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4.2 MATHEMATICS ALT. B (I22)

4.2.1 Mathematics Alt. B Paper 1 (122/1)

SECTION I (S0 marks)

Answer all the questions in This section in the spaces provided.

1.Without using a calculator, evaluate:

‘3(-5 — +7)/ +2(-3 + -6). _ (3 marks)

2.The first four prime numbers are written in descending order to fom1 a number. (a) Write down the number. (1 mark)

(b) Find the total value of the hundreds digit in the number. (2 mark)

3. Without using a calculator evaluate:

( 2/3 of5/2/5-23/10)/(3 marks)************************** 4.Tito owned Ksh 600 to Nekesa, Ksh 750 to Mwita and Ksh 650 to Auma. He had Ksh I200 to repay to the three people in proportion to what he owed them. Calculate the amount of money Mwita received more than Nckesa. (3 marks)

5.Given that r = 2 and h = 3r — l, evaluate 7r +2‘-h . d4h — 2r

6.The surface area of a cube is 1176 cmz. Determine the length of one of its sides. (3 marks)

( 7. By construction, divide the line PQ below into six equal parts. (3 marks) l 1 P Q Given that tan x = % and .r is an acute angle, without using mathematical tables or a calculator, find the value of 2 sin x —- cos x. (3 marks) A box contains five shillings coins and ten shillings coins. The number of ten shillings coins are 6 times as many as the five shillings coins. The total value of all the coins in the box is Ksh 2600. Determine the total number of coins in the box. (4 marks) _2 1 Simplify LE, leaving your answer in index form. Hence evaluate the expression. 4-3 + 8§ (4 marks) A retailer bought a mobile phone for Ksh 5750. The marked price at the retailer’s shop was l2% higher than the buying price. Afler allowing a certain discount, the retailer sold the mobile phone for Ksh 61 I8. Calculate the percentage discount. (3 marks) l6 Factorise 9112 — F (2 marks) c Three types of books A. B and C were each piled on a table to attain the same height. The thickness of the books were l2 mm, 28 mm and 54 mm for types A, B and C respectively. Find: (a) the least height attained; (3 marks) (b) the number of type A books piled. (l mark) The sum of the interior angles of a regular polygon is l260°. Find the size of each interior angle. (3 marks) The corresponding lengths of two similar triangles are 5 cm and 7.5 cm. lf the area of the larger triangle is 22.5 cm’, calculate the area of the smaller triangle. (3 marks) The area of a sector of a circle is 77 cml. The arc of the sector subtends an angle of 45° at the centre of the circle. Find the circumference of the circle. (Take it = 2%). (4 marks) SECTION ll (50 marks) Answer only five queslions in this section in the spaces provided. The figure below represents a solid prism with a semi-circular groove. The dimensions are as shown. 