CSEC Maths Workbook Answers - Collins

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Collins CSEC Maths Workbook answersSection 1 Computation1. a)b)c)182315 0.125[2] 0.667[2] 0.2[2]375100075 1002 1002. a) 0.375 b) 0.75 c) 0.02 3. a)58 3[2]34150[2]857137701.071.225 000[1][1][1][1]5. a)b)c)d)2 500 000 2.5 1060.003 251 3.251 10 3362 000 3.62 1050.000 009 9.0 10 6[1][1][1][1]6. a)Angle A 2 180 60 6Angle B 3 180 90 6Angle C 1 180 30 6[1]c) Melissa receivesd) 40 20 m[2]5Kerry received 1 of x which is 455[2]7. a) Duration 9 hr 30 min 7 hr 25 min 2 hr 5 min [2](605 ) hr 2.08 hoursAverage speed distance 92 km 44.2 km/hr1 1 2 5 310 time1510 15103102.08 310 15 10 5103[2][4]2b) 14.25 (1.24) 14.25 1.5376 12.7124 12.7 (to 3 significant figures)[3]9. a) 2.14(3 1.26) 6.42 2.6964 3.7236 3.72 (to 3 significant figures)[2]b)2.150.82 0.22 2.15 2.150.64 0.22 0.42(to 3 significant figures) 15[1]301000 0.03 km[1][1][1]Section 2 Number theoryLet x represent the total sum sharedb) Time taken (hours) 2 hr [1] 100 20%c) 1 litre 1000 ml2 litres 2000 ml45Then, x 45 (5) 225.[1]1260105030 m [1]510Therefore, 100% 25 100 12513. a) 2.5 1000 2500 mb) 3000 cm 30 mKerry receives 18. a)[1]c)[1]b) Length of longest piece 5 3 2 512. a) 0.125 80 10b) 20% 2520[2]14. a)b)c)d)1 32 2 5[1][1][1]20[2] 5 100 71.4%711. a) 20 000 cmb) 0.2 kmc) 0.9 km1% 25[2]b) 0.135 0.135 100 13.5 %c)[2][2] 5 100 62.5%810. 1 cm 10 000 000 cm (from scale given)1 cm 100 km4.2 cm 420 km 5.119 5.12[3]31. a) Whole numbers – Bb) Integers – Cc) Natural numbers – A[1][1][1]2. a) Factors of 12 – 1, 2, 3, 4, 6, 12b) Factors of 10 – 1, 2, 5, 10c) Factors of 21 – 1, 3, 7, 21[2][2][2]3. a) Factors of 12 – 1, 2, 3, 4, 6, 12Factors of 18 – 1, 2, 3, 6, 9, 18HCF – 6b) Factors of 30 – 1, 2, 3, 5, 6, 10, 15, 30Factors of 15 – 1, 3, 5, 15HCF – 15c) Factors of 15 – 1, 3, 5, 15Factors of 25 – 1, 5, 25Factors of 40 – 1, 2, 4, 5, 8, 10, 20, 40HCF – 5[2][2][1][2][2][1][2][2][2][1]4. a) The first four multiples of 5 – 5, 10, 15, 20b) The first four multiples of 6 – 6, 12, 18, 24c) The first four multiples of 12 – 12, 24, 36, 48[2][2][2]5. a) Multiples of 9 – 9, 18, 27, 36Multiples of 12 – 12, 24, 36LCM – 36b) Multiples of 5 – 5, 10, 15, 20, 25, 30, 35, 40Multiples of 8 – 8, 16, 24, 32, 40LCM – 40c) Multiples of 6 – 6, 12, 18, 24, 30, 36, 42Multiples of 7 – 7, 14, 21, 28, 35, 42LCM – 42[2][2][1][2][2][1][2][2][1]6. a) 8, 13, 21b) 22, 26, 30c) 25, 36, 49[2][2][2]

