Mathematics Grade 9 ANA Exemplars 2014 Memo (English)

Downloaded from hlayiso.com Page 1 of 10 ANNUAL NATIONAL ASSESSMENT 2014 GRADE 9 MATHEMATICS EXEMPLAR QUESTIONS MEMORANDUM KEY M Method mark CA Consistent Accuracy mark A Accuracy mark 1. MULTIPLE-CHOICE QUESTIONS 1.1 C 1.2 C 1.3 B 1.4 C 1.5 B Give 1 mark for each 1. correct answer. 1.6 C 1.7 D 1.8 C 1.9 D 1.10 B [10] 2. NUMBERS, OPERATIONS AND RELATIONS 2.1.1 0,0067  A 1 mark (1) 2.1.2 3,56 × 10  1 mark (1) 2.1.3 2,7, 2√2, 8  A because 2√2 ≈ 2,83. 1 mark (1) 2.1.4 −3√3 , -5,25 , -16  A because −3√3 ≈ −5,20 1 mark (1) 2.1.5 9 < 13 < 16 Answer: 1 mark (1) 3 < √13 < 4 A 2.2.1 0,125 ÷ √25 = 0,125 ÷ 5  M 5: 1 mark = 0,025  CA Answer: 1 mark (2) 2.2.2 1 2 + (0,5) 2 25 = + 0,25  4 = +  : 1 mark = : 1 mark 1 =6  2 (3) Answer: 1 mark or 6,25 + 0,25 6,25: 1 mark = 6,5  Answer: 2 marks (3) 2.2.3 √169 + 3 × 5 ÷ 2 √169 + 3 × 5 ÷ 2: 1 mark = (13 + 15) ÷ 2 13 + 15: 1 mark = 14 Answer: 1 mark (3) 2.2.4 10 × 0,01 10: 1 mark = 10 × 0,1  0,1: 1 mark = 1 Answer: 1 mark (3) Grade 9 Mathematics Exemplar Memorandum 1

Downloaded from hlayiso.com Page 2 of 10 2.3 96:120M = 8:10 Ratio: 1 mark = 4:5A Answer:1 mark (2) 2.4 : : M Simplifying : 1 mark Answer: 1 mark (2) = 5: 3: 2 A 2.5 : M ÷ : 1 mark (2) = 5: 8 A Answer: 1 mark 2.6 5: 3: 4: First mass = × 240 = 100 g A Second mass = × 240 = 60  A Third mass = × 240 = 80  A or Third mass = 240 − (100 + 60 ) = 80 CA Answer: 1 mark each (3) 2.7 Decreased amount = × 1 250 : 1 mark = 500A Answer: 1 mark (2) 2.8 Increased number = × 280 : 1 mark = 700A Answer: 1 mark (2) 2.9 . . =  Formula/substitution: 2 marks 3 000( )(0,8) = 960  =4 A Answer: 1 mark (3) or = (1 + ) 3 960 = 3 000(1 + 0,08 ) 1,32 = 1 + 0,08 0,32 = 0,08 = 4A 2.10 A = (1 + )  Formula/substitution: 2 marks A = 6 500(1 + 0,075)  A = R8 074.93 A Calculation: 1 mark Interest=A-P Answer: 1 mark (4) = R1 574,93 CA or Year 1: 6 500 × 7,5% = 487,50 Year 2: 6 987,50 × 7,5% = 524,06 Year 3: 7 511,56 × 7,5% = 563,37 Total interest is R1 574,93 2.11 = (1 + )  Formula/substitution: 3 marks = R10 000(1 + 0, 1)  = R13 310,00 A Answer: 1 mark (4) or Year 1: 10 000 × 10% = 1 000,00 Year 2: 11 000 × 10% = 1 100,00 Year 3: 12 100 × 10% = 1 210,00  The amount = R10 000 + R3 310,00 =R13 310,00 CA Grade 9 Mathematics Exemplar Memorandum 2

