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4.0 4.1 4.1.1 I 2 3 4 5 6 7 8 PART TWO: THE YEAR 2013 KCSE EXAMINATION QUESTION PAPERS MATHEMATICS ALT. A (121) Mathematics Alt. A Paper 1 ( 121/1) SECTION I (50 marks) Answer all the questions in this section in the spaces provided. _- Z " _ Evaluate Q — (2 marks) 6 X -2 72 +' 8 X 3 The production of milk, in litres, of 14 cows on a certain day was recorded as follows: 22, 26, 15, 19, 20,16, 27,15, 19, 22, 21, 20, 22 and 28. Detennine: (a) the mode; (1 mark) (b) the median. (2 marks) Use logarithms, correct to 4 decimal places, to evaluate: 3,1794 X 0.038 1.243 Simplify the expression: (4 marks) 161111-9112 4m2 - mu — 3n’ (3 111111115) A wholesaler sold a radio to a retailer making a profit of 20%. The retailer later sold the radio for Ksh l 560 making a profit of 30%. Calculate the amount of money the wholesaler had paid for the radio. (3 marks) A point P on the line AB shown below is such that AP = %AB. By construction locate P. (3 marks) c a A B Chelimo’s clock loses 15 seconds every hour. She sets the correct time on the clock at 0700h on a Monday. Determine the time shown on the clock when the correct time was l900h on Wednesday the same week. (3 marks) Given that sin (x + 20)° = —0.7660, find x, to the nearest degree, for 0° 3 x S 360°. (3 marks) A number m is formed by writing all the prime numbers between 0 and 10 in an ascending order. Another number n is formed by writing all the square numbers between 0 and 10 in a descending order. (a) Find m — n; (2 marks) (b) Express (m — n) as a product of its prime factors. (1 mark) The figure below shows a net of a solid. (Measurements are in centimetres). 3 5 3 4 5 3 3 5 Below is a part of the sketch of the solid whose net is Sh0WIl above. Complete the sketch of the solid, sho\m'ng the hidden edges with broken lines. (3 marks) The interior angles of an octagon are Zx, -%x, (x + 40)°, 110°, 135°, 160°, (Zx + 10)” and 185°. Find the value of x. (2 marks) Astraight line passes through points (-2, 1) and (6, 3). Find: (a) the equation of the line in the form y = mx + c; (2 marks) (b) the gradient of a line perpendicular to the line in (a). (1 mark) A triangle ABC is such that AB = 5 cm, BC = 6cm and AC = 7cm. (a) Calculate the size of angle ACB, correct to 2 decimal places. (2 marks) (b) A perpendicular drawn from A meets BC at N. Calculate the length AN correct to one decimal place. (2 marks) A cylindrical pipe 2% metres long has an intemal diameter of 2l millimetres and an extemal diameter of 35 millimetres. The density of the material that makes the pipe is 1.25 g/cm‘. Calculate the mass of the pipe in kilograms. (Take 1r = 3%). (4 marks) The figure below represents a pentagonal prism of length 12cm. The cross-section is a regular pentagon of side 5 cm. _ _ _ _ - - - - ' ' ' ' > ' ’ _ ’__._‘\ _ - » - ' ' ' ' ' ' _ _- — \ _ _ - — - ” \ . \ ‘\ . \ \ ___‘\ _ _ _ _ _ _ _ _ _ _ _ __ 12cm 5cm Calculate the surface area of the prism