ITEM BANKS FOR REVISION
DEPARTMENT OF EDUCATION
MST ACADEMY
MATHEMATICS GRADES 8&9
1
TABLE OF CONTENTS
GRADE 8 FEBRUARY TEST ........................................................................................................... 3
GRADE 8 FEBRUARY TEST MEMORANDUM ............................................................................ 5
GRADE 8 APRIL TEST ..................................................................................................................... 7
GRADE 8 APRIL TEST MEMORANDUM ...................................................................................... 9
GRADE 9 APRIL TEST ................................................................................................................... 11
GRADE 9 APRIL TEST MEMORANDUM .................................................................................... 14
GRADE 8 MAY TEST ...................................................................................................................... 16
GRADE 8 MAY TEST MEMORANDUM ....................................................................................... 19
GRADE 9 MAY TEST ...................................................................................................................... 22
GRADE 9 MAY TEST MEMORANDUM ....................................................................................... 26
GRADE 8 AUGUST TEST .............................................................................................................. 28
GRADE 8 AUGUST TEST MEMORANDUM ............................................................................... 31
GRADE 9 AUGUST TEST .............................................................................................................. 34
GRADE 9 AUGUST TEST MEMORANDUM ............................................................................... 37
GRADE 8 SEPTEMBER COMMON TASK .................................................................................. 40
GRADE 8 SEPTEMBER COMMON TASK MEMORANDUM ................................................... 44
GRADE 9 SEPTEMBER COMMON TASK MEMORANDUM ................................................... 52
GRADE 8 JUNE EXAM ................................................................................................................... 57
GRADE 8 JUNE EXAM MEMORANDUM .................................................................................... 61
GRADE 9 NOVEMBER 2017.......................................................................................................... 65
GRADE 9 NOVEMBER 2017 MEMORANDUM .......................................................................... 74
GRADE 8 NOVEMBER 2018.......................................................................................................... 84
GRADE 8 NOVEMBER 2018 MEMORANDUM .......................................................................... 91
GRADE 8 NOVEMBER 2019.......................................................................................................... 96
GRADE 8 NOVEMBER 2019 MEMORANDUM ........................................................................ 107
GRADE 9 NOVEMBER 2019........................................................................................................ 115
GRADE 9 NOVEMBER 2019 MEMORANDUM ........................................................................ 124
2
GRADE 8 FEBRUARY TEST
QUESTION 1
1.1. What is the LCM of 24, 30 and 36 using factorisation
B 22 x32 x 5
A 2x3
C 23 x32 x 5
D 23 x33 x 5
1.2. R600 is invested for 3 years at 9% p.a. how much simple interest will be earned?
A R27,00
B R162,00
C 609,00
D R762,00
1.3. What will be the answer of: - 62 + (-3)3 - 64 - 3 125
A -186
B -22
C 8
D 50
3
3 1
1.4. What will be the answer of: 4  5  1
4
5
2
A
177
20
B
177
40
C 8
5
11
D8
5
40
1.5 .What will be the answer of: 2,34 + 7,39 - 0,9 - 45,2
A -44,47
B -36,84
C -36,37
D 36,37
QUESTION 2
2.1
Simplify the following:
2.1.1. (-12)2 - 169
2.1.2. (
(3)
1 2
4
) +5
2
9
(4)
2.1.3. (ab2)3 x a2b3
(3)
2.2. Complete the table below by providing correct conversions.
(3)
Decimal
Fraction
Percentage
0,125
1
8
2.2.1
0,4
2.2.2
2.2.3
3
2.3. Increase 80 in the ratio 3:2
(2)
2.4. A car travelling at a constant speed travels 60 km in 30min. How far will it travel
in 2hrs, if it continues at the same constant speed.
(5)
2.5. Calculate the final amount if R3 500 is invested for 5 years at a compound interest
rate of 6% p.a . (round off the answer to two decimal places).
(4)
2.6. Use the conversion rate (10,6 Kenyan shillings is equal to 1 South African rand) to
convert R540 into shillings.
(3)
QUESTION 3
3.1
Given the numeric pattern : 1; 8; 27; ____ ; ____; _____
3.1.1. Extend the pattern by three more times.
(3)
3.1.2. Write the general rule of the pattern.
(2)
3.2.
Given the geometric pattern below.
1
3
2
3.2.1. Draw the next pattern.
(2)
3.2.2. Write the general rule of the pattern.
(2)
3.2.3. How many squares will figure number 23 have?
(4)
4
GRADE 8 FEBRUARY TEST MEMORANDUM
QUESTION 1
QUESTION
ANSWER
1.1. 1.2. 1.3. 1.4. 1.5
C B B C A
QUESTION 2
2.1.1
(-12)2 - 169
=144 – 13
=131
2.1.2.
2.1.3
(
1 2
4
) +5
2
9
=
1 49


4 9
=
9  196

36
=
205

36
5
25

36
(ab2)3 x a2b3
=a3b6 x a2b3 
=a5b9 
2.2.1
12,5% 
2.2.2.
2

5
2.2.3
40% 
2.3.
3
x 80 
2
=140 
2.4.
V=
s

t
5
=60km/30min 
=2km/min 
S = vxt 
= 2km/min x (2x60min) 
=2km/min x 120min
=240km 
2.5.
A= p( 1+
r n
) 
100
=R3500 (1+
6 5
) 
100
= R3500 (1,06)5 
=R3500 x 1,34
= R4,690 
2.6.
10,6 shillings = R1 
X = R540 
X= 5 724 shillings 
QUESTION 3
3.1.1.
66; 125; 216
3.1.2.
Tn = n3 
3.2.1.

3.2.2.
Tn = 2n+1
Tn = 2n+1
3.2.3
 2(23)  1 
 46  1 
 47 
6
GRADE 8 APRIL TEST
QUESTION 1
1. Given the expression: 18 x3  2 x 2  4 x  1
1.1.
How many terms are in the expression?
(1)
1.2.
Write down the coefficient x 2
(1)
1.3.
What is the value of the constant term?
(1)
1.4.
Calculate the value of the expression, when x  1
(3)
1.5.
Write down the degree of the expression
(1)
2. Simplify the following:
2.1.
2.2.
2.3.
 9 x( x 2  3)  5 x(2 x 2  5)
(2)
3(2a  7) 2
(3)
(3x  3)(2 x  2)
(2)
3. Factorise the following:
3.1.
 3t 2  15t
(2)
3.2.
5x(2 x  1)  5d (2 x  1)
(2)
3.3.
2 x( y  p)  3c( p  y)
(3)
3.4.
81  y 2
(2)
QUESTION 2
Simplify the following:
2.1.
10 x 4  15 x 3
 5x
(2)
2.2.
3 xy  y 2
3x  y
(2)
x 2  2x  8
x2
2.3.
(3)
x2  9
2
2.4. x  5 x  6
(5)
2.5.
8x 3 y 4  4 x 5 y 5
2x 2 y 2
(3)
7
2.6. 3
64
512
(2)
QUESTION 4
Solve for x
4.1.
𝑥 − 19 = 4
(2)
4.2. 3𝑥 − 10 = −22
4.3.
(2)
𝑥 2 + 9𝑥 = −14
(3)
4.4. 2 x  0,25
(3)
8
GRADE 8 APRIL TEST MEMORANDUM
QUESTION 1
No.
1.1
1.2
1.3
1.4
1.5
2.1.
Answer
4
-2
-1
18 x3  2 x 2  4 x  1
=18(-1)3 -2 (-1)2 + 4(-1) -1
=-18 -2-4-1
-25
Third degree
 9 x( x 2  3)  5 x(2 x 2  5)
 9 x 3  27 x  10 x 3  25 x 
 19 x 3  2 x
2.2.

3(2a  7) 2
 3[( 2a  7)( 2a  7)]

 3(4a  14a  14a  49) 
2
 3(4a 2  28a  49)
 12a 2  84a  147
2.3.
3.1.

(3x  3)(2 x  2)
6x2  6x  6x  6 

6x2  6
2
 3t  15t
 3t (t  5)  
3.2.
5 x(2 x  1)  5d (2 x  1)
3.3.
 (5 x  5d )(2 x  1)  
2 x( y  p)  3c( p  y )
 2 x( y  p)  3c( y  p) 
 (2 x  3c)( y  p)
3.4.
81  y 2
 (9  y )(9  y )
 

9
QUESTION 2
No. Answer
2.1. 10 x 4  15 x 3
 5x
3
5 x (2 x  3)

 5x
  x 2 (2 x  3)
2.2.
2.3.
2.4.
2.5.
 2 x 3  3 x 2
3 xy  y 2
3x  y
y (3 x  y )

(3 x  y )
y


x2  2x  8
x2
( x  2(( x  4)

( x  2)
 x4

2
x 9
2
x  5x  6
( x  3)( x  3)

( x  3)( x  2)
x 3

x2
8x3 y 4  4 x5 y 5
2x2 y 2
4 x y 4 (2  x 2 y )

2x2 y 2
 
 
 
 

 


3
 2 xy 2 (2  x 2 y )
 4 xy 2  2 x 3 y 3
2.6.






64
512
4


8

1

2 
3

10
QUESTION 4
No. Answer
4.1. x  19  4
x  23
 

4.2. 3𝑥 − 10 = 
−22
3x  22  10 

3x  12

x  4
2
x  9 x  14 
4.3.
x 2  9 x  14  0
( x  7)( x  2)  0
x0

or



x  7

x
2  0,25 
1
2x 

4

2 x  22
 
x  2
4.4.


GRADE 9 APRIL TEST
QUESTION 1
1. Choose the correct answer
Two angles are ____________ when they add up to 180o
1.1.
A. Complementary B. Equilateral
1.2.
C. Equal
D. Supplementary
The sum of angles of a quadrilateral is equal to _________
A. 150˚
1.3.
B.180˚
C. 360˚
D. 480˚
The value of 𝑥 is...
2x
40o
A.15˚
1.4.
B. 20˚
C. 25˚
The value of 𝑦 is...
A
B
y + 20
y+8
D
11
C
D. 42˚
A. 76˚
1.5.
B. 52˚
C. 28˚
D. 56˚
What is the name of the quadrilateral below?
A. Rhombus
B. Parallelogram
C. Kite
D. Rectangle
QUESTION 2
1. Find the sizes of the unknown angles with reasons.
2.1.
a
(2)
120o
b
2.2.
(2)
c
2.3.
(1)
106o
d
2.4.
(2)
e
88o
2.5.
(4)
X+30o
87o
y
QUESTION 3
3.1.
In ∆EDF, DF is produced to C. What is the size of  E, give reasons
for your answer.
(4)
E
4
D
5
3
F
12
C
3.2.
Find the size of y , if ∆ABC is an isosceles ∆, with  A=65o
(4)
B
y
y
65o
A
C
y
3.3.
In the figure below, CS//HN.  EAW =70o, AE =AW and  CAE = x .
Determine the value of x .
(4)
3.4.
ABCD is a parallelogram calculate the size of  B
(4)
(4)
3.5.
Find the sizes of x, y and z, with reasons.
(4)
Q
P
30
y
z
O
Find the value of x
F
A
C
o
x
R
3.6.
42
o
S
(4)
45o
B
G
D
X+15o
P
3.7.
Name 3 properties of a kite
(3)
3.8.
Name 2 properties of a trapezium
(2)
13
GRADE 9 APRIL TEST MEMORANDUM
QUESTION 1
Question 1.1.
1.2.
1.3.
1.4.
Answer
D
C
C
A
1.5.
B
QUESTION 2
No. Answer
2.1. a + 120o =180o (sum of  ’s on a strt line) 
a = 60o
2.2. b+ 90o =180o(sum of  ’s on a strt line) 
b = 90o
2.3. c=106o (vertical opp.  ’s) 
2.4. e = 88o ( alt.  ’s) 
d= 88o (corrs.  ’s) or vertical opp.  ’s) 
2.5. y + 87 = 180o (  ’s on a strt. line) 
y = 93o 
x +30o =87o ( vertical opp.  ’s) 
x = 57o 
QUESTION 3
No. Answer
3.1. 3x + 5x = 180o (  ’s on a strt line) 
8x =180o
X=22,5o 
5x = 4x +  E (ext  of a ∆ = sum of opp int.  ’s) 
 E = 5x -4x
E = x
  E = 22,5o
3.2.  A =  C ( base  ’s of isco ∆) 
  C =  A =65o
 A +  B +  C = 180o ( sum of  ’s on a ∆) 
65o +  B + 65o = 180o
 B=180o-130o
 B = 50o 
3.3.  E2 +  W1 +70o = 180o (sum of  ’s of ∆) 
 E2 =  W1 (Base  ’s of isco. ∆) 
2  E2 +70o = 180o
2  E2 =110o
 E2 =  W1 = 55o 
x =  E2 (alt  ’s) 
3.4. 2x-20o + x +50o = 180o (co-inter/Suppl.  ’s) 
3x+30o =180o
3x=150o
14
x= 50o 
 B=  D =2x-20 (opp  ’s of quard) 
 B=2(50o) -20o
 B=100o-20o
 B= 80o 
3.5. x=42o (alt  ’s) 
z = 30o (alt  ’s) 
y + x = 180o (sum of  ’s on ∆) 
y + 42o =180o
y =138o 
3.6.  OFB =  DGP (corrs.  ’s) 
 OFB =  DGP = 45o 
But
 DGP =  CGP (vert. opp.  ’s) 
  CGP =x+15o =45o
X+15o=45o
X = 30o 
3.7.
 Two pairs of sides are of equal length
 One pair of opposite angles is equal
 Only one diagonal is bisected by the other
 The diagonals meet at 90o
3.8.
 One pair of parallel sides
 No sides, angles nor diagonals are equal
 Sum of angles equal to 360o
15
GRADE 8 MAY TEST
QUESTION 1
1. Solve the following equations.
1.1.
2 x  7  15
(2)
1.2.
6 x2  8x  0
(3)
1.3.
x2  7 x  6  0
(3)
1.4.
2 x  0,25
(3)
QUESTION 2
2.1.
Construct the following angles without using a protractor. Use a pair of
compasses. (Construct on the page provided)
2.1.1
Construct a 90˚ angle on point N on line MN
(3)
2.1.2
Construct a 60˚ angle on point K on line KL
(2)
2.1.3
Bisect the angle in 2.1.2
(2)
2.2.
Construct XYZ with XY = YZ = 4cm and state what type of triangle
(3)
2.3.
Construct rectangle VBMN with VB = 6cm, MN = 4cm and diagonals
2.4.
Figure
VM and BN intersecting at G.
(3)
Complete the table below:
(6)
Name of polygon
Sum of interior angles
2.4.1. _______
2.4.3.
2.4.2._______
2.4.4._______
180o
2.4.5._______
2.4.6._______
16
QUESTION 3
3.1. Consider the parallel lines below. AB and CD are cut by a transversal line GH
below. Find the sizes of the following angles with reasons.
G
B
r
s
110˚
A
t
D
y
w
z
C
x
H
3.1.1.  w
(2)
3.1.2.  y
(2)
3.1.3.  s
(2)
3.2. Find the value of a
(2)
3.3. Given the kite with AB =AC and BD =CD. Prove that  ABE   ACE
(4)
B
1
A
1
2
2
2
1
E
1
D
1
C
17
3.4. Given ABC with PQ//AB. Prove that ∆ ABC /// ∆ PQC
(4)
B
A
P
3.5. Given the parallelogram.
Q
C
3.5.1. Find the size of  x
(2)
3.5.1. Find the size of  y
(2)
18
GRADE 8 MAY TEST MEMORANDUM
QUESTION 1
Answer
1.1. 2 x  7  15
2 x  15  7 
2 x  22
x  11

1.2. 6 x 2  8 x  0
2 x(3x  4)  0 
2 x  0or 3x  4  0
4
x  0orx 

3
1.3.
x2  7 x  6  0
( x  6)( x  1) =0
x  6orx  1 
1.4. 2 x  0,25
1
2x  
4
x
2  2 2 
x =-2
QUESTION 2
Answer

2.1.1

2.1. 1.