Downloaded from hlayiso.com Page 3 of 10 2.12 Speed Time / ) ( (h) 80 3 50 50 = 80(3) (indirect proportion) 80(3) =  Formula/substitution: 2 marks 50 = 4,8 ℎ : 1 mark or Answer: 1 mark (4) = = 80 × 3 ℎ = 240  = 240 = 50 /ℎ = 4,8 hours or 4 hours 48 min 2.13 Length in 3,5 Shadow in 5,2 29,2 , = ,  (direct proportion) Formula/substitution: 2 marks , = , × ,  : 1 mark , Answer: 1 mark (4) = 19,65  3. PATTERNS, FUNCTIONS AND ALGEBRA 3.1.1 (2 ) + 3 = 4 + 3 M 4 : 1 mark = 7  CA Answer: 1 mark (2) 3.1.2 (a b ) . ab − (ab) a b ab : 1 mark = a b ab − a b M a b : 1 mark = a b − a b M Applying exponential laws: 1 mark = a b (b − 1) A Answer: 1 mark (4) 3.1.3 × M Reciprocal: 1 mark =  A 9: 1 mark 4a b: 1 mark (3) 3.1.4 × M Applying exponential laws: 1 mark = or A Answer: 1 mark (2) 3.1.5 ( ) = M 4 : 1 mark = 4 CA 4 : 1 mark = 64 A Answer: 1 mark (3) Grade 9 Mathematics Exemplar Memorandum 3

Downloaded from hlayiso.com Page 4 of 10 3.1.6 ( ) ( + 2): 1 mark × M ( ) ( )( ) ( − 2): 1 mark = A ( + 2)( − 2):1 mark Answer: 1 mark (4) 3.1.7 −2 −3 – 2 3 ( ) ( ) = M Common denominator: 1 mark 3( − 2) − 2( − 3): 1 mark = M 3 − 6 − 3 + 6: 1 mark = A Simplification: 1 mark Answer: 1 mark (5) = A 3.1.8 3 × 24 9 = M Simplification: 1 mark 8b : 1 mark = A a : 1 mark (3) 3.1.9 −1 3 +3 = ( )( ) M ( − 1)( + 1): 1 mark ( ) 3( + 1): 1 mark = A Answer:1 mark (3) 3.2.1 3 (3 − 4 − ) =9  − 12 −3 A 1 mark for each term (3) 3.2.2 (2 − 3)( + 1) 2 : 1 mark = 2  −  − 3 A − : 1 mark −3: 1 mark (3) 3.2.3 ( − 3) − ( + 4) − 6 + 9: 1 mark = − 6 + 9 − − 4 M − − 4 : 1 mark = −10 + 9 Answer:1 mark (3) 3.3.1 10 − 5 5t: 1 mark = 5 (2 − 1)  A 2t − 1: 1 mark (2) 3.3.2 81 − 100 9 − 10a: 1 mark = (9 − 10 )(9 + 10 )A 9 + 10a: 1 mark (2) 3.3.3 ( + )(2 + )A Answer: 2 marks (2) 3.3.4 6 ( − ) + ( − ) − ( − ): 1 mark = 6 ( − ) − ( − ) : 1 mark = ( − )(6 − 1) (6 − 1): 1 mark ( − ):1 mark (4) 3.3.5 ( + )(4 − )M Common factor: 1 mark = ( + )(2 + )(2 − )A (4 − ): 1 mark (2 + )(2 − ): 1 mark (3) 3.3.6 +5 +6 ( + 3): 1 mark = ( + 3)( + 2) ( + 2): 1 mark (2) 3.3.7 2 − 18 + 36 = 2( − 9 + 18) 2( − 9 + 18): 1 mark = 2( − 6)( − 3) A Answer:2 marks (3) Grade 9 Mathematics Exemplar Memorandum 4