7. a) Distributive lawb) Associative lawc) Commutative law4. a) TT 6.30 US 1.00[1][1][1]TT 5000 US (50006.30 1.00) US 793.65b) Amount left US 793.65 US 650 US 143.65US 1.00 TT 6.30US 143.65 TT (143.65 6.30) TT 905[3]8. a) 110112 (1 24) (1 23) (0 22) (1 21) (1 20) 16 8 0 2 1 27[2]b) 2324 (2 42) (3 41) (2 40) 32 12 2 46[2]c) 12425 (1 53) (2 52) (4 51) (2 50) 125 50 20 2 197[2]5. a) Percentage profit selling price cost price 100%cost price 420 000 350 000 100350 000 20%[2]b) Loss 75 000 40 000 35 000Percentage loss losscost price9. a) 2 152 7R12 3R12 1R10R1Depreciation after 1 year 0.10 180 000 18 000Value of car after 1 year 180 000 18 000 162 000[2]ii) Depreciation after 2 years 0.10 162 000 16 200Value of car after 2 years 162 000 16 200 145 800[2][2]b) 5 27R25 1R00R16. a) i)1025c) 8 9028 1R30R1100(132810. a)Total interest repaid P R T 120 000 10 5100() 10 000 1 [2]2.51005) 11 314.08Compound interest 11 314.08 10 000 1314.08 [2]1101102 1100127. a) Discount 10% 6500 650Amount paid 6500 650 5850b) Tax 15% 3000 450Amount paid 3000 450 34501001111Answer – 10011112100 60 000[2]ii) Total amount of money repaid 60 000 120 000 180 000[2]180 000iii) Monthly instalment 5 12 3000 per month[2]nb) Amount P 1 R[2]8 11 R 100% 35 000 100 46.7%75 000[2]c) i)111125 5[2][3]b) 11112 10012[2][2]Section 4 Sets0110Answer – 11021. a) n(U) 19b) A {6, 9, 12, 15, 18, 21}c) B {5, 7, 9, 11, 13, 15, 17, 19, 21}d)U[3]Section 3 Consumer arithmetic1. a) Total hire purchase price 800 (300 10) 800 3000 3800b) Money that would be saved 3800 3000 800[2][1]82. a) Amount received by the grocer 120 32 3840 [1]b) Profit 3840 3360 480[1]c) Percentage profit 4803360 100 14.3%3. a) Hourly rate 600 15 per hour40b) Overtime rate 1.5 15 22.50Overtime wage 8 22.50 180c) Wage for 40 hours 40 15 600Wage for 10 hours overtime 10 22.50 225Total wage 600 225 ]2. a)b)c)d)[2][3]4n(A) 8n(B) 5A B {9, 11, 15}A B {2, 3, 4, 5, 7, 9, 11, 13, 15, 17}[3][1][1][1][1]

U3. a)A7. a)Bn(u) 36n(B) 2525 – xn(C) 18x18 – x[1]2Ub)A[3]Bb) 25 x x 18 x 2 3645 x 36x 45 36x 98. a)n(u) 40n(B) 23[1]Uc)A[2]n(C) 156–xB8 x4 xx9–x5–x4 x[1]2Ud)A[6]b) 8 x 6 x x 9 x 4 x 5 x 4 x 2 40x 38 40 [2]c) x 38 40x 40 38x 2[2]d) n(Biology only) 8 x 8 2 10[1]e) n(Chemistry and Biology only) 6 x 6 2 4 [1]B[1]4. a) Number of subsets 23 8b) { }, {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6}, {2, 4, 6}[1][2]5. a)b)c)d)InfiniteFiniteFiniteInfinite[1][1][1][1]6. a)b)c)d)e)f)g)h)B and ED is a subset of BB and C OR C and DAB, D OR EC4Number of subsets 24 16[1][1][1][1][1][1][1][1]n(P) 18Section 5 Measurement1. a) C 2πr 2 3.14 6 37.68 cmb) A πr2 3.14 62 113 cm2[2][2]θ A 120 113 37.7 cm2 [2]360360Area of triangle AOB 1 ab sin C 1 6 6 sin 120 22c) Area of minor sector d) 15.6 cm2e) Area of shaded region 37.7 15.6 22.1 cm2f) Length of minor arc g) Length of major arc 5[2][2]θ C 120 37.68 12.56 cm [2]360360θ C 240 37.68 25.12 cm [2]360360

