SENIOR PHASE GRADE 9 NOVEMBER 2018 MATHEMATICS MARKING GUIDELINE

[Pages:13]SENIOR PHASE GRADE 9

NOVEMBER 2018 MATHEMATICS MARKING GUIDELINE

MARKS: 140

This marking guideline consists of 13 pages.

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MATHEMATICS

(EC/NOVEMBER 2018)

INSTRUCTIONS AND INFORMATION

1. Give full marks for answers only, unless stated otherwise. 2. Accept any alternate correct solutions that are not included in the marking guideline. 3. Underline errors committed by learners and apply Consistent Accuracy (CA). 4. THE FINAL MARK MUST BE CONVERTED TO 100.

M

Method

CA

Consistent Accuracy

A

Accuracy

S

Statement

SF

Substitution in Formula

R

Reason

S/R Statement and Reason

KEYS

QUESTION 1 [10 marks] Ques.

1.1 C 1.2 A 1.3 B 1.4 A 1.5 D 1.6 B 1.7 C 1.8 A 1.9 C 1.10 D

Mark allocation

Total (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) [10]

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(EC/NOVEMBER 2018)

MATHEMATICS

QUESTION 2 [25]

Ques.

2.1

2.2.1

2.2.2

Solution

1 042 000 000 = 1, 042109 A

3z2

4

2 3

z3

7z 2

3z2

4

2 3

z3

7z 2

M

3z2 4z2 M

3

5z2 / 5 z2 CA

33

2(x 3)2 3(x 1)(2x 5)

2(x2

6x

M

9)

3(2x2

3x

M

5)

2x2 12x 18 6x2 9x 15

4x2 3x 33 CA

2.2.3

2 x 1

y

2

3y2

2y 3 xy 2

2M

3 xy 2 2y

2

M

9x2 y2 CA

4 OR

2x1 y 2

3y2

22 x2 y 2 32 y 4

M

1 x2 y2

4 M

1

9

9 x2 y2 CA

4

OR

3

Mark allocation

Total

Answer: 1Mark (1)

14 z3

3

2 7z

:

1

Mark

4z 2 : 1 Mark

3

Answer: 1 Mark (3)

(x2 6x 9) : 1 Mark

(2x2 3x 5) : 1 Mark

Answer: 1 Mark (3)

2

2y

3 xy 2

: 1 Mark

3 xy 2

2

2y

: 1 Mark

Answer: 1 Mark

OR

22 x2 y2 32 y4 : 1 Mark

1 x2 y2 4 : 1 Mark

1 9 Answer: 1 Mark

OR

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Ques.

2.2.4 2.3.1 2.3.2

Solution

2

x 1

y

2

3y2

1 2x1 y

2

M

3y2

1 22 x2 y2

M

32 y4

9x2 y2 CA

4

169x6

y p99q

0

3 x12

M

13x3 1

x4M

13A

x

ax2 5ax 6a

M

a(x2 5x 6)

CA

a(x 3)(x 2)

(2x 3y) (3y 2x)x2

(2x 3y) (2x 3y)x2 M (2x 3y)(1 x2)CA (2x 3y)(1 x)(1 x)CA

2.4.1

x 2x 3 1 23

6

x 2

6

2x 3

3

6

1

M

3x 4x 6 6

7x 0

x 0 CA

MATHEMATICS

(EC/NOVEMBER 2018)

Mark allocation

Total

1 2x1 y 2

:

1

Mark

3y2

1 22 x2 y2

:

1

Mark

32 y4

Answer: 1 Mark

(3) 13x3 : 1 Mark

x4 : 1 Mark Answer: 1 Mark

(3) a(x2 5x 6) : 1 Mark

(x 3) : 1 Mark (x 2) : 1 Mark

(3)

(2x 3y) (2x 3y)x2 : 1 Mark

(2x 3y)(1 x2) : 1 Mark

(2x 3y)(1 x)(1 x) : 1 Mark

(3)

by LCM: 1 Mark

Answer: 1 Mark

(2)

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(EC/NOVEMBER 2018)