correct to 4 significant figures. (4 marks) Given the inequalities x — 5 5 3x — 8 < 2x — 3. (a) Solve the inequalities; (2 marks) (b) represent the solution on a number line. (1 mark) SECTION ll (50 marks) Answer only five questions in this section in the spaces provided. A farmer had 540 bags of maize each having a mass of l I2 kg. Afier drying the maize, the mass decreased in the ratio 15: l 6. (a) Calculate the total mass lost afier the maize was dried. (3 marks) (b) A trader bought and repacked the dried maize in 90 kg bags. He transported the maize in a lorry which could carry a maximum of 120 bags per trip. (i) Determine the number of trips the lorry made. (3 marks) (ii) The buying price of a 90 kg bag of maize was Ksh 1 500. The trader paid Ksh 2 500 per trip to transport the maize to the market. He sold the maize and made a profit of 26%. Calculate the selling price of each bag of the maize. (4 marks) (a) Solve the equation, L2’? = (4 marks) X _ (b) The length of a floor of a rectangular hall is 9m more than its width. The area of the floor is 136 ml. (i) Calculate the perimeter of the floor. (4 marks) (ii) A rectangular carpet is placed on the floor of the hall leaving an area of 64 mi. If the length of the carpet is twice its width, determine the width of the carpet. (2 marks) A trader bought 2 cows and 9 goats for a total of Ksh 98 200. If she had bought 3 cows and 4 goats she would have spent Ksh 2 200 less. (a) F om1 two equations to represent the above information. (2 marks) (b) Use matrix method to determine the cost of a cow and that of a goat. (4 marks) (c) The trader later sold the animals she had bought making a profit of 30% per cow and 40% per goat. (i) Calculate the total amount of money she received. (2 marks) (ii) Determine, correct to 4 significant figures, the percentage profit the trader made from the sale of the animals. (2 marks) Two towns, A and B are 80 km apart. Juma started cycling from town A to town B at 10.00 am at an average speed of 40 km/h. Mutuku started hisjoumey from town B to toum A at 10.30 am and travelled by car at an average speed of 60 km/h. (a) Calculate: (i) the distance from town A when Juma and Mutuku met; (5 marks) (ii) the time of the day when the two met. (2 marks) (b) Kamau started cycling from town A to town B at l0.2lam. He met Mutuku at the same time as Juma did. Determine Kamau's average speed. (3 marks) s=t’—5t +3t+l0. (a) Find s whent = 2. (b) Determine: (i) the velocity of the particle when t = 5 seconds; (ii) the value of t when the particle is momentarily at rest. (c) Find the time, when the velocity of the particle is maximum. 21 The displacement, s metres, of a moving particle from a point O, afier t seconds is given by, (2 marks) (3 marks) (3 marks) (2 marks) ln the figure below, OABC is a trapezium. AB is parallel to OC and OC = SAB. D is a point on OC such that OD: DC = 3:2. A q B P O > C D (a) Given that OA = p and AB = q, express in tem1s of p and q: (i) OB; (ii) AD; (m) cs. k and r are scalars. Determine the values of k and r. (l mark) (2 marks) (2 marks) (b) Lines OB and AD intersect at point X such that AX = kAD and OX = rOB, where (5 marks) (a) On the grid provided, draw the square whose vertices are A (6, -2), B (7, -2), C(7,—l)andD(6,—l). 