19
2.1.2
and
2.1.3.




K
2.2

4cm
4cm

Isosceles triangle 
2.3.



2.4.
2.4.1 square 
2.4.2. 360o 

2.4.3.

2.4.4. triangle 
2.4.5.parallelogram
2.4.6. 360o
QUESTION 3
Answer
3.1.1.
110˚+w=180˚(co-interior s )
w=180˚-110˚
20
w=70˚
3.1.2. 𝑦 = 110˚ (Alt. s )
3.1.3.
𝑠 = 110˚ (vertically opp. s )
3.2. 52o +79o +a=180o (sum of s in triangle) 
a=49o
3.3. In Δ ABE and Δ ACE
AE is common
AB =AC (given) 
E1=E2 (Diagonals of a kite intersecting at 90o) 
 ABE   ACE (90o, H, S) 
3.4.
In Δ ABC and Δ PQC
𝐶̂ = 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛
𝐴𝐵̂𝐶 = 𝑃𝑄̂ 𝐶 (corresponding angles) 
or
𝐵𝐴̂𝑃 = 𝑄𝑃̂𝐶 (corresponding angles)
∴ Δ ABC /// Δ PQC
3.5.1. 2(112) +2x =360o (sum of angles of parm) 
224o +2x = 360o
2x=360o-224o
2x=136o
x= 68o
3.5.2. y=112o (opp. angles of parm) 
21
GRADE 9 MAY TEST
QUESTION 1
1.1.
Which property is NOT TRUE for a parallelogram?
A. Opposite angles are equal.
B. Opposite sides are equal and parallel.
C. The diagonal are equal.
D. Interior angles sum up to 360°.
1.2.
Which statement is true about the figure below:
A.
B.
C.
D.
∠𝐴1 = ∠𝐵1
∠𝐴1 = ∠𝐵3
∠𝐴1 = ∠𝐴3
∠𝐴1 + ∠𝐴2 = 180°
Corresponding ∠𝑠 ∥ 𝑙𝑖𝑛𝑒𝑠
Exterior alternate ∠𝑠 ∥ 𝑙𝑖𝑛𝑒𝑠
Vertically opp. ∠𝑠
co-interior ∠𝑠 𝑜𝑓 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
1.3.
A.
B.
C.
D.
An equilateral triangle is __________
A triangle with two sides equal
A triangle with all sides equal
A triangle with all sides not equal
A triangle with one angle equal to 90o
1.4.
If the height of a triangle is TRIPLED, the area of the new triangle would be
________
3 + 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
3 × 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
4 × 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
½ × 𝑏 × ℎ3
A.
B.
C.
D.
22
1.5.
Find the length of AD (rounded off to whole number).
A
15 cm
C
8 cm
D
A. 11
B.16
B
8 cm
C. 23
D.64
QUESTION 2
2.1. In the figure below, ABC is a straight line, calculate the value of 𝑥 (with
reasons).
(2)
E
D
𝑥
A
2.2.
2𝑥
3𝑥
C
B
In ∆𝐿𝑂𝑉, 𝑂𝑉 𝑖𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑡𝑜 𝐸 𝑎𝑛𝑑 𝐿𝑉 = 𝐿𝑂.
2.2.1. Calculate the size of ∠𝐿 with reasons.
(3)
2.2.2. Calculate the value of 𝑦 with reasons.
(3)
23
2.3. In the quadrilateral below 𝐷𝑆 ∥ 𝑉𝑆 and 𝐷𝑉 ∥ 𝑆𝑇.
2.3.1. What type of a quadrilateral is DSTV?
(1)
2.3.2. Calculate the value of 𝑎. (with reasons)
(3)
2.3.3. Calculate the value of 𝑏. (with reasons)
(4)
2.4. Given the kite with AB =AC and BD =CD. Prove that  ABD   ACD
(4)
QUESTION 3
3.1. Calculate the PERIMETER of the figure below if 1 𝑏𝑙𝑜𝑐𝑘 = 1𝑐𝑚 .
(2)
3.2. If the area of the rectangle below is 60 𝑐𝑚2 . Calculate the value of 𝑥.
(4)
24
3.3. Consider the figure below.
3.3.1. Calculate the length of 𝑥.
(4)
3.3.2. Calculate the length of 𝑦
(4)
3.4. Calculate the area of the figure below if 1 𝑏𝑙𝑜𝑐𝑘 = 1 𝑐𝑚
(7)
25
GRADE 9 MAY TEST MEMORANDUM
QUESTION 1
QUESTION
ANSWER
1.1.
C
1.2
1.3
C B
1.4
1.5
B A
QUESTION 2
Answers
2.1.
𝑥 + 2𝑥 + 3𝑥 = 80°
6𝑥 = 180°
6𝑥
∠𝑠 𝑜𝑛 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒
1800
= 6 
6
𝑥 = 30° 
2.2.1 ∠𝑉1 = 35° ∠𝑆 𝑜𝑝𝑝 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠 𝐼𝑠𝑜 ∆
80° 𝐼𝑛𝑡 ∠𝑠 𝑜𝑓 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 
∠𝐿 = 180° − 70°
∠𝐿 = 110° 
2.2.2. 𝑦 = ∠𝐿 + ∠𝑂
𝐸𝑥𝑡 ∠𝑜𝑓 𝑎 ∆
𝑦 = 110° + 35° 
𝑦 = 145° 
2.3.1 𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝑜𝑟𝑃𝑎𝑟𝑚 
2.3.2
3𝑎 + 𝑎 = 100°

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∠𝑠 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑚
4𝑎 = 100°
4𝑎
4
=
1000
4

𝑎 = 25° 
2.3.3.
𝑎 + 3𝑎 + 3𝑏 + 2𝑏 = 180° 𝐶𝑜 − 𝑖𝑛𝑡 ∠𝑠 // lines 
25° + 3 × 25° + 5𝑏 = 180°
5𝑏 = 180° − 100°
5𝑏
5
=
80°
5
𝑏 = 16°
2.5.
In  ABD and  ACD
AB=AC(given) 
BD =CD (given) 
AD is common
 ABD   ACD (S,S,S) 
26
∠𝐿 + 350 + 35° =
QUESTION 3
3.1.
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 1 + 3 + 2 + 2 + 2 + 3 + 1 + 8 = 22𝑐𝑚
3.2.
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 
𝐴=𝑙×𝑏
60𝑐𝑚2 = 15𝑐𝑚 × (𝑥 + 3)𝑐𝑚 
60𝑐𝑚2
15𝑐𝑚
= 𝑥 + 3 cm
4𝑐𝑚 − 3𝑐𝑚 = 𝑥 
𝑥 = 1𝑐𝑚 
3.3.1. DB2 = AB2 + AD2 
𝑥 2 = (9𝑐𝑚)2 + (12𝑐𝑚)2

𝑥 2 = 81𝑐𝑚2 + 144𝑐𝑚2
𝑥 2 = 225 𝑐𝑚2 
√𝑥 2 = √225𝑐𝑚2
𝑥 = 15𝑐𝑚 
3.3.2. BC2= CD2+BD2 
(17cm) 2  y 2  (15cm)2 
𝑦 2 = (17𝑐𝑚)2 − (15𝑐𝑚)2
𝑦 2 = 289 𝑐𝑚2 − 225𝑐𝑚2
𝑦 2 = 64 𝑐𝑚2 
√𝑦 2 = √64𝑐𝑚2
𝑦 = 8𝑐𝑚 
3.4.
𝐴𝑟𝑒𝑎 = 𝐴(𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒) + 𝐴(𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒)
1
= 𝑙 × 𝑏 + 2 𝑏 × ℎ 
1
= 5𝑐𝑚 × 2𝑐𝑚 + 2 × 5𝑐𝑚 × 3𝑐𝑚 
= 10𝑐𝑚2  + 7,5𝑐𝑚2 
= 17,5𝑐𝑚2 
27

GRADE 8 AUGUST TEST
QUESTION 1
1.1.
Calculate the following without the use of a calculator. You need to show all
your calculations. Leave your answers in its simplest form.
1.1.1.
1 2

5 5
(2)
1.1.2.
4 2

7 3
(3)
1.1.3.
1
2
3
4
(3)
1.1.4. 3
1
3
x2
4
5
(3)
1.1.5. 18,01 − 1,73
(2)
1.1.6. 0,927 x 12,04
(3)
1
1.1.7. √64 − (√9)
3
1.1.8.
1.2.
3
125 -
2
(3)
64
4
(3)
Write down the values of the marked points (A, B and C) in decimal form. (3)
QUESTION 2
2.1. Use the theorem of Pythagoras in the diagram below and complete the
following:
𝑐
A
(1)
𝑎 C
2.2.
Use the figure below to calculate a and b:
A
(6)
10𝑐𝑚
8cm
𝑎
𝑎
𝑏
𝑏 2 = ...............................
B
B
D
𝑏
28
9cm
C
2.3.
Find the value of y.
(a)
3 cm
A
y
(3)
B
5 cm
C
2.4.
Matthews has a rectangular lawn with 8 𝑚 long and 9 𝑚 wide. How much
distance will Matthew walk diagonally across the lawn from one corner to the
other.
(3)
QUESTION 3
3.1. Find the perimeter of the following figures:
(2)
9 cm
3.1.1
3.1.2.
(2)
Radius = 28mm (𝜋 = 3,14)
29
3.2. Find the area of the shaded region of the following shape which is made up of a
rectangle and a triangle:
(5)
2cm
(5)
3cm
6cm
3.3. Calculate the area of ABC below:
(3)
A
C
B
30
GRADE 8 AUGUST TEST MEMORANDUM
QUESTION 1
No.
Answer
1.1.1. 1 2

5 5
1 2


5
3
 
5
1.1.2. 4 2

7 3
12  14


21
26


21
1.1.3. 1
2
3
4
1 4
 x 
2 3
4
 
6
1.1.4.
1
3
x2
4
5
13 13
= x 
4 5
169
=

20
9
=8 
20
1.1.5. 18,01
- 1,73_
16,28
1.1.6.
0927
x1204
3708
0000 
1854
0927____
11,16108
3
31
1.1.7.
1
2
√ − (√9)
64
1
=  (3) 2 
8
1 9
= 
8 1
1  72
=

8
 71
=

8
1.1.8.
3
3
125 -
64
4
4

2
=5 - 2
=3
=5 -
1.2.
A=
B=
C=
29
100
33
100
40
100
= 0.29
= 0.33
2
= = 0.4
5
QUESTION 2
No.
2.1.
Answer
𝑏 2 = 𝑎2 + 𝑐 2 
2.2.
𝐴𝐷2 = 𝐵𝐷2 + 𝐴𝐵 2
102 = 𝑎2 + 82 
𝑎2 = 102 − 82
𝑎 = √100 − 64 
𝑎 = √36
𝑎 = 6𝑐𝑚
𝐴𝐶 2 = (𝐵𝐷 + 𝐷𝐶)2 + 𝐴𝐵 2
2
𝑏 2 = (6 + 9) + 82 
𝑏 2 = 225 + 64
𝑏 2 = 289
𝑏 = √289
𝑏 = 17𝑐𝑚 
2.3.
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
2
𝑦 2 = 3 + 52 
𝑦 2 = 9 + 25
𝑦 2 = 34
𝑦 = √34
32
𝑦 = 5,8𝑐𝑚
2.4.
𝑟2 = 𝑥
2
+ 𝑦2
𝑟2 = 8
2
+ 92 
𝑟 2 = 64 + 81
𝑟 2 = 145
r =12.04cm 
QUESTION 3
No.
3.1.1
Answer
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 4𝑠
= 4(9)
= 36𝑐𝑚
3.1.2. 𝐶 = 2𝜋𝑟 
𝐶 = 2𝜋(28) 
𝐶 = 175,93𝑚𝑚 
3.2.
𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
𝐴𝑟𝑒𝑎 𝐴 𝑜𝑓 ∎ = 𝑙 × 𝑏
= 6 × 3
= 18𝑐𝑚2 
𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒
1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 ∆= 2 𝑏 × ℎ
1
= 2 (6)(2) 
= 6𝑐𝑚2 
Area of shaded region
3.3.
= 𝐴∎ + 𝐴∆
= 18𝑐𝑚2 + 6𝑐𝑚2
= 24𝑐𝑚2 
1
𝐴∆= 2 𝑏 × ℎ
1
=2 (10)(12) 
=60cm2  
33
GRADE 9 AUGUST TEST
QUESTION 1
1.1 What is the product of (4y-y) 94x+3y)
A. 16𝑥 2 + 8𝑥𝑦 − 3𝑦 2
B. 8𝑥 2 + 16𝑥𝑦 − 3𝑦 2
C. 8𝑥 2 + 8𝑥𝑦 − 3𝑦 2
D. 16𝑥 2 − 16𝑥𝑦 − 3𝑦 2
1.2 Given the sequence; -13; -10; -7; __; __; __the next three terms are
A. -10; -13; -16
B. -4; -3; -2
C. -4; -3; -1
D. -4; -1; 2
1.3 The shape of the base of a triangular prism is?
A. Triangle
B. Square
C. Rectangle
D. Circle
1
𝑥
1.4 Given the equation: 4 = 260. The value of 𝑥 will be:
A. 1
B. 4
C. 65
D. 80
1.5 A 3D-shape is likely to have a___
A. Length, Breadth and Height
B. Length, Width and Breadth
C. Base, Face and Height
D. Rectangle, Triangle and Circle
QUESTION 2
2.1 Determine the values of a, b, c, and d.
34
(4)
2.2 Given the table:
Position in the sequence 1
2
3
4
5
g
Term
8
15
e
f
57 …..
2.2.1.
1
Tn
Determine the values of e, f and g
(3)
2.2.1 Determine the general rule of the pattern in the form, Tn=...
(2)
2.3 Write down the equation which will best describe the relationship between output
values in the table below in the form Tn=...
(2)
Input : x
1
2
3
Output: y
4
2
0
4
n
-2 …
QUESTION 3
3.1
3.1.1.
3.1.2
Simplify the following expressions
12𝑥 2 𝑦−6𝑥𝑦 2
(3)
3𝑥𝑦
−2(4𝑥 − 2𝑥 2 ) − 6 + 5𝑥(3𝑥 2 − 𝑥 + 5)
(4)
QUESTION 4
4.1. The rectangular prism has L = 6cm, B = 4cm and H = 3cm. Calculate its surface
area.
(4)
4.2. Given the cylinder as shown below:
4.2.1. Calculate the surface area of the cylinder, If 𝝅=3,14
35
(4)
4.2.2. If the dimensions of the cylinder are halved what would be the volume of the new
cylinder?
(3)
4.3. Owami has three glass tanks. She wants to use one as a decorative fish tank
for Mylo - the rainbow fish.
4.3.1. Which fish tank will give Mylo the most water to swim in? Show your
calculations.
(10)
4.3.2. Write the volume of the cubic fish tank in mm3
36
(1)
GRADE 9 AUGUST TEST MEMORANDUM
QUESTION 1
QUESTION
ANSWER
1.1
A
1.2
D
1.3
A
1.4
1.5
C A
QUESTION 2
No.
2.1.
2.2.1.
Answer
a = 8, b = 14, c = -2,
d = -3
𝑇𝑛 = 7𝑛 − 6
𝑒 = 22
𝑓 = 29
𝑔 = 9
2.2.2.
𝑇𝑛 = 7𝑛 − 6
2.3.
𝑇𝑛 = −2𝑛 + 6
QUESTION 3
No.
3.1.1
ANSWER
12𝑥 2 𝑦
3𝑥𝑦
−
6𝑥𝑦 2
3𝑥𝑦