Downloaded from hlayiso.com Page 5 of 10 3.4.1 2 − 5 = 5 + 16 3 : 1 mark −3  = 21  M or −21 = 3  M 21: 1 mark = −7  A −7 =  A Answer:1 mark (3) 3.4.2 −1 − =3 2 × 2: Multiplying by 2: 1 mark 2 − + 1 = 6 M Simplifying:1 mark + 1 = 6 M Answer: 1 mark (3) = 5A 3.4.3 ( − 2) (2 + 1) 5 + = 4 3 3 × 12: 3( − 2) + 4(2 + 1) = 4 × 5 M 3( − 2): 1 mark 3 − 6 + 8 + 4 = 20  M 4(2 + 1): 1 mark 11 = 22 4 × 5: 1 mark = 2 A Simplifying: 1 mark Answer: 1 mark (5) 3.4.4 ( − 3)( + 4) = 0 − 3 = 0 or + 4 = 0 = 3or = −4A Answer: 2 marks (2) 3.4.5 −1 = 0 ( − 1)( + 1) = 0 Factorising: 1 mark − 1 = 0 or + 1 = 0 (3) = 1 or = −1 Answer: 2 marks or = 1M ∴ = 1 or = −1 3.4.6 3 =3  Applying exponential law: 1 mark + 1 = 4 Equating exponents: 1 mark = 3 Answer: 1 mark (3) 3.4.7 = −27 = √−27 = −3 Answer: 2 marks (2) 1 3.4.8 2 = 64 2 =2  2 : 1 mark = −6 Answer: 1 mark (2) 3.5.1 2 −3 +9 +2 Substitution: 1 mark = 2(−2) − 3(−2) + 9(−2) + 2 Simplifying: 1 mark = −16 − 12 − 18 + 2 Answer: 1 mark = −44A (3) 3.5.2 5 −1 5(2)( ) = 2  Substitution: 1 mark (−3) =  Simplifying: 2 marks (3) Grade 9 Mathematics Exemplar Memorandum 5

Downloaded from hlayiso.com Page 6 of 10 2 =1 3 3.5.3 3 −2 − = 3(2) − 2(2)(−3) − (−3)  Substitution: 1 mark = 12 + 12 − 9 Simplifying: 2 marks Answer: 1 mark (3) = 15 3.5.4 2×3 ( ) = 2×3 = 2×3  Simplifying: 1 mark = 54 Answer: 1 mark (2) 3.6.1 Figure 1 2 3 4 Number of 1 2 3 4 black tiles Number of 6 10 14 18 white tiles Answer: 2 marks (2) 3.6.2 =4 +2 4 : 1 mark 2: 1 mark (2) 3.7.1 Triangular numbers Answer: 2 marks (2) 3.7.2 ( ) : 2 marks =  ( ) =  Substitution: 1 mark = 210 Answer: 1 mark (4) 3.8.1 −2 −1 0 1 -11 -8 -5 -2 1 mark each (4) 3.8.2 −3 −1 0 1 1 1 mark each (4) 1 − -1 -1 or - 3 3.9.1 = 2A Answer: 1 mark (1) 3.9.2 = 2 A Answer: 2 marks (2) 3.9.3 = 6 unitsA Answer: 1 mark (1) Grade 9 Mathematics Exemplar Memorandum 6

Downloaded from hlayiso.com Page 7 of 10 3.10.1 y= x-1 y=− x+1 . .4 .3 = 3 + 1 .2 = 3 − 2 .1 -5 -4 -3 -2 -1 .0 1 2 3 4 5 2 .-1 3 .-2 .-3 .-4 .-5 X-intercept: 1 mark per graph+A Y-intercept: 1 mark per graph+A Correct labelling of graph: 1 mark per graph +A (6) 3.10.2 The lines are perpendicular.A (1) 3.11 (3; 3) Answer: 2 marks (2) 3.12.1 4 −2x : 1 mark Gradient of = = −2. −2 4:1 mark Equation of = −2 + 4 4 Gradient of = = −2 −2 Equation of = −2 − 4 −2x : 1 mark −4:1 mark (4) 3.12.2 ||  because the gradient of = gradient of  || : 1 mark Reason :1 mark (2) Grade 9 Mathematics Exemplar Memorandum 7

