## Header Ads NECO 2018 Mathematics Obj And Essay Answer – June/July Expo

## Mathematics OBJ:1CDAAEABAEC11AEDDCDCDCC21CEBDEDCBBC31CBEEECBDCC41DBCBCDDBCA51BCBDCDCCECALWAYS SUBSCRIBE IF YOU WANT YOUR ANSWERS BEFORE THE EXAMCLICK HERE FOR no.1 THE IMAGE1a)Log 10(20*-10)-log10(*+3)=log105(20*-10/*+3)=log10 =520*-10/*+3=55(*+3)=20*-105*+15=20*-1015+10=20*-5*25=15**=25/15*=5/3=1 2/31b)Discount percent =15%Discount amount =#600Actual amount paid on the article =?Original amount on the article =*15%*=#60015/100* =60015*=600*10015*=60000*=60000/15*=#4,000Therefore actual amount paid on the article=#4,000-#600=#3,400Actual amount paid on the article =#3,400CLICK HERE FOR no.2 THE IMAGE2a)(X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5= (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/Z^5= X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5=X^3/2+1 * Y^-9/4-4 * Z^3/4-5=X^5/2 * Y^-25/4 * Z^-17/4=X^10/4 * Y^-25/4 * Z^-17/4=(X^10/Y^25 Z^17)^1/42b)√2/k + √2 = 1/k – √2Multiply both sides by (k+√2)(k-√2)√2(k-√2) = k+√2√2k-√2 = k+√2√2k-k = 2+√2K(√2 -1) = 2+√2K = 2+√2/√2-1K = -(2+√2)/1-√2RationalizingK = -(2+√2) * 1+√2/1-√2K = -(2+√2)(1+√2)/1 – 2K = (2+√2)(1+√2)K = 2+2√2 + √2+2K = 4+3√23)V = Mg√1 – r²Square both sidesV² = m²g²(1-r²)V²/m²g² = 1-r²r² = 1 – v²/m²g²r = √1-(v/mg)²If v = 15, m = 20, and g = 10r = √1 – (15/20*10)²r = √1 – (0.075)²r= √(1.075)(0.925)r = √0.994375r = 0.9972CLICK HERE FOR no.4 THE IMAGE4i)length of Arc of the sectorTitter= 72degree, r = 14cmL= titter / 360 x 2 pie r==> L= 72/360 x 2 x 22/7 x 14=44352/2520=17.6cm4ii) perimeter of the sectorPerimeter = titter/360 x 2 pie r + 2r=17.6 +(2×14)=17.6+28=45.6cm4iii) Area of the sectorArea = Titter/360 x pie r^2=72/360 x 22/7 x (14)^2=72 x 22 x 196/2520Area= 310464/2520=123.2cm^2CLICK HERE FOR no.5 THE IMAGE5a)Mode = mass with highest frequency = 35kgMedian is the 18th mass= 40kg.5b)In a tabular formUnder Masses(x kg)30,35,40,45,50,55Under frequency(f)5,9,7,6,4,4Ef = 35Under X-A-10, -5, 0, 5, 10, 15Under F(X-A)-50, -45, 0, 30, 40, 60Ef(X – A) = 35Mean = A + (Ef(X – A)/Ef)= 40 + 35/35= 40 + 1= 41kgCLICK HERE FOR no.6 THE IMAGE6a)log2 = 0.3010Log3 base 10 = 0.4771(i) Log10 3.6 = Log10 36/10= log10 36 – log10 base 10= log10 (9×4) -1=log10 9+log10 4 – 1=log10 3² + log10 2² – 1=2log10 3 + 2log10 2 – 1= 2(0.4771) +2(0.3010) -1= 0.9542 + 0.6020 – 1= 0.55626aii)Log10 0.9= log10 9/10 = log10 9-log10 10= 2log10 3 – 1= 2(0.4771)-1= -0.0458= 1.95426b)(3√5 – 4√5)(3√5-4√5)/(3√5+4√5)(3√5-4√5)= 45 – 60 + 80 = 6045-60+60-80= 5/35 = 1/7CLICK HERE FOR no.7ai THE IMAGECLICK HERE FOR no.7aiii THE IMAGECLICK HERE FOR no.7b THE IMAGECLICK HERE FOR no.7bii THE