= 4𝑥 − 2𝑦 
3.1.2
−2(4𝑥 − 2𝑥 2 ) − 6 + 5𝑥(3𝑥 2 − 𝑥 + 5)
= −8𝑥 + 4𝑥 2 − 6 + 15𝑥 3  − 5𝑥 2 + 25𝑥
= 15𝑥 3 − 𝑥 2 + 17𝑥 − 6 
37
QUESTION 4
No.
4.1.
ANSWER
𝑺𝑨 = 𝟐𝑳𝑩 + 𝟐𝑩𝑯 + 𝟐𝑳𝑯√
= 𝟐(𝟔)(𝟒) + 𝟐(𝟒)(𝟑) + 𝟐(𝟔)(𝟑)√√
= 𝟒𝟖 + 𝟐𝟒 + 𝟑𝟔
= 𝟏𝟎𝟖𝒄𝒎𝟐 √
4.2.1.
𝑺𝑨 = 𝟐𝝅𝐫 𝟐 + 𝟐𝝅𝐫𝐇 √
= 𝟐𝛑(𝟔)𝟐 + 𝟐𝛑(𝟔)(𝟏𝟓)√
=226, 08+565,2 √
= 791.28𝑐𝑚2 √
4.2.2.
V=𝛑𝒓𝟐 . 𝑯 √
=𝛑 𝟔𝟐 . 𝟕, 𝟓√
=847,8cm3√
4.3.1.
CUBE:
𝑽 = 𝒔𝟑 √
= (𝟖)𝟑 √
= 𝟓𝟏𝟐𝒄𝒎𝟑 √
Volume of a rectangular prism:
𝑽 = 𝑳 × 𝑩 × 𝑯√
= 𝟏𝟏𝒄𝒎 × 𝟏𝟎𝒄𝒎 × 𝟓𝒄𝒎
= 𝟓𝟓𝟎𝒄𝒎𝟑 √
Triangular prism
:𝑽 = 𝟏⁄𝟐 𝒃 × 𝒉 × 𝑯√
= 𝟏⁄𝟐 × 𝟖𝒄𝒎 × 𝟔𝒄𝒎 × 𝟐𝟐𝒄𝒎√
= 𝟓𝟐𝟖𝒄𝒎𝟑 √
The rectangular prism will contain most water√
38
4.3.2.
𝑽 = 𝟓𝟓𝟎 × 𝟏𝟎𝟑 𝒎𝒎𝟑
= 𝟓𝟓𝟎 𝟎𝟎𝟎𝒎𝒎𝟑 √
39
GRADE 8 SEPTEMBER COMMON TASK
QUESTION 1
1.1.
4
The sum of 67 and
3
10
10
B. 497
C. 6 7
D. 714
B. 1,267
C. 12,67
D. 126,7
The surface area of a cube with sides equal to 9cm is____
A. 81cm2
1.4.
is _____
2,138 – 0,871=
A.1267,0
1.3.
7
3
A. 77
1.2.
6
B. 729cm2
C. 486cm2
D. 18cm2
Pythagoras theorem states that ______
A. The square of the hypotenuse side is equal to the sum of the other two
sides.
B. The square of the hypotenuse side is equal to the sum of squares of the
other two sides.
C. The sum of the sides of the right-angled triangle is equal to 1.
D. The longest side of the right-angled triangle is equal to the sum of the
height and the base of the triangle.
1.5.
The median of the given set of data is _____
12; 14; 16; 17; 10; 13; 12; 13; 12
A. 13
B. 12
C. 17
D. 10
QUESTION 2
2.1. Calculate the following:
2.1.1.
3 4
 
7 7
(2)
2.1.2.
5 3
 =
12 4
(3)
2.1.3.
9
12
5
7
(3)
=
3
1
2.1.4. 8 x 4 =
5
8
25
2.1.5. √144 −
3
(3)
64
=
4
(3)
40
2.1.6. 12,93 + 8,11 =
(2)
2.1.7. 111,027 x 62,129= (round answer to one decimal place)
(2)
QUESTION 3
3.1. Use the figure below to calculate a and b:
(6)
A
8cm
10𝑐𝑚
B
3.2. Find the
value𝑎of r. D
𝑏
C
9cm
(3)
A
r
6cm
m
B
C
8cm
3.3. Prove whether the triangle is a right-angled triangle.
(5)
D
13cm
8cm
E
F
5cm
QUESTION 4
4.1. Calculate the periemter of the rectangle below.
(2)
7cm
6,2cm
4.2. Angel baked a 24cm square cake. What length of ribbon will he need to put
outside around the cake?
(2)
41
4.3. Calculate the area of ABC below.
(3)
B
6cm
A
C
D
18cm
4.4. Calculate the area of a circle below. (use  =3,14)
14cm
(3)
14cm
4.5.1. Calculate the area of the figure below.
(5)
8cm
5cm
14cm
4.5.2. Write the area of a figure above in m2
(2)
QUESTION 5
5.1. Calculate the surface area and volume of a rectangle below:
5.1.1.
(7)
6cm
3cm
5cm
5.1.2. Calculate the surface area and volume of a triangular prism below:
4cm
m
3cm
12cm
m
42
(10)
5.2. A swimming pool is 8m long, 6m wide and 2m deep. Calculate the volume of the
swimming pool. Write your answer in cm3.
(4)
QUESTION 6
6.1. Distinguish between a sample and population and provide examples.
(4)
6.2. Organise the following set of data using stem-and-leaf display.
(4)
564; 534; 544; 568; 588; 590; 510; 523; 533; 537; 555
6.3. The data shows the scores obtained when a six-sided dice was thrown 20 times.
3; 4 ; 2; 6; 1; 3; 2; 1; 6; 6; 2; 5; 5; 3; 2; 1; 6; 3; 2; 4
6.3.1. Find the mode of the data set.
(2)
6.3.2. Record data using tally/frequency table.
(2)
6.3.3. Use the set of data to draw a bar graph.
(5)
6.4. The pie chart represent percentage of learners participating on different sporting codes.
The total number of learners at the school is 800.
6.4.1. Which sporting code has the least percentage?
(1)
6.4.2. What number of learners participated in Rugby?
(2)
43
GRADE 8 SEPTEMBER COMMON TASK MEMORANDUM
QUESTION 1
QUESTION
1.1.
1.2.
ANSWER
A B
1.3.
1.4.
1.5.
C
B
A
QUESTION 2
NO.
ANSWER
2.1.1. 3 4

7 7
7
7
= 
=1
2.1.2.
5 3

12 4
𝟓−𝟗
= 𝟏𝟐 
−𝟒
= 𝟏𝟐 
𝟏
=− 𝟑
2.1.3.
9
12
5
7
=
9
7
= 12 𝑥 5
63
=60
3
=160
2.1.4.
3
1
8 x4
5
8
43
33
=5 x
8