h)m6cAd) Volume of cylinder πr 2h 3.14 22 6 75.36 cm3 [2]120 30 6ce) TOTAL volume of the two hemispheres (or one sphere) 4 πr 3 4 π(2)3m3X7. a) BDC 30 (BDC is an isosceles triangle)b) DBC 180 (30 30) 180 60 120 ABD 150 120 30 AX 6 cos 30 5.2 cmAB 2 5.2 10.4 cmPerimeter of shaded region length of minor arc AB 12.56 10.4 22.96 cm [2]2. a) A l b 6 8.2 49.2 cm2b) V A h 49.2 12 590 cm3c) Total surface area 2 (12 6) 2 (12 8.2) 2 (6 8.1) 144 196.8 98.4 439.2 cm23 33.49 cm3f) TOTAL volume of perfume bottle 75.36 33.49 108.85 cm3B[2][2]28. a) Area of trapezium 1 (8 10) 6 54 cm22b) 110 x 120 60 360x 290 360x 360 290x 70 [4][1]1. a)Score (x)TallyFrequency (f )x f1 332 483 4124 4165 2106 848[4]7 174. a) Area of shaded cross-section πr 2 3.14 22 12.56 cm2[2]2b) Volume of cylinder πr h 12.56 8 100.48 cm3 [2]c) Area of curved part of cylinder h 2πr 8 2 3.14 2 100.48 cm2[2]8 3249 19 () (4 4) (π(2)22) 4 16 6.28[3] 26.28 cm2b) Total area Area ofrectangle ABCD Area of semi-circle (8 6) (π(4)22) 48 25.12 22.88 m25. a) i)ii)b) i)ii)Length of one side of square 196 14 cmPerimeter of square 4 14 56 cmCircumference 56 cm2πr 56r 562227 8.91 cmb) Mode 6c) Median 4 5 4.5d) Mean [2][1][1] 4.6[3]e) Probability (score 5) 2. a)[2]713730[2]6. a) Curved surface area of the cylinder 2πrh[2] 2 3.14 2 6 75.36 cm2b) TOTAL surface areas of the two hemispheres(or one sphere) 4πr 2 4 3.14 22 50.24 cm2[2]c) TOTAL surface area of perfume bottle 75.36 50.24 125.6 cm2 [1]6[5][1][1]2( )iii) Area of circle πr 2 22 (8.91)2 249.5 cm2[2]Section 6 Statistics3. a) Total area Area of triangle Area of square Area of semi-circle1 2 42[1][2]c) ADB 180 30 75 [2][2][2]8 1 3 130 1330Height (cm)Number ]

b)b) (See graph)fuel10018ºutilities9048º 54º60º180º80rentsavings70cumulative frequencyfood[3]604. a) Mean score 505. a)b)c)d)e)f)302010 10 6.1 [2]14–1614–1610.516.57.5 4.5 3See graph[2][1][1][1][1][1][1]1410203040height / cm5060Lower quartile 22 cmMedian 32 cmUpper quartile 38 cmInterquartile range 38 22 16 cmSemi-interquartile range 16 8 cmd) i)P (less than 35 cm) [5][1][1][1][1][1]262100 0.62ItemBudgeted amount15100 0.15rent 1000food 9009006000 360 54 fuel forher car 3003006000 360 18 utilities 8008006000 360 48 Total 3000 6000[1]30006000 360 60 86Angle of sector inpie chart10006000savings10[1]ii) P (greater than 42 cm) 100 85 10012frequency0c) i)ii)iii)iv)v)3. a)1061b) Arrange the scores in ascending order to find themedian – 5, 5, 6, 6, 6, 6, 6, 7, 7, 7Median score 6c) Modal score 6 (most frequent score)4006 5 7 7 6 5 6 6 6 7420010age6. a) Number of letter Ms 2Number of letters 112P (selecting an M) 11 360 180 360 515[4][2]5b) Number of yellow balls 8 40 25[4]20[2]5 1 1 30 16 [2]c) P (student scores more than 6) 1 2 307