MATHEMATICS

Ques. 2.4.2

2.4.3

Solution

x2 x 12 x2 x 12 0 (x 4)(x 3) 0M

CA

x 4 or x 3 5x2 1

5x2 522 5M

x 2 2

x 4CA

5

Mark allocation

Total

(x 4)(x 3) : 1 Mark

Both solutions: 1 Mark

(2) 52 : 1 Mark Answer: 1 Mark

(2) [25]

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6 QUESTION 3 [26]

MATHEMATICS

(EC/NOVEMBER 2018)

Ques. Solution 3.1 3.1.1

Shape

1 2 3 4

Number of rectangles 5 9 13 17

A

q 17 and r 25

3.1.2

A A

Tn 4n 1

3.1.3

Tn 4n 1

205 4n 1SF

n 51 Shape number

51

has

205

rectangles.CA

3.2.1 3.2.2 3.3.1

3.3.2

Tn n2 A1A

Tn n2 1

T10 Tn

(10)2 1

101CA

SF

A 1x2

2

A 1 2SF2

A

2 1

CA

1 B S2F 4

2 B

4

CA

... 25 ... 101

Mark allocation

Total

q = 17 and r = 25 : 1 Mark

(1) 4n : 1 Mark 1 : 1 Mark (2)

Tn =205 : 1 Mark Answer: 1 Mark

(2)

n2 : 1 Mark 1 : 1 Mark (2) SF(n 10) : 1 Mark Answer: 1 Mark

(2)

SF x = -2 : 1 Mark Answer: 1 Mark

(2) SF x = B : 1 Mark

Answer: 1 Mark

(2)

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(EC/NOVEMBER 2018)

MATHEMATICS

Ques.

Solution

3.4.1 Common difference = 7 (10) 3

y intercept = -1 Since x = 0 A

y 3x 1A

OR m y2 y1

x2 x1 m 7 (10)

2 (3)

m 3 A

3.4.2

y 3x 1 A

y 3x 1

8 3q 1 SF q 3CA

3.5.1 3.5.2

y x 3 0 x 3

x 3A

A A

A

7

Mark allocation

Total

Explanation: 1 Mark

Answer: 1 Mark

If ANSWER ONLY Full Marks

OR

m = 3: 1 Mark Answer: 1 Mark

If ANSWER ONLY Full Marks

(2) SF both values : 1 Mark

Answer: 1 Mark

(2)

Answer: 1 Mark

(1)

y x 3 x int ercept : 1 Mark y int ercept : 1 Mark Straight Line : 1Mark

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MATHEMATICS

Ques. Solution

3.6.1 Monday to Tuesday A

OR

Saturday to Sunday A

3.6.2 15 packets of sweets sold A

OR

10 packets of sweets sold A

3.6.3 Thursday to Saturday A

3.6.4 The decrease varied.A

3.6.5 The sales were constant.A

OR

No increase or decrease in the sales.A

QUESTION 4 [12] Ques. Solution

4.1 SI P.i.n M 720 1800.i.5SF

i 720 1800 5

r 720 100 1800 5

r 8% CA

4.2 Let the breadth of the original playground x

The length of the original playground x 1

The perimeter of the original playground 2 x 1 x M

4x 2 M

The length of the new playground x 4

The breadth of the new playground x 1

The perimeter of the new playground 2 x 4 x 1 M

4x 6M

The difference in perimeter 4x 6 4x 2

4meters CA

4.3 Total distance travelled 210km

Total time travelled 2,5hours M

Average Speed Dis tan ceM

Time

210km SF

2, 5hours

84km / h CA

(EC/NOVEMBER 2018)

Mark allocation Answer: 1 Mark

Total

(1) Answer: 1 Mark

(1)

Answer: 1 Mark (1)

Answer: 1 Mark

(1)

Answer: 1 Mark

(1) [26]

Mark allocation Formula: 1 Mark Substitution: 1 Mark Answer: 1 Mark

Total

(3)

2 x 1 x : 1 Mark

4x 2 : 1 Mark

2 x 4 x 1 : 1 Mark

4x 6 : 1 Mark

Answer: 1 Mark

(5)

Distance & Time : 1 Mark Formula : 1 Mark Substitution : 1 Mark

Answer: 1 Mark (4) [12]

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