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www.mathsatsharp.co.za
November Exam Memorandum
Grade 9 Mathematics
Marks: 150
Time: 2 hours
Notes
1. This answer paper consists of 13 pages and 2 sections.
2. There are other possible ways to answer some of the questions than those answers given
here. However, the final answer should still be the same.
3. Should you find any errors in the test during your moderation please do let us know via email:
[email protected] so that we can correct it as soon as possible. We really
appreciate your help towards making this a fantastic free resource for all maths teachers
across South Africa.
Section 1 - Algebra
Question 1
1.1. 1.1.1. 360 2 252 2 240 2
180 2 126 2 120 2
90 2 63 3 60 2
45 3 21 3 30 2
15 3 7 7 15 3
5 5 1 5 5
1 1
360 = 23 x 32 x 5 β 252 = 22 x 32 x 7 β 240 = 24 x 3 x 5 β (3)
1.1.2. HCF = 22 x 3 = 12 β
LCM = 24 x 32 x 5 x 7 = 5 040 β (2)
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1.2. 1.2.1. β(β9)(β4) +42
β7 (3)
= β36 β 6 β
= 6 β 6 β
= 0 β
1.2.2. ββ273
β β25 (2)
= β3 β 5 β
= β8 β
1.2.3. 32
5 Γ· 3
4
7 Γ· 0.35 (3)
=17
5 Γ·
25
7 Γ·
35
100 β
=17
5 Γ
7
25 Γ
100
35 β
=68
25ππ 2
18
25 β
1.3. 1.3.1. The bank owns = 1 β1
4β 0.4 β 35% (2)
= 1 β 0.25 β 0.4 β 0.35 β
= 0 β
The bank does not own a part of the business.
1.3.2. Henrietta = π 250 000 Γ1
4= π 62 500 β (3)
Yvonne = π 250 000 Γ 0.4 = π 100 000 β
Kimberley = π 250 000 Γ 35% = π 87 500 β
1.4. 3 741 360 000 = 3.74136 x 109 β (1)
1.5. 5.627 x 10-5 = 0.00005627 β (1)
Please note: The question specifically asks that no calculator is used, this means that the steps must be shown and marks awarded for the steps. If only the answer is given only one mark should be given.
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1.6. (π₯3π¦β4)
2
(π₯2π¦β3)β2 Γπ₯3π¦0
π¦5 (3)
=π₯6π¦β8
π₯β4π¦6 Γπ₯3
π¦5 β
=π₯6π₯4
π¦6π¦8 Γπ₯3
π¦5 β
=π₯13
π¦19 β
[23]
Question 2
2.1. Given the following table:
π 1 2 3 4 b 9 d
π 15 12 9 a 0 c -24
2.1.1. a = 6 Β½ mark b = 6 Β½ mark (2)
c = -9 Β½ mark d = 14 Β½ mark
2.1.2. ππ = β3π + 18 ββ (2)
2.1.3. π100 = β3(100) + 18 (1)
π100 = β282 β
2.2. 2.2.1. String Γ· 2 + 1.5cm +1.5cm Answer
β β β (3)
2.2.2. 18 Γ· 2 = 9 9 + 1.5 = 10.5 10.5 + 1.5 12cm (2)
β β
2.2.3. 7 β 1.5cm = 5.5cm β
5.5 β 1.5cm = 4cm β
4 x 2 = 8cm β (3)
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2.2.4. (2)
Input 12cm 20cm 50cm 84cm
Output 9cm 13cm 28cm 45cm
2.3. 2.3.1. m = -3 β (3)
y-intercept = 9 β
x β intercept: 0 = β3π₯ + 9
β9 = β3π₯
π₯ = 3 β
2.3.2. (3)
Mark for y-intercept and for x-intercept β
Mark for drawing graph correctly β
Mark for labeling axes. β
