100% WAEC GCE MATHEMATICS OBJ & ESSAY ANSWERS AVAILABLE HERE

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^ means raise to power NO (1) If Given, m^2 - 2mn + n^2 - 9r??2 m^2 - 9r^2 + n^2- 2mn m^2 - (3r)^2 + n(n-2m) (m-n)^2 - 9r^2 =(m - n - 9r)(m -n +9r) NO (1B) 5x-4y=6...,....1 3^3(y-x) = 1/27....2 From......2 3^3y-3x= 1/3^3 3^3y-3x = 3-3 3y-3x=-3 Y=-1+x.........2 Substitute 2 in 1 5x-4 (-1+x)=6 5x-4(-1+x)=6 X+4=6 X=6-4 X=2 From.....2. y=-1+2=1 Y=1 (2A) All taxable income is = 10% + 15% + 20% + 25% = 70% of 850 = 70/100*850 = #595 (2B) 1: 70/100*120 =#84 2: 70/100*30= #21 3: 70/100*100 =#70 4: 70/100*40 = #28 total= 84+21+70+28=#203 =203 +595 =#798 NO (3A) QT^2 =QS^2 + ST^2 5^2 =4^2 + ST^2 ST^2 =25 - 16 ST^2 =9 ST=sqrt (9)= 3cm Cos(tita)=3/5 cos(tita)= 0.6 tita= cos^-1(0.6) = 53degree cos53(degree)=SR/6 0.602 =SR/6 SR= 0.602*6 = 3.6cm /TR/=3.6/Sin53(degree)=4.5cm NO (3B) Adj=12 Opp=5 Hyp=13 cosx=12/13 Sinx=5/13, Tanx=5/12 Cosx - 2Sinx/2Tanx = 12/13-2(5/13) / 2(5/12) = (12/13 - 10/13) / 5/6 = (2/13) / (5/6) = 2/13*6/5 = 12/65 (4A) Log10(75/10) - Log10(5/9)^2 + Log10(100/243) = Log10(75/10) - Log10(25/81) + Log10(100/243) = Log10(75/10*81/25) + Log10(100/243) = Log10(3*81/10) + Log10(100/243) = Log10(243/10) + Log10(100/243) = Log10( 243/10* 100/243) =Log10(10)=1 (4B) X=?(10,11,12,13,14,15?) Y=? (2,4,6,8,10,12,14,16?) U=? (10,11,12,13,14,15,16,17,18,19,20?) (i.) XnY = ?(10,12,14?) (ii.) X?/?/?/' = ? (16,17,18,19,20?) X?/?/?/'nY=?(16?) n(X?/?/?/'nY)=?(16?) (5A) Considering change in WZX ZX^2 = Zw^2 + WX^2 ZX^2 = 6^2 + 8^2 = 36+ 64 ZX^2=100 ZX= sqrt(100) =10cm ZX= 2ZM ZM= ZX/2 Zm=10/2 =5cm Mx=5cm OM is the height of the pyramid OX^2= OM^2 + MX^2 13^2= OM^2 + 5^2 169= OM^2 + 25 OM^2 = 169 -25 OM^2 = 144 OM= sqrt(144) =12cm height of pyramid = 12cm (5B) COS < 0XZ = 12/13 ar^3/ ar^5= 1/4 1/r^2=1/4 r^2= 4 r = sqrt(4) r=2 (i) Common ratio(r)= 2 (ii) Substitute r = 2 into eqn(1) a + ar^2 = 40 a(1+2^2) = 40 a(5) = 40 a= 40/5= 8 Fifth term (F5) = ar^4 = 8*2^4 =8*16 =128 (10A) (i) 1/2x - 5x/6 - 5/3 -2 (10A) (ii) (-2, 0,2) diagram (10B) /PQ/ = 10m Let KR be the height of pole tan 45(degree)= KR/ PR KR= tan 45(degree) * (10-x) KR= (10 -x) --------------eqn(1) tan 58(degree)= KR/x KR= tan 58(degree) * X KR= 1.6 *X KR= 1.6x ---------------eqn(2) equation(1) and(2) 10 - x= 1.6x 10= 1.6x + x 2.6x = 10 x= 10/2.6 x= 3.85m (i) Distance fom P to the pole is (10 - x) = (10 - 3.85)= 6.15m (ii) Height of pole = KR= 1.6x = 1.6 * 3.85 = 6.16m (11A) Pr(Manful pass) = 2/3 Pr(John)= 5/8 Pr(Ernest) = 3/4 Pr(all pass)= 2/3 * 5/8 * 3/4 = 5/16 (11B) Tabulate x: 1,2,3,4,5,6 f: 4,2,3,2,6,3 E(f) = 20 f1x1=4 , f2x2=4, f3x3=9 , f4x4 =8, f5x5=30, f6x6=18 Efx = 4+4+9+8+30+18 = 73 Cf = 4,6,9,11,17,20 Q1= Ef/4 = 20/4 = 5th Q3= 3Ef/4 = 3*20/4 =15th Q1 corresponds to 2 Q3 corresponds to 5 inter-quatile range = Q3 - Q1 = 5 - 2= 3 (12A) (i) OBJECTIVE
VERIFIED MATHS OBJECTIVE: 1-10: BAABBCDBCC 11-20: ADDBCACACC 21-30: BDCBDABACA 31-40: CACCDCDDDB 41-50: DDDABADCCD