SENIOR PHASE GRADE 9 SENIOR PHASE GRADE 9 SENIOR PHASE · SENIOR PHASE GRADE 9 NOVEMBER 2012 ARTS AND CULTURE MEMORANDUM MARKS: 100 ... Answer all the questions. 2. Write neatly and

Province of the EASTERN CAPE EDUCATION

SENIOR PHASE

GRADE 9

NOVEMBER 2012

ARTS AND CULTURE MEMORANDUM

MARKS: 100

This memorandum consists of 4 pages.


SENIOR PHASE

GRADE 9

NOVEMBER 2012


MARKS: 100


Name and Surname: ___________________________!!____________________________________________!!____________________________________________


SENIOR PHASE

GRADE 9

NOVEMBER 2012

ECONOMIC AND MANAGEMENT SCIENCES MEMORANDUM

MARKS: 100



SENIOR PHASE

GRADE 9

NOVEMBER 2012

MATHEMATICS

MARKS: 100 TIME: 2 hours

This question paper consists of 16 pages.


SENIOR PHASE

GRADE 9

NOVEMBER 2012

MATHEMATICS

MARKS: 100 TIME: 2 hours

This question paper consists of 16 pages.

2 MATHEMATICS (NOVEMBER 2012)

INSTRUCTIONS AND INFORMATION 1. Answer all the questions. 2. Write neatly and legibly. 3. Do not change the numbering of the questions. 4. Show all your calculations, correct your answer to TWO decimal places where

necessary. 5. A non-programmable calculator may be used.

(NOVEMBER 2012) MATHEMATICS 3

QUESTION 1 There are TEN multiple-choice questions in QUESTION 1. For each question FOUR possible answers are given and only one answer is correct. Write the number then select the letter for the correct answer and write it next to the corresponding question number. Do not rewrite the question. EXAMPLE : e.g. 1.11 If 3 loaves of bread are equally divided among 6 people, each will get: A 1

2 B 13 C 2

3 D 2

The correct answer is which is letter A. Answer: 1.11 A 1.1 Which of the following is not a property of rational numbers? A Terminating decimals B Recurring decimals C Square root of a perfect square D Cube root of a prime number (1) 1.2 (2 2 ) 3 x (2 2 ) 3 Simplified is: A 2 10 B 2 1 C 2 D 1 (1) 1.3 An output that is given by the 28th term in the sequence 5;; 9;; 13;; 17;; … is. A 112 B 113 C 116 D 117 (1) 1.4 a = 4 ; b = 6 and c = 5 then the value of 2a + bc = A 77 B 70 C 54 D 38 (1)


1.5 A pattern used to find the number of tiles used to surround square flower beds:

A 4(n + 1) B 2(2n + 1) C n 2 + 1 D n + 1 (1) 1.6 1.6 A 3D figure which has 18 edges, 8 faces and 12 vertices is a … A decagonal prism. B pentagonal prism. C hexagonal prism D square based pyramid. (1) 1.7 Triangle A is transformed to triangle A'. This type of transformation is a …

50º 50º

A translation – reduction. B rotation – reduction. C reflection – reduction. D reflection – rotation (1)

A A'


1.8 Select the graph(s) that best represent(s) the height of water in a regular cylinder bucket being filled from a tap with constant flow of water.

1 2 3

A 1 and 2 B 3 only C 1 and 3 D 2 only (1) 1.9 The spinner below is rotated. The probability that the arrow will point to a prime

number is:

10 1 2 9 3 8 4

7 6 5

A 14

B

35

C 25

D 12 (1)


1.10 The correct tally table for the following data

1; 2; 3; 1; 2; 3; 4; 1; 3; 2; 2; 1; 1 is:

[10] QUESTION 2 2.1 Angela spent of her money for entertainment. If she now has R30 left, how

much did she have at first? (2) 2.2 2.2.1 What would be the height of a stack of 200 000 sheets of paper of the

same size, if the thickness of ONE sheet of paper is 0,08928 mm? (1) 2.2.2 Write your answer of QUESTION 2.2.1 in scientific notation. (1) 2.3 Andiswa bought a R1 500 hi-fi sound system on hire purchase. The deposit

was R150 and the balance is payable monthly over 3 years at 18% p.a. simple interest.