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A f _-_.- .._. 1 Y ,,_..: . _ ' *'i1‘.1"Tii~ 'TliL"1 1;" LT“ ‘if’ LI; ';l;l_i;I L __.~ - mt. .._.,., ..,_ .._.-. 1,__, ._.1 _ ._--. .r*,_;__T ,_._- 5., 1-rm . ;.., - ~*‘*** 11; .1. ,1»: :::: »-* 7:4‘Y‘1V\>'7‘%\4\7’ +'- »_-_L -it --= ,%.-= %fim”§ i IIIIIII n - ~""\ ii '\ IIIIIIIII III III \ ‘ > I--I 7 I II In ::II ) HIIIIIIH I IIIII II II .J 1; _j_;_. ; I E. >1? (b) On the same grid, draw: (i) A’B’C'D', the image of ABCD, under an enlargement scale factor 3, centre (9, -4); (3 marks) "5: :5" .. i f"j * 111, ¢-_ >0 t +4»- ,:;1j;‘ ‘ 1;. IA I ‘*1 444 t _,.‘.i w 9—+‘ t fr" " w i .‘_L_..7 ‘L ll -I-ll 1747 'I""t" "llH' iIHQI“** ""¢'.‘;;__'f'§f.‘ .,. A _ . _ L . .. >\ IIIII IIII IHIE Hill 1' .::: mi It ‘-=i!i 1+» i:: ii; fif} ‘ II mii gigs Iii (ii) A"B"C”D”, the image of A’B’C'D' under a reflection in the line x = 0; (2 marks) (m) A'"B"’C"'D’”, the image of A"B"C"D" under a rotation of +90° about (0,0). (2 marks) (c) Describe a single transformation that maps A’B’C'D' onto A”'B’”C’”D"’. (2 marks) 24 The figure below represents a cone of height 12 cm and base radius of 9 cm from which a similar smaller cone is removed, leaving a conical hole of height 4 cm. l2cm \ \ \ \ \\ .-__--_____; / -A / 0 / , / / / / _____./,_-__ ____ \ \ ITI ‘L ___-- /1 : "_\\ ~~____ _ — ' ' /,_.----———r-~——~~~__‘\ " ‘ - c ~ > \ ‘1i9CIl1—i-vi (a) Calculate: (i) (ii) (b) (i) (ii) the base radius of the conical hole; (2 marks) the volume, in terms of 11:, of the smaller cone that was removed. (2 marks) Determine the slant height of the original cone. (l mark) Calculate, in terms of 1:, the surface area of the remaining solid afler the smaller cone is removed. (5 marks) 4.1.2 Mathematics Alt. A Paper 2 (121/2) SECTION l (S0 marks) Answer all the questions in this section in the spaces provided. 1 The sum of n terms of the sequence; 3, 9, 15, 21, is 7500. Determine the value of n. (3 marks) 2 A quadratic curve passes through the points (—2, 0) and (l, 0). Find the equation of the curve in the form y = ax’ + bx + c, where a, b and c are constants. (2 marks) 3 Make d the subject of the fomula, d1 P = %mn1 — qT (3 marks) 4 Solve the equation 2 logx — log (x — 2) = 2 log 3. (3 marks) 5 (a) Using a pair of compasses and ruler only, construct an escribed circle to touch side XZ of triangle XYZ drawn below. (3 marks) Z Y X (b) Measure the radius of the circle. (l mark) 6 The equation of a circle is given by xi + 4x + yl —— 2y — 4 = 0. Determine the centre and radius of the circle. (3 marks) 7 (a) Expand (l — x)’. (l mark) (b) Use the expansion in (a) up to the term in x3 to approximate the value of (0.98)5. (2 marks) 8 The position vectors