0.8m 2m l.4m . (a) Calculate: (i) the volume of the prism; (4 marks) (ii) the total surface area of the prism. (4 marks) (b) All the rectangular faces are painted. Calculate the percentage of the surface of the prism that is painted correct to I decimal place. (2 marks) (8) (b) (C) (=1) (b) Ma Three vertices ofa parallelogram ABCD are A(——7, 3), B(l, — 1) and C(5, l). On the grid provided, draw the parallelogram ABCD. (2 marks) Detennine: (i) the gradient of the line AB; (2 marks) (ii) the equation of line AB in the form y = mx + c, where m and c are constants. (2 marks) Another line L is perpendicular to CD and passes through point (l , 3)i Determine: (i) the equation of L in the fonn ax + by = c where a, b and c are constants; (3 marks) (ii) the coordinates of the y-intercept of line L. (l mark) The roots of a quadratic equation are % and -1. Write down the quadratic equation in the form axz + bx + c = 0, where a, b and c are integers. (3 marks) (i) Barasa bought (2y + 1) mangoes at y shillings each. The total cost of the mangoes was Ksh 55. Find the cost of each mango. (4 marks) (ii) Karau spent Ksh 95 more than Barasa to buy the same type of mangoes. For every 6 mangoes he bought, he was given one extra mango. Calculate the total number of mangoes Karau got. (3 marks) The angle of elevation of the top T, of a vertical mast from a point P, 100 m away from the foot F, of the mast is 14°. (8) (b) (c) (<1) Using a scale of l cm to represent 10 m, make a scale drawing to represent the above infonnation. (3 marks) Using the scale drawing, determine the height of the mast. (2 marks) A support cable, 27 m long, is fixed tightly at a point D on the mast 5 m below T and at a point C on the ground. Points P, F and C lie on a straight line with P and C on opposite sides of F. On the scale drawing, show the position of the cable. (2 marks) Use the scale drawing to detennine: (i) the angle of depression of C from D; (l mark) (ii) the distance of C from P. (2 marks) 2| ln the figure below, P, Q, R and S are points on the circumference of the circle centre O. TP and TR are tangents to the circle at P and R respectively. POQ is a diameter of the circle and Giving reasons in each case, find the size of: (a) (b) (c) (d) (e) LROP; LPSR; LORP; LTRP; L RTP. (2 marks) (Z marks) (2 marks) (2 marks) (2 marks) 22 The figure below represents a cone whose vertical height is I2 cm and slant height is l5 cm. % A (i) the radius, OA, of the cone; (2 marks) (a) Calculate: (ii) the volume ofthe cone. (2 marks) (b) A smaller cone of radius 6 cm is cut off from the cone above to leave a frustum. Calculate: (i) the height of the smaller cone; (2 marks) (ii) the volume of the smaller cone: (2 marks) (m) the volume of the fi'ustum. 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It decelerated constantly and came to rest after 25 seconds. (a) On the grid provided, draw the velocity—time graph for the motorcycle. (4 marks) (b) Use the graph to detem1ine: (i) the deceleration of the motorcycle; (2 marks) (ii) the total distance travelled; I (2-marks) (m) the average speed for the motorcycle, correct to 3 significant figures. (2 marks) 113 4.2.2 Mathematics Alt. B Paper 2 (122/2) SECTION l (50 marks) Answer all the questions in this section in the spaces provided