1419
=

40
19
=35 40
2.1.5.
25
√
−
144
3
64
=
4
5 4
= 12-2
5−24
=
12
19
=− 12
7
=1 12
44
2.1.6. 12,93 + 8,11 =21,04
2.1.7. 111,027 x 62,129= 6,9
QUESTION 3
NO.
3.1.
ANSWER
AD2 = BD2+AB2
102=a2 +82
a2=100-64
a2=36
a=6cm
b2 = BC2+AB2
=(6+9)2 + 82
=225 +64
=289
b=17cm
3.2.
AC2 = BC2+AB2
r2=82 +62
r2=64+36
r=10cm
3.3.
LHS
DF2 =132 
DF2 = 169
RHS
EF2+DE2 = 52 +82
= 25 +64
=89
LHS  RHS
It is not a right angled triangle
QUESTION 4
NO.
4.1.
ANSWER
7cm+7cm+6,2cm+6,2cm =26,4cm
4.2.
24cm+24cm+24cm+24cm=96cm
45
4.3.
1
Area = 2bxh
1
=2x18x6
=54cm2
4.4.
Area = r 2 
=3,14 x72
=153,86cm2
4.5.1. Area =lxb +1bxh
2
1
=8x5 +26x5
=40+15
=55cm2
4.5.2. 55cm2  (100 ) 2 
=0,0055m2
QUESTION 5
NO.
ANSWER
5.1.1. SA=2(lxb) +2(bxh)+2(lxh) 
=2(5x3) +2(3x6) +2(5x6) 
=30+36+60
=126cm2
Volume = lxbxh
=5x3x6
=90cm3
5.1.2. r2 = x2+y2
=32+42
=25
r=5
1
SA=2bxh +lxb+lxb+lxb
1
=23x4 +12x3+12x5+12x4
=6+36+60+48
=150cm2
1
Volume =2bxhxH
1
5.2.
=23x4x 12
=72cm3
Volume = lxbxh
= 8x6x2
46
=96cm3
QUESTION 6
NO.
6.1.
ANSWER
Sample is a set of data selected from a
population e.g. Grade 8 class
Population is the broader group of people
to whom one is intending to generalise the
results of the study e.g. learners at
Maswameni primary school
6.2
51
0
52
3
53
347
54
4
55
5
56
48
58
8
59
0
6.3.1 Mode =2
6.3.2. score Tallies
Frequency
1
111
3
2
1111
5
3
1111
4
4
11
2
5
11
2
6
1111
4
6.3.3.
Scores
Scores of a six sided dice
thrown 20 times
6
5
4
3
2
1
0
1
2
3
4
5
Dice Sides
6.4.1 Netball 
6.4.2. 34 x800 =272
100
47
6
GRADE 9 SEPTEMBER COMMON TASK
QUESTION 1
4x3 and 4x2 are ____
1.1.
A. Like terms
1.2.
B. Unlike terms
C. Constants
D. Variables
Which number satisfies the equation: x(x-1)=x+8
A. 0
1.3.
B. 2
C. -2
D. 3
The rule for the pattern: 14; 19; 24; 29; 34 is _____
A. 7n+7
1.4.
B. 4n+10
C. 5n-9
D. 5n+9
The gradient between the two points: (2;-1) and (3,0) is______
1
A.4
1.5.
B.2
C.
−1
D. -1
3
The volume of a rectangular prism with length =6,3cm, Breadth=4,3cm and
height =9cm is____
A. 243,81cm3
B. 19,6cm3
C. 27,09cm3
D. 20cm3
QUESTION 2
2.1. Determine the values of a, b and c
(6)
a
1
b
42
y= 5x2-3
5
2.2. Determine
the rule.
(2)
c
-5
15
9
-3
3
-1
-3
1
2.3. If 2x+3y=6; complete the table below.
X
y
0
4
6
0
48
(4)
2.4. Given: y= -3x2 +9; determine the value of y if x= -4
(3)
QUESTION 3
3.1. Given:
14p3 - 4p2 - 5 – 8(6p2 + p) +2p
3.1.1. How many terms does the expression have? (Without simplifying)
(1)
3.1.2. What is the degree of the expression?
(1)
3.1.3. What is the coefficient of p3?
(1)
3.1.4. Calculate the value of the expression if p=-1
(3)
3.2. Simplify: 3a2-5a+7 – a2+2a-19
(3)
3.3. Determine the product of: 6c (9c2 + 11c -12)+2c2
(2)
3.4. Simplify: (x-2)2 - (x+2) (x-6)
(4)
3
3.5. Simplify; 4(√𝑥 2 - √𝑥 6 - x 2)
(3)
QUESTION 4
4.1. Given: y=-33x-2; Determine the variables and constant of the equation.
(3)
4.2. Solve the following equations:
4.2.1. y + 22 =15
(2)
4.2.2. 3(x-9) = 0
(3)
4.2.3. z2 -59= 110
(3)
4.2.4. 64x -4=0
(4)
4.3. If
𝑎+3
𝑏
5
= 6; Determine b when a=5
(3)
QUESTION 5
5.1. Say whether the following graphs are (increasing, decreasing or constant)
5.1.1.
5.1.2.
5.1.3.
49
(3)
5.2. Determine the coordinates of A; B and C
(3)
5.3. Given: 𝑦 = 4𝑥 − 3, complete the table below and plot the graph.
(5)
𝒙
−3
−2
−1
0
1
2
3
𝒚
5.4. The graph below shows the number of push-ups that Angel did within a week.
Number of push-ups
Number of push-ups
24
22
20
18
16
14
12
10
8
6
4
2
0
Monday
Tuesday Wednesday Thursday
Friday
Saturday
Sunday
Days of the week
5.4.1. Which day has a least number of push-ups?
(1)
5.4.2. Which day has the highest number of push-ups?
(1)
5.4.3. What is the total number of push-ups Angel did for the week?
(2)
5.4.4. Is the graph linear or non-linear?
(1)
50
QUESTION 6
6.1. Calculate the surface area and volume of the cube below:
(8)
5cm
5cm
5cm
6.2. Calculate the surface area and volume of a triangular prism below:
(8)
5cm
4cm
m
3cm
6.3. Given the cylinder.
6.3.1. Calculate the volume of a cylinder. Write your answer in m 3 (use   3,14 ) (4)
r= 1,5mm
h= 5,3mm
6.3.2. If only the radius of the can is doubled, what will be the new volume of the
cylinder?. (Write answer in mm3)
(3)
51
GRADE 9 SEPTEMBER COMMON TASK MEMORANDUM
QUESTION 1
QUESTION
1.1.
1.2.
1.3.
1.4.
1.5.
ASNWER
B
C
D B
QUESTION 2
NO.
ANSWER
2.1.
y=5x2-3
=5(1)2-3
a=2
y=5x2-3
42+3 =5x2
45=5x2
9=x2
b=3
y=5x2-3
c=5(5)2-3
c=5x25-3
=125-3
c=122
2.2.
2.3.
2.4.
Tn = -3n+2
x
-3
0
3 6
y
4
2
0
-2
y= -3x2 +9
=-3(-4)2 +9
=-3x16+9
=-48+9
=-39
52
A
QUESTION 3
NO.
ANSWER
3.1.1. 5 terms
3.1.2. Third degree
3.1.3. 14
3.1.4. 14p3 - 4p2 - 5 – 8(6p2 + p) +2p
=14(-1)3-4(-1)2-5-8[6(-1)2+(-1]+2(-1) 
=-14-4-5-48+8-2
=-65
3.2.
3a2-5a+7 – a2+2a-19
=3a2-a2-5a+2a+7-19
=2a2-3a-12
3.3.
6c (9c2 + 11c -12)+2c2
=54c3 +66c2-72c +2c2
=54c3 +68c2-72c
3.4.
(x-2)2- (x+2) (x-6)
=(x-2) (x-2)- (x+2) (x-6) 
=x2-4x+4 –[x2-4x-12] 
= x2-4x+4- x2+4x+12
=16
3.5.
3
4(√𝑥 2 - √𝑥 6 - x 2)
=4(x-x2-x2) 
=4(x-2x2) 
=4x-8x2
QUESTION 4
NO.
ANSWER
4.1.
Variables: x and y
Constant: -2
4.2.1
. y + 22 =15
Y=15-22
Y=-7
53
4.2.2
3(x-9) = 0
3x-27 =0
3x=27
27
x= 3
x=9
4.2.3. z2 -59= 110
z2=169
z2=132
z=13
4.2.4. 64x -4=0
(44)x =4
44x =41
4x=1
1
x=4
4.3.
𝑎+3
5
=
𝑏
6
(5) +3 5
=6
𝑏
8 5
=
𝑏 6
5b=48
48
b= 5
3
b=95
QUESTION 5
NO.
ANSWER
5.1.1.
Decreasing
5.1.2.
increasing
5.1.3.
Constant
5.2.
A(0;-2) ; B(2;0)  and C(4;2) 
5.3.
𝒙
−3
−2
−1
0
𝒚
-15
11
-7
-3
1
1
2
5
3
9
54
𝑦 = 4𝑥 − 3
5.4.1.
Monday
5.4.2.
Thursday
5.4.3.
10+13+19+22+11+12+15=102
5.4.4.
Non-linear
QUESTION 6
NO.
ANSWER
6.1.
SA = 6(lxb)
=6(5x5)
=6x25
=150cm2
Volume=lxbxh
=5x5x5
=125cm3
6.2.
1
SA= bxh +lxb+lxb+lxb
1
2
=23x4 +12x3+12x5+12x4
=6+36+60+48
=150cm2
1
Volume =2bxhxH
1
=23x4x 12
=72cm3
6.3.1.
Volume= r 2 xh
=3,14x1,5x5,3
=24,96mm3
24,96
=1000000000
55
=0,000000025m3
6.3.2.
Volume= r 2 xh
=3,14x(1,5x2)x5,3
=49,93mm3
56
GRADE 8 JUNE EXAM
QUESTION 1
1.1.
100 x10 -100 10 
A. 990
1.2.
B. 900
C. 910
D. 901
C. R64 925, 00
D. R16 250,
C. Isosceles
D. Right-
What is 75% of R65 000,00
A. R48 750, 00
B. R65 075, 00
00
1.3.
Which triangle has two equal sides
A. Equilateral
B. Scalene
angle
1.4.
What is the size of K
A. 76o
1.5.
B.104o
C. 38o
D. 142o
What is the value of y
60°
8y-4
A. 8
B. 64
C. 56
D. 68
QUESTION 2
2.1. Write 470000000 in scientific notation
(2)
2.2. Write 792 as the product of its prime factors.
(2)
2.3. Determine the LCM and HCF of 28 and 42
(4)
2.4. Simplify the following:
2.4.1. 2 xy  17 x  8 y  25 xy  xy  9 y
(2)
2.4.2. (a-b) (2a+b)
(2)
57
2.4.3. ( x  y)  5( x  7)
(2)
2.4.4. (5 x3 ) 2 .6 x 2
(2)
a 2b3 .4ab3
a 2b 2
(3)
2.4.5.
2.5. Calculate the following:
2.5.1. 117 + (-104) – (-33) + 62 =
(2)
2.5.2. 121  3 64  33 
(2)
2.5.3. 23.53  72 
(2)
2.6. Solve the following equations:
2.6.1. y  22  4
(2)
2.6.2. x 2  2  14
(3)
QUESTION 3
3.1. The temperature recorded at Bloemfontein increased from -2oC to 13oC.
What is the difference in temperature?
(2)
3.2. Happiness uses 2 packets of jelly to make a pudding for 6 people. How many
packets of jelly does she need to make a pudding for 18 people?
3.3. Increase 72 in the ratio 4:3
(3)
(2)
3.4. A car travels a distance of 220km in 2 hours on a tarred road. How many
kilometers can it travel in 3 hours at the same speed?
3.5. Calculate the discount of 8% on the TV set that cost R8 600, 00
(3)
(2)
3.6. Ralph invests R17 000, 00 for 3 years at 9% per annum. What will the value
of the investment be at the end of 3 years?
(3)
3.7. Sipho has $11 USD and wants to buy computer game which costs $10 AUD.
Does he have enough money to buy the game? if 1,00 USD =1,15 AUD. Give a
reason
(3)
QUESTION 4
4.1. Given the sequence: 2; 4; 8; 16;___; ___; ___
4.1.1. Extend the sequence by three more terms
(3)
4.1.2. Describe the general rule of the sequence in words
(2)
58
4.1.3. Write the general rule of the sequence in algebraic language
(2)
4.1.4. What will the value of term 8 be?
(2)
4.2. Given the geometric pattern:
1
3
2
4.2.1. Draw figure number 5
(2)
4.2.2. Write the general rule of the pattern in algebraic language.
(2)
4.3. Complete the flow diagram below.
(4)
-3
2x2 -6
4.3.2.____
__
4.3.1.____
__
2
4.4. Complete the table below:
x
y
1
4
2
9
(4)
3 7
4.4.2___
14 4.4.1___ 249
4.5. Given the equation: y=
3x 2
. Find the value of y; if x  4
6
(3)
QUESTION 5
5.1. Find the value of the unknown angles with reasons.
5.1.1. Find the size of  a.
(2)
5.1.2. Find the size of  b.
(2)
59
5.1.3. Find the size of  x and y
(4)
5.1.4. Find the size of x.
(2)
5.2. Given  ABC, D marked on BC with AD  BC. Prove that  ABD   ACD
(4)
A
B
C
D
5.3. Given  ACD, B marked on AC and E marked on AD. BE is drawn. BE//CD.
Prove that  ACD ///  ABE
(4)
A
B
C
E
D
60
GRADE 8 JUNE EXAM MEMORANDUM
QUESTION 1
QUESTION
ANSWER
1
2
3
4
5
A A C B D
QUESTION 2
ANSWER
2.1. 4,7x108
2.2. 
2 792
2 396
2 198
3 99
3 33
11
3
2
2 x3 x11
2.3. 28 and 42
HCF: 28(1;2;4;7;14;28) 
42( 1;2;3;6;7;14; 21;42)
LCM: 28(28; 56; 84;112;...) 
42(42; 84 ;126;...)
HCF is 14
LCM is 84
2.4.1. 2 xy  17 x  8 y  25 xy  xy  9 y
=-22 xy -17 x +17 y 
2.4.2. (a-b) (2a+b)
2a2+ab-2ab-b2
2a2+ab-b2
2.4.3. ( x  y)  5( x  7)
= x  y  5x  35 
=  4 x  y  35 
2.4.4.
(5 x3 ) 2 .6 x 2
= 25 x 6 .6 x 2 
=150 x 6 2
=150 x 8 
a 2b3 .4ab3
2.4.5.
a 2b 2
= 4.a 2 1 2b3 3 2 
=4 ab4 
2.5.1.
117 + (-104) – (-33) + 62
61
=117-104+27+36 
=76 
2.5.2. 121  3 64  33 
=11+4-27
=12
2.5.3. 23.53  72 
=8x125+49
=1049
2.6.1. y  22  4
y  26 
2.6.2. x 2  2  14
x 2  16 
x 2  42
x4

QUESTION 3
ANSWER
3.1. 13oC –(-2oC)=15oC
3.2. 2 packets ---6 people
x ----18 people
18.2

x=
6
x =6 packets
4
3.3. 72x =96
3
3.4. 220 km--- 2 hours
x --- 3 hours
220x3

x=
2
x =330 km
3.5.
8
 R688 
8 600,00 x
100
3.6. A =P(1+rt) 
=R17 000, 00 (1+0,09 x 3) 
= R21 590, 00
3.7. $1,00 USD =1,15 AUD
$11 USD=11,5 AUD
Yes he has 11,5 AUD which is enough to buy the computer game. 
62
QUESTION 4
ANSWER
4.1.1. 32; 64;128
41.2. Multiply the previous term by 2 to get the next term 
4.1.3. Tn = 2.2n-1
4.1.4. Tn = 2.2n-1
T8 = 2.28-1
T8 = 2.27
T8 = 256
4.2.1.

4.2.2. Tn = 3n+1
4.3.1 Tn =2x2 -6
=2(-3)2-6 
=18-6
=12
4.3.2. 2 =2x2 -6 
2x2-8=0
x2-4=0
(x-2) (x+2)=0
X=2 or -2 
4.4.1 Tn = 5n-1
Tn=5(7)-1
Tn=34
4.4.2. Tn = 5n-1
249 =5n-1
5n=249+1
5n=250
n=50
3x 2
4.5. y=
6
3( 4) 2
y=

6
48
y= 
6
63
y= 8
QUESTION 5
ANSWER
5.1.1.
a=89o (sum of ' s in a  )
5.1.2.
b=102o  (opp. ' s of quard) 
5.1.3. x= 120o (corr. ' s ) 
y=60o  (vert. opp. ' s )
5.1.4. 4x+x =90o  (complimentary ' s )
5.2. In  ABD and  ACD
AB=AC (given) 
AD is common
 ADB =  ADC=90o (given) 
 ABD   ACD (90o, H, S) 
5.3. In  ACD and  ABE
A is common 
 ABE=  C(corr.  ’s)
 AEB=  D (corr.  ’s) 
 ACD ///  ABE (A,A,A) 
64
GRADE 9 NOVEMBER 2017
GENERAL EDUCATION AND TRAINING
MATHEMATICS
NOVEMBER 2017
GRADE 9
MARKS: 100
TIME: 2 HOURS
This question paper consists of 9 pages and a formulae sheet
65
INSTRUCTIONS TO CANDIDATES
Read the following instructions carefully before answering the questions
1. This question paper consists of ten (10) questions. Answer ALL of them.
2. Question one (1) consists of multiple choice questions. Write the letter
corresponding to the correct answer.(e.g. 1.6 A).
3. Clearly show all the calculations and answers must be rounded off to TWO
decimal places unless otherwise stated.
4. An approved calculator (non-programmable and non-graphical) may be used.
5. The examination duration is 2 hours.
6. The examination is out of 100 marks.
7. Number your answers EXACTLY as the numbering system indicated on this
question paper.
8. The diagrams are NOT drawn to scale.
9. You are reminded of the need for clear presentation in your answers.
66
QUESTION 1
1.1
(2)
Simplify √4𝑥 2
A
2𝑥 −2
B
2𝑥
2𝑥 2
C
D
−2𝑥 2
1.2
Two Twice the number less 15 is equal to the number. Then, find the number?
A
1.3
5
B
2
C
3
D
(2)
15
Below is a square ABCD with a perimeter of 16cm.Calculate the length of the
(2)
diagonal AC
A
1.4
1.5
8 𝑐𝑚
B
16𝑐𝑚
4√32 𝑐𝑚
C
D
4√2 𝑐𝑚
A polygon can be defined as:
(2)
A
a closed three-dimensional shape with straight sides.
B
a closed two-dimensional shape with three or more straight sides
C
a closed plain figure with straight sides.
D
a closed figure with length, width and height.
The correct tally table for the following data:
(2)
2 ; 3 ; 1 ; 2 ; 3 ; 4 ; 1 ; 3 ; 2 ; 2 ; 1 ; 1; 4; 1; 2; 3 is?
Number
1
A
Tally
B
Number
1
C
Number Tally
1
Tally
D
Number Tally
1
2
2
2
2
3
3
3
3
4
4
4
4
[10]
67
QUESTION 2
2.1
Write 0.0010034 𝑐𝑚 in scientific notation.
2.2
Calculate the following without using a calculator .Show in each case all calculations.
2.2.1
2.2.2
2.2.3
2.2.4
2.3
3
(1)
(3)
1
√0.16 +√8
(2𝑥 2 + 5)0 − 1
(2)
6
(2)
52
x 53
(𝑚 − 1)(−4𝑎2 )(𝑚 + 1)
(3)
Solve the following:
2.3.1
6(𝑦 + 2) − 𝑦 = −8
(3)
2.3.2
𝑥−2
(3)
2.3.3
Tutsi is currently 2 years old and his sister Camilla, is half his age. By the time
3
+2=
𝑥+1
5
Tutsi will reach 42 years, how old will be Camilla?
(2)
[19]
QUESTION 3
3.1
Factorise fully:
3.1.1
𝑥 2 + 3𝑥 − 4
(2)
3.1.2
8𝑦 2 − 72
(2)
3.1.3
𝑥(𝑛 + 3) + 𝑡(𝑛 + 3)
(2)
[6]
68
QUESTION 4
4.1
Marlene has won R120 000 from the scratch card and must invest the money. She must first
pay off her debt of 60% of the winning amount before investing the rest of the money. She
has to choose between the two following options:
Option 1
Simple interest of 7.5% per annum for 3 years.
Option 2
Compound interest of 6.8% per annum for 3 years.
4.2
4.1.1
How much was Marlene’s debt?
(2)
4.1.2
Calculate how much was invested?
(2)
4.1.3
What is the actual interest obtained if she chooses to invest through option 1?
(2)
4.1.4
What is the actual interest obtained if she chooses to invest through option 2?
(2)
4.1.5
Which investment option is best for Marlene? Give a reason.
(2)
The distance between Acornhoek town and Nelspruit city is 140km.The motorist left
Acornhoek town at exactly 7: 00 𝑎𝑚 intending to attend the meeting in Nelspruit that
1
was scheduled to commence in 1 hours later. Upon arrival he phoned his wife and
2
informed her that he arrived in Nelspruit 10 minutes before the commencement of the
meeting.
4.2.1
4.2.2
At what time did the motorist arrived in Nelspruit? Write your answer in 24
hour notation.
(1)
What was the motorist’s average speed?
(2)
[13]
69
QUESTION 5
5.1
Study the graph below
5.1.1
Calculate the gradient of this graph
(2)
5.1.2
Use the gradient to determine the equation of the graph above
(2)
5.1.3
Use the equation obtained in QUESTION 5.1.2 to find the y- value when 𝑥 =
(2)
0.5
5.2
Below is a triangle ABC drawn on a Cartesian plane. Reflect the given shape along
(3)
the y-axis such that 𝐴 → 𝐴′ , 𝐵 → 𝐵 ′ 𝑎𝑛𝑑 𝐶 → 𝐶 ′ . Now write down the new coordinates
𝐴′ , 𝐵 ′ and 𝐶 ′
[9]
70
QUESTION 6
6.1
Below is a rectangle ABCD. AB=14 units , AD=12 units, DE=9 units and HC=5 units.
6.1.1
Calculate the length AH.
(3)
6.1.2
Calculate the area of the triangle ADE
(2)
6.1.3
Calculate the perimeter of the triangle AEH.
(4)
[9]
QUESTION 7
7.1
In the diagram below OABC is a straight line and ∠EBD= 𝑦 + 10° ; ∠𝐸𝐵𝐴 = 3𝑦 − 20°
and GH//EF.
7.1.1
Determine the size of y and give reasons for your answer. Show all your
(4)
working.
7.1.2
Calculate the size of
GAC, give reasons for your answer.
71
(4)
7.2
Study the diagram below:
7.2.1
Prove that
AEB III
ADC.
7.2.2
If AB=5𝑐𝑚, BC=3 𝑐𝑚, CD=9 𝑐𝑚, BE=𝑥 𝑐𝑚, ED=y 𝑐𝑚 and AE=4 𝑐𝑚.
(4)
Determine the length of the line segments 𝑥 𝑎𝑛𝑑 𝑦 (correct to two decimal
(5)
places).
[17]
QUESTION 8
8.1
Consider the rectangular prism given below and answer the questions that follow:
7 𝑐𝑚
5 𝑐𝑚
10 𝑐𝑚
8.1.1
Calculate the surface area of the rectangular prism.
(2)
8.1.2
Determine the volume of the rectangular prism above.
(2)
8.1.3
What will be the new volume if the length of the rectangular prism is
doubled?
What is the capacity of the original container (rectangular prism) when filled
with water? Give your answer in litres.
(3)
8.1.4
(2)
[9]
72
QUESTION 9
9.1
The following stem and leaf shows the marks (%) obtained by a certain group of 17
students after writing an examination. The total sum of their marks is 707.
Stem
Leaf
1 1
7
2 2
4
6
3 1
2
4
4 6
9
5 0
3
6 5
7
3
7
7 0
key
9.2
5 4 = 54
9.1.1
Determine the modal mark?
(1)
9.1.2
Find the median of the marks.
(1)
9.1.3
Write down the range.
(1)
9.1.4
Calculate the average of the marks.
(2)
Use the above data and answer the following questions:
9.2.1
What is the probability of obtaining a mark greater than 50%?
(2)
9.2.2
What is the probability of getting a mark that is a perfect square?
(1)
[8]
73
GRADE 9 NOVEMBER 2017 MEMORANDUM
GENERAL EDUCATION AND TRAINING
MATHEMATICS
MEMORANDUM
NOVEMBER 2017
GRADE 9
MARKS:10
0
TIME:2
HOURS
This memorandum consists of 10 pages
74
QUESTION
ANSWER
1.1
1.2
B D
1.3
C
1.4
1.5
B B
TOTAL MARKS
10
QUESTION 2
No
Answer
Mark Allocation
2.1
0,0010034 = 1,0034 × 10−3 𝑐𝑚
1 mark for the
Marks
1
correct answer
2.2.1
1 mark for writing
1
3
√0.16 +√8
16
√
=√100 + √2×2×2 
4
1
=10 + 2
=
10
9
2.2.2
simplifying