of points F, G and H are f, g and h respectively. Point H divides FG in the ratio 4:— 1. Express h in terms off and g. (2 marks) 10 ll 12 13 14 15 6 0/W Two machines, M and N produce 60% and 40% respectively of the total number of items manufactured in a factory. lt is observed that 5% of the items produced by machine M are defective while 3% of the items produced by machine N are defective. lf an item is selected at random from the factory, find the probability that it is defective. (3 marks) Two taps A and B can each fill an empty tank in 3 hours and 2 hours respectively. A drainage tap R can empty the full tank in 6 hours. Taps A and R are opened for 5 hours then closed. (a) Determine the fraction of the tank that is still empty. (2 marks) (b) Find how long it would take to fill the remaining fraction of the tank if all the three taps are opened. (2 marks) 4 Simplify the expression &, leaving the answer in the fonn a \/b + c where a, b and c are integers. \/-5 + \/3 (3 marks) A point P moves inside a sector of a circle, centre O, and chord AB such that 2cm < OP 5 3 cm and angle APB = 65°. Draw the locus of P. (4 marks) The table below shows income tax rates in a cenain year. Monthly income in Kenya shillings 1 Tax rate in each shilling Up to 9 680 10% from968l to 18800 15% from 18 801 to 27 920 20% from 27 921 to 37 O40 25% over 37 040 30% In that year, a monthly personal tax relief of Ksh 1 056 was allowed. Calculate the monthly income tax paid by an employee who eamed a monthly salary of Ksh 32500. (4 marks) Solve the equation 6 cos’ x + 7 sin x — 8 = 0 for 0° S x S 90°. (4 marks) The positions of two towns are (2° S, 30° E) and (2° S, 37.4° E). Calculate, to the nearest km, the shortest distance between the two towns. (Take the radius of the earth to be 6 370 km) (2 marks) The vertices of a triangle T are A(l , 2), B(4, 2) and C(3, 4). The vertices of triangle T’, the image ofT are A'(%, 1), B'(2, l) and C’(%, 2). Determine the transformation matrix M = (3 that maps T onto T’. (3 marks) SECTION ll (50 marks) Answer only five questions from this section in the spaces provided. The Hire Purchase (H.P.) price of a public address system was Ksh 276000. A deposit of Ksh 60000 was paid followed by I8 equal monthly instalments. The cash price of the public address system was 10% less than the H.P. price. (=1) (b) (C) Calculate: (i) the monthly instalment; (2 marks) (ii) the cash price. (2 marks) A customer decided to buy the system in cash and was allowed a 5% discount on the cash price. He took a bank loan to buy the system in cash. The bank charged compound interest on the loan at the rate of 20% p.a. The loan was repaid in 2 years. Calculate the amount repaid to the bank by the end of the second year. (3 marks) Express as a percentage of the Hire Purchase price, the difference between the amount repaid to the bank and the Hire Purchase price. (3 marks) In the figure below, PR is a diameter of the circle centre O. Points P, Q, R and S are on the circumference of the circle. Angle PRQ = 72°, QS = QP and line USV is a tangent to the circle at S . Giving reasons, calculate the size of: (8) (b) (C) (<1) (6) L QPR; (2 marks) L PQS; L OQS; L RTS; L RSV. (2 marks) (2 marks) (2 marks) (2 marks) (a) Complete the table below for y = x‘ + 4x3 — 5x — 5. (2 marks) X 1-sl—4|—3|-2l—1|0||[2{ =x-‘+4.