l Round off each of the numbers in the expression to 3 significant figures, hence evaluate the expression. O-2528 _ 0-0149 (2 marks) 2 4 2 3 - 2 Given the matrices A = [3 0] and B = [1 lj,find AB — 5B_ (3 marks) 3 Three types of coffee, A, B and C are mixed such thatA:B = 4:3 and B:C = 1:2. Detennine the mass of type C in a mixture of 52kg. (3 marks) 4 In a Geometric Progression (G.P), the 4th term is 24, and the 6th term is 96. Detennine: (a) the common ratio of the G.P; (2 marks) (b) the first term of the G.P. (2 marks) 5 Two fair dice are rolled together and the sum of the numbers showing on the top faces noted. (a) Represent all the possible outcomes in a probability space. (2 marks) (b) Determine the probability that the sum is greater than 6 but less than l0. (1 mark) 6 Two points A and B are such that OA = and AB = M is a point on AB such that AM : MB = 3:1. Determine: (a) OB; (2 marks) (b) the coordinates of M. (2 marks) l 14 In the figure below, TPR is a tangent to the circle at P. Angle APB = 75° and angle BPR is twice angle APT. T P R ‘ B Determine the size of angle BAP. (2 marks) Given that 2 cos(x —— 3O)° = -0.9, determine the value of x for 0° S x S 180° correct to 2 decimal places. (3 marks) The vertices of a triangle RST are R( l, 3), S( l _ 7) and T(— l, 4). Triangle RST is mapped onto O 1 —l 0 triangle R'S’T' by transformation matrices P =£l 0] followed by Q =[ 0 _l]. Find the coordinates of R’S'T’. (3 marks) Using the method of completing the square, solve the equation 214 + 8x = l5. correct to one decimal place. (3 marks) (a) Using a ruler and a pair of compasses only, constnict an inscribed circle in triangle ABC given below. (2 marks) C A B (b) Measure the radius of the circle. (l mark) 115 ln a camp, there was enough food to feed 2000 people on equal rations for 90 days. Afier 20 days 500 more people joined the camp. Calculate the number of days that the remaining food would be used to feed the people. (4 marks) Figure PQRS below represents a garden in which, PQ = 80m, PS = 75 m, SQ = 40m, LSQR = 68° and LSPQ = LSRQ. S R 68° P 80m Q Calculate, to 2 significant figures, the length of SR. (3 marks) The table below shows part of income tax rates in a certain year. Monthly lncome in Ksh Tax Rate in each Shilling Up to 10 164 10% From 10 16530 19 740 15% From 19741 to29 316 20% In a certain month of that year, Abdala’s income was Ksh 21 820. He was entitled to a monthly personal tax relief of Ksh H62. Calculate the income tax paid by Abdala that month. (4 marks) Ali and Kinjo bought the same type of pencils and rubbers from the same shop. Ali bought 2 pencils and 3 mbbers for Ksh 66, Kinjo bought 7 pencils and 2 rubbers for Ksh I29. Find the cost of a pencil. (3 marks) The table below shows marks scored by students in a mathematics test. Marks |20-29 l 3o_39 40"-49|50-s9|eo-69|70-79] 80-89190- Number of students On the grid provided below, draw an ogive to represent the data. (3 marks) 1 16 ' SECTION n (50 marks) Answer only five questions from this section