b) y x2y kx216 k (2)2Section 7 Algebrax 31. a) i) 42x 133(x 3) 4(2x 1)12 3x 9 8x 412 11x 512ii) 2x 1 x 3323(2x 1) 2(x 1) 66x 3 2x 2 64x 5 67x 1 2x 1 453 b)k k 4Therefore, y 4x2When x a, y 6464 4a2[3]a2 4x2y(1 3xy2)(1 3x)(1 3x)(3x 1)(x 2)6p2 3pq 2p q 3p(2p q) (2p q) (3p 1)(2p q)3. a) mx c ymx y c2. a)b)c)d)x y c[3]b)When x a, y 33 [2]a xTkT2k2x a xaxk2 aT 2x aT 2k22[2]2 224. a) 3x y 6x y 3xyb) 2x 2y 6x2 3x 5x 2y 6x22c) 4x 5y5. a) 2 * 3 (2 3) 3(2)(3) 25 18 7b)(1)( 2) 3( 2)2 1[2][2] 5 15[2][2]6. a) y xy kx12 k(4)k 124k 3Therefore, y 3xWhen y 9, x a9 3a[3][3]2x 39Or x 3 0x 3221When x 3 , y 7 2 3 8 3a 3a 3When x 7, y bb 3(7) 21 3x 1Substituting x 1 into equation (2)y 3(1) 1 3 1 2b) y 2x 7 .(1)x2 xy 6 .(2)From equation (1), y 7 2xSubstituting equation (1) into equation (2)x2 x (7 2x) 6x2 7x 2x2 63x2 7x 6 0(3x 2)(x 3) 0Either 3x 2 03x 2[2]21247. a) 2x 3y 8 .(1)3x y 1 (2)From equation (2), y 3x 1Substiting equation (2) into (1)2x 3(3x 1) 82x 9x 3 811x 3 811x 8 311x 1111x 11[2] a123a 4When x 4, y bb T k12a3a 12aa[3]1y xky xk6 2k 6 2k 1212Therefore, y x[3][2][2][2]pq xpqx ax pqc)c)[2]m644a2 16a 16a 4When x 5, y bb 4(5)2b 1003(7x 1) 5(2x 1) 15 421x 3 10x 5 6031x 2 6031x 60 231x 6262x 31x 2164( )When x 3, y 7 2(3) 1[3]8[6]

c) 6x2 13x 5 0(2x 1)(3x 5) 0Either 2x 1 02x 18. a) 2x 8 122x 12 82x 20x 202x 102x3b)1x 2[2]xOR 3x 5 03x 5 2 64x 3x 367x 3636x 75x 313. a) 2x 5x 1 0a 2, b 5, c 1[3]c) 8 2x 2 2x 2 8 2x 62x 66x 29. a) i) 50 xii) 50x 35(50 x)b) i) 50x 1750 35x 220015x 2200 175015x 450x [1][1]ii) 50 30 201x 4002First piece xSecond piece x 2Third piece 3xii) x (x 2) 3x 5x 2iii) 5x 2 425x 42 25x 40c)a 3 b4ab b b 2 4ac2a[3][1]x 7 72 4(1)( 2)2(1)[2]x 7 572 7 57Either x 2OR x [2][1][4][2][2][2][1][1]c) y mx cy 2x cWhen x 2, y 3 3 2(2) cc 4 3c 3 4c 7Therefore, the equation of the line is y 2x 7.[1]2[2]2. a) Gradient b) Midpoint 4x 3OR x 2 0x 2 7.271. a) Gradient 21b) Gradient 2[2]12. a) x 7x 10 0(x 5)(x 2) 0Either x 5 0x 5OR x 2 0x 2b) 3x2 10x 8 0(3x 4)(x 2) 0Either 3x 4 03x 4 5 572 0.27Section 8 Relations, functionsand graphs[3][1][1]d) y2[4]14. a) 2 18x2 2(1 9x2) 2(1 3x)(1 3x)b) (2x 3) (3x 4)c) (2p 3q) (4r s)405 a2b3 5 52 4(2)(1)2(2)x b) i)11. a) x8y4b) x3yx b) x2 7x 2 0a 1, b 7, c 245015x 30x 8 b b 2 4ac2a 5 174 5 17Either x 4 0.22 5 17OR x 4 2.28[2]x x x x 310. a)[2]2y 2 y1x2 x1(11 110 3 ( 2) 5 2x1 x2 y 1 y 2 2 3 1 11, 2 2 , 22) ([3][1]) (12, 6)[2]c) Length (x2 x1)2 (y2 y1)2 (3 ( 2))2 (11 1)2[2] (5)2 (10)2 11.2 units9[2]