\"-5x—5l | |19| l |—5[ l (b) On the grid provided, draw the graph ol'y = x‘ + 4x3 — 5.r — 5 for —- 5 5x 5 2. (3 marks) (c) (i) Use the graph to solve the equation x’ + 4.\-1 — 5x — 5 = 0. (2 marks) (ii) By drawing a suitable straight line on the graph, solve the equation x"‘ + 4r‘ — Sx — 5 = - 4x — l. (3 marks) The figure ABCDEF below represents a roofofa house. AB = DC = l2m, BC = AD = 6m, AE (8) (b) =BF=CF=DE=5mandEF=8m. E . 8 m F Sm A C l2m 6m B Calculate, correct to 2 decimal places, the perpendicular distance of EF from the plane ABCD. (3 marks) Calculate the angle between: (i) the planes ADE and ABCD; (2 marks) (ii) the line AE and the plane ABCD, correct to 1 decimal place; (2 marks) (m) the planes ABF E and DCFE, correct to l decimal place. (3 marks) (a) Complete the table below, giving the values correct to l decimal place. (2 marks) X“ 0 | 40 so | 120 160 I 200 240 2 sin (X + 20)“ 0.7 2.0 0.0 - 2.0 \/§ cosx | 1.7 1.3 | — 0.9| -1.6 (b) On the grid provided, using the same scale and axes, draw the graphs of y = 2 sin (x + 20)° andy = \/5 cos x for 0° Sx S 240°. (5 marks) (c) Use the graphs drawn in (b) above to determine: (i) the values of x for which 2 sin (x + 20) = x/5 cos x; (2 marks) (ii) the difference in the amplitudes of y = 2 sin (x + 20) and y = \/3 cos x. (l mark) Three quantities R, S and T are such that R varies directly as S and inversely as the square of T. (a) Given that R = 480 when S = 150 and T = 5, write an equation connecting R, S and T. (4 marks) (b) (i) Find the value of R when S = 360 and T = l,5. (2 marks) (ii) Find the percentage change in R if S increases by 5% and T decreases by 20%. (4 marks) The equation of a curve is given by y = 5x — éxz. (a) On the grid provided, draw the curve of y = 5x — fr’ for 0 5 x 5 6. (3 marks) (b) By integration, find the area bounded by the curve, the linex = 6 and the x-axis. (3 marks) (c) (i) On the same grid as in (a), draw the line y = 2x. (l mark) (ii) Determine the area bounded by the curve and the line y = Zx. (3 marks) 105 24 The table below shows marks scored by 42 students in a test 35 69 57 S8 40 56 36 62 49 41 68 73 65 64 64 54 74 73 4] 61 S6 I 86 47 ' s1 I (a) Starting with the mark of 25 and using equal class intervals of 10, make a frequency distribution table. (2 marks) (b) On the grid provided, draw the ogive for the data. (4 marks) (c) Using the graph in (b) above, estimate: (i) the median mark; (2 marks) (ii) the upper quartile mark. (2 marks)
