# Mathematical Literacy Grade 11 P1 2018 Memo

> Grade 11 Mathematical Literacy Paper 1 memorandum from the KwaZulu Natal June 2018, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-literacy-june-p1-qpmemo/62f6e7b7-80e0-4fe4-b2f1-4afce0f9e34a
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 11
- Document type: Memorandum
- Year: 2018
- Exam period: KwaZulu Natal June
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematical Literacy · Grade 11 · KwaZulu Natal June · 2018. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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### Page 1

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education

Department:
Education
PROVINCE OF KWAZULU-NATAL

NATIONAL

_ SENIOR CERTIFICATE

GRADE 11

AAAS. AEBS S888 228

Cee ie

: MATHEMATICS P1

COMMON TEST

JUNE 2018

MARKS: — 100

Thyuk: 2 hours

This question paper consists of 8 pages.

(Copyright reserved Please turn over

### Page 2

Mathematies PT

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2 June 2018 Common Test
Grade 11 - NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions:

1.

to

we

an

6.

9.

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, et cetera, which you have used in
determining the 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 to TWO decimal places, unless stated otherwise.
Diagrams are NOT necessarily drawn to scale.

Write neatly and legibly.

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Mathematics P| June 2018 Common Test

Grade 11 - NSC

QUESTION I

1.1 Solve forx in each of the following:

LLL (Qv4D(v-)=0 Q)
1.1.2 2y° +411=3x+421 (correct to TWO decimal places) (4)
I4
ii 02e=1e8e-— “sy
fae]

1.2. Solve simultaneously for. andy:

ye2v4l and 3x? -Sxy44y° =24 (5)
1.3. Consider the equation ky* + kv +2=0
1.3.1 Solve for x in terms of k. (2)

1.3.2. For which values of & will the equation Av* + kv +2 = 0 have non-real
roots? (3)

QUESTION 2

i 1
62 x12"! x27 2

2.1 Evaluate, WITHOUT using a calculator: ; (4)
322)
2.2. Calculate the value of ————————. WITHOUT using a calculator.
102" _ Voz"
Show ALL your calculations. (4)
[8]

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June 2018 Common Test

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Mathematics P|
Grade Tb - NSC

QUESTION 3

Jabu manutactures wall clock faces in the shape of a regular octagon (cight-sided shape
having sides of equal length). He cuts off identical isosceles triangles from the corners
of a square sheet of glass having sides 16 cm long.

Ifthe length ofa side of the clock face is x, calculate the value of v correct to TWO
decimal places.

(7
QUESTION 4
4.1 Consider the quadratic number pattern: 0;~-9:-16; -21:-24
4.1.1 Determine then term (T,,) of the pattern. (4)
4.1.2. Calculate the 30” term of the pattern. 2)
4.1.3. Determine which term of the pattern will be equal to 200. (4)

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Mathematies P1 June 2018 Common Test

Grade 1] - NSC

4.2 The first three figures in a pattern of grey and white squares are shown below.

Figure | Figure 2 Figure 3

4.2.1 Considering Figure 4 in the pattern, determine:

(a) The number of grey squares in this figure. (1)
(b) The number of white squares in this figure C1)
(c) The number of dots in this figure. (1)

Considering the n’’ figure, determine an expression in terms of #7 for:

(a) The number of grey squares. (1)
(b) The number of white squares. (1)
(c) The number of dots. (2)

4.2.3. There are 320 dots in one of the figures. Determine the number of grey
squares in this figure. Q)

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Mathematics P| June 2018 Common Test

Grade | 1 - NSC

QUESTION 5

5] Given: f(x) = +2.

xt3

aay

.L.1 Draw a neat sketch graph of f, indicating the asymptote(s) and

intercept(s) with the axes. Show all your calculations. (6)

n
~)

Anaxis of symmetry of f has the equation + = mx + 5. Write down the
value of 77. (1)

5.2 In the diagram below, g represents the function g(x) = a*,a>0. The graph / is
symmetrical to g about the y-axis. The point B [? ; 2; lies on the curve of

g and C is the y-intercept of both gand fr.

cay

h
5.2.1 Write down the range of g. (1)
5.2.2 Calculate the value of a. (2)

