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Maths Answer Ppsmi2006 F4 P2
 

Maths Answer Ppsmi2006 F4 P2

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    Maths Answer Ppsmi2006 F4 P2 Maths Answer Ppsmi2006 F4 P2 Document Transcript

    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 1. a) ∈ Q P1 P b. ∈ P P2 Q R 3 2. 1 x 15 x 15 x 21 K1 3 22 x 7 x 7 x 9 K1 7 2 2 1 x 15 x 15 x 21 + 22 x 7 x 7 x 9 K1 3 7 2 2 1228.5 N1 4 3. 4m + 2n = 6 @ 6m + 3n = 9 K1 5n = -5 @ 10m = 20 K1 n = -1 , m = 2 N1 N1 4 https://panelmathsaddmathswordpress.com/ 1
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 4. a) 270o x 2 x 22 x 21 @ 90o x 2 x 22 x 14 K1 360o 7 360 o 7 270o x 2 x 22 x 21 + 90o x 2 x 22 x 14 + 7 + 7 K1 360o 7 360 o 7 135 N1 b) 270o x 22 x 21 x 21 @ 60o x 22 x 14 x 14 K1 360o 7 360 o 7 270o x 22 x 21 x 21  60o x 22 x 14 x 14 K1 360o 7 360 o 7 936.8 N1 6 5. 2x2  5x + 3 = 0 K1 ( x - 1 ) ( 2x - 3 )= 0 K1 x = 1, 3 N1 N1 2 4 6. a) Statement P1 b) If 4k < 20, then k < 5 If k < 5, then 4k < 20 P2 c) n + 1 is not an even number P2 5 https://panelmathsaddmathswordpress.com/ 2
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 7. a) y = 4 P1 b) Identify mRS = 3 P1 y − 20 K1 = 3 or equivalent x − 13 y = 3x − 19 or equivalent N1 c) 0 = 3x − 19 19 1 K1 x= or 6 N1 3 3 8. < AGD P1 Tan θo = 6 K2 5 θo = 50.19 o @ 50° 11’ N1 4 9. a) 6 P2 11 b) 1 = 12 K2 3 x x = 36 N1 5 https://panelmathsaddmathswordpress.com/ 3
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 10. a) mRQ = 5 P1 3 b) 10 = 5 ( 3) + c K1 3 c = 5 K1 y = 5 x+5 N1 3 c) 0 = 5 x + 5 K1 3 x = -3 N1 6 11. a) i. > P1 ii. < P1 b) PQRST is a pentagon P2 c) 3n2 where n = 0, 1, 2, 3, … P2 6 https://panelmathsaddmathswordpress.com/ 4
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 12. a) i. m OA = 4 P1 ii. c = -18 K1 y = 4x – 18 N1 iii. mAB = 2 K1 5 c = 18 N1 5 b) i. Identify ∠ NLJ P1 JL = 12 2 + 12 2 or 16 ⋅ 97 K1 tan θo = 5 K1 16.97 θo = 16.42o or 16o 25’ N1 ii. Identify ∠NMJ P1 tan θo = 5 K1 13 θo = 22⋅62o or 22o 37’ N1 12 https://panelmathsaddmathswordpress.com/ 5
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 13. a) i. { 2, 3, 5, 6, 8, 9, 11, 12 } P1 ii. { 2, 6, 8, 12 } P1 4 1 iii. @ P2 8 2 6 3 iv. @ 10 5 P2 b) i. 60o x 2 x 22 x 28 360o 7 K1 60o x 2 x 22 x 28 + 28 + 14 + 14 + 14 +14 360o 7 K1 113 1 3 N1 ii. 60o x 22 x 28 x 28 @ 150o x 22 x 14 x 14 360o 7 360o 7 K1 60o x 22 x 28 x 28 + 14 x 14 - 150o x 22 x 14 x 14 360o 7 360o 7 K1 350 N1 12 https://panelmathsaddmathswordpress.com/ 6
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 14. a) 3n2 – 11n – 4 = 0 K1 ( n – 4 ) ( 3n + 1 ) = 0 K1 n = 4, - 1 N1, N1 3 2 b) x – 2x + 24 = 0 K1 ( x – 6 ) ( x + 4 )= 0 K1 x = 6, - 4 N1, N1 c) 1 x 22 x 7 x 7 x 10 K1 2 7 1 x 1 x 22 x 7 x 7 x 5 K1 2 3 7 1 x 22 x 7 x 7 x 10 + 1 x 1 x 22 x 7 x 7 x 5 K1 2 7 2 3 7 898 1 N1 3 15. a). Age Frequency Mid point Upper boundary (I) ( II ) ( III ) 1-5 1 3 5.5 6-10 4 8 10.5 11-15 9 13 15.5 16-20 9 18 20.5 21-25 5 23 25.5 26-30 2 28 30.5 All values in column ( I ) correct P2 All values in column ( II ) correct P1 All values in column ( III ) correct P1 b) Mean= 1(3) + 4(8) + 9(13) + 9(18) + 5(23) + 2(28) or 485 K2 30 30 = 16.17 N1 https://panelmathsaddmathswordpress.com/ 7
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks c) Histogram Using the uniform scales for 0.5 ≤ x ≤ 30.5 and 0 ≤ y ≤ 9 K1 6 bars* drawn correctly. K2 Note: 5 or 4 bars* drawn correctly. K1 The 6 bars with correct frequencies are drawn. N1 d) 16 x 100 = 53.33% 30 P1 * referring to candidates frequencies 12 https://panelmathsaddmathswordpress.com/ 8
    • PPSMI ASSESMENT 2006 Panel Pakar Runding GC MMMT MATHEMATICS 1449/2 FORM 4 No Marking Scheme Marks 16. a) Age Frequency Cumulative Upper (years) frequency boundary (I) ( II ) 10 – 14 0 0 14.5 Row I 15 - 19 10 10 19.5 20 – 24 20 30 24.5 25 – 29 50 80 29.5 30 – 34 60 140 34.5 35 - 39 36 176 39.5 40 - 44 18 194 44.5 45 - 49 6 200 49.5 All values in Row I correct. P1 All values in Column ( I ) correct excluding Row I correct. P2 All values in Column ( II ) correct excluding Row I correct. P1 b) Mean = 10(19.5) + 20(24.5) + 50(29.5) + 60(34.5) +36(39.5) + 18(44.5) +6(49.5) 200 Or 6750 200 K2 33.75 N1 c) x-axis is drawn with the right direction and uniform scaled from P1 14.5 ≤ x ≤ 49.5 y-axis is drawn with the right direction and uniform scaled from P1 0 ≤ y ≤ 200 All eight points* plotted correctly. P2 Note : Seven or six points* plotted correctly. P1 All the right eight points plotted correctly and ogive is drawn smoothly passing N1 through all the points. * referring to candidates frequencies https://panelmathsaddmathswordpress.com/ 9