in the spaces provided. Kurao borrowed Ksh 300 000 from a financial institution. The institution charged compound interest at the rate of 18% per annum on the outstanding balance at the end of each year. At the end of the first and second years, he made equal repayments of Ksh 134 000. He repaid the remaining amount of money plus interest at the end of the third year. Calculate: (a) the interest charged at the end of the first year; (2 marks) (b) the principal at the beginning of: (i) the second year; (2 marks) (ii) the third year. (2 marks) (c) the amount of money Kurao paid at the end of the third year. (2 marks) (d) the total interest charged on the loan. (2 marks) (a) The nth term of a sequence is given by U" = n? —n + 3 Determine: (i) the l0th term of the sequence; (2 marks) (ii) the difierence between the 30th and the 20th terms of the sequence; (2 marks) (m) the value of n given that U“ = 243. (3 marks) (b) ln a research, it was found that the number of bacteria tripled every hour. Given that the number of bacteria at the start of a certain hour was I80: (i) write an expression for the number of bacteria alter r hours; (l mark) (ii) determine the number of bacteria, to the nearest million, afler 12 hours. (2 marks) The time in minutes each student in a group took to solve a certain mathematics question is shown in the table below. TimeinMinutes |0-1[1'-2|2-3l3-4l4- l -6 6-7 | 7-8 Number of students (a) State the modal class. (1 mark) (b) The above data was represented in a pie-chart, determine the angle that represented the number of students who answered the question in 4 — 5 minutes. (2 marks) (c) Calculate the mean time taken to solve the question. (4 marks) (d) On the grid provided, draw a bar graph to represent the data. (3 marks) A curve is represented by the equation y = —2x2 + 7x — 3. (a) Make a table of the values of x for: —l ix 5 4 and the corresponding values of y for the curve. (2 marks) (b) On the grid provided, draw the graph of y = —-2x2 + 7x — 3 for —l Ex 5 4. (3 marks) (c) Use the graph to determine: (i) the roots of the equation —211 + 7x — 3 = 0; (2 marks) (ii) the instantaneous rate of change of the curve at x = 1. (3 marks) Four vectors are such that OA = —-2|‘ +j, OB = 3i + Sj, OC = —8i — l2j and OD = 2i — 4j. (a) Express in tenns of i and j, the vectors: (i) AB; (2 marks) (ii)‘ CD. (2 marks) (b) Detem1ine the co-ordinates of the mid-point of AD. (3 marks) (c) Calculate to 3 significant figures, the magnitude of BC. (3 marks) Two points P and Q lie on the equator. The position ofP is (0°, l2° E) and that 0fQ is (0°, 60° W). (Take the radius of the earth to be 6370 km and 1t = g). ' (a) (i) Calculate the distance from P to Q in kilometres. (3 marks) (ii) Detennine the local time at Q when the time is 9.00 pm at P. (3 marks) (b) A point T is due North of Q. An aeroplane flying from Q at I001 km/h takes 2h to reach T. Detennine the position of T. (4 marks) A relation connecting three variables R, C and T is such that R varies directly as the square