4+5
10
=10 
1×1×1
1 mark for
4(1)+1(5)
=
3
+√2×2×2
100
1×1×1
3
16
3
1 mark for the
correct answer
OR 0,9
(2𝑥 2 + 5)0 − 1
1 mark for
= 1−1
simplifying
= 0
(2𝑥 2 + 5)0
2
1 mark for the
correct answer
2.2.3
6
52
1 mark for writing
× 53
= 6 × 53−2 OR
6×5×5×5
=6×5
= 6×5
5×5

53−2 or for
expanding
1 mark the correct
= 30= 30
answer
75
2
2.2.4
(𝑚 − 1)(−4𝑎2 )(𝑚 + 1)
1 mark for
= (𝑚 − 1)(𝑚 + 1)(−4𝑎2 )
(𝑚2 − 1)( −4𝑎2 )
= (𝑚2 − 1)( −4𝑎2 )
1 mark for
= −4𝑎2 𝑚2 + 4𝑎2 
simplifying
= 4𝑎2 (1 − 𝑚2 )
1 mark for the
3
correct answer
2.3.1
6(𝑦 + 2) − 𝑦 = −8
1 mark for collecting
6𝑦 + 12 − 𝑦 = −8
like terms
3
6𝑦 − 𝑦 = −8 − 12
5𝑦 = −8 − 12
1 mark for
5𝑦 = −20
simplifying
−20
5
𝑦=
1 mark for the
𝑦 = −4
correct answer
2.3.2
𝑥−2 2 𝑥+1
+ =
3
1
5
1 mark for
3
multiplying all terms
Find the LCM of 3;1 and 5 and multiply each term by that
by 15
LCM
𝑥−2
15 (
3
2
) + 1 (15) = (
𝑥+1
5
) 15
1 mark for
5(𝑥 − 2) + 30 = 3(𝑥 + 1)
simplifying
5𝑥 − 10 + 30 = 3𝑥 + 3
2𝑥 = −17
1 mark for the
𝑥 = −17

2
correct answer
OR
𝑥 = −8,5
2.3.3
1
2
1 mark for the
of 2
2
method
1
× 2= 1
2
∴ 𝐶𝑎𝑚𝑖𝑙𝑙𝑎 𝑖𝑠 1 𝑦𝑒𝑎𝑟 𝑜𝑙𝑑 𝑛𝑜𝑤
When Tutsi reaches 42 years, the sister will be 1 year
younger than him i.e, Camilla will be 42 − 1 = 41 𝑦𝑒𝑎𝑟𝑠
1 mark for the
correct answer
[19]
76
QUESTION 3
3.1.1
𝑥 2 + 3𝑥 − 4
1 mark for each factor
2
8𝑦 2 − 72
1 mark for the factor
2
8(𝑦 2 − 9)
(𝑦 − 3)
8(𝑦 − 3)(𝑦 + 3) 
1 mark for the factor
Multiply 1st and last terms i.e. 𝑥 2 (−4) = −4𝑥 2
Find the factors of −4𝑥 2 𝑤ℎ𝑜𝑠𝑒 𝑠𝑢𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑑𝑑𝑙𝑒 𝑡𝑒𝑟𝑚:
𝑇ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 − 𝒙 𝑎𝑛𝑑 𝟒𝒙.
𝑇ℎ𝑒𝑛 𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑡ℎ𝑒 𝑚𝑖𝑑𝑑𝑙𝑒 𝑡𝑒𝑟𝑚 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒𝑠𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠.
= 𝑥 2 + 3𝑥 − 4
= 𝑥 2 − 𝒙 + 𝟒𝒙 − 4
= 𝑥(𝑥 − 1) + 4(𝑥 − 1)
= (𝑥 − 1)(𝑥 + 4)
OR
𝑥 2 + 3𝑥 − 4
= (𝑥 − 1)(𝑥 + 4)
3.1.2
𝑦+3
3.1.3
𝑥(𝑛 + 3) + 𝑡(𝑛 + 3)
1 mark for the factor
= (𝑥 + 𝑡)(𝑛 + 3)
(𝑥 + 𝑡)
2
1 mark for the factor
(𝑛 + 3)
[6]
QUESTION 4
4.1.1
60
1 mark for method
Debt = 100 × 𝑅120 000
2
1 mark for the answer
=𝑅72 000
OR
40
= 𝑅120 000 − 100 (120 000)
= 𝑅72 000
4.1.2
40
100
1 mark for method
× 𝑅120 000= 𝑅48 000
77
2
1 mark for the correct
ALTERNATIVE SOLUTION
answer
= 𝑅120 000 − 𝑅72 000
= 𝑅48 000
4.1.3
OPTION 1
1 mark for substitution
2
𝐼 = 𝑃𝑅𝑇
7,5
1 mark for the answer
𝐼 = 𝑅48 000 (100) 3
𝐼 = 𝑅10 800
4.1.4
OPTION 2
1 mark for calculating
𝐴 = 𝑝(1 + 𝑖)𝑛
the accumulated
6,8
2
amount
𝐴 = 48 000(1 + 100)3
𝐴 = 𝑅58 472,95
1 mark for the answer
𝐼 = 𝑅58 472,95 − 𝑅48 000
= 𝑅10 472,95
4.1.5
Option 1because the interest is relatively higher by
1 mark for choosing
R327,05 compared to option 2
option 1
2
1 mark for the reason
4.2.1
07:00 hours + 1 hour 20 minutes
1 mark for the answer
1
1 mark for formula
2
08: 20 𝐻𝑜𝑢𝑟𝑠
4.2.2
𝑠𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒

20
= 140 ÷ (160)
1 mark for answer
= 105𝑘𝑚/ℎ𝑟
[13]
QUESTION 5
5.1.1
Consider any two points i.e (0; 5) and (4;0)
𝑦1 − 𝑦2
∴ 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑥1 − 𝑥2
1 mark for correct
substitution
5−0
= 0−4
1 mark for the correct
= −1,25
answer
78
2
5.1.2
Equation:
𝑦−5
𝑥−0
=
1 mark for simplifying
−5
4
(gradient)
1 mark for the correct
4𝑦 − 20 = −5𝑥
answer
𝑦 = −5
𝑥 + 5
4
5.1.3
𝑦=
𝑦=
−5
𝑥+5
4
−5
4
2
1 mark for substitution
(0,5) + 5
2
1 mark for the answer
𝑦 = 4,375
𝑦 = 4,38
5.2
𝐴′ (−2; 4)  , 𝐵 ′ (−2; 3) and 𝐶 ′ (−4; 3)
1 mark for each pair of
3
coordinates
[9]
QUESTION 6
6.1.1
(𝐴𝐻)2 = (𝐴𝐵)2 + (𝐵𝐻)2 
1 mark for formulae
3
(𝐴𝐻)2 = (14)2 + (12 − 5)2 
(𝐴𝐻)2 = (14)2 + (7)2
1 mark for substitution
𝐴𝐻 = √(14)2 + (7)2
1 mark for the answer
𝐴𝐻 = 7√5 units
𝐴𝐻 = 15,65 units
6.1.2
1
Area of a triangle ADE = 2 𝑏 × ℎ
1 mark for substitution
2
1
= 2 × 9 × 12
= 54 𝑢𝑛𝑖𝑡𝑠 2 
6.1.3
1 mark for the answer
(𝐴𝐸)2 = (𝐴𝐷)2 + (𝐷𝐸)2
1 mark for calculating
(𝐴𝐸)2 = (12)2 + (9)2
AE
𝐴𝐸 = √144 + 81
𝐴𝐸 = 15 𝑢𝑛𝑖𝑡𝑠
1 mark for calculating
(𝐸𝐻)2 = (𝐸𝐶)2 + (𝐶𝐻)2
EH
(𝐸𝐻)2 = (14 − 9)2 + (5)2
1 mark for adding
(𝐸𝐻)2 = (5)2 + (5)2
𝐴𝐸 + 𝐸𝐻 + 𝐻𝐴
79
4
𝐸𝐻 = √25 + 25
𝐸𝐻 = √50 units
1 mark for the correct
∴ 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝐴𝐸 + 𝐸𝐻 + 𝐻𝐴
answer
= 15 + √50 +7√5
= 37,72 𝑢𝑛𝑖𝑡𝑠
[9]
QUESTION 7
7.1.1
3𝑦 − 20 + 𝑦 + 10 = 90 (Complimentary ∠s add up 90°) 1 mark for
4𝑦 = 90 + 10
statement.
4𝑦 = 100
1 mark for the
𝑦 = 25°
reason
OR
1 mark for
3𝑦 − 20 + 𝑦 + 10 + 90 = 180°(𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 ∠s)
simplifying
4𝑦 + 80° = 180°
1 mark for the
4𝑦 = 180° − 80°
correct answer
4𝑦 = 100°
7.1.2
4
𝑦 = 25°
∠𝐺𝐴𝐶 ?
1 mark for
∠EBC =25° + 10° + 90° = 125°
(Given)
4
calculating ∠EBC
∠𝐸𝐵𝐶 =∠GAC= 125°(corresponding ∠s, EF//GH
1 mark for the
Alternative Solution
reason
∠ABF=∠EBC
(Vertically opposite ∠s)
1 mark equating
=25° + 10° + 90°
∠𝐸𝐵𝐶 =∠GAC=
= 125°
125°
∠GAC=∠ABF=125°(Alternate interior ∠s)
1 mark for the
reason
OR
1 mark for finding
∠ABF=∠EBC
1 mark for the
reason
1 mark equating
∠GAC=∠ABF=125°
80
1 mark for the
reason
7.2.1
1 mark for each
4
statement and
reason
NB// : No mark is
awarded if the
STATEMENT
In
AEB and
REASON
ADC
∠A=∠A
Common
∠AEB=∠ADC
Corresponding angles; EB//DC
∠ABE=∠ACD
Corresponding angles; EB//DC
AEB III
reason is not given
ADC.
AAA
NB: No marks if reasons are not given

7.2.2
Statement
AEB III
ADC.
𝐴𝐷 𝐷𝐶 𝐴𝐶
=
=
𝐴𝐸 𝐸𝐵 𝐴𝐵
𝐷𝐶
𝐸𝐵
Reason
1 mark for the
Proved in 7.2.1
statement and
Proportional sides
reason
𝐴𝐶
= 𝐴𝐵 
1 mark for
𝐷𝐶 𝐴𝐶
=
𝐸𝐵 𝐴𝐵
9 8
=
𝑥 5
45 = 8𝑥
𝑥 = 5.625 𝑐𝑚
1 mark for
𝑥 = 5.63 𝑐𝑚
𝑥 = 5.63 𝑐𝑚
𝐴𝐷
1 mark for
𝐴𝐸
𝐴𝐶
= 𝐴𝐵
𝐴𝐷 𝐴𝐶
=
𝐴𝐸 𝐴𝐵
4+𝑦 8
=
4
5
1 mark for the
20 + 5𝑦 = 32
correct value of 𝑦 =
5𝑦 = 32 − 20
2,4 𝑐𝑚
5𝑦 = 12
𝑦 = 2,4 𝑐𝑚
81
5
[17]
QUESTION 8
8.1.1
Surface area of a rectangular prism:
1 mark for
= 2𝑙𝑏 + 2𝑙ℎ + 2𝑏ℎ
substitution
= 2(10𝑐𝑚 x 5𝑐𝑚) + 2(10𝑐𝑚x7𝑐𝑚)+2(5𝑐𝑚x7𝑐𝑚)
1 mark for the
=100𝑐𝑚2 + 140𝑐𝑚2 + 70𝑐𝑚2
answer
2
= 310𝑐𝑚2 
8.1.2
Volume=𝑙 × 𝑏 × ℎ
1 mark for
= (10𝑐𝑚) × 5𝑐𝑚 × 7𝑐𝑚
substitution
= 350𝑐𝑚3 
1 mark for the
2
answer
8.1.3
𝑉𝑜𝑙𝑢𝑚𝑒𝑁𝑒𝑤 = 2𝑙 × 𝑏 × ℎ
1 mark for the
= 2(10𝑐𝑚) × 5𝑐𝑚 × 7𝑐𝑚
deriving the formula
= 700𝑐𝑚3 
1 mark for
3
substitution.
1 mark for the
answer
8.1.4
1000𝑐𝑚3 = 1 𝑙𝑖𝑡𝑟𝑒 = 1000𝑚𝑙
1 mark for formula
2
350𝑐𝑚3 =?
350
× 1𝑙𝑖𝑡𝑟𝑒
1000
1 mark for the
answer
= 0.35 𝑙𝑖𝑡𝑟𝑒𝑠
[9]
QUESTION 9
9.1.1
Mode=53
1 mark for the
1
answer
9.1.2
Median=46
1 mark for the
1
answer
9.1.3
Range= highest mark − 𝑙𝑜𝑤𝑒𝑠𝑡 𝑚𝑎𝑟𝑘
= 70 − 11
= 59
82
1 mark for answer
1
9.1.4
𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑚𝑎𝑟𝑘𝑠
1 mark for
Average=mean =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠
=
substitution
707

