<p>SUBUKIA DISTRICT JOINT EXAM- 2013 MATHEMTICS 121/1 MARKING SCHEME</p><p>1. - = M1</p><p>X8 = M1</p><p>X = A1</p><p>3 2. M1</p><p>M1</p><p>20</p><p>M1</p><p>A1 4</p><p>3. Gradient L1 = </p><p>Eqn L1 = M1</p><p>2y- 12 =3x M1 Co-ordinates of p: 2y-3x=12 y-2x=-2 A1 x=16, y=30 (16, 30)</p><p>3 4. Width = 3x-2 M1 ∵(3x + 1) (3x-2) = 28</p><p>1 9x2 – 3x – 30 = 0 X = 3 ± 18 M1 X =2 or – 1.67 ∵Length = 7cm, width = 4cm A1 3 5. M1</p><p>M1</p><p>A1</p><p>3 B1</p><p>6.</p><p>Opp = = M1 Tan(90-x)0 = = 2 A1 3 7. AB= M1</p><p>-2(K+12) -9 (2K-16)=10 -2K-24-18K+144=10 M1 -20K = -110 ∵K=5.5 A1 3</p><p>2 8. B1 Angle of 1200</p><p>B1 Trianlge</p><p>B1 Locating E</p><p>B1 locating D</p><p>4 9. Vol of water: ¾ x 2.5 = 1.875l M1 Vol of ethanol ¼ x2.5 = 0.625l Mass of water 1875 x 1 = 1875g M1 Mass of ethanol 625 x 1.2 = 759g Mass of mix= 1875 + 750=2625g</p><p>A1 3 10. 9.74 x 105,000=Ksh.1,022,700 M1 1,022,700-403897=Ksh.618,803 M1 = 832,752.66 yen A1 3 11.</p><p>B1 = 2 ∵ x = 3 M1</p><p>3 A1</p><p>= 0 ∵y=1</p><p>= 1.5 ∵z=2</p><p>OB = 3i + j + 2k 3 12. 3-5<2x -2<2x -1<x B1 8-4≤-3x -4≤-3x x≤ B1</p><p>B1</p><p>3 13. Log (x +24)-log32=log(9-2x)+log102 M1 Log = log [(9-2x)100] =900-200x M1</p><p>X = 4.484 A1 3 14. B1Accurate diagram B1 for accurate measurement B1 for accurate labelling</p><p>4 3 15. SP= x 450,000= ksh 418500 M1 418,500=113% ? 100% M1 =ksh 370,358.98 A1 3 M1</p><p>16. Let ext angle be x ∵ int angle= 4x M1</p><p>A1 X+4x=180 ∵x=360 No of sides= = 10 sides 3 17. i) COB = 270 B1 B1 (angle substained by Same arc) B1</p><p>5 B1</p><p>0 ii) DBC= 36 B1 (angle subtended by same arc) B1 B1 B1 iii) DOC = 720 </p><p>(angle in a Δ ) B1 B1</p><p> iv) OCA= 630</p><p>(base angles of isosceles Δ)</p><p> v)DEB-630 </p><p>(angles subtended by same arc)</p><p>10 18. a) B3 (for all values B2 for at least 6 values B1 for any 4) 2cos( ½ - 1.9 1.9 1.4 1 0 X-300) 1.7 2COS2 2 1 -1 -1 1 1 X</p><p> b) see graph S1 P1 For 2cos( ½ X-300) C1 For 2cos( ½ X-300) P1 For 2COS2X C1 For 2COS2X</p><p>6 B1</p><p>B1 (B1 for at least 2) B1</p><p> c)i) x=1710 ± 30 ii) x=90, 1560,2250±30 10 19. a) distance from NRB = 500 – (60 x ) = 35km X 150KM M1 </p><p>NRB m MB A1 = + </p><p>= 20X-12X = 3000 M1 X= 375KM M1</p><p>7 b) time taken by bus remaining distance M1 = 2hr 5min A1 1hr 40min M1 Avg speed is = 75km’hr B1 M1 A1 10 20. See graph ( a, b, c) B1 OBJECT B1COORDINATES IMAGE 1 B1IMAGE 1 PLOTED B1COORDINATE IMAGE II B1 IMAGE II PLOTED B1 COORDINATE IMAGE III B1IMAGE III PLOTED</p><p> d. Reflection on the line y=x B1 e. Area = ½ (8) 7 M1 = 28 sq units A1 10 10 2 21. a) y- ½ x – 2 B2 X 0 1 2 3 4 5 6</p><p>8 y -2 -1.5 0 2.5 6 0.5 16</p><p>A = ½ .1 [(2+0) + 2x1.5)] + ½ .1 [(0+16) + 2 (2.5 + 6 + 10.5)] M1 = 2.5 + 27 = 29.5 sq units A1</p><p>BI b) [ x3 – 2x] + [ x3 – 2x ] + [ (36-12) – ( - 4)] + = 28sq units % error = x 100 = 5.357%</p><p>M1</p><p>M1</p><p>M1</p><p>M1</p><p>A1 10 22. i) Vol. – r2h</p><p>9 = x 0.72 x 1 M1 = 1.54 m3 A1</p><p> ii) vol. = X area x h = ½ x 0.52 sin 60 x 3.2 M1 = 0.346m3 A1 b) i) vol. of pillar= vol. of solid pillar- vol. of cylinder hole = (1.54 + 0.346)- π 0.22.4.2 M1MI = 1.886 – 0.528 M1 = 1.358m3 A1 ii) mass = 2.7 x 1358000 M1 = 3, 666,600g. A1</p><p>10 23. a) bearing of R from T = 450 B1 distance RT= 84KM B1 B1 B1 B1 B1</p><p> b) Bearing of S from R = 670 B1 SR = 81km ± 5km B1 c) Bearing of S from P = 3040 B1 SP = 91km B1</p><p>10 10 24. a) i) P in 1h = = 1.05m M1 = 105 cm deep A1 ii) depth of water = 2.1m + 0.35 = 2.45m R: 6hrs – 2.1m M1 4hrs = = 0.6m New depth = 2.45 + 0.6 – 1.4 M1 = 1.65m b) 1hr the three pipes fill ½ + - = M1 depth of the tank = x 6.5 = 4.33m A1 1hr = x 6.5m M1 1hr = 3.09m M1 4.33m = = 1.4 hrs M1</p><p>A1 10</p><p>11</p>