of C and inversely as T. When R = 30, C = .6 and T = 2.4. (a) Find: (i) the constant of proportionality; (3 marks) (ii) the equation connecting R, C and T. (l mark) 1 18 (b) Given that R = 40 and C = 8, determine: (i) the value of T; (2 marks) (ii) the percentage change in R when C decreases by 10% and T increases by 8%. (4 marks) In the figure below the area bounded by the curve, the y-axis, the x-axis and the line x = 5 represents a map of a piece of land. _l...‘._;_l_l....,; ‘_l_ -4»>;_#i>4‘_|_A-y~[;_;_LA_ I“ l l . l I l ‘J l , ‘w Ml 7“,-7 7; 7» ,- ;¢~~— w ‘-1 ; -¢+~- -- a - 1,; » 9 4 '1', ~t~»l~,~1~-", - ‘- —rr—y _LA44p w+* 4: l ‘F Tlc v 1 ~4~P 1 7 » _ ~» », —— _t __ _ __ __ _ _rA.A~l- _,.Y_ | - l _..-.-r--§— ~ ,_l_ll_,‘_,_,.,. “ct, _‘..‘_.t_ .;._l.._. ,<_,..j_ s_._,,_.;_;_tl ,,_..‘._ _LA~4— _'_t.»l_ ‘_,_ _'..k_,,_ rw 8 l ‘ - ii. V ' _,L_,., l_ rt _‘P~O~4—‘> A. 4 _;,.I¢j[._;i;. 1.8;} 11;. »»~7,}~_~l_ ._i_._; ‘~49... 7 ,' L4 _ ._ _, t ,_j_.j_l_ 4-. _ . 4 O\ “C m, c . 4l_L _ _ _,, , , -qg L ._.,.. . J_‘,4_4_ _ ~Li1—P LL 4.<»~,- ' ——l4~v- <4 _,,,. i. ._. _ ,__ _L : A >- V )._|_ . ,,,_‘_ _l _ __ .. ..,_',.4|.t _ _ e- _ _l_A_l»o~——r~1>~‘>vL e r 7 * e —* "rv'v',"—~f~~r _, _,,.l t._._.w+.d.._,, ,u‘ ka--1')_+_ . “J _.. ,' t_s.. _‘_;.l_l_ .., ___._l_. . _,T_ _t ma "Y ._,""'w;' i ll ’ -1-L-I *7 .3-;~~»~, , 7-— vw— '—~¢—§ vr.‘1- T I V ~<~fY l~ T,‘ s ,> flue? '**Fi+-113' f -7; 'v'i‘r." _ ._,,,1 Vt l,.,,_,_..T,v,_l,,, _,_‘_..,,,‘,. tl,_,._;_t¢_. __r+.;..l_ . __t.»t._ . ~ _\_l_l_-N-l— J_.._,_ 4, .,,;_tl_ 4 l ~—~—-|_——*-‘—-l l l . . ~—-- _ _r, . -t.,_».._ _‘~>v>—L4 __l_tl,._ >v'l—,y<:~'~4—Z—‘r——i>‘—[ .;_;+t A l_4 ‘:_ ‘ ' we _L -l_ B) M) H~ »l _r _ ctlc ,1 _,, c s_ L _r set ,3 r —* uer + 4,4,, L4~>¢ +t.<_ ,T_ < L4 _ _‘_L--4-v~ » 177"‘. A .,. ‘.tt.__,. _ ,_,_;__ _ ,_;_-_iu. __.;4 IJAF '__m.. V T.-,~ If »‘t¢_;_.._;>l‘+.~_H »_;~~- >¢——~ _. _.__...._,,_,_ ,,1_} 1, ‘V ‘I ) ' l l 1 ‘ V,‘ -_,_r . ‘T, _‘_‘_‘_.__rT>_‘<_...._ >.,_ _ _,_,‘_t . Wt 7§Q*{*‘h ~:~\~4*P- l l IV ..._ _. .__ _ ._,,_ ~~» _ 4. '.,, 5.4. It /1,4, 7 w‘~-‘Lb-»-~‘ » .~‘ ‘.1 ,‘ ' . '* <1‘.'i:'iJ“2‘::;::';::':t:.':::‘:“m 1 [~"'——‘——~l.; ‘ , .~..,l__ 0 1 2 3 4 5 (a) Estimate the area of the map in cml by: (i) the counting technique; (2 marks) (ii) using the trapezium rule with 5 strips of equal width. (3 marks) (b) Given that the actual area of the map is 30% cml, calculate: (i) the percentage error, correct to 2 significant figures, when the trapezium rule is used to estimate the area of the map; (2 marks) (ii) the actual area in hectares of the piece of land if the scale used was 1212000. (3 marks)