2.3.1 Calculate the total amount she would pay for the hi-fi. (4) 2.3.2 Determine the monthly installment Andiswa will pay if the insurance

premium of R10,50 is added monthly. (1) [9]


QUESTION 3 Mr Nkuti, a young man is staying on the fifth floor of a multistory building. He prefers to use steps as a form of exercise rather than using a lift. The steps of the building are constructed as shown in the structures below.

Structure 1 Structure 2 Structure 3 Structure 4 3.1 Draw the next structure. (1) 3.2 The table below shows the relationship between the structure number and the

number of blocks. Structure number ( n) 1 2 3 4 ….. 6 n Number of blocks (b) 1 3 6 10 ….. 3.2.1 Write the general rule for any structure (i.e. nth structure). (2) 3.2.2 How many blocks can be used to form structure 6? (1) 3.3 Design a flow diagram using y = 3x – 5 where x lies between 0 and 5. (3)


3.4 Read the graph below and then answer the questions that follow.

3.4.1 Determine the equation of the graph above. (2) 3.4.2 Use the equation obtained in QUESTION 3.4.1 to find the value of 𝑦 when

x = 3 (1) 3.5 The admission policy of Jenge Junior Secondary School states that the current

year’s admission must be twice the previous year’s admission. In the fourth year the total number of learners is 1 500. How many learners were admitted in the first year (i.e. four years ago)? (4)

[14]


QUESTION 4 4.1 Factorise: 9p 2 q – 81p 2 q 3 4.2 Simplify: 4.2.1 (3x – 2 )(5x +1) (2) 4.2.2

223

432

812

zyxzyx

x xyyx

168 32

(4)

4.3 Solve for x in the equations below. 4.3.1

34

)8(32

6

xxx (4)

4.3.2 2 x2 = 64 (3) [17] QUESTION 5 5.1 The sum of the angles of any polygon is 180º (n – 2) where n stands for the

number of sides. If the sum of the angles of a regular polygon is 1 260º, calculate the number of sides. (2)

5.2 In the figure below, PQRS is a parallelogram with diagonal QS. P Q

S R

Prove that ∆ 𝑆𝑃𝑄 ≡ ∆ 𝑄𝑅𝑆. (4) 5.3 The sides of a triangle are 6 cm, 7 cm and 10 cm. Find the length of the longest

side of a similar triangle whose shortest side is 12 cm. (2)


5.4 On the set of axes given below consider quadrilateral LMNO with its coordinates and then answer the questions that follow.

5.4.1 Determine the coordinates of the image under the transformation rule

(x) → (x ; y – 7) (2) 5.4.2 Use ANNEXURE 1 to draw an image of the quadrilateral LMNO under

the transformation rule in QUESTION 5.4.1. (2) 5.4.3 On the same ANNEXURE 1, slide the image 4 units to the left. (2) [14]


QUESTION 6 6 The figure below has PQ // ST, P𝑆V = 125º and Q𝑇𝑈 = 95º. R

P Q 125º 95º V S T U

6.1 Calculate with reason(s) the size of P𝑄R. (3) 6.2 In the sketch below, AB is horizontal and CD is vertical. C

x 2 A x 1 P x 3 B x 6 x 5 x 4

D

6.2.1 Elevation (1) 6.2.2 Depression (1) [5]


QUESTION 7 7.1 Study the graphs below and then answer the questions that follow.

7.1.1 Work out the speed of train B (Hint: if speed = ) (1) 7.1.2 Which of the two trains is faster? Motivate your answer. (2) 7.2 The figure below represents a triangular prism.