16. a) x 3 0x 3Therefore, when x 3, f (x) is undefined.b) g(2) 2 (2) 3 7d) Gradient of perpendicular bisector 2y mx c1When x 2 , y 61 1 c216 4 c1c 6 425c 4125Therefore, y 2 x 46 2()f (7) c) f (x) 2x 1x 3Step 1 – Interchange x and y[3]x 2y 1y 3Step 2 – Make y the subject of the formulax(y 3) 2y 1xy 3x 2y 1xy 2y 1 3xy(x 2) 1 3x[1]y 1 3xx 2Therefore, f 1(x) 1 3x[1][1]y321A27. a) 2x 5x 3 a(x b) c a(x2 2bx b2) c ax2 2abx ab2 cEquating coefficients:a 22ab 52(2)b 52 3 x(2, 0)5b 4ab2 c 35 22 4 c 3()–3c 3 –449–6B5. a) f (2) 5(2) 2 10 2 8b) f ( 1) 5( 1) 2 5 2 7[1][2]11g (4) 3(4) 1215fg(x) 5 3x 2 3x 211gf (x) 3(5x 2) 15x 6( ) 1Therefore, f (x) 498[3]b)5 49c) Coordinates of minimum point 4 , 82d) 2x 5x 3 0(2x 1)(x 3) 0Either 2x 1 02x 1x 25()[1][1]1x 2Or x 3 0x 3[1][2]e) f)[2]y[2]54f) y 5x 2Step 1 – Interchange x and yx 5y 2Step 2 – Make y the subject of the formula5y x 2x 25)5Axis of symmetry x 4[1][1][1][1][2]y 5(So f (x) 2 x 4 2 (0, –6)4. a) i) One-to-oneii) Many-to-oneiii) One-to-manyiv) Many-to-manyb) Functions – one-to-one, many-to-onee)258c 8–5d)[3]x 22–2c)[3]x 361154Let y 2x 1x 30–2 –1–1154 Therefore, fg(2) 3. a) When line meets x-axis, y 03x 6 03x 6x 2Therefore, A (2, 0)b) When the line meets the y-axis, x 0y 3(0) 6 6Therefore, B (0, 6)1c) Area of triangle OAB 2 2 6 6 units squared2(7) 17 3[1]3(–3, 0)21 ,0210–5 –4 –3 –2 –1–1[2]–2123 xminimum pointaxis of symmetry10[4]

8. a)b)c)d)e)f)g)9. a)x 3, x 1f (x) (x 3)(x 1) x2 3x x 3 x2 2x 3f (x) 5( 1, 4)x 1x 0, x 2 2 x 0xyb)161412108 2 10234565 3 4 305124f (x)01[2][2][2][2][2][2][2][10]2y0–5 –4 –3 –2 –1–21212345 x–410–6–88–10–126–14–164[4]c) From the graph, solutions are:( 2, 10) and (3, 0)d) x 4, x 3e) 4 x 3f) x 3, x 22–2–10123x5420 0211. a) a 30 0 3 ms 2–2b) a –4c) d [2]40 20 20 1 ms 250 30 2040 040 2 ms 2100 80 20[2][2] 2b) a 2b 5[2]c) x 1x 3[2]d) (1, 4)[1]e) Choose any two points on the tangent drawn at thepoint x 2.(5, 3) and (1, 5)Gradient 5 3 8 21 510. a)x 4 5 4 3 2 1d) a 0 mse) Distance Area of trapezium1 1600 m3432121060 8g(x) 1614121086420 2[2]13. a) x 2212[2][1]8c) Average speed 30 0.267 ms 111010b) 3.5 1 2.5 minutes06[2]d12. a) Average speed t 60 0.167 ms 1y0[1] 2 (30 50) 40[3]f (x) 8[2][2][2][2]1[6]0–3 –2 –1–1123 x–2–311x 2[1]