5.2 5.2.1 MATHEMATICS ALTERNATIVE B (122) Mathematics alternative b (122/1) l . - 3A 5~+ 7h++2/\ 3+- 6h =— 3/*12h+2J\ 9h M1 =36+— 18 M1 =_ 2 Al 3 2. (a) Number is 7532 Bl Bl 2 3. (b) Total value of hundreds digit = 500 \I'|LM\AJY\) 4:- --U‘ + _. ulw—- o mm % \I'| ow + -.0 mmo _. o §{E_A=£ Ml Ml _ _ _:___=§ 15 _- O _. U1 _.| O I\J O\ -PL» ‘L3’ 2_6=Q#£=_ A1 T 4. Nckesa: Mwita: Auma = 600 : 750 : 650 = 12 : 15 : 13 Bl Amount Mwita got more than Nckcsa =£ _ Q Ml =i 40#12oo 4O#1200 4O#12oo = 450- 36o= 90 A1 =90 3 5. h=3r~1( h= 3#2— 1= 5 » 7r’+2rh:7#2’+2#2#5 \,"‘4h- 2r \/4#5— 2# 2 Z 28 -I-%20 J16 M1 _fl _ 4 M1 =12 Al 3 Area of each face = % = 196 M] Length of side \/ 196 = 14 Ml L 3 A R 2sinx— cosx = 2 X =§_ 5 \ B2 P Q smx 3 and cos — 5 Bl U1v|>~ LIIU-I U1-§ Z -_- _i=l A A1 5 5 i 3 B1 Line, PR, drawn and dlvlded mto six (6) equal parts Joining QR and drawmg five lmes parallel to QR rntersectmg wrth 3PQ- 5x+ 6x( 10) = 2600 5x + 60x = 2600 _ 2600 " ' 65 =40 Total number of coins: =4O+6X40= 280 Ml Ml Al L 4 3*><3“ > M1 *1; + ' 1. 4 .8 FT2 =3‘><2’ = 10368 Ml Al L 4 \/ powers of 3 \/ powers of 2 Marked price = 5750 >< 1.12 = 6440 % discount = i = 5% Al 3 Ml Ml 12 . '— =(3a)*— b c (bc)2 M1 = <3“ i)(3“- i) L 2 l3 (a) 12 28 \IbJ£»L»->l\Jl~J >------L,gq\ -—~:\1\1\1; The height (LCM) = 2= >< 3’ >< 7 = 756 A1 (b) Number of books = $23 = 63 Bl 4 54 27 27 s->-g,.)\Q M1 \/ factorization M1 Let number of sides ben .'.(2n— 4)>< 90 = 1260 2n >< 90 = I260 + 360 ,, = % = 9 Al Size of each angle = %,60 = 140° L 3 Ml {~@ 7.5 .. =——~= 1.5 LSF 5 .'.A.S.F= 1.s*= 2.25 B1 . _ 22.5 M1 Area of smaller mangle - 2'2 5 =10cm1 A1 3 1 Q 45: r#7#T60 77 r: / M1 ' 45#22 =14 Al Circumference = 2 # 14# % M1 Al 4 =88 cm (a) (i) Volume of prism = Area of crosssection # L =;1.4#o.s- l# Q#»o.7h|;: 2 M1 2 7 M1 = 0.35 # 2 M1 Multiplication by length = 0.7 m’ A1 (ii) Total S .A = o.s# 2# 2 + 2 # 1.4 + 07¢ %# 2 M1 mtangular Ml triangular + 0.35 # Z = 5+ 4-4+ 0-7 M1 cross section = 11.1 m’ A1 0,) = L Ml 6+ 4.4+ 2'0.3Sh = 54.05405405 % 10 J A lac-4;) ;_ 1+ z_ a < C. (5) ') r'5'£'5“#"§4' @$§ -| (b) (i) grad AB = L 2 _|\l ~_l (ii, y=- —><- v=- — __l . .. 2m: equation of line g = 2 (ii) y- intercept: when x= O y= 2 # 0+ 1 = 1 co-ordinates —°rT='5 N-M >< N-~l\| w (c) (i) Let grad L be m 1(m=2 3 '(|,-I) -z& B1 B1 plotting vertices A, B and C. identifying vertex D (-3, 5) and competing parallelogram. Ml Al Y“ 1 1 M1 or y=- %x+ %- 1 Al B1 Ml y- 2>t= 1 Al /0,1 h Bl 10 19 I B 1 or equivalent (a) (x—5)(x+ l)= 0 x2+x—%x—%=() Ml x:+%x—%=O 2x=+x-1=0 A1 0) <2y+1>=5s B1 (2y+11)=0 M1 y=—5i or y= 5 A1 2 price of one mango Sh 5 B1 (ii) no. of mangoes Karau got mangoes bought = 95_'%5:-' = 30 Ml extra mangoes = % = 5 Al Total mangoes = 30 + 5 = 35 B1 10 2 0.| T P 0 D _.____ If lovm (a) J use of scale Bl angle of elevation 14° \/ drawn B1 completion of scale drawing Bl (b) height of mast —- 2.5 i 0.1 B1 = 2.5 >< 10 = 25 m Bl Bl Bl (c) position of cable drawn (d) (i) L of depression of C from D 48°i1° Bl (ii) Distance from P to C (l0+1.8iO.l)> i :':i%i|;:J1§jftf", i W t § K i "1 Z \ l s l 4; "t—' 75'} ___,.. I .7. mt ._._.. ‘ 4 "?i+* 31.. _,.t. ii»; \ __ ___.__, .,_..._ y _ \ __..5...._.. ‘ ‘cc. ,._. ._ -._ 1 ___.,. L.: _'_..'_..l__._._ Q :;'.l_-‘ . . , _l_ V. ._.. l 1 1 ‘r t -»~l...l...l., 1 1 . _i,._‘._l... _._i., _ l 1 4 _._.--_._lTn,__ ______,:,_,.,_.‘. - - F_ _ ll 1""‘; 1.1‘ *1‘ _....§ M ..__. ~ >..