Verified Neco Mathematics Theory (1a) 142n=34base10 1*x^2+4*x^1+2*n ^o=34 X^2+4x+2=34 X^2+4x-32=0 X^2+8x-4x-32=0 (X-4)(x+ =0 X-4=0 or x+8=0 X=4 or x=-8 But x can only be 4 (1b) 86.13/2.7 8613*10^2/ 27*10^-1 8613/27*10^-2+1 319*10^-1 3.19*10^2-1 3.19*10^1 (2a) 8/3+4/3=12/3 8/3/4*750 8/3*1/4*750/1 6000/12 500 7/3/4*750 4/3*1/4*750/1 3000/12 250 (2b) A=P+I I=PRT/ 100 where p=1250,R=3%,T=8yrs =1250*3*8/100 =300 Amount is 1250+300 =1550 (3a) Tn=a+(n-1)d -118=32+(n-1)-3 -118=32-3n +3 -118=35-3n 3n=153 n=153/3 n=51 Sn=n/ 2(2a+(n-1)d) S51=51/ 2(64+(50)-3) =51/2*-86 =51*-43 =-2,193 (3b) Tn=ar^n-1 T30=2*7^30-1 =2*7^29 (4a) n(f)=72 n(m)=64 n(c)=62 n(fnm)=18 n(fnc)=24 n(mnc)=20 n(fnmnc)=8 U=38+34+26+10+1 6+12+8 =144 candidates (4b) 5root18-3root72 +4root50 5*3root2-3*6roo t2+4*5root2 15root2-18root2 +20root2 = 17root2 NOTE THAT ROOT IS square root sign eg: 5root2 is 5_/2 (5a) I-Nasiru is not tall II-Victor is fat and nasiru is tall III-Nasiru is tall or victor is fat IV-Nasiru is tall and victor is not fat (5b) a=(2,-3) b=(-1, 2), c=(-2 3) 4a-3b+2c 4(2,-3) -3(-1, 2), +2(-2, 3) (8+3-4 -12) (7, -12) (9a) AO/AM=DO/BM X+6/x=7/5 5x+30=7x 5x-7x=-30 -2x=-30 X=-30/-2 X=15m (I) Volume=1/ 3pieR^2H-1/ 3pier^2h 1/3*22/ 7*7*7*21-1/3*22/ 7*5*5*15 1078-392.86 =685.14m^3 (II)Total surface =pieRL-pierl 22/7*7*22.1-22/ 7*5/7*15.8/1 486.2-248.3 =237.9m^3 (9b) Total surface area 237.9*200 47,580. Sent at 10:27 AM on Monday me: (6a) x-y/x=1, x/2+y/2=17/6 x=1+y/2 substitute equation(3) into equation(2) x/2+y/2=17/6,x/ 2=17/6-y/2 1+y/2=2(17/6-y/ 2) 2+y/2=2(17-3y/ 6) 2+y=4(17-3y/6) 6(2+y)=4(17-3y) 3(2+y)=2(17-3y) 6+3y=34-6y 6y+3y=34-6, 9y=28 y=28/9 from x=1+y/2 x=1+(28/9)/ 2=1+28/18 x=18+28/18=46/ 18=23/9 x=23/9, y=28/9 (6b) (m^2-2mn+n^2)/ (m^2-n^2) (m^2-mn-mn+n^2) /(m-n)(m +n) m(m-n)-n(m-n)/ (m-n)(m+n) = (m-n)(m-n)/ (m-n)(m+n) =m-n/m+n (6c) (x-2)(x-3)=12 x^2-3x-2x+6=12 x^2-5x-6=0 x^2-6x+x-6=0 x(x-6)+1(x-6)=0 (x+1)(x-6)=0 x+1=0 or x-6=0 x=-1 or x=6 (11a) Y^2=x^2+z^2-2xz cosY =100+36-120*-co s80 =136+120*0.1736 =156.83 Y=square156.83 Y=12.52km Distance XZ=12.52km Bearing of Z from X X/sinX=y/sinY 10/sinX=12.52/ sin100 12.52sinX=10sin 100 Sinx=10sin100/ 12.52 Sinx=10*0.9846/ 12.52 Sinx=0.7866 X=sin^-1 (0.7866) X=51.9degree (11b) 2pier(r+h)=A 2*22/ 7*7(7+h)=528 44(7+h)=528 308+44h=528 44h=220 h=220/44 h=5cm (13) Purchase br /> 20 crate of 30 eggs per crate @500 each=10,000 Transport=1000 Marketing ticket(50*6)=30 0 Total=11,300 Sales br /> 191/2 crate of @30 eggs per crate=585eggs Sold @ 30 per eggs=585*30=175 50 Remaining eggs=1/2*30=15 eggs Sold @ 20 per egg 15*20=#300 Total= 17850 Gross profit=17850-10 000 7850 Net profit=17850-11 300 =6550 Percentage profit=profit/ cost price*100/1 6550/11300*100 57.9% Approximately 58% (14a) Ade= 1/2(musa), A=1/2M M=1/2(0) A=1/2(1/2 0)= 1/4 O A+M=114 Ade=57 years Musa =32 years obi=27 years (14b) SP=N10,000 Discount= 10% SP after discount =10/ 100*(10,000) =N1000 35%=(Sp-cp)/cp 35/ 100= (1000-cp)/cp 0.35cp=1000-cp 1.35cp=1000 cp=1000/1.35= N740.74 (15a) (I)Market value=1.70 No shares bought=520 He spent 520*1.70=884 (II) A bonus of one to ten 520/10 Bonus share=52 shares Total share held=520+52 572 shares Dividend @ 5% of 50k 2.5k per share Total income=2.5k*572 1430k =#14:30 (15b) Appreciation @ 5% Cost, Appreciation , Value First year: 7500, vallue 157,500 2nd year: 7875, 165,375 3rd year: 8268.75, 173643.75 4th year: 8682.19, 182,325.94 Therefore it will appreciate to #182,325.94 at the 4th year
Posted on: Mon, 07 Jul 2014 10:58:42 +0000
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