7.2.1 Determine the height (h) of the base of the prism. (2) 7.2.2 Calculate the total surface area of the prism. (4) [9]


QUESTION 8 8.1 The table below shows the class interval of the exam marks of 120 learners in

Grade 9. Marks 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 100 No. of

Learners 3 5 2 9 18 28 30 12 11 2

If the pass mark is 40%, how many learners failed the exam? (1) 8.2 Talita wrote 8 Mathematics tests in 2012. For her to get Level 7 in the CASS

mark for the subject, she must get a minimum average of 80 marks for her 8 tests. What is the minimum total mark she must get in order to obtain Level 7? (1)

[2] QUESTION 9 9.1 A survey was conducted to test the relationship between the hand length and

shoe size. The table below shows 10 measurements of different hand lengths and shoe sizes.

Hand length 5 7 2 9 6 7 4 9 8 5 Shoe size 12 13 10 15 12 15 11 16 15 11 9.1.1 Use ANNEXURE 2 to draw a scatter graph using the information in the

table. (5) 9.1.2 What conclusion can you draw about the relationship between the hand

length and the shoe size? (1) 9.1.3 Find the median hand length. (2) 9.1.4 Find the mode of the shoe size. (1) 9.1.5 Calculate the mean of the shoe size. (2) 9.1.6 Determine the range of the hand length. (1) 9.2 Any game played has the following three possibilities; win, draw and loss. Ama-

Bokoboko played two friendly games. Draw a two-way table to list all possible outcomes. (3)

9.3 What is the probability of: 9.3.1 Winning both games? (1) 9.3.2 Winning 1 game and losing 1 game? (1) 9.3.3 Winning at least 1 game? (1)


9.4 The two graphs below represent the same information about the share prices in 2009.

Share Prices (R) Share Prices (R) _ 350_ 220_ 300_ . 210 _ . 250_ 200 _ 200_ . . . . 190 _ 150_ . . 180_ 100 _ 170 | | | | | | 50 | | | | | | Jan Feb Mar Apr May Jun Jul Jan Feb Mar Apr May Jun Jul Months Months

Graph 1 Graph 2 9.4.1 Which ONE of the graphs represents the information more clearly? (1) 9.4.2 Why does Graph 1 look different from Graph 2? (1) [20] TOTAL: 100


ANNEXURE 1 SURNAME : NAME : PROVINCE : DATE : QUESTION 5.4.2 and 5.4.3


SENIOR PHASE

GRADE 9

NOVEMBER 2012


MARKS: 100



SENIOR PHASE

GRADE 9

NOVEMBER 2012


MARKS: 100



SENIOR PHASE

GRADE 9

NOVEMBER 2012

ECONOMIC AND MANAGEMENT SCIENCES MEMORANDUM

MARKS: 100



SENIOR PHASE

GRADE 9

NOVEMBER 2012

MATHEMATICS MARKING GUIDELINES

MARKS: 100

This marking guideline consists of 12 pages.


QUESTION 1 1.1 D √ (1) 1.2 D √ (1) 1.3 B √ (1) 1.4 D √ (1) 1.5 A √ (1) 1.6 C √ (1) 1.7 C √ (1) 1.8 D √ (1) 1.9 C √ (1) 1.10 C √ (1) [10]

QUESTION 2

2.1 Let the amount she had be = x

Amount she spent = x

Amount left = 53 x = 30 √

53 x = 30

53 x (

15

) = 30 (15

)

3x = 150 x = 50 √ Hence Angela had R50 at first.

OR 3 : 5 = 30 : x √ 3/5 = 30/x 3x = 5 (30) x = 150/3 x = 50 √

(2)

Forming equation Answer

2.2 2.2.1 200 000 (0,08928) = 17 856 mm √ (1) Answer


2.2 2.2.2 17 856 = 1,7856 x 10 4 √ (1) Correct answer 2.3 2.3.1 Principal is R1 500 – R 150 = R1 350 √

S I = P x R x T √

= R1 350 x 10018

x 3

= R729 √ Amount paid = Principal + Interest + Deposit = R1 350 + R729 + R150 = R2 229 √ (4)

