# KCSE Past Papers Maths B 2013

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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)

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:::. 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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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