c) See shaded regionb) y 4y4B16y 43[2]A 41421210–4 –3 –2 –1–11234 x108–2–36[1]c) y 2x 3yy 2x 3021d) i)123x–3[1]d) y 3x 7y6y 3x 74321234 x4(5A 4B 5A B 156810121416 AP (4, 11)Q: y 252)17. a) y 2 x 4 –214. a) Inequality 1 :Inequality 2:Inequality 3:b) See graph215. a) i) y 0ii) x 0iii) y 4b) x 0y 0y 4y 6 x16. a) P: x 10,b) x 10y 2x y 15y x710Profit (4 6000) (11 7000) 101 000[1]Q (10, 5)Profit (10 6000) (5 7000) 95 000[1]R (4, 5)Profit (4 6000) (5 7000) 59 000[1]Quantity of refrigerator A 4Quantity of refrigerator B 11ii) Maximum profit 101 000[2]–20–2 –1–1A B 15230–4 –3 –2 –1–1B 1][1][1][1][2]b) x 4c) Two1d) x 2 and x 3[1][1][2]e) Maximum point. Coordinates of maximum point5 49 4, 8[2]()18. a) Let y 3x 2Interchanging x and yx 3y 23y x 2x 2y 3Therefore, f 1(x) x 23b) Let y 6 xInterchanging x and yx 6 yy 6 xTherefore, g 1(x) 6 x12[2][2]

x 32. a)c) Let y 2x 1Interchanging x and yy 3x 2y 1Cx(2y 1) y 32xy x y 32xy y 3 xy(2x 1) 3 x3 xy 2x 13 xTherefore, f 1(x) 2x 1[3]Section 9 Geometry andtrigonometry1. a)90º60º60ºA[3]B6 cmb) AC 10.4 cmb)[4][1]3. a) b)C30º[3]5.5 cmc)60ºA45ºX7.5 cmB[3]d)[5][1]Angle BCX 45 4.C120ºD[3]7 cmA138.1 cm6.4 cmB[5]

5. a) 2x x 1803x 180x 1803x 60 b) x 60 90x 90 60x 30 c) x 90 120 100 360x 310 360x 360 310x 50 d) d 60 (vertically opposite)a 180 60 120 c 120 (vertically opposite)e 120 (alternate)b 60 [2][1][2][1][1][1][1][1]6. a) 90 The angle in a semi-circle is a right angle.[1]b) 60 The angles subtended by a chord at the circumferenceof a circle and standing on the same arc are equal. [1]c) 180 80 100 The opposite angles of a cyclic quadrilateral aresupplementary.[1]d)6029. AB2 AC2 BC272 AC2 4.52AC2 72 4.52AC 28.75AC 5.4 cm[2]10. AB2 AC2 BC2AB2 42 5.22AB2 43.04AB 43.04AB 6.6 cm[2]11.tan 60 RQ RQ 6.9 cm7. a) QPR 80 [2]The angle formed by the tangent to a circle and achord, at the point of contact, is equal to theangle in the alternate segment.b) QOR 160 [2]The angle subtended by a chord at the centre of acircle is twice the angle that the chord subtends at thecircumference, standing on the same arc.c) QSR 180 (80 80) 180 160 20 [2]Tangents QS and RS are equal lengths. Triangle RQS isisosceles, making RQS 80 .8. a) ROC 60 [2]The angle subtended by a chord at the centre of acircle is twice the angle that the chord subtends at thecircumference, standing on the same arc.b) ABC 30 CAB 90 30 60 [2]c) OCQ 30 QCP 90 30 120 QPS 180 (120 30) 30 [2]d) QOC 120 COR 60 OCR 6tan Q 8Q tan 1(68)Q 36.9 [2]RQ13. sin RPQ PQRQsin 30 12RQ 12 sin 30 RQ 6 cm14. sin PQR sin PQR [2]PRPQ69PQR sin 1(69)PQR 41.8 [2]RQ15. cos PQR PQRQcos 42 8.2RQ 8.2 cos 42 RQ 6.1 cm[2]PR16. cos RPQ PQ3.8cos RPQ 9.4RPQ cos 1(3.89.4 )RPQ 66.2 [2]17. Using the sine ruleABsin C4sin CBC sin A 6sin 42 6 sin C 4 sin 42 sin C 4 sin 42 6C sin 1(0.446)C 26.5 [2]18. Using the sine ruleBCAB sin A sin CBC 9sin 20 sin 120 9 sin 20 BC sin 120 180 – 60 60 2RCA 90 60 30 [2]PR12. tan Q RQ 30 The angle subtended by a chord at the centre of acircle is twice the angle that the chord subtends at thecircumference, standing on the same arc.[1]e) 70 The angle formed by the tangent to a circle and achord, at the point of contact, is equal to the angle inthe alternate segment.[1]tanQ PRRQ12RQ12tan 60 BC 3.6 cm[2]14[2]