Calculating the principal Formula Interest Answer

2.3.2

Monthly instalment = + insurance

= 362079R

+ insurance

= R57,75 + R10,50 = R68,25 √ (1) Answer

[9] QUESTION 3 3.1

√ Structure 5 (1) Correct drawing


3.2 3.2.1 1st term : 1(2) = 1 2 2nd term : 2(3) = 3 2 3rd term : 3(4) = 6 2 4th term : 4(5) = 10 2

Therefore )1(2

nn √√

(2) Answer 3.2.2

b = )1(2

nn

If n = 6

b = )16(26

= 3 x 7

b = 21 21 blocks can be used to form structure 6 √ (1) Answer

3.3 Input Output

Process -2 1 √ 2 1 3x – 5 √ 3 4 √ 4 7 (3)

Minus 1 for any wrong output value

3.4 3.4.1 The y-intercept is 3 i.e. c

Take points (0 ; 3) and (-2 ; 0) If x = -2 ; y = 0 mx + c = y

-2m + 3 = 0 -2m = -3

m = 23

√

Hence y = 23

x + 3 √ (2)

substituting into formula and calculating m Answer


3.4.2 If x = 3

𝑥 + 3 = y

(3) + 3 = y

29

26

= y

2

15 = y

y = 721

√

(4)

Multiplying by LCM Removing the brackets Grouping like terms Answer

3.5 Let the first year’s admission be represented by x

(any letter can be used) √ x + 2x + 4x + 8x = 1 500 √ 15x = 1 500 √ x = 100 √ Therefore 100 learners were admitted in the first year. (4)

Representing the unknown Forming the equation Simplifying left hand side Answer

[14] QUESTION 4 4.1 9p 2 q – 81p 2 q 3

= 9p 2 q (1 – 9q 2 ) √√ = 9p 2 q [(1 – 3q) (1 + 3q)] √√ (4)

Common factor and difference of 2 squares Correct factors of difference of 2 squares

4.2 4.2.1 (3x – 2) (5x + 1)

= 15 x 2 + 3x – 10x – 2 √ = 15 x 2 – 7x – 2 √ (2)

Removing brackets Answer


4.2.2

223

432

812

zyxzyx

x xyyx

168 32

= 234

464

12896

zyxzyx

√√

= 4

3 23 zy √√

(4)

Product of numerator and Product of denominator Answer

4.3 4.3.1

2

6x +

4)8(3 x

= x + 3

2

)6(4 x +

4)8(12 x

= 4(x + 3)

2(x – 6) + 3(x + 8) = 4(x + 3) √√ 2x – 12 + 3x + 24 = 4x + 12 √ 5x + 12 = 4x + 12 5x – 4x = 12 – 12 x = 0 √ (4)

Simplifying the left hand side and the right hand side Grouping like terms Answer

4.3.2

2 x2 = 64 2 x2 = 2 6 √ 2x = 6 √ x = 3 √ (3)

Converting 64 into power Equating exponents Answer

[17] QUESTION 5 5.1

180º(n – 2) = 1 260º

180

)2(180 n =

1801260

n – 2 = 7 √ n = 7 + 2 √ n = 9 Hence, if the sum of the angles of a polygon is 1 260º, it has 9 sides. (2)

Simplification Answer


5.2 In ∆ SPQ and ∆ QRS PQ = RS (opposite sides of a parm.) √ SQP = QSR (alt s, PQ // RS ) √ SQ = SQ (common) √ ∆ SPR = ∆ QRS (SS) √ OR PQ = RS opposite sides of a parm. √ SP = QR opposite sides of a parm. √ SQ = SQ common √ Hence ∆ SPQ ≡ ∆ QRS (SSS) √ (4)

Statements with reason Answer

5.3

x

10126 √

6x = 120

6

6x =

6120

x = 20 cm √ The length of the longest side is 20 cm. (2)

Setting up the proportional sides Answer

5.4 5.4.1 L (3 ; -5 )

√ M (5 ; -6 ) N (6 ; -5 ) √ O (5 ; -3 ) (2)

2 marks for 4 coordinates correct 1 mark for 1or 2 wrong coordinate No mark for 1 correct co-ordinate


5.4 5.4.2 √√ (2) Award for correct position of image.