~ acceleration pans constant speed deceleration (b) (i) deceleration = % = 2 m/s’ (ii) Total distance =%(1s><1s)+%(15+so)><10+1o><5o+%(2s>-¢7>- . J - W ‘ ‘ F“ i | 1*?‘ tag ? »=-e Ml A1 Ml A1 M1 L 10 S1 Bl Bl Bl 4 ~S. ’o»iss'o§ea 1, or equivalent 5.2.2 Mathematics Alternative B Paper 2 (122/2) l. 4.957 _ 4.96 0.2638 - 0.0149 ' 0.263 - 0.015 B1 20 Bl T :<: 12> 2% /~£=<: 3>.=.> s».»-P.= =24 Al C 6 combined ratio A:B:C = 413:6 B1 massoftypeC=%X52 Ml 3 4' (a) "*1 (b) when when _ = E ar' 24 y'1=4_.r= 12 r=2=>a><2“=24=a=%=3 r=-2=>ax(-2)’=24=a=2—‘:;=-3 B1 4 Mi A1 Bl (=1) O\UlJ>uaI~J'—-+ \|o\u|-|>w-- oo\:o\u-l=~uo~ \ooo\|o\u-:>u-> \ooo\1o~ul-A :5\ooo\1o\u| :E‘,\ooo\1o\ 10 12 B2 J probnbility space (b) P(6 < x <10) =Q=i 36 12 B1 3 (=1) QB= (§)+(;‘) M‘ =<:> ~ (b) co-ordinates of M OM = QA + %/w = (§)+%(i) M1 = (§)+(§)= (Z) coordinates of M are (5,8) A1 4 Let angle APT = x“ 3x+ 75 = 180° x = 35° Bl angle BAP = angle BPR = 2 >< 35° = 70° Bl 2 2cos(x — 30)’ = — 0.9 c0s(x — 30)° =— 0.45 Ml (x— 30)’ = cos" — 0.45 = 116.74“ Al x= 146.74° B1 3 9 0 l (1 o)(—é -(i) . M1 = (ff -<1) 0_ =(—1 coordinates: R’(—3,— 1), S(—7,— 1) and T(—4,1) (-1 OX 3 7 4) Ml 3 1 1‘) 0/"7 l1l—1 Al 3 2x’+8x= 15 xz+4x= 7.5 2 $1: 51 Ml x+4x+(2) 7.5+(2) x+2= t/11.5 Ml i3.4 1.4 or — 5.4 A1 3 C Bl Bl A I B radius = 2.4 i 0.1 Bl 3 bisecting 2 or 3 angles constructing radius and com- pleting circle A 2000 x 90 Fraction for 2000 persons for 20 days _ 20 _ 2000 X 2000 >< 90 Fraction of food per person per day Ml _ Z " 9 A1 Remaining fraction of food = % No of days to feed 2000 + 500 persons = 1; 1 X 2500 M1 9 ' 180000 l B: A1 9x] 56 4 _ 751+ 801- 40* M1 °°SP_ 2><75><%= 1436.4 M1 3"‘ bracket ~(21s2o - 19740) ><12% = 416 Net tax =(1o16.4+1436.4+ 416)- 1162 M1 = 1706.8 Al 4 2p + 3r = 66.....(i) 7p + 2r = l29...(ii) M1 4p + 6r = 132..(m) 21p + 6r = 3l7.....(iv) 17p =255 M1 p =15 A1 f 4}? 43 5: 10 10 II III IIII IIIII IIIII III _ IIII IIII1 IIII JIIIIIIIIIII:IIIII: lIIIIIII:III IIIII IIIIIIII IIIIIIIIII IIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIII:III:III IIIIIIIIIIIIIIIIII III IIII IIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIII:IIIIIIIIII IIIIIIIIIIIIIIIIII IIIIIIIIII IIIIII:IIIIII::IIIII:IIIIIIIIII IIIIII IIIIII IIIII II:IIIIIIIIII: IIIIIIIIIIIIIIIIIIIIIII IIIIIIIIII """"IllllI:lI'lIIllllllilllfillln ':lI:I:i':I:'-SHIRE:Zmliilillflll IIIIIIIIIIIIIIIIIIIIIIII:IIIIIIII:I IIIIIIIIIIIIIIIIIIUIIIII IIIIIIII III IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII::::::: .:mm= IIIIIIII:II IIIIIII IIIII IIIIIIIIIIIIII III: IIIIr II II!4I I:IIIIIII:::. I m: ii III IIIIIIII IIIIIII Sllllll IIIIII IIIIIII 1 IIIIIIIIII:IIIIIIII:IIIIIIlIIIIIII IIIII:IIIlI IIIIIIII IIIIIUIIIIIII IIIIII IIIIIIIIIIIIIIIIIIIISIIIIIII IIIIIIIIIIIIIIIIIII IIIU IIIIIII IIIIIIIIIIIII::II IIIIIIIIIII II IIIIIIII IIIIIIIII IIIIIIIIIII IIIIIIIIIIII IIIIIIIIIIIIIIIIIII IIIII:IIIIIIIIIIIIII IIIII IIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIII IIIIIIIII IIIIIIIII IIIIIIII IIIIIIII IRIIIIII IIIIIIII IIIIIII IIIIIII IIIIF IIIII IIII IIII IIII III IIII II IIII III III III III III III IR II III II II II II II II II