19. Using the cosine rulec2 a2 b2 2ab cos CBC 2 8.52 6.22 2(8.5)(6.2) cos 40 BC 2 29.95BC 29.95BC 5.5 cmii) Using the sine ruleasin A65sin 100 cos ABC 223. T Translationcos ABC 0.646ABC cos 1(0.646)ABC 49.8 [2]21. a)T1m40 tan 40 [3]iii) Bearing of O from B 180 30 49.2 259.2 [1]7 6.2 – 5.62(7)(6.2)b) tan 40 50 sin 100 65sin OBA 0.758OBA sin 1(0.758)OBA 49.2 2O50 sin OBAsin OBA [2]20. Using the cosine rulec2 a2 b2 2ab cos C5.62 72 6.22 2(7)(6.2) cos ABC2b sin BP30 Q(53)[3]24. Enlargement of scale factor 3Centre of enlargement ( 4, 8)[3]25. Reflection in the line x 4[2]26. a) i) (0, 0)ii) 90 iii) Anticlockwiseb) Congruent trianglesA' ( 2, 2) , B' (0, 2), C' (0, 1)c) Image after the transformationA" ( 1, 0), B" (1, 0) , C" (1, 1)d)[1][1][1][1][3]y[2]OTOP11OP311OP tan 40 13.1 m[2]2A’AB’11c) tan 30 OQ111OQ tan 30 C’COQ 19.1 m–2PQ OQ OP 19.1 13.1 6m–10B123x–1[3]22. a)–2–3110ºO65 kmB27. a) Sum of the interior angles of a triangle 180 .Triangle BCD is an isoseles triangle.30º50 kmCDB Ab) i)OAB 70 30 100 110º[5]65 km70º[1]NNO[3]BN50 km30º30º70ºA15180 642 58 [2]b) BAD 64 . The angles subtended by a chord at thecircumference of a circle and standing on the same arcare equal.[2]c) ADB 90 . The angle in a semi-circle is a right angle.ABD 180 (64 90) 26 [2]d) BDT 64 . The angle formed by the tangent to a c

9. a) 215 7 R 1 1 R 1 0 R 1 3 R 1 2 2 2 1111 2 [2] b) 527 5 R 2 0 R 1 1 R 0 5 5 102 5 [2] c) 890 11 R 2 0 R 1 1 R 3 8 8 100132 8 [2] 10. a) 110110 2 11001 2 1001111 Answer – 1001111 2 [3] b) 1111 2 1001 2 0110 Answer – 110 2 [3] Section 3 Consumer arithmetic 1. a) Total hire purchase price 800 (300 10) 800 3000 3800 [2 .

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CSEC Home Economics Free Resources LIST OF CONTENTS CSEC Home Economics Syllabus Extract 4 CSEC Home Economics Syllabus 5 CSEC Home Economics – Family and Resource Management Specimen Papers and Mark Schemes: Paper 01 227 Mark Scheme 239 Paper 02 241 Mark Scheme 253

3 Collins CSEC Physics Workbook answers A1 Scientific method 1. a) Time taken for 20 oscillations t/s 15.5 20.1 23.7 26.9 29.8 32.4 Length l/m 0.150 0.250 0.350 0.450 0.550 0.65 Period T/s 0.775 1.005 1.185 1.345 1.490 1.620 T 2/s 0.601 1.010 1.404 1.809 2.220 2.624 (6)

Collins CSEC Chemistry Workbook answers A1 States of matter 1. a) i) Ammonium chloride (1) ii) Diffusion Diffusion is the movement of particles from an area of higher concentration to an area of lower concentration until the particles are evenly distributed. (2) iii) The ammonia solution gave off ammonia gas and

Collins CSEC Chemistry Workbook answers A1 States of matter 1. a) i) Ammonium chloride (1) ii) Diffusion Diffusion is the movement of particles from an area of higher concentration to an area of lower concentration until the particles are evenly distributed. (2) iii) The ammonia solution gave off ammonia gas and