2016 NECO:MATHEMATICS OBJ AND THEORY QUESTIONS AND ANSWERS NOW AVAILABLE HERE

Obj from our second solver use the first one or this one. Preferable this one. Any one is ok..nobody is perfect when it comes to obj MATHS OBJ: 1-10: ADADDDCEBE 11-20: CCDCAACECA 21-30: DCDAECAEED 31-40: EAADBEEADC 41-50: BBDCCBCECD 51-60: CADBAECDEC NEW CHANGES IN THEORY MATHS THEORY NO 1,3,4,5,6,7,8,9,10 ================================== (1a) Tabulate: Draw both vertical and horizontal lines across the digits. I drew only verticals cos it's txt messaging. x|1|2|3|4 1|1|2|3|4 2|2|4|0|2 3|3|0|3|0 4|4|2|0|4 (1b) S.I=PRT P=N15,000, R=10%, T=3yrs S.I=#15,000, R=10%, T=3yrss S.I=15,000* 10/100 *3 S.I=N4,500 A=P S.I A=N15,000 N4,500 A=#19,500 ============================ (3) Let x represent Donald's age and let y represent his son's age x=5y---(1) (x-4(y-4)=448---(11) substitute for x in (11) 5y^2-20y-4y 16=448 5y^2-24y 16-448=0 5y^2-24y-432=0 y=-b _sqroot(b^2-4ac)/2a a=5, b=-24 c=-432 y=24 _sqroot[(-24^2)-4*5*-432]/2*5 y=24 _sqroot(576 8640)/10 y=24 _sqroot(9216)/10 y=24 _96/10 y=24 96/10 or 24-96/10 y=120/10 or -72/10 y=12 or -7.2 since age cannot be negative y=12years but x=5y x=5*12 x=60years ================================ (4a) Convert 30000 litres to metres = 30000/1000 metres = 30metres Depth of fuel = h 7.54.2 = 30m^3 = 31.5hm^3 = 30m^3 = h = 30m^3/31.5m = h = 0.9m (4b) Depth of the tank = l*b*h where l= 7.5, b = 4.2, h = 1.2 =(7.5 4.2 1.2)m^3 =37.8m^3 Convert metres to litres = 37.8* 1000litres = 37800litres Litres of fuel needed to fill the tank=37800litres/30800litres =7800litres ================================ (5a) sector for building project =48000/144000*360=120degree sector for education = 32,000/144000*360=80degree sector for saving = 19200/144000*360=48degree sector for maintenance = 12000/144000*360=30degree sector for miscellaneous = 7200/144000*360=18degree sector for food items = 360-(120 80 48 30 18) =360-296 =64degree THEN DRAW A PIE CHART WITH THE ANSWERS YOU GOT ABOVE. (5b) Food = 48000 32000 19200 1200 7200 x = 144000 Food => 118400 x = 144000 Food => x = 144000 — 118400 Food = x = #25,600 ================================= (6) 3sqroot[(41.02*sqroot0.7124)/(42.87*0.207*0.0404)] Tabulate No|log 41.02|16130 *sqroot0.7124|(1^-.8527)/2 |(2^- 1.8527)/2 |1^-9264 41.02*sqroot0.7124|1.6130 | 1^-.9264 |=1.5394 42.87|1.6322 *0.207| 1^-.3160 |0.9482 *0.0404| 2^-.6064 |=1^-.5546 |1.5394 |1^-.5546 |=(1.9848)/3 antilog4.588|0.6616 Answer=4.588 ================================== (7a) 3^2n 1 - 4(3^n 1) 9=0 3^2-3 - 4(3^n -3) 9=0 (3^n)^2-3 - 4(3^n -3) 9=0 let 3^n = p p^2 -3 - 4(p-3) 9=0 3p^2/3 - 12p/3 9/3 = 0 p^2 - 4p 3 = 0 p^2 - 3p - p 3 = 0 p^2p(p-3) - 1(p-3) = 0 (p-1)(p-3) = 0 p-1 = 0 or p-3 = 0 p = 1 or 3 Recall 3^n = p when p=1 3^n = 3^0 n = 0 when p = 3 3^n = 3^1 n = 1 (7b) log(x^2 4) = 2 logx - log^20 log(x^2 4) = log^100 = log^x - log^20 (x^2 4) = log(xx) x^2 4 = 5x x^2-5x 4 = 0 x^2-4x - x 4 = 0 x(x-4) - 1(x-4) = 0 (x-1)(x-4) = 0 x-1 = 0 or x-4 = 0 x = 1 or 4 ========================== (8ai) In <> ABC Cos B=(a^2 c^2-b^2)/2ac Cos B=(13^3 13^2-10^2)/(2*13*13) Cos B=(338-100)/338 Cos B=238/338 Cos B=0.7041 B=cos^-0.7041 B=45.2degrees therefore =90.4degrees In <>AOC By sine rule r/sine44.8=10/sine90.4 r=10sin44.8/sin90.4 r=7.05cm (8aii) Circumference =2*pie*R =2*22/7*7.05 =44.31 =44.3cm(1d.p) (8b) p(-1,2) , (2,6) (y-y1)/(y2-y1)=(x-x1)/(x2-x1) (y-2)/(6-2)=(x 1)/(2 1) (y-2)/4=(x 1)/3 3(y-2)=4(x 1) 3y-6=4x 4 3y-4x=10 ============================= (9a) Let the lens digit x and unit digit be y, therefore x-y=5 ---(1) 3xy-(10x y)=14 ----(2) 3xy-10x-y=14 -----(3) frm eq(1); x=5 y --- (4) therefore, 5(5 y)(y)-10(5 y)-y=14 (15 3y)y-50-10y-y=14 3y^2 4y-50-14=0 3y^2 4y-64=0 3y^2 -12y 16y-64=0 3y(y-4) 16(y-4)=0 (3y 16)(y-4)=0 y=-16/3 or 4 therefore from eqn(1); x 4=5 x=5 4=9 the number is 94 (9b) (3-2x)/ 4 (2x-3)/3 (3(3-2x) 4(2x-3))/12 (9-6x 8x-12)/12 =(2x-3)/12 ============================ (10a) y=(2x^2 3)^5 let U=2x^2 3 Y=u^5 du/dx = 4x dy/du = 5u^4 dy/du = (2x^2 3)^4 dy/dx = du/dx dy/du dy/dx = 4x.5(2x^2 3)^4 dy/dx = 20x(2x^2 3)^4 (10b) y=3x^2 2x 5 dy/dx =6x 2 dy/dx =6(3) 2 dy/dx =18 2 dy/dx =20 (10c) R-W=Wv^2/gx Wv^2=gx(R-W) Wv^2=gRx-Wgx Wv^2 Wgx=gRx W(v^2 gx) =gRx W=gRx/V^2 gx R=2, g=10, x=3/2, V=3 W= 10*2*3/2/3^2 10*3/5 W=30/9 15 W=30/24 W=5/4 FROM PROF KEN EKWUEME