5.0 5.1 5.1.1 THE YEAR 2013 KCSE EXAMINATION MARKING SCHEMES MATHEMATICS ALT. A (121) Mathematicsa Iternative a Paper l (121/1 l. 36 _ -103 -12 -27 =-3-4 Ml =_7 Al 2 2. (a) Mode = 22 Bl (b) Median l5,15.16.19.19.20.20.21.22.22.22.26,27.28 20+21 medi:m=?—- MI = 20.5 Al 3 3. N0. LO 1.794 0.2538 Ml 0.038 2.5798 2.8336 Ml 1.243 0.0945 7.7391+3 M‘ *3 ‘/ 0.3799 15797 L 4 all log \/ + and - operations V’ 4. (4m + 3n)(4m — 3n) (4m + 3n)(m — n) = 4m~3n m—n M1 faclorizing numerator v’ Ml faclorizing denominator ¢ AI 3 5. Retailer l30% —- 1560 __. 1560 X I00 M1 l00% ——il30 = 1200 Wholesaler 120% —~ I200 00 M 1200 X 1 l lO0% -no = 1000 Al 3 207 D sl Bl A P Bl Bl 3 construction of equal pans on AC draw DP//CB such that AP= % AB locating point P From 0700 h Monday to I900 h Wednesday =24><2+l2h =60h Time lost = 60 X 15 = 900 sec = I5 min Time shown on clock: l900h -15 min = 184 Sh Ml M1 Al 3 x+20=23O°orx+20=3lO' x = 210° or x = 290° Bl Bl Bl 3 for 230° or 310° (a) 2357_ 941 1416 (b) l416 = 2’X3X59 Bl Bl Bl 3 for 2357 and 94l J for 1416 E Bl C A B 3 Bl Bl lines AF, ED equal and parallel to BC lines AB, FC equal and parrallel or lines AE and FD equal and parallel or lines CD, EB equal and parallel. completing the solid showing dotte lines. d ZX+%.\'+.\'+4O+l|0+l35+l60+2Jt+l0+185 -'h= 440=x=440><-2—=s0° Al 2 11 -T = I080 Ml 12 (a) Gradient of line: % =% line equation L3 = L x — 6 4 _ = L _ y 3 4 (x y = :11-x + 1% (b) Gradient of perpendicular line —Lm' = —l 4 m’ =-4 6) Ml Al L. 3 13 (a) 5’=7’+6’—2><6><7cosC Ml (b) h cos C C = 44.42’ = 7 sin 44.42 = 4.9 cm = 49+3s-25 24 Al M1 i 4 Volume of pipe material Q(1.7s=-1.0s=)><2s0¢m Ml 7 Ml I540 Cm“ mass of pipe 1540 X 1.25 1000 1.925 kg Ml Al 4 209 h = 2.5 tan 54° = 3.44] cm Bl Area of pentagonal faces =2(%><5><3.441><5) M1 = 86.025 Total area Ml = 86.025 + 5( l2 X 5) Al = 386.0 4 (a) x—5S3x-—8 —?_xS—3 x?.l.5 Bl 3x—8<2x—3 x<5 Bl l.5Sx<5 (b) 0:1-**5“ Bl 7. (a) Mass after decrease 112 X % Ml or equivalent = 105 kg Total decrease (112-105)>< 6 = l5000 Total 945000 + l5000 Ml Selling price per bag: 960000 X L26 M1 :0 =1920 A1 l0 21 l 8. (a) (b) (i) (x+3)(x-—2)=24 x’+x—3O=O (x+6)(x—5)=0 Ml x=—6 0rx=5 Al Ml Ml (x+9)x = 136 x1+9x-— l36=0 (x+ l7)(x—9)= 0 x =-17 or x= 8 '."x=8 A1 penmeter =2(8+l7)=50m B1 Ml Ml (ii) 2xXx= l36—64 2.1-:= 72 x*= 36 x= 6m Al T Ml 2c + 9,; = 98200 Bl (8) Bl 3c + 4g = 96000 (b) Det.of(§ Z)=—l9 M’ =_Tl9'(-43 .29) Bl w—L(-§ "§)(§ i)(Z>'=-1%(_§ 'Z)(ZZ§3Z) M‘ -001” _‘i9>(§)=~%<24800X1.3+9X5400Xl.4 Ml = l32S20 Al (ii) % profit _ 132520 — 98200 - 98200 X lO0% Ml = 34.95% Al l0 (a) (i) Time taken by Juma = fih Time taken by Mutuku = T Let x km be distance from A 2(5x— 160): I20 10x=440 x= 44km (ii) Time they met 10.00 am + %h = 10.00 +l h 6min =ll.O6am (b) Speed if Kamau delayed by 2| minutes Kamau‘s time = — %)h speed needed: = 58% km/h Al 10 _Q:-1=l 40 so 2 3x—%(80—x) __i 120 '2 Bl 80-x Bl Ml Ml Al Ml Al M1 4 0 _l _4h Ml 21 (a) Displacement.s,whent=2 2"—S><2‘+3><2+lO Ml =4 Al (b) (i) velocity when I = 5 seconds V=~‘1i=3r‘—l0r+3 BI dr whenl=5,V=3><5’—l0><5+3 M1 =23 Al (ii) 3:’ - 101+ 3 = 0 M1 (3r—1)(r—3)=0 M1 = i = 1 3, r 3 A1 (c) time when velocity of particle is at its maximum acceleration = = 6; __ 1Q = Q Ml ,=%=1%S Al 10 (a) (i) QB=I3+q (b) also E(r—l)+rz1=—kE+3kQ —k=r—l and r=3k —k=3k—l -4/( =—1=/< =2 subxrirure r = 3 X— (ii) Ap=—g+%><5q (m) QB=—5Ql +g+q A_X=/<(A.D) =/<<-2+32> =-—kE+ 3kg A_X=—@+'(QB) =—n+r(11+q) =E(r——l)+ Bl Ml =—Lv+3Q Al Ml _ “Hp Al Bl Bl '21 Ml Ml I ._- zi Al -¥>~ A .