAL: 75

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