17
1 mark for the
= 41.59%
9.2.1
2
answer
6 students scored marks above 50%
∴ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑚𝑎𝑟𝑘𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 50% =
1 mark for number
6

17
2
of students
1 mark for the
answer
9.2.2
49 is the only mark that is a perfect square.
1
1 mark for the
1
answer
∴ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 17
[8]
GRAND TOTAL : 100 MARKS
83
GRADE 8 NOVEMBER 2018
MST ACADEMY SCHOOLS
MATHEMATICS FINAL EXAM
GRADE 8
NOVEMBER 2018
QUESTION PAPER
TOTAL MARKS: 100
TIME: 2 Hours
Number of pages including cover page: 7
84
INSTRUCTIONS AND INFORMATION
1. This question paper consists of FIVE questions. Answer ALL the questions.
2. Question 1 consists of 5 multiple choice questions. Learners must write the
letter of the correct answer in their answer book. (e.g.: 1. 1 B)
3. Clearly show ALL the calculations, diagrams, you have used in determining
the answers as marks will be awarded according to that.
4. An approved calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.
5. If necessary, answers should be rounded off to TWO decimal places,
unless stated otherwise.
6. Number the answers EXACTLY as the questions are numbered.
7. It is in your own interest to write legibly and to present the work neatly.
8. Drawings are not necessarily drawn to scale.
85
QUESTION 1
1.6.
The LCM of 6 and 9 is ___
B. 6
1.7.
B. 9
B. Tn = 10n+13
C. Tn = 10 +3n
D. Tn = 3n
The transformation that has taken place on ABC is ___
A. Translation
1.9.
D. 54
The general rule for the sequence: 3; -7; -17; ... is ___
A. Tn =-10n+13
1.8.
C. 18
B. Reflection
C. Rotation
D. Reduction
Suzan has made a 13cm square cake. What length of ribbon will she need to
put around the edges of the cake?
A. 26cm
B. 13cm
C. 52cm
D. 39cm
1.10. The range of the given set of data: 2;3;4;5;7;9. is ___
A. 7
B. 2
C. 30
D. 9
[5X2=10]
QUESTION 2
2.2.
Calculate the following:
2.2.1. 13 – 8 + 27 x 3 =
(2)
2 3
- =
5 7
(3)
2.2.2. 3
2.2.3. 0,213 + 12, 01 – 7,87 x 0,5=
2.3.
(2)
Simplify the following:
b5c 7
2.3.1. 4 5 
b c
2.3.2. ( x 2 )3 .
(2)
2x 5
=
x3
(3)
2.3.3. 4 x( x 2  3)  3x(2 x 2  5
(3)
86
2.3.4.
2.4.
3
512
 33 - 49
64
(3)
Solve the following equations
2.4.1. 2 x  28  6
(3)
2.4.2. 2 4 x  9  25
(3)
2.5.
There are 132 learners in Grade 8. The ratio of boys to girls is 1: 2.
2.5.1. How many boys are there?
(2)
2.5.2. How many girls are there?
(2)
2.6.
Ralph deposits R450 in a new savings account. No further deposits or
withdrawals are made. Calculate the interest he will earn after 6 years at
4,75% simple interest.
2.7.
(4)
Sophy went to Italy. She changed £325 into euros (€). The exchange rate
was £1= €1, 68. How much euros did she get?
(2)
[34]
QUESTION 3
3.4. Extend the number patterns by three more terms.
(3)
1; 4; 9; ___; ___; ___
3.5.
Given the geometric pattern.
1
2
3
3.5.1. Draw the fifth figure of the geometric pattern
(2)
3.5.2. Write down the general rule to help calculate the number of squares of
the figures.
(2)
3.5.3. How many squares will figure 22 have?.
3.6.
Determine the rule on the flow diagram
56
7
9
11
________
72
88
87
(2)
(3)
3.7.
Draw the graph of y  x  5
(4)
[16]
QUESTION 4
4.1. Name four properties of a parallelogram
(4)
4.2. Given the kite with AB =AC and BD =CD. Prove that  ABD   ACD
(4)
4.3. Given ABC with PQ//AB. Prove that ∆ ABC /// ∆ PQC
(3)
B
A
Q
P
4.4. Find the sizes of x, y and z, with reasons.
C
(6)
Q
P
33
47o
o
y
z
x
R
4.5. What is the sum of the interior angles of a kite?
88
S
(2)
4.6. Given the platonic solid.
4.6.1. What is the name of the platonic solid?
(2)
4.6.2. How many vertices, faces and edges does it have?
(3)
[24]
QUESTION 5
5.1. Find the area of the figure below.
(4)
3cm
5cm
10cm
5.2. Calculate the
surface area of the rectangular prism.
(4)
2cm
6cm
12cm
5.3 Calculate the mean of the set of data.
7, 8, 9, 4, 5, 10,12, 23, 44
89
(2)
5.4. Given the bar graph
Number of learners in Grade 8
35
30
25
20
15
10
5
0
8A
8B
8C
Boys
8D
8E
Girls
5.4.1. Which data sets are being compared?
(1)
5.4.2. Which class has the least number of boys?
(1)
5.4.3. Which class has the highest number of girls?
(1)
5.4.4. How many classes are there in Grade 8?
(1)
5.5. A jar containing seven yellow marbles, ten green marbles and five red marbles.
What is the probability that a green marble will be picked?
(2)
TOTAL: 100
90
GRADE 8 NOVEMBER 2018 MEMORANDUM
MST ACADEMY SCHOOLS
MATHEMATICS FINAL EXAM
GRADE 8
NOVEMBER 2018
MEMORANDUM
TOTAL MARKS: 100
TIME: 2 Hours
Number of pages including cover page:
5
91
QUESTION 1
QUESTION 1.1.
ANSWER
C
1.2.
A
1.3.
B
QUESTION 2
NO.
ANSWER
2.1.1. 13 – 8 + 27 x 3 = 86 
2.1.2.
2 3
3 5 7
17 3
=  
5 7
119  15
=

35
104
34
=
2 
35
35
2.1.3. 0,213 + 12, 01 – 7,87 x 0,5=
8,29
2.2.1 b 5 c 7
b 4c5
=b5-4.c7-5 
=b.c2
2.2.2
2x 5
( x 2 )3 . 3
x
=x2x3 . 2x5-3
=x6.2x2
=2x8
2.2.3. 4 x( x 2  3)  3x(2 x 2  5
=4x3+12x-6x3+15x
=-2x3+27x
2.2.4.
512
3
 33 - 49
64
8
=  27  7 
4
=22
2.3.1. 2 x  28  6
2x=34
2 x 34


2
2
X=17
2.3.2. 2 4 x  9  25
2 4 x  16 
24x  24 
4x=4
X=1
2.4.1. 1:2
92
1.4
C
1.5
A
ALLOCATION OF MARKS
2 marks for the answer
1 mark for changing mixed
numbers to improper fraction
1 mark for simplification
1 mark for the answer
MARKS
2
3
2 marks for the answer
2
1 mark for 2nd law of
exponents
1 mark for the answer
2
1 mark for application of 3rd
and 2nd laws of exponents
1 mark for simplification
1 mark for the answer
3
2 marks for expansion of
binomials
1 mark for the answer
3
1 mark for simplification
2 marks for the answer
3
1 mark for additive inverse
1 mark for simplification
1 mark for the answer
3
1 mark for additive inverse
1 mark for exponential form
1 mark for the answer
3
2 marks for the answer
2
Boys: girls
1
 x132

3
 44
2.4.2. 1:2
Boys: girls
2
= x132
3
=88
2.5
I= Prt 
4,75
=R450x
x6 
100
=R128,25
2.6
£1= €1, 68
£325 x1,68 = £546
2 marks for the answer
2
1 mark for the formula
1 mark for substitution
2 marks for the answer
4
2 marks for the answer
2
QUESTION 3
NO.
ANSWER
3.1.
1; 4; 9;16; 25; 36
3.2.1.
ALLOCATION OF
MARKS
1 mark for each term
2 marks forthe answer
MARKS
2 marks for the answer
1 mark for substitution
1 mark for the answer
2
2
1 mark for any correct
method used
2 marks for the rule
3
3
2

3.2.2. Tn= n+3
3.2.3. Tn= n+3
=22+3
=25
3.3.
Input
Output
7
56
9
72
11
88
Relationship
8x7
8x9
8x11
8xn
Tn = 8n
93
3.4.
1 mark for the x
intercept
1 mark for y the
intercept
1 mark for the straight
line passing through
intercepts
1 mark for the scale
QUESTION 4
NO.
ANSWER
4.1.
Sum of interior angles equal to
360o
Opposite angles are equal
Opposite sides are parallel
Diagonals bisect each other
4.2.
. In Δ ABD and Δ ACD
AD is common
AB =AC (given) 
BD=CD (given) 
4
ALLOCATION OF MARKS
1 mark for each property
MARKS
4
1 mark for each statement
and reason
1 mark for conclusion
4
1 mark for each statement
and reason (two)
1 mark for conclusion
3
2 marks for each statement
and reason
6
2 marks for the answer
1 mark for the answer
1 mark for each answer
2
2
3
ALLOCATION OF MARKS
1 mark for the formulae
1 mark for substitution
1 mark for simplification
1 mark for the answer
MARKS
4
 ABD   ACD (S,S,S) 
4.3.
4.4
In Δ ABC and Δ PQC
𝐶̂ = 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛
𝐴𝐵̂𝐶 = 𝑃𝑄̂ 𝐶 (corresponding angles)

or
𝐵𝐴̂𝑃 = 𝑄𝑃̂𝐶 (corresponding angles)
∴ Δ ABC /// Δ PQC
X =33o (alt. ' s) 
180o-(33o+47o)=100o (sum
Y=
of
' s in a  )
Z=180o-100o = 80o (sum of ' s
on a strt. Line) 
4.5.
360o
4.6.1. Octahedron 
4.6.2. 8 faces
12 edges
6 vertices
QUESTION 5
NO.
ANSWER
5.1.
𝐴𝑟𝑒𝑎 = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∎ + 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆
1
= 𝑙 × 𝑏 + 2 𝑏𝑥ℎ
94
=10x5 +
1
x10x3
2
=50+15
=65cm2
5.2.
5.3.
SA = 2(lxb) + 2 (lxh) + 2 (bxh) 
= 2(12x6) + 2 (12x2) + 2( 6x2) 
=144 +48 +24
=216 cm2
7  8  9  4  5  10  12  23  44
Mean =