- O or equivalent ' / i I 6/1 ll \ _ \\ I l B .r>” A"'l"+ , . “°l <»¢,¢,~ -val l1s“4$"e‘?"’ 9* ‘Er ii 4 s C/H 811/ 3 . 9 " 2 ’ 4 \/ (a) ABCD J drawn Bl (b) (i) Centre identified and used J Bl Bl AA‘, BB’, CC‘ and DD‘ drawn J Bl completion of square A'B‘C'D' and labelled (ii) A"B"C"D" B2 A"B"C"D" drawn J (m) A...B."c...D,.. B2 A...B...C...D...drawn (c) Reflection on line y = —x Bl reflection Bl line y =—x l0 v H) (i) L=A_ 9 u 9x4 r=T=3c‘m Al (ii) volume of material drilled out =%nx?x4 Ml =l27I (b) Slant height of cone =flUTF=wm» m (c) Surface area of solid after conical has been drilled rr><9><15+1r><(9’-3’)+rr><3x5 Ml Ml Al for7ZX9><3><5 Ml summing up Al TF' 5.1.2 Mathematics Alternative A Paper 2 (121/2) 1. 1“term,a _l 7500- 2 3n’ =75O0 n = \/2500 =50 = 3; common difference,d=6 B1 {2><3+(n—1)><6} Ml Al ? 2. y=(x+2)(x-1) y=x’+x-2 Ml L 2 3. qdz _i Li P-Zmn n qd’_1 i___ 1_p n 2mn dz = %mn"‘ — nP q dz /%mn’— nP q Ml Ml AI T 4. Log = log 3’ (0%) 9 _I’_= x—2 X2—9X+18=0 (x-6)(x-3)=0 x=6orx=3 Ml Al Ml ‘ ? 5 (=1) I W \ \ \ \ \ B1 extending YX and YZ X \ Z Bl bisecting 4; vxz and XZW Bl escribed circle drawn / Y Y z’ }? z V Bl (b) radius = 3.1 4 allow 1 0.] Completing square on L.H.S. xZ+4x+4+y2-2y+l=4+4+1 B1 (x+2)2+(y—l)2=9 Bl centre of circle : (-2, 1) B1 radius of circle: 3 units 3 (a) (1 - X)5 = l + 5(-x) + 10(-X)’ + l0(-x)’ + 5(-x)‘ + (-x)5 =l-5x+l0x2-10x3+5x“-x’ B1 (b) (0.98)’ = (1 - 0.02)‘ = x = 0.02 (O.98)5 = 1 — 5(0.02) + lO(O.O2)2 » lO(O.O2)3 Ml =1- 0.1+ 0.004 - 0.00008 = 0.90392 Al ? 220 By: _“@2 0:-'23 -150 -115 _ -1 4 Ml I3‘ 4+(-1)f+ 4+(-1)? _-_1 A '3f+3$ A1 ? P(dcfective) : M - 0.6 >< 0.05 = 0.03 N - 0.4 x 0.03 =0.012 P(defective) 0.03 + 0.02 = 0.042 Ml ? 0.03 0‘ -0; A1 04 Q91 4! .g3 Ml For 0.6 X 0.05 or 0.4 X good defective good defective (a) Fraction filled if A and R are open for 5h i_i=i 5X<3 6) 6 Fraction of tank still empty=1—%=% B1 Bl (b) Fraction filled ifA,BandRare0pen forlh L L_i=l 3+2 6 3 - 1.2 I 3 T1metakentofillthetank=€¢€=gx3 M1 = %h or 15 min Al 4 »/E = 4~/§(~/3-1/5) M1 /§+/5 (1/5+./'3')(\/§—~/5) =4/§(/5-E) M1 5-3 =2./§(~/3-/5) :2‘/E_6 A1 T 221 N A A AAOB = 130° arc AB - solid curve arc A'B' - broken curve region shown Locus of P I 1 C 130' q Bl B1 B1 B1 T 9680 X 0.1 = 968 9120 X 0.15; 9120 X 0 =1368 =1824 Net tax = (968 + 1368 + 1824 = 4249 Ml .2; 4580 X 0.25 Ml = 1145 +1145)-1056 Ml A1 4 6(1-sin2x)+7sinx-8=0 Ml 6-6sin2x+7sinx-8 6sin2x-7sinx+2=0 (3sinx-2)(2sinx-1) sinx=% 0rsinx=% x=41.81° orx=30° =0 =0 M1 Ml Al 4 Distance between town = 21: X 6370 cos 2° >< = 822.212128l = 822 km s K and S L M1 360 Al <“ "><1 4 3>~<‘ c d 2 2 4 _ a+2b=§ 4a+2b=2 3a =%=>a=% |~_)-- %+2b= =b=0 c+2d=l 4c+2d=1 3c =0=>c=0 0+2d =1==d=% M=(’5 Z) --N- --[Q gum.