5.2.3. Determine the equation of / in the form y= 0b". 2)
5.2.4 B’ is the reflection of B in the y-axis. Calculate the average gradient
between B’ and C. (4)

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Mathematics PI June 2018 Common Test

Grade 11 - NSC
53 The diagram below shows the graph of a parabola /, with turning point
A(2:—3)and y— intercept B(O ;-S).
ay
.
>
oO
A(2 ;-3)
B(O 5-5)

in
bo

Show that the equation of the parabola can be written as

ibe (5)
5.3.2 Use the graph to determine the value(s) of & for which the equation
1» - .
a ae 2xy-—5=k will have real and unequal roots. (2)
5.3.3. The parabola / is shifted vertically until the new y — intercept is at the
origin. Determine the new turning point of the parabola. (2)
[25]

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Mathematics PI June 2018 Common Test

Grade 1] - NSC

QUESTION 6

i 9
P $3h-& and g(x) =x +c.

Phe diagram below represents the functions: /(.) = =

e A,B and C are the points at which f intersects the axes.
¢ Dis the turning point of /.

eg passes through A and B.

e [isapointon g.

uy
o '

: x

>
6.1 Determine the coordinates of A,B and C. (5)
6.2 Determine the coordinates of D. GQ)
6.3 Use the graphs to determine the values of yv for which f(x) = g(x). (2)
64 Determine the equation of g in the form g(x) = mx +e. (3)
6.5 Determine the length of DE, if DE is perpendicular to the y-axis. (4)

[17]

TOTAL MARKS: 100

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### Page 10

3
GRADE IL
Marking Guideline

sc 2018 Common Test

Mathematics P 1

June 2018 Common Test

13.1 | be? hv +2=0
eek 7 _ apy
| each tk = 42) Y substitution into quadratic
| 2k formula
| _aktvk? -8k
“= 2k Y answer
_ ~ Q)
13.2 [kh -8K <0 Y bP -8k <0
K(k -8) <0 Y factors
Y answer >)
[21]

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3x2)" (3x2?

Y writing as prime bases

Y simplification using laws

Y simplification using laws

Y answer

NOTE: Ifa calculator is used,

then NO marks will be awarded.
(4)

Copyright Reserved

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2.2
i ¥ determining square roots
-10!
10"
* Tg (10-1) ¥ factorising denominator
__10
10°-1
210
99 Y answer
3)
OR OR
vl0"
hom® — Ji0%
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= Y simplification
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99 (4)
QUESTION 3
The length of the side of the clock is x
« 16 16-2
the length of each corner cut out is = v i x

(16— } CS } -

—"|+ Bx
2 2

256-32x+x° >

4
4

512-64x42x7 = 4x?
2x? +64x-512=0
x? +32x-256 =0
—btvb -Aac
2a
—32+.4(32)° — 4(1(-256)
21)
X= 6,63 “Fx # 38,63
de of the clock is 6,63 em

[Theorem of Pythagoras]

x=

the length of the

¥ equation

Y simplification

Y standard form

¥ substitution into quadratic
formula

Y both values of x

Y¥ answer

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### Page 11

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Mathematics PI z
GRADE II
Marking Guideline

QUESTION 5

+. 2018 Common Test

3
3.11 For x-intercept,

substitute p= 0

For y-intercept,
substitute x= 0:

~6

043 OR

| The graph therefore goes through the origin; (0 ; 0)

x

substitution

Y answer

¥ horizontal asymptote: y = 2
Y vertical asymptote: x = -3 |

Y passing through the origin

¥ shape

(6)

fa

¥ answer dd)

RIE

¥ answer |
OR
Y answer a)

Y substitution

Y answer

Q)

Y change sign of x

¥ answer

Q)

w
iS
&

Y coordinates of Bi
Y¥ coordinates of C
Y substitution into average

gradient formula
v answer (4)

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Mathematics PL

8
GRADE II
Marking Guideline

June 2018 Common Test

33.1. | ysatet py tg
y=atx—2)-3

-5=a(0-2)'-3
-2=4a

Y substitution of turning point, A
Y substitution of point B

Y value of a
Y substitution of @

Y simplification

1 5
y= 3a +2x-5
(5)
5.3.2 | k<-3 vY answer
(2)
533 [(2;2) Y x-coordinate
v y-coordinate
@)
[25]

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