5.4.3 Answer on diagram above with the following coordinates

L'' (-1; -5) M'' ( 1; -6) N''(2; -5) O'' (1; -3) √√ (2)

Correct gliding on ANNEXURE is to be awarded

[14]


QUESTION 6 6.1

P𝑄R + Q𝑇S = 180º sum of angles on a straight. line Hence Q𝑇S = 180º – 95º = 85º √ but P𝑄R = Q𝑇S corresponding angles, PQ//ST √ Therefore P𝑄R = 85º √ (3)

Obtaining angle QTS Equating with reason angles PQR and QTS Answer

6.2 6.2.1 x 1 √ (1) Answer 6.2.2 x 4 √ (1) Answer [5] QUESTION 7 7.1 7.1.1

S = 25,1

20kmtD

= 16 km/h √ Speed of train B is 16 km/h (1) Answer

7.1.2

Speed of train A tD

= 75,0

30

= 40 km/ h √ Therefore Train A is faster than Train B because it runs at a speed of 40km /h whilst Train B runs at a speed of 16 km/h. √ (2)

Answer Reason

7.2 7.2.1 In ∆ ADB OR ∆ADC

DB = DC = 2,1 cm AD 2 = AB 2 – DB 2 Pythagoras Theorem √ = (3 cm) 2 – (2,1 cm) 2 = 9 – 4,41 = 4,59 AD = 59,4

= 2,14 cm √ (2)

Theorem Answer


7.2.2 Total Surface Area = 2(triangular base area) + area rectangle +2(Area of other rectangular face)

= 2(21

b x h) + (l x b) + 2 (l x b) √

= 2(21

x 4,2 x 2,14) + (4,2 x 15) + 2(3 x 15) √√

= 8,99 + 63 + 90 = 161,99 cm 2 √ (4)

Formula Correct substitution Answer

[9] QUESTION 8 8.1 No. of learners that will fail = 3 + 5 + 2 + 9 = 19 √

(1) Addition and correct answer

8.2 Minimum total marks she can obtain = 80 x 8 = 640 √

(1) Multiplication and answer

[2]


QUESTION 9 9.1 9.1.1 Hand length and Shoe Size Correlation Graph

Labelling x-axis 1 mark; Labelling y-axis 1 mark; Title of graph 1 mark Plotting points 2 marks (5) 9.1.2 The bigger the shoe size the longer the length of the hand

and vice versa. √ (1) Correct reason for relation.

√√√√√


9.1.3 2; 4; 5; 5; 6; 7; 7; 8; 9; 9

Median = 2

76 √

= 2

13

= 621

√ (2)

Identifying middle numbers and dividing Answer

9.1.4 Mode is 15 √ (1) Answer 9.1.5 Mean =

.

= 130 / 10 = 13 √ Therefore the mean is 13. √ (2)

1 mark for correct sum Answer

9.1.6 Range = 9 – 2 = 7 √ (1) Answer 9.2 Win (W) Draw (D) Loss (L)

√√√ (3)

1 mark per row / column Correctly completed table

Win (W) WW WD WL

Draw (D) DW DD DL

Loss (L) LW LD LL

9.3 9.3.1 √ (1) Answer 9.3.2 √ (1) Answer 9.3.3 √ (1) Answer 9.4 9.4.1 Graph 1 (1) Answer 9.4.2 In graph 1 the scale units on y-axis are small i.e. 10 units

This results in graph being stretched (enlarged). On the other hand the scale units on y-axis of graph 2 are bigger i.e. 50 units this results in the graph being compressed. √ (1) Answer

[20] TOTAL: 100

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SENIOR PHASE GRADE 9 SENIOR PHASE GRADE 9 SENIOR PHASE · SENIOR PHASE GRADE 9 NOVEMBER 2012 ARTS AND CULTURE MEMORANDUM MARKS: 100 ... Answer all the questions. 2. Write neatly and