2.3.3. Decreasing β (1)
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2.4. For graph a: gradient =π¦2βπ¦1
π₯2βπ₯1
=4β3
0β2
=1
β2 ππ β
1
2 β
So: π¦ = β1
2π₯ + π Substitute in a point
4 = β1
2(0) + π
π = 4 β
β΄ π¦π = β1
2π₯ + 4
For graph b: gradient =π¦2βπ¦1
π₯2βπ₯1
=3β0
2β(β2)
=3
2+2
=3
4 β
So π¦ =3
4π₯ + π Substitute in a point
3 =3
4(2) + π
π = 11
2 β
β΄ π¦π =3
4π₯ + 1
1
2 (4)
[26]
Question 3
3.1. 3.1.1. a and b β (1)
3.1.2. β9π3 β 11. ββ (2)
3.1.3. 12 β (1)
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3.1.4. 7π4π3 + 6π2π β 9ππ3 β 11π + 12 with π = β2 πππ π = 3 (3)
= 7(β2)4(3)3 + 6(β2)2(3) β 9(β2)(3)3 β 11(β2) + 12 β
= 7(16)(27) + 6(4)(3) β 9(β2)(27) + 22 + 12 β
= 3 024 + 72 + 486 + 22 + 12
= 3 616 β
3.2. 3.2.1. (9π₯ + 4)(8π₯ β 7) +π₯2+10π₯+21
π₯+3 (4)
= 72π₯2 β 63π₯ + 32π₯ β 28 +(π₯+3)(π₯+7)
(π₯+3) ββ
= 72π₯2 β 31π₯ β 28 + π₯ + 7 β
= 72π₯2 β 30π₯ β 21 β
3.2.2. (π₯ β 6)2 + (π₯ β 6)(π₯ + 6) (3)
= (π₯ β 6)(π₯ β 6) + (π₯ β 6)(π₯ + 6)
= π₯2 β 12π₯ + 36 β +π₯2 β 36 β
= 2π₯2 β 12π₯ β
3.3. 3.3.1. π₯+3
4+
π₯β6
5= 12 (4)
5(π₯+3)
20+
4(π₯β6)
20=
12(20)
20 ββ
5π₯ + 15 + 4π₯ β 24 = 240 β
9π₯ = 240 + 24 β 15
9π₯ = 249
π₯ = 272
3 β
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3.3.2. π₯2 + 10 = β7π₯ (2)
π₯2 + 7π₯ + 10 = 0
(π₯ + 5)(π₯ + 2) = 0
π₯ + 5 = 0 or π₯ + 2 = 0
π₯ = β5 or π₯ = β2 ββ
3.3.3. (π₯ + 3)(π₯ β 8) = β4π₯ (4 +13
π₯) (3)
π₯2 β 5π₯ β 24 = β16π₯ β 52 β
π₯2 β 5π₯ + 16π₯ β 24 + 52 = 0
π₯2 + 11π₯ + 28 = 0 β
(π₯ + 4)(π₯ + 7) = 0
π₯ = β4 ππ π₯ = β7 β
3.4. (π₯ + 3)(π₯ + 7) = 5ππ β (4)
π₯2 + 10π₯ + 21 β 5 = 0
π₯2 + 10π₯ + 16 = 0
(π₯ + 2)(π₯ + 8) = 0
π₯ = β2 ππ π₯ = β8 β
Not possible (as it would make the length and breadth negative which is
impossible) β
β΄ Length = -2 +3cm = 1cm and breadth = -2 + 7cm = 5cm β
[27]
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Section 2 - Geometry
Question 4
4.1. Base is 5cm β
2 angles both 30Β° β
Other two sides equal β
Drawn neatly in pencil β
Clearly and neatly labeled. β (5)
4.2. All three sides 3.5cm β
All three angles 60Β° β
Drawn neatly and in pencil and labelled. β (3)
4.3. 4.3.1. In βBEC and βBDC.
1. BE = DC Given β
2. πΈοΏ½ΜοΏ½πΆ = π΅οΏ½ΜοΏ½π· Alt angles, BA || DF β
3. BC is common β
β΄ βBEC β‘ βBDC (SAS) β (4)
4.3.2. Rectangle β
opposite sides equal and parallel AND all four angles are 90Β° β (2)
4.3.3. π·οΏ½ΜοΏ½πΆ = π΅οΏ½ΜοΏ½πΈ β Alternating angles, BD || EC β (4)
π΅οΏ½ΜοΏ½πΈ = πΆοΏ½ΜοΏ½πΉ Alternating angles, BC || EF β
β΄ π·οΏ½ΜοΏ½πΆ = πΆοΏ½ΜοΏ½πΉ Both equal to π΅οΏ½ΜοΏ½πΈ β
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4.4. 4.4.1. In βABC and βACD.