IMAGE7ai)T3=>a+2d=6(eqi)T7=>a+6d=30(eqii)Eqii minus eqi gives6d-2d=30-64d=24d=24/4d=6Common difference=67aii)Putting d=6 into eqia+2(6)=6a+12=6a=6-12a=-6(7aiii)10th term T10=a+9d=-6+9(6)=-6+54=487bi)T3=>ar²=9/2(eqi)T6=>ar^5=243/16(eqii)Dividing eqii by eqiar^5/ar²=243/16 divided by 9/2r³=243/16*2/9r³=27/8r³=3³/2³r=3/2Putting this into eqia(3/2)²=9/2a(9/4)=9/2a=9/2*4/9a=4/2=27bii)Common ratio r=3/2 as aboveCLICK HERE FOR no.8 THE IMAGE8)x=a+by(eqi)when y=5 and x=1919=a+5b(eqii)when y=10 and x=3434=a+10b(eqiii)solving eqii and eqiiia+10b=34a+5b=19=>5b=15b=15/5=3putting b=3 in eqii19=a+5(3)19=a+15a=19-15a=48i)Putting a=4 and b=3 in eqix=4+3yThis is the relationship between xand y8ii)When y=7x=4+3(7)x=4+21x=258b)3x/x+2 – 5x/3x – 1 + 1/3Find the L. C. M3(3x-1)(3x)-3(x+2)(5x)+(x+2)(3x-1)/(x+2)(3x-1)(3)27x²-9x-15x²-30x+3x²-x+6x-2/3(x+2)(3x-1)Collect like terms15x²-34x-2/3(x+2)(3x-1)CLICK HERE FOR no.10a THE IMAGE10a)Obtuse 105 + reflexReflex =255°Now 2w = reflex2w =255°W = 255/2 =127.5°Also 2x = obtuse2x = 105°X = 105/2 = 52.5°Now EDF = y(base angles of an isosceles triangle)BED=X=52.5°(angles in the same segment)EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)Y+y = 52.5°2y = 52.5°Y = 52.5°/2=26.25°CLICK HERE FOR no.10b THE IMAGE10b)Draw the diagramOpp/adj = TanR|TB|/|BR| = TanR100/|BR| = Tan60°|BR| = 100/tan60|BR| = 100√3|BR| = 100√3 * √3/√3=100√3/3m OR 57.7mCLICK HERE FOR no.11 THE IMAGE11a)x+y/2 =11x+y= 11*2x+y= 22 —(1)x-y= 4 —-(11)x+y = 22—-(1)–x-y= 4—-(11)____2y = 18y= 18/2y=9Substitute y=9 in equ 1x+9=22x=22-9x=13x=13, y=9x+y= 13+9= 22Sum of the two number11b)(6x + 3) dx(6x + 3)dx(6x +3)^6 – (6x + 3)^1(6 x + 3)^5(7776x^5 + 243)38,880x/6 + 2436480 x^6 + 243x9(720x^6 + 27x)11c)y = x² + 5x – 3 (x = 2)y = 2² + 5(2) – 3y = 4 + 10 – 3y = 14 – 3y = 11Gradient of the curve = 1112a)Pr of Abu to pass = 3/7Pr of Abu to fail = 1 – 3/7 = 7-3/7 = 4/7Pr of kuranku to pass = 5/9Pr of kuranku to fail = 1 – 5/9 = 9 – 5/9 = 4/9Pr of musa to pass = 12/13Pr of musa to fail = 1 – 12/13 = 13 – 12/13 = 1/13Pr of only one of them passing is=(3/7*4/9*1/13)+(5/9*4/7*1/13)+(12/13*4/7*4/9)=12/819+ 20/819 + 192/819=12+20+192/819 = 224/819= 32/11712b)10Red + 8green + 7blue = 2512bi)pr of different colour isProf(RG)+(RB)+(GB)+(BG)+(BR) +(GR)=(10/25*8/24)+(10/25*7/24)+(8/25*7/24)+(7/25*8/24)+(7/25*10/24)+(8/25*10/24)=80/100 + 70/600 + 56/600 + 56/600 + 70/600 + 80/600= 80+70+56+56+70+80/600= 412/800 = 103/20012bii)pr of atleast one must be=Pr[RB+BR+GB+BG+BB]= (10/25*7/24)+(7/25*10/24)+(8/25*7/24)+ (7/25*8/24) + (7/25*7/24)=70/600+70/600+56/600+56/600+49/600=70+70+56+56+49/600=301/600

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