9
1 mark for the formula
1 mark for substitution
1 mark for simplification
1 mark for the answer
1 mark for adding and diving
by the number of data set
1 mark for the answer
1 mark for the answer
1 mark for the answer
1 mark for the answer
1 mark for the answer
1 mark for the answer not
simplified
2 marks for simplified answer
=13,56
5.4.1
5.4.2.
5.4.3.
5.4.4.
5.5.
Number of boys and girls
8B
8B
Five 
10 5
P (Green) =
= 
22 11
TOTAL: 100
95
4
2
1
1
1
1
2
GRADE 8 NOVEMBER 2019
GENERAL
EDUCATION AND TRAINING
MST ACADEMY
GRADE 8
MATHEMATICS
FINAL EXAMINATION
NOVEMBER 2019
MARKS
: 100
DURARION : 2 hours
This question paper consists of 10 pages
96
INSTRUCTIONS AND INFORMATION
1.
This Question Paper has two sections, Section A and Section B.
2.
Section A has 10 multiple choice questions each with 4 possible
answers.
3.
Answer Section A on the answer-sheet provided by circling the letter of
the correct answer (A – D).
4.
Section B has 8 questions. Answer ALL questions.
5.
Read through the questions carefully and make sure that you allocate
enough time for each question.
6.
Show all working unless otherwise stated.
7.
Round off your answers to two decimal places unless otherwise stated.
8.
A non-programmable calculator may be used unless otherwise stated.
9.
Write as neatly and clearly as possible.
10. Tear off the answer sheet (Section A) and the grid (Question 5.1.2)
from your question paper and submit them with your answer book.
11. Write your name on each sheet of paper submitted.
97
SECTION A
QUESTION 1
1.
Circle the letter of the correct answer from the four possible answers on
the mark-sheet provided.
1.1.
The HCF of 18; 30 and 48 is:
A 3
1.2.
B 4
C 6
D 8
(1)
The transformation that moves each point of a figure in the same
direction and same distance is
A Enlargement. B Rotation C Reflection D Translation
1.3.
4,8 − 2,042 = ⋯
A 2,38
1.4.
B 2,420
C 2,756
D 2,758
(1)
The sum of interior angles of a quadrilateral is
A 180°
1.5.
(1)
B 360°
C 90°
D 270°
(1)
If the temperature is −7℃ and then it rises by 15℃, what will the
temperature be?
A −22℃
1.6.
If
3
4
B 22℃
C 8℃
D −8℃
(1)
of the 4 500 000 people in a city are between the ages of 15
and 40, how many people is this?
A 3 375 000
1.7.
B 281 250
C 337 500
D 33 750 000 (1)
Which quadrilateral is best described by this characteristic? A
rectangle with adjacent sides equal.
A Trapezium. B Kite
1.8.
C Rhombus D Square.
(1)
Which equation best expresses the statement:
“The sum of the squares of 𝑡 and 𝑝 is 25”.
A 2𝑡 + 2𝑝 − 25
B √𝑡 2 + √𝑝2 − 25
D 𝑡 2 + 𝑝2 = 25
C (𝑡 + 𝑝)2 − 25
(1)
98
1.9.
In ∆𝐴𝐵𝐶 below 𝑥 = …
A
52°
3𝑥 − 15°
44°
B
C
A 43°
B 36°
C 37°
D 33° (1)
1.10. Six counters in a bag are numbered 3; 4; 7; 9; 10; 11.
One counter is drawn at random from the bag. The probability that
the number drawn is a prime number is ….
A
1
B
6
1
2
C
1
3
D
4
6
(1)
[10]
SECTION B
QUESTION 2
2.1. Simplify without using a calculator.
3
2.1.1. √125 − √
2.1.2.
1
2
1
(3)
4
1
1
1
4
3
4
+ ÷( − )
(4)
2.1.3. (−5) − (−8) − (−7) − (+2)
2.2.Write 0,00125 in scientific notation.
(3)
(2)
[12]
QUESTION 3
3.1 Mr Catch saves money for his intended relocation to Britain. He keeps himself
updated with the exchange rates by watching the daily business news on TV. On
a particular day the Rand/ Pound exchange rate was £1= R18, 40. How
many pounds will he get in exchange for his savings of R500 000?
(2)
3.2 A pair of jeans priced at R550 is put on sale for 25 % discount. How much is
the new price?
(2)
3.3 Mrs Tate saves a lump sum of R50 000 for her daughter’s university fees. If
99
her money is invested for five years on simple interest option of 5 % per
annum, how much pay-out will she receive at the end of five years?
(3)
[7]
QUESTION 4
4.1.
The first four terms of a number pattern is 2 ; 7 ; 12 ; 17 ; …
4.1.1. Find the next three terms of the pattern.
(3)
4.1.2. Find the general term of the pattern in the form Tn = …
4.2.
(3)
Study the flow diagram below and answer the questions that follow.
4.2.1. Calculate the output values of a and b
(4)
4.2.2. Draw a graphical representation of the relationship
(3)
[13]
QUESTION 5
5.1. Add 3𝑥 − 7𝑥 2 + 4 and 3 + 2𝑥 − 𝑥 2
5.2.Simplify:
5.2.1. 2𝑥(1 − 𝑥 + 𝑦) − 𝑥(𝑦 − 3 + 2𝑥)
5.2.2.
5.2.3
(4𝑎2 )(−3𝑎3 )
4𝑥
−
5.3. Find the value of
10𝑥 2 −15𝑥
𝑥
2
(3)
(3)
−6𝑎4
12𝑥 2 −4𝑥
(2)
(3)
5𝑥
+
𝑦
6
if 𝑥 = 2 and 𝑦 = −3
(2)
[13]
QUESTION 6
6.1
Solve for 𝑥
100
6.1.1. 2x – 1 = –5
(2)
6.1.2. 3x – 2 = x + 4
(2)
6.1.3.
𝑥
−3
+ 2 = −2
(3)
6.2
The sum of two numbers is 165 and their difference is 27. Find the
numbers.
(3)
[10]
QUESTION 7
7.1. In the diagram below calculate the size of x .
x
0
20
(2)
0
60
7.2. Calculate the size of m.
T
U
35
0
0
50
Y
m
W
X
(3)
7.3.1 A is transformed object to image A'. Mention two types of
transformation that took place.
101
(2)
7.3.2 The diagram show rotation of rectangle ABCD. Describe the rotation in
words.
(2)
7.4 Complete the prove that ▲ABC ≡ ▲EDC
A
400
B
1
C
2
400
D
E
102
𝐴̂ = …….. = 400
= ……..
[ ………………..]
[ Vertically opp. angles]
𝐵̂ = ……..
[ Remaining angles]
……………………… [……………….]
(6)
[15]
QUESTION 8
8.1
Study the shape below and answer the questions that follow.
20cm
10cm
6cm
9cm
8.2
8.1.1 Calculate the perimeter of the shape.
(2)
8.1.2 Calculate the area of the shape.
(3)
Below is a closed rectangular box.
2m
3m
6m
8.2.1 Calculate the volume of the box.
(2)
8.2.2 Draw a net of the box.
(1)
8.2.3 Calculate the surface area of the box.
(3)
[11]
QUESTION 9
9.1 Below are marks of a grade 9 class after writing a mathematics test out
of 40. Answer the questions that follow based on the data. All answers
must be rounded off to one decimal place.
27 25
28 31
27 29 31
25 27 28
24
28
25
28
103
27
25
28
28
29
31
24
24
26
30
30
25
9.1.1 Calculate the range.
(2)
9.1.2 What is the mode of the data?
(1)
9.1.3
(2)
Find the median.
9.1.4 Calculate the mean.
(2)
9.2 Write down the probability of getting an odd number when you roll a six sided
die.
(2)
[9]
GRAND TOTAL [100]
104
SECTION A
QUESTION 1
Name: _____________________________ Class: ________
Marks:
10
Circle the letter of the correct answer. Submit this with your answer sheets.
Question
Answer
1.
A
B
C
D
2.
A
B
C
D
3.
A
B
C
D
4.
A
B
C
D
5.
A
B
C
D
6.
A
B
C
D
7.
A
B
C
D
8.
A
B
C
D
9.
A
B
C
D
10.
A
B
C
D
105
Name: _____________________________ Class: ________
QUESTION 5
5.1.2
Submit this with your answer sheets.
106
GRADE 8 NOVEMBER 2019 MEMORANDUM
MST ACADEMY
MATHEMATICS
GRADE 8
MARKING GUIDELINE
FINAL EXAM
NOV 2019
MARKS
: 100
DURATION
: 2 Hours
107
SECTION A
QUESTION 1
1.1
C

1.2
D

1.3
D

1.4
B

1.5
C

1.6
A

1.7
D

1.8
D

1.9
D

1.10
B

[10]
SECTION B
QUESTION 2
2.1
Simplify
2.1.1
125 –
3
1
4
= 5 – 
=4 
(3)
+
2.1.2
2+1
=
4
÷(
)
4−3
÷ ( 12 )
= ÷(
=
–
×(
)
)
= 9
(4)
108
2.1.3
(–5) – (–8) – (–7) – (+2)
= –5 + 8 + 7 –2
(3)
= 8
2.2
2
0,00125 = 1, 25 × 10-3 
[12]
QUESTION 3
3.1

He will get
= £ 27 173, 91 
3.2
New price = R550 ×
(2)

= R412,50  or any other correct method used.
(2)
A= P(1 + in) 
3.3
= 50 000(1 + 0,05 × 5) 
= R62 500 
(3)
[7]
QUESTION 4
4.1.1 22 ; 27 ; 32
(3)
4.1.2 Find the general term of the pattern in the form Tn = …
d=5
T1 = 2 = 5(1) – 3
T2 = 7 = 5(2) – 3
T3 = 12 = 5(3) – 3  (method)
4.2 .1 a = 3(0)+4
(4)
=4
b = 3(2)+4 
= 10 
 (y-int)
 (x-int)
 (line)
(3)
4.2.2
109
[13]
QUE STION 5
- 7x 2+3x + 4
5.1
(2)
+(- x2+2x+3)
−8𝑥 2 + 5𝑥 + 7
5.2 5.2.1.
2x(1 – x + y) – x (y – 3 + 2x)
= 2x – 2x2 + 2xy - xy +3x – 2x2 
= - 4x2 +xy+5x
5.2.2
(3)
(4𝑎2 )(−3𝑎3 )
−6𝑎4
=
−12𝑎5
−6𝑎4

(3)
110
=2a
12𝑥 2 −4𝑥
5.2.3
4𝑥
10−15𝑥
−(
5𝑥
)
= 3 x – 1 - 2 x + 3 
= x +2 
𝑥
5.3
2
𝑦
2
−3
6
2
6
+ = +
(3)

= = 
(2)
[13] .
QUESTION 6
6.1 6.1.1
6.1.2
2x – 1 = - 5
2x = - 5+1 
2x =- 4
x=-2
3x – 2 = x + 4
(2)
3x– x = 4+2
2x = 6
x=3
(3)
6.1.3
−𝑥
3
= −4
−𝑥 = −12
∴ 𝑥 = 12
(3)
111
6.2
x + y=165
+(x-y=27)
2x=192
x=96
y=165-96
(3)
y=69
[10]
QUESTION 7
7.1
x+200= 600 (Alternate angles parallel lines)
x=600 - 200
(2)
x = 400 
7.2
(3)
𝑇̂+ 500 + 350 =1800 (Sum of angles in a ∆)
𝑇̂ = 180 0 - 850
= 950 
𝑇̂ = m = 950  (Opposite angles of a parm)
Rotation and Translation.
(2)
7.3.1
7.3.2 Rotation
90
degrees
anticlockwise.
(2)
7.4
𝐴̂ = 𝐸̂  = 40

̂
𝐵̂ = 𝐷

0
[ Alternate Ls, AB॥DE] 
[ Vertically opp. angles]
[Alternate Ls, AB॥DE]
∆ABC ≡ ∆EDC  [AAA] 
(6)
[15]
[10]
QUE STION 8
8.1.1 Perimeter 10 cm + 9 cm + 6 cm + 11 cm + 4 cm + 20 cm 
=
= 60 cm 
(2)
8.1.2 Area = 10 𝑐𝑚 × 9 𝑐𝑚 + 4 𝑐𝑚 × 11 𝑐𝑚 
(3)
112
= 90 𝑐𝑚2 + 44 𝑐𝑚2 
=134 𝑐𝑚2
8.2.1
(2)
𝑉 = 𝑙𝑏ℎ
= 6 𝑚 × 3 𝑚 × 2𝑚
= 36 𝑚3
8.2.2
(1)
8.2.3 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 2(6 𝑐𝑚 × 3 𝑐𝑚) + 2(3 𝑐𝑚 × 2 𝑐𝑚) + 2(6 𝑐𝑚 × 2 𝑐𝑚) 
= 36 𝑐𝑚2 + 12 𝑐𝑚2 + 24 𝑐𝑚2
= 72 𝑐𝑚2 
(3)
[11]
QUE STION 9
9.1
9.1.1
Range = 31 – 24 
9.1.2
9.1.3
(2)
= 9
Mode = 28 
(1)
24 24 24 25 25 25 25 26 26 27 27 27 27
28 28 28 28 28 28 29 29 30 30 31 31 31
27+28
Median =
2