- T Ml Ml L 3 J f0l'lTl8llOl'l and solution of simultaneous equations J formation and solution of simultaneous equations (a) 0) 2760001; 600 =l2000 (ii) 276000 >< 0.9 = 248400 (b) 248400 X 0.95 = 235980 235980 X 1.22 = 3398ll.2 (c) 3398112 - 276000 63811.2 276000 X100 =23.12% 00 Ml A1 M1 A1 Ml M1 Al Ml Ml Al T 223 0'/'23 450 425 (b) 4PQS = I80” - 2(72) = 36° Bl (c) LOQS =36° - l8°= 18° Bl (d) ARTS = 180- (36+ 18) = l26° B1 (e) ARSV = 90° - 36° = 54° Bl a) AQPR=90°-72°=l8° B1 LPQR = 90° - angle subtended by diameter Bl APSQ = 72° - angle subtended at the circumference by chord PQ equal and base is of isosceles AQPS = 72° B1 base angles of isosceles A OPQ = 18° Bl extension angle RTS equal to sum of opposite interior B1 angles TSP and TPS 4RSV = 4RPS - angle in altemate segment. Bl 10 or equivalent 19. (a) x -5-4-3-2-1012 y -5 15 13 3 -5 9 =x»‘+4x1-5x-5 (b) (c) (i) X = -4.8, -0.7, 1.5 B2 B2 A \ ze- I Sl / to. Pl / >\ c1 I-5! -4» -3 -2 -ii?’-5k 2 3) * ‘k -10* . -J ‘J= -41-: (ii) y=-4x-1 Pl Solutions Ll x=-4,-l, l. Bl 10 allow Bl for 4 correct Suitable scale All correctly plotted 10.1 allow Bl for 2 values \/ plotting for line 224 (a) = distance of EF from place ABCD slant height from F to BC = J5’- 3’ M1 E 9§w\ F =4 ‘ ‘\“\\5/t1 I = distance of EF from planeABCD ) It i _ ' 1,_ _ — - * ‘ =J4’- 2’ M1 ’ ‘L _ _ l / 1/ ' ,, mm 15 fl =/fi=3.46m Al (b) (i) angle between planes ADE and ABCD _ tan_, /§ Ml or equivalent ‘ 2 =60° Al (ii) angle between line AE and plane ABCD = sin‘ ‘Z? Ml or equivalent = 43 .9° A 1 (m) angle between planes ABFE and DCFE = 2¢tan' ' im M1 tan' ' ;,— 0r equivalent /E ¢ 12 = s1.s° M1 doubling Al 10 21. (a) x 0 40 80 120 160 ZOO 240 y= 1.7 1.3 -1.3 2sinx+20 B1 y = 0.3 -1.6 -0.9 /gcos x Bl (b) “73€é§§f5?‘ L s5sy¢y?1 1° +'k f. .7 1 ;_-_ A A ., . ,__._ ._.__, .» ff‘: ' 1;: '. ,..:;,l*"fi 1 . 1.5 ~— ~ s » .=.,,,¢- .. .,.. .. ..i,~ IIi_Z'IlVl§._‘ “If I_, __ . .f‘ I“ _ .1 II: II _' .. 1.0 ’.T.:i’:":_‘.f¢_‘:’:;: :1." :.;| .1.:":t _:;' _:_I1 0.5 , 51 11; §_§‘::::‘..;:;;,-=;__;:_;: .1: I .,:‘:;"_iQ ' :: 1,1: 0.0 ? -0.5 V ‘vb _,,__ o 511201130? 1_ 1.01» \...,_ -1.0 ¢;;:1?:;:j5§ 59% my ;. Q YE 1; ;, ";.:_ gs, J V1 k . -1.5 _._. . .1 : .,.,, _..,._. . %qk.,,.,. . ,. . w __ _ _ __ .4.:‘>__,.l . ,..,._._ 4 . ., _. _ , ~ l 4 -- - é— - ,_ _ __ *1 % i V l._. 1,... __ Z __ _ _ _ --—-----+~--4 ~»---fli A- 4 »__~ _.k_A~.i .%.__€__ ___..____ ""*—-:—_~:*:'f_—-">%' --ff-_~—~f--~-* '-> <'r—--— —~ *—< — —i—v————~~/ii J § _ , _ 'T‘T‘_’§.i. ‘M * ‘ _ _ ‘ _ _ ,,_ -A _:‘.f::tTttL;“t;'t1; ..'. _ ._ . 1 , 3 _ '1 _~;;: ‘H: :, --,...1_._._-... Ha. _ a _ » ~ ~:'~l—~_;~ ~ £4"? (c) (1) Identification of median = 575 1 O 5 (ll) Identification of upper quartile mark = 665 1: 0 5 Bl \/ marks class column Bl J frequency column ___ ___ _._ __ __ _ ‘___ ‘_ _ _ ._ ._¢ _, 4.- ‘ >_._t_.a.,__a - czrgs t" *,:::E1: 2;; 3:::l:*—;“
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