III III III III III III III I:IIIII II IIIIII IIIIIIIIII III:IIIIII IIII IIIII IIIIIIIIII IIIIIIIINI IIIIIIIIII I: I IIIIIIIIII I IIII IIIBII II IIIIII IIIII IIIIIIIIIIII IIIII IIIIIIIIIIII IIIIIIIIIIIIIIII IIIIIIIIIII IIIIIIIII IIIIIIII IIIIIII Jilli- 9= Iilllllh 0.IIIIIIIIII II IEEISIHPIIEIIEH 1 IIIIIIIIIIIIIIII 1 IIIIIIIIIIIIIIII II ‘“ IIIIISS" ‘ll: - -El N IIII IIII III _DIIII III III _ IIII II III o»IIII I: III II I III II . II II ::' Q :I -I I I IIIII IIIII IIIIIIII IIIIIIII IIIIIIIII IIIIIIIIII IIIIIIIIIII W5 5'. s IIIII III:I III I IIIII IE’--I E IIII \X\ IIIII IIIII » IIRII J IIIII IIIIQ U IIIII IIII . IIII 3: IIII . IIII V; IIII III II cf 4, IO, 18, 28,37,44,48,5O Bl canbe lmplled Pl Cl 3 300000 >< 0.18 = 54000 (i) 300000+ 54000 — 134000 = 220000 (ii) 220000 X 1.18 —134000 = 125600 125600 >< 1.18 = 148208 Total interest charged: (300000 + 22000 + 125600) >< 0.18 = 54000 + 39600 + 22608 = 116208 A1 = 116208 M1 A1 M1 A1 M1 Al M1 A1 M1 10 or equivalent 134000 >< 2 + 148208 — 3000C0 18 (a) (1) U,..=101—10+3 =93 (ii) U_=1»— U1. = (30:— 30+ 3)— (20Z— 20+ 3) = 873- 383 = 490 (m) n: — n + 3 = 243 n:—n—240= 0 (n+15)(n—16)= 0 n=—l5 or n= 16 n=16 (b) (i) Number after rhours =180><3' (ii) Number to the nearest million after 20 hours 180 >< 3‘: = 95659380 = 96000000 M1 A1 Ml A1 M1 M1 A1 B1 M1 Al 10 0:-'23 -150 415 19. (a) Modal class: 4 - 5 Bl (b) % >< 360" M1 = 30° Al (c) mid values 0.5,1.5,2.5,3.5,4.5,5.5,6.5,7.5 fx= 1,6, 7.5,l7.5,36,33,32.5,22.5 Zfx= 1+ 6+ 7.5+ 17.5+ 36+ 33+ 32.5+ 22.5 Ml Ml _ _ 156 ..mean—Y Ml = L 43 A1 (<1) '11: 1) . . L;7‘;i1 ;"'I-‘¥i";"l_i;;:%}"Y“"ffl "*'T -1 ‘ I , Q: ~l_ I4 - , ,, 13 pf ' S1 A l B2 ‘_.. (‘*1 ,7... . '}-jtglf | 10 J scale and labelling 8barsJ (allow Bl for5 - 7bars J) 0:-'23 -150 415 20 (a) (b) x-101234 -12-3230-7 B2 J51 ' = \ ‘I . / I ‘ ‘i/7#+ — i 3\‘ Z); \\\ 1 10 —s S1 Pl Cl (c) (i) Roots of equation x = 0.5 °Y B1 x=3 B1 (ii) tangent line ¢ drawn B1 gradient: i52__—01 M1 = 3 Al 10 (=1) (i) Afi (11) QD= (b) mid point =%{(”’})+<_i§~)}=%(_°3,-) » 1 ° > _ — 1.5j coordinates of mid point is (0,-1.5) (c) B_C= QC .'.|B§\ = QB—QA=3i+5j—(—2i+j) 1v11 31+ 5j+ 2i—j 51+ 4; 0p- QC= 21- 4j—(—8i— 121') 21-41+ 8i+12j 101+ 8j Al Ml Al of vector AD Ml Al Bl —QB=—8i—l2j-—(3i+5j) M1 =111"-17/ 1/111+ 17* 1/121+ 289 1 20.2 Ml i 10 (a) Distance PR = % >< 2 >< % >< 6370 (11) Time difference = % 11 Local time at Q: =9.00pm-4h48 min =4.13 (b) Distance travelled in 2 h = 1001 X ._9 Q _ .. 360 ><2>< 7 ><6370- 2002 6: 2002><3eo><1 2><22><(0.9><8)’ M1 (") N°WR _ 1.0s><3.2 = 30 Al % change in R : 40 — 30 Ml 40 >< 100 = 25% AI IO 9 8 1 1/ 7f “J X (vi i v‘/ §‘\/ ‘ ¢ \/ / 4% ‘o(./ \/ ./ / 3. J‘./ \/ \/ / 1 / /\/ ./t/\/\/ 0| 34- Ml A1 Bl M1 (1) 24+ %(13) = 30% (ii)%><1{2+2+2(6+8+8+6)} = $60) = 30 cm’ A1 i_ (i) % error = Mx 100 30; _ A _237 Ml = A1 (ii) 1 cm E 120m 1 cm’ E 14400 mz . B1 30% cm’ E %40400o0g) x % =44.4 ha M1 i whole square and part square ordinates 2, 6, 8, 8, 6, 2 substitution into formula simplification
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