= 27,5 
(2)
9.1.4
Mean =

= 27,35 
(2)
113
9.2
(2)
1
P(Odd Number) =2 
Total
114
[09]
[100]
GRADE 9 NOVEMBER 2019
GENERAL
EDUCATION AND TRAINING
GRADE 9
MATHEMATICS
NOVEMBER 2019
MARKS: 100
TIME: 2 HOURS
This question paper consists of 9 pages
115
INSTRUCTIONS TO CANDIDATES
Read the following instructions carefully before answering the questions
1. This question paper consists of six (6) questions.
Answer ALL of them.
2. Question one (1) consists of multiple choice questions. Write the letter corresponding to
the correct answer.(e.g. 1.5 A).
3. Clearly show all the calculations and answers must be rounded off to TWO decimal
places unless otherwise stated.
4. An approved calculator (non-programmable and non-graphical) may be used.
5. The examination duration is 2 hours.
6. The examination is out of 100 marks.
7. Number your answers EXACTLY as the numbering system indicated on this question
paper.
8. The diagrams and graphs are NOT necessarily drawn to scale.
9. You are reminded of the need for clear presentation in your answers.
116
QUESTION 1
Choose the letter of the correct answer from the four possible answers given.
1.1
Which of the following numbers is rational?
A
√−1
B
𝜋
C
√5
D
1,23̇
(2)
1.2
1.3
1.4
3
Calculate the value of √0,027𝑥 3
A
0,16𝑥 3
B
3𝑥 3
C
0,3𝑥
D
27 9
𝑥
1000
(2)
The 3-D object with 5 faces,5 vertices and 8 edges is
A
Triangular based pyramid
B
Triangular prism
C
Cylinder
D
Square base pyramid
(2)
In triangle ABC, BC is produced to D such that ∠𝐵 = 3𝑥° , ∠𝐵𝐶𝐴 = 𝑥° and ∠𝐴𝐶𝐷 = 5𝑥°.
Calculate ∠𝐵𝐴𝐶
A
180°
B
60°
C
30°
D
4𝑥°
(2)
1.5
If 𝑝 is a point on the line defined by 𝑦 = 𝑥,then the coordinates of 𝑝 are …
A
(−3; 3)
B
(2; −2)
C
(−2; 2)
D
(−3; −3)
(2)
[10]
QUESTION 2
2.1
Decrease 72 in the ratio of 4: 5
(2)
2.2
Determine the highest common factor(HCF) of 36 and 48
(3)
117
2.3
Mr Kgarudi purchased a radio for R12 000.He used a hire purchase loan with an agreement of 10% per
annum for a period of 2 years. The loan requires that a 11,5% cash deposit is paid.
2.4
2.5
2.6
2.3.1
How much is the cash deposit?
(2)
2.3.2
How much will Mr Kgarudi pay over the 2 years excluding the cash deposit?
(3)
2.3.3
In total how much will Mr Kgarudi pay for his radio?
(2)
Simplify the following:
2.4.1
√0.27𝑥 2 − 0.02𝑥 2
(2)
2.4.2
3𝑦 . 3𝑦+2
3𝑦+1
(2)
2.4.3
2𝑎−3 𝑏−2 × 𝑎4 𝑏 2
(2)
Factorise fully the following:
2.5.1
−49 + 𝑘 2
(2)
2.5.2
5(−𝑥 + 𝑦) + 𝑥 − 𝑦
(2)
Solve the following equations:
2.6.1
2.6.2
𝑡 2 + 𝑡 = 12
3𝑦 − 1 =
2.6.3
20𝑏 = 1
2.6.4
3𝑥+1 =
(3)
𝑦−3 𝑦+1
+
2
3
(3)
(2)
1
27
(2)
[32]
QUESTION 3
3.1
Study the geometric pattern below and answer the questions that follows:
Figure
Number of lines
1
6
2
11
3
(1)
3.1.1
Complete the table by writing down the number of lines on figure 3
3.1.2
Write down the general rule of the pattern(𝑇𝑛 ) =
118
(2)
3.2
3.3
3.1.3
Hence, determine which figure will have 71 lines.
(2)
3.1.4
Use the rule or otherwise to determine the number of lines on 6th figure
(2)
Study the graph below and answer the questions that follow:
3.2.1
What name is given to this type of a graph?
(1)
3.2.2
Determine the slope of the graph above.
(2)
3.2.3
Write down the coordinates where the graph cuts the y-axis.
(1)
3.2.4
Hence find the equation of the given graph in the form, 𝑦 =
(2)
The diagram below shows the transformation where shape 𝐷𝐸𝐹𝐺 has been transformed into 𝐷 ′ 𝐸 ′ 𝐹 ′ 𝐺 ′
3.3.1
Show by means of calculations why this type of transformation is called reduction?
(3)
3.3.2
If the perimeter and area of shape 𝐷𝐸𝐹𝐺 is 100𝑐𝑚 and 550𝑐𝑚2 respectively, calculate the area
(2)
of 𝐷 ′ 𝐸 ′ 𝐹 ′ 𝐺 ′
[18]
119
QUESTION 4
4.1
Refer to the diagram below where WZ= 9𝑐𝑚,WY= 23𝑐𝑚 and ZX= 6𝑐𝑚. Calculate:
W
23 cm
9 cm
Z 6 cm X
4.2
Y
4.1.1
the length of XY.
(4)
4.1.2
the area of∆𝑊𝑋𝑌.
(2)
A cylinder with a height of 9m and a radius of 3.5m are filled with water (𝜋 = 3.14):
4.2.1
Determine the volume of water in the cylinder.
4.2.2
Calculate the surface area of the cylinder above:
(3)
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 2𝜋𝑟(𝑟 + ℎ)
(2)
[11]
QUESTION 5
5.1
Calculate, with reasons the values of 𝑥 and 𝑦 in the diagram below:
(4)
120
5.2
On the diagram below, It is given that ∠𝐴𝐹𝐸 = 2𝑦 − 10° and ∠𝐷𝐺𝐹 = 3𝑦 + 25°.Now for which value of
(4)
𝑦 are AB and CD parallel? Show all steps of your working and give reasons for your statements.
5.3
In the figure below, 𝐴 and 𝐶 are points on a circle with centre O. 𝐵 is a point on chord 𝐴𝐶 such that OB ⊥
(6)
AC. Prove that AB = BC.
𝐂
𝐁
1
2
𝐀
1 2
𝐎
5.4
Prove that ∆𝑀𝐿𝐽 ≡ ∆𝐾𝐿𝐽
(3)
5.5
Study the diagram below and Prove that ∆𝐶𝐵𝐷 lll ∆𝐶𝐴𝐸
(4)
121
[21]
QUESTION 6
6.1
The table shows the results of a particular survey that was carried out to determine the goals of different
genders. Suppose that the survey was based on 500 respondents from each of the two groups.
Goals
Gender
Poor
Total
Males
150
𝒂
500
Females
400
100
500
𝒃
450
1 000
Total
6.2
Rich
6.1.1
Calculate the values of 𝑎 and 𝑏 in the table.
(2)
6.1.2
Calculate the probability of picking a person with a goal of getting rich?
(2)
A fruit farmer collected the plums and measured their masses, the results are shown on the diagram below
122
40
35
Frequency
30
25
20
15
10
5
0
0
10
20
30
40
50
60
70
80
90
Mass of plums
6.2.1
What is the total number of plums measured?
(1)
6.2.2
Determine the modal class
(1)
6.2.3
Which method of representing data is shown above? Give a reason
(2)
[8]
GRAND TOTAL 100
123
GRADE 9 NOVEMBER 2019 MEMORANDUM
GENERAL
EDUCATION AND TRAINING
GRADE 9
MATHEMATICS
NOVEMBER 2019
MARKING GUIDELINES
MARKS: 100
This marking guideline consists of 8 pages
124
QUESTION 1.1
ANSWER
1.2
D C
1.3
1.4
1.5
D
B
D
TOTAL MARKS
10
QUESTION 2
No
Answer
2.1
4
5
Mark Allocation
4
× 72
1 mark 5 × 72
= 57.6
Marks
2
1 mark for the correct
answer
2.2
2.3.1
36 = 2 × 2 × 3 × 3
1 mark for listing
48 = 2 × 2 × 2 × 2 × 3
factors of 36 and 48
∴ 𝐻𝐶𝐹 = 2 × 2 × 3
1 mark for 2 × 2 × 3
= 12
1 mark for the answer
11.5
1 mark for
100
× 𝑅12 000
3
2
11.5
× 𝑅12 000
100
= 𝑅1 380. 𝑂𝑂
1 mark for the correct
answer
2.3.2
𝐴 = 𝑃(1 + 𝑖𝑛)
1 mark for formulae
10
= (𝑅12 000 − 𝑅1 380)(1 + 100 × 2)
3
1 mark for correct
substitution
= 𝑅12 744.00
1 mark the answer
OR
OR
𝐼 = 𝑃𝑅𝑇
10
= 𝑅10 620 ×
×2
100
1 mark for formulae
= 𝑅2 124
1 mark for interest
∴ 𝑂𝑣𝑒𝑟 2 𝑦𝑒𝑎𝑟𝑠 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 ∶
value
𝑅10 620 + 𝑅2 124
1 mark for the answer
for interest
𝑅12 744.00
2.3.3
Total for the radio:
1 mark for 𝑅12 744 +
= 𝑅12 744 + 𝑅1 380
𝑅1 380
= 𝑅14 124
1 mark for the correct
answer
125
2
2.4.1
√0.27𝑥 2 − 0.02𝑥 2
1 mark for simplifying
= √0.25𝑥 2
1 mark for the
2
answer
25 2
=√
𝑥
100
(5)2
= √(10)2 𝑥 2 
=
5
𝑥
10
1
2
= 𝑥
2.4.2
3𝑦 . 3𝑦+2
3𝑦+1
=3
𝑦+𝑦+2−(𝑦+1)
1 mark for Simplifying
2
1 mark for the answer

3𝑦+1 
2.4.3
2𝑎−3 𝑏−2 × 𝑎4 𝑏 2
1 mark for simplifying
= 2𝑎−3+4 𝑏−2+2
1 mark for the answer
2
2𝑎
2.5.1
−49 + 𝑘 2 
2 marks for the
= (𝑘 − 7)(𝑘 + 7)
answer
2
𝑶𝑹
= (−7 + 𝑘)(7 + 𝑘)
2.5.2
5(−𝑥 + 𝑦) + 𝑥 − 𝑦
1 mark for
= −5(𝑥 − 𝑦) + 1(𝑥 − 𝑦)
−5(𝑥 − 𝑦) + 1(𝑥 − 𝑦)
= (−5 + 1)(𝑥 − 𝑦)
1 mark for the answer
2
= −4(𝑥 − 𝑦)
OR
5(−𝑥 + 𝑦) + 𝑥 − 𝑦
−5𝑥 + 5𝑦 + 𝑥 − 𝑦
−4𝑥 + 4𝑦
−4(𝑥 − 𝑦)
2.6.1
𝑡 2 + 𝑡 = 12
1 mark for the
𝑡 2 + 𝑡 − 12 = 0
standard form
𝑡 2 − 3𝑡 + 4𝑡 − 12 = 0
1 mark for factors
126
3
𝑡(𝑡 − 3) + 4(𝑡 − 3) = 0
1 mark for both values
(𝑡 + 4)(𝑡 − 3)= 0
of 𝑡
𝑡+4=0
and 𝑡 − 3 = 0
𝑡 = −4
2.6.2
𝑡 = 3
3𝑦 − 1 =
𝑦−3 𝑦+1
+
2
3
1 mark for multiplying
all terms by the LCM
Find the LCM of 2 and 3 and multiply each term by that
LCM
6(
3
6
1 mark for simplifying
3𝑦−1
𝑦−3
1
2
) = (6)
𝑦+1
+ (6) (
3
)
6(3𝑦 − 1) = 3(𝑦 − 3) + 2(𝑦 + 1)
1 mark for the correct
18𝑦 − 6 = 3𝑦 − 9 + 2𝑦 + 2
answer
18𝑦 − 3𝑦 − 2𝑦 = 6 − 9 + 2
13𝑦 = −1
𝑦=
2.6.3
−1
13

20𝑏 = 1
1 mark for writing as
20𝑏 = 200 
an exponent
𝑏 = 0
1 mark for the correct
2
answer
2.6.4
3𝑥+1 =
3
𝑥+1
1
27
1 mark for
2
3𝑥+1 = 3−3
−3
=3 
1 mark for the answer
𝑥 + 1 = −3
𝑥 = −4
[32]
QUESTION 3
3.1.1
1 mark for the answer
1
𝑇𝑛 = 𝑇0 + 𝑛 × 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2 marks for the
2
= 1 + 𝑛(5)
answer
Figure
1
2
3
Number of lines
6
11
16

3.1.2
= 5𝑛 + 1
127
3.1.3
3.1.4
𝑇𝑛 = 5𝑛 + 1
71 = 5𝑛 + 1
70 = 5𝑛
𝑛 = 14
1 mark for the
𝑇𝑛 = 5𝑛 + 1
1 mark for substitution
𝑇𝑛 = 5(6) + 1
1 mark for the answer
2
equation
1 mark for the answer
2
= 31
3.2.1
Linear graph
1 mark for the answer
1
3.2.2
𝑆𝑙𝑜𝑝𝑒 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
1 mark for substitution
2
𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 ((−4; 0)𝑎𝑛𝑑 (0; −2)
𝑦1− 𝑦2
=
𝑥1− 𝑥2
1 mark for then
=
answer
0−(−2)
−4−0

1
= − 2
3.2.3
(0; −2)
1 mark for answer
1
3.2.4
𝑦−(−2)
1 mark for equating
2
𝑥−0
1
= − 2
with the gradient
2𝑦 + 4 = −𝑥
1 mark for the answer
1
𝑦 = − 2 𝑥 − 2
3.3.1
𝑙𝑒𝑡 𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑘
1 mark for the
𝑆𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
∴ 𝑘 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑟𝑖𝑔𝑖𝑜𝑛𝑎𝑙
3
formulae
1 mark for the value
4
=
20
for 𝑘
1
1 mark for the reason
= 5
0 < 𝑘 < 1, 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑒 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛
3.3.2
𝐴𝑟𝑒𝑎 = (𝑘)2 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐷𝐸𝐹𝐺
1 mark for
1
= (5)2 × 550
Substitution
= 22𝑐𝑚2 
1 mark for the answer
(2)
[18]
QUESTION 4
4.1.1
(𝑊𝑌)2 = (𝑊𝑍)2 + (𝑍𝑌)2 …….Pythagoras theorem
1 mark for the
(23)2 = (9)2 + (𝑍𝑌)2 
Statement and
reason
128
4
529 − 81 = (𝑍𝑌)2
1 mark for substitution
𝑍𝑌 = √448
1 mark for ZY value
𝑍𝑌 = 21.17𝑐𝑚
1 mark for the answer
∴ 𝑋𝑌 = 𝑍𝑌 − 𝑍𝑋
𝑋𝑌 = 21.17 − 6
= 15.17𝑐𝑚
4.1.2
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑊𝑍𝑌 =
1
(21.17 × 9)
2
= 95.27𝑐𝑚2 
1 mark for the area of
2
∆ 𝑊𝑍𝑌& and
∆𝑊𝑍𝑋
1
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑊𝑍𝑋 = (6 × 9)
2
1 mark for the area of
= 27𝑐𝑚2 
∆𝑊𝑋𝑌
∴ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑊𝑋𝑌 = 95.27 − 27
= 68.27𝑐𝑚2 
4.2.1
4.2.2
𝑉 = 𝜋𝑟 2 ℎ
1 mark for formulae
= 3.14 × (3.5)2 × 9
1 mark for substitution
= 346.19𝑐𝑚3 
1 mark for the answer
Surface area= 2𝜋𝑟(𝑟 + ℎ)
1 mark for substitution
= 2 × 3.14 × 3.5(3.5 + 9)
1 mark for the answer
3
2
= 274.75𝑐𝑚2 
[11]
QUESTION 5
5.1
3𝑥 = 𝑥 + 50°………..Alt ∠s; AB//DC
1 mark for statement
2𝑥 = 50
and reason
𝑥 = 25°
1 mark for the value of 𝑥
∠ABC = 3𝑥 = 75° … … ∠s 𝑜𝑝𝑝 = sides
1 mark for 75° + 75° +
75° + 75° + 𝑦 = 180° … … … . . 𝑖𝑛𝑡 ∠s of a ∆
𝑦 = 180°. 𝑖𝑛𝑡 ∠s of a ∆
𝑦 = 180° − 150°
1 mark for the value of 𝑦
4
𝑦 = 30°
5.2
If 𝐴𝐵 // 𝐶𝐷
1 mark for statement
∠CGF = ∠AFE = 2y − 10°… … … … . 𝐶𝑜𝑟𝑟 ∠s; AB // CD
1 mark for reason
2𝑦 − 10° + 3𝑦 + 25° = 180°… … ∠s on a straight line
1 mark for statement
5𝑦 = 165°
1 mark for the answer
129
4
𝑦 = 33°
OR
∠CGF = 180° − (3𝑦 + 25°)
= 155 − 3𝑦
∠CGF = ∠AFE … … . Corr ∠s; AB // CD
155 − 3𝑦 = 2𝑦 − 10
155 + 10 = 2𝑦 + 3𝑦
165 = 5𝑦
𝑦 = 33°
5.3
Statement
1 mark for statement
Reason
6
1 mark for reason
In ∆OBA and ∆OBA
𝑂𝐶 = 𝑂𝐴
Radii
1 mark for statement
𝑂𝐵 = 𝑂𝐵
Common
and reason
∠𝐵2 = ∠𝐵1 = 90

∴ ∆𝑂𝐵𝐶 ≡ ∆𝑂𝐵𝐴
90°𝐻𝑝𝑦 𝑠𝑖𝑑𝑒
∴ AB = BC
Corresponding sides of
congruent ∆𝑠
∆𝑠 congruent 
1 mark for statement
and for reason
1 mark for statement
and reason
1 mark for statement
and reason
5.4
∠𝑀𝐿𝐽 = ∠𝐾𝐿𝐽 = 90° ………………..Given
1 mark for statement
𝐿𝐽 = 𝐿𝐽 … … … … … … … … … … … … … … … 𝐶𝑜𝑚𝑚𝑜𝑛
and reason
𝑀𝐽 = 𝐾𝐽 … … … … … … … … … … … … … … … . 𝐺𝑖𝑣𝑒𝑛
1 mark for statement
∴ ∆𝑀𝐿𝐽 ≡ ∆𝐾𝐿𝐽90°𝐻𝑝𝑦 𝑠𝑖𝑑𝑒
and reason
3
1 mark for conclusion
5.5
∠C is common
1 mark for statement
∠CAE = ∠CBD
Corresponding ∠s ; (DB//EA)
∠CEA = ∠CDB
Corresponding ∠s ; (DB//EA)
OR
∴ ∆𝐶𝐴𝐸 III ∆𝐶𝐵𝐷
4
1 mark for statement
and reason
Sum ∠s ∆
1 mark for statement
AAA
1 mark for reason
[21]
130
QUESTION 6
6.1.1
150 + 𝑎 = 500
1 mark for 𝑎 𝑣𝑎𝑙𝑢𝑒
2
𝑎 = 350
1 mark for 𝑏 𝑣𝑎𝑙𝑢𝑒
OR
𝑎 + 100 = 450
𝑎 = 350
𝑏 + 450 = 1000
𝑏 = 550
OR
𝑏 = 150 + 400
𝑏 = 550
6.1.2
550
1000
2 marks for the answer
2
1 mark for the answer
1
1 mark for the answer
1
Histogram
1 mark for the name
2
Because there is no space between the bars
1 mark for reason
11
=20
6.2.1
3 + 14 + 26 + 36 + 21
= 100
6.2.2
50 < 𝑥 < 60
OR
50-60
6.2.3
[8]
GRAND TOTAL 100
131