Downloaded from hlayiso.com Page 8 of 10 4. SPACE AND SHAPE 4.1 Statement Reason = + ext ∠ of ∆= sum of int. opp ∠  Statement: 1 mark 75° = + 44° or ext ∠ of ∆ Correct reason:1 mark ∴ = 31° Answer for : 1 mark Statement: 1 mark Correct reason:1 mark = = 31° alt ∠s and AD || BC (5) ∴ = 31° 4.2 Statement Reason In ∆AEW : Statement: 1 mark + = 110° sum of ∠s of ∆ = 180° Correct reason:1 mark ∠s opp.equal sides of ∆ Statement: 1 mark but = = 55° alt ∠s and CS || HW Correct reason:1 mark ∴ = = 55° Answer for : 1 mark Correct reason:1 mark (6) 4.3 Statement Reason = 95° − 30° M ext. ∠ of ∆ A Statement: 1 mark = 65° A Correct reason:1 mark or Answer: 1 mark = 180° − 95° adj. suppl. ∠s or is a str. ∠ = 85°M/A or ∠s on a str. line + 85° + 30° = 180° sum of ∠s of ∆= 180° A + 115° = 180° = 65° A Statement: 1 mark + = 180° M co - interior ∠s and AB || CDA Correct reason:1 mark = 115° A (6) Answer: 1 mark 4.4 Statement Reason Statement: 1 mark In ∆ ∆ Correct reason:1 mark =  common A Statement: 1 mark = = 90° given A Correct reason:1 mark A =  given A Statement: 1 mark ∴∆ ≡∆  90°H s A Correct reason:1 mark Statement: 1 mark Correct deduction:1 mark (8) Grade 9 Mathematics Exemplar Memorandum 8

Downloaded from hlayiso.com Page 9 of 10 4.5.1 Statement Reason + + + = 180°M co-int ∠ and || A Statement: 1 mark Correct reason:1 mark But = and = °M given Statement: 1 mark Correct reason:1 mark ∴2 +2 = 180°A sum of ∠ of ∆= 180° Statement: 1 mark Correct reason:1 mark + = 90°A Correct deduction:1 mark sum of ∠ of ∆= 180° Correct deduction:1 mark (8) ∴ = 90°A 4.5.2 Statement Reason In ∆ and ∆ Statement: 1 mark 1. = M givenA Correct reason:1 mark 2. = M both 90°A Statement: 1 mark Correct reason:1 mark 3. = M sum of ∠ of ∆= 180° Statement: 1 mark ∠ ∠ ∠A Correct reason:1 mark ∴∆ ||| ∆ M Statement: 1 mark Correct reason:1 mark (8) 4.5.3 Statement Reason = = M prop. sides of similar ∆ A Statement: 1 mark = M Correct reason:1 mark =2 A Statement: 1 mark = A Correct reason:1 mark Correct deduction:1 mark ∴ =8 A Correct deduction:1 mark (6) 4.6.1 Statement Reason In ∆ and ∆ Statement: 1 mark =  givenA Correct reason:1 mark =  given A Statement: 1 mark =  common sideA Correct reason:1 mark ∴∆ ≡ ∆  A Statement: 1 mark Correct reason:1 mark Correct deduction:1 mark Correct reason:1 mark (8) 4.6.2 = (corr∠s of congruent ∆s) ∴ bisects BÂC (2) Grade 9 Mathematics Exemplar Memorandum 9

Downloaded from hlayiso.com Page 10 of 10 5. MEASUREMENT 5.1 Let the length of the ladder be . = 12 + 5 ( ℎ) M Formula/substitution: 2 marks = 169 A 12 Calculation: 1 mark = 13 A Answer: 1 mark (4) 5 5.2.1 = − Formula/ substitution: 1 mark = (25 − 4) M Pyth  Reason: 1 mark = 4,58  Answer: 1 mark (3) 5.2.2 = 3× = 12  Answer: 1 mark (1) 5.2.3 Kite. Answer: 1 mark (1) 5.2.4 × Formula: 1 mark Area of the kite = × ×  or 1 = × 12 × 4  Substitution: 1 mark 2 Answer: 1 mark (3) = 24  5.3.1 Perimeter = 2 × 100 + 2 Formula/ substitution: 2 marks = 200  + 2 × 3,14 × 30  Answer: 1 mark ≈ 388,4  ∴ the no. of rounds = 4 000 ÷ 388,4  Calculation: 1 mark = 10,298661 Answer: 1 mark (5) Peter must run 11 times round the track.  5.3.2 Area = × + = 60 × 100 + 3,14 × 60  Formula/ substitution: 2 marks = 6 000 + 11 304 = 17 304  Answer: 1 mark (3) Grade 9 Mathematics Exemplar Memorandum 10