1. π΅οΏ½ΜοΏ½π΄ = πΆοΏ½ΜοΏ½π· Alternating angles Β½ mark
2. π΅οΏ½ΜοΏ½πΆ = π΄οΏ½ΜοΏ½π· Alternating angles Β½ mark
3. π΄οΏ½ΜοΏ½πΆ = π΄οΏ½ΜοΏ½πΆ Last angle in triangle Β½ mark
β΄ β CAB ||| β ACD Β½ mark (2)
4.4.2. Any two of the following angles:
πΊοΏ½ΜοΏ½π· = π΄οΏ½ΜοΏ½πΆ = 30Β° Alt angles, AD || BG β (2)
πΊοΏ½ΜοΏ½π· = π΄οΏ½ΜοΏ½πΆ = 30Β° Corrsp angles, BE || CD β
π΄οΏ½ΜοΏ½πΆ = πΈοΏ½ΜοΏ½π· = 30Β° Corrsp angles, AD || BG β
4.4.3. π·οΏ½ΜοΏ½π΅ + π΄οΏ½ΜοΏ½πΆ + π΅οΏ½ΜοΏ½π· + π΄οΏ½ΜοΏ½πΆ = 360Β° Sum of angles in quad = 360Β° β (3)
πππ π·οΏ½ΜοΏ½π΅ = π΅οΏ½ΜοΏ½π· = π¦ Opp. Angles of rhombus β
β΄ π¦ + 30Β° + π¦ + 30Β° = 360Β°
2π¦ = 300Β°
β΄ π¦ = 150Β°
β΄ π΄οΏ½ΜοΏ½π· =1
2π¦ = 75Β° Diags bisect angles in rhombus β
[25]
Question 5
5.1. 5.1.1. π΄π΅2 = π΅πΊ2 + π΄πΊ2 Pythag β
π΄πΊ2 = 10.842 β (18.78
2)
2 ββ
π΄πΊ2 = 29.3335
π΄πΊ = β29.3335
π΄πΊ = 5.42π β (4)
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5.1.2. P = 6 x 10.84m β (2)
P = 65.04m β
5.13. A = area of rectangle + 2 x area of triangles (2)
A = l x b + 2 ( Β½ x b x h)
A = 18.78 x (21.68 β 2 x 5.42) + 2 ( Β½ x 18.78 x 5.42)
A = 203.58m2 + 101.79m2
A = 305.37m2
5.1.4. Area of circle = ππ2
Area = π (21.68
2)
2 Β½ mark
Area = 369.15m2 Β½ mark
Circle area seen = 369.15 β 305.37
= 63.78m2 (2)
5.2. 5.2.1. (4)
β for 24cm (3 x 8cm width)
β for 10cm
β for 32cm (4 x 8cm length)
β for correct net shape
5.2.2. SA = 2 (l x b + l x h + h x b) β (3)
SA = 2( 32cm x 24cm + 10cm x 32cm + 24cm x 10cm) β
SA = 2 656cm2
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5.2.3. Volume = l x b x h (4)
Volume of box = 24cm x 32cm x 10cm
Volume of box = 7 680cm2 β
Volume of one toilet roll = ππ2β
= π Γ (8
2)
2Γ 10 β
= 502.65ππ2
Left over volume = volume of box β 12 x volume of toilet roll β
= 7 680cm2 β 12 x 502.65cm2
= 1648.2ππ2 β
5.3. 5.3.1. Dodecahedron β 5.3.2. Octahedron β
5.3.3. Hexagonal prism β (3)
[24]
Question 6
6.1. (5)
Original Point Image Rule
(-8; 1) (1; -8) Reflected about the line y = x β
(-4; -1) β (-1, -3) Translate 3 units right and 2 units down
(3; 2) (12; 8) β Enlarge by a factor of 4
(5; -5) (5; 5) Reflect around the x-axis β
(2; 4) β (-2; 4) Reflect about the y-axis
6.2. 6.2.1. The perimeter will become 3 times smaller β (1)
6.2.2. The area will become 9 times smaller ββ (2)
[8]
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7%
18%
14%
22%
21%
18%
Frequency of Different Coloured Cars
Red
Blue
Silver
Green
White
Yellow
Question 7
7.1.
Colour Tally Frequency
Red // 2
Blue //// 5
Silver //// 4
Green //// / 6
White //// / 6
Yellow //// 5
(2)
7.2. Green and white ββ (2)
7.3. (4)
Red =2
28 Γ 360Β° = 26Β° Green =
6
28 Γ 360Β° = 77Β°
Blue =5
28 Γ 360Β° = 64Β° White =
6
28 Γ 360Β° = 77Β°
Silver =4
28 Γ 360Β° = 51Β° Yellow =
5
28 Γ 360Β° = 64Β°
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7.4. 7.4.1. Average = 28+79+35+16+34
5 β (2)
=192
5
= 38.4 cars β
7.4.2. Range = 79 β 16 = 63 β (1)
7.4.3. Tuesday, it has the highest number of cars, which would be over peak hour traffic. (2)
[13]
Question 8
8.1. Shirt 1 1
3 S1 + P =
7
30
Pants 0.7 Shirt 2 1
3 S2 + P =
7
30
Shirt 3 1
3 S3 + P =
7
30
Shirt 1 1
3 S1 + S =
3
30
Skirt 0.3 Shirt 2 1
3 S2 + S =
3
30
Shirt 3 1
3 S3 + S =
3
30
(2)
8.2. P (pants and Shirt 1) = 7
30 ββ (2)
[4]
Total [150]