NECO 2018/19 Mathematics (Maths) Runs - Essay and Objectives Questions/Answers (Expo/Runz) - Scholarships, Nigeria Universities, Polytechnics And Admissions News
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# NECO 2018 - MATHS ANSWERS

## Monday 4th June 2018OBJ – General Mathematics – 10:00am – 11:45amEssay – General Mathematics – 12:00noon –2:30pm+++++++++++++++++++++++++++++++++++++=============================COMPLETED========================================================MATHS OBJ:1-10: CDAAEABAEC11-20: ACDDCDCDAC21-30: CEADEDCABC31-40: CBEECCBDCC41-50: DDCBCDDBBA51-60: BCECCBBCEE===========================(1a)Log10(20x - 10) - log10(x+3) = log10^5Log10(20x-10/x+3)= log10^520x - 10/x + 3 = 5Cross multiplying 20x - 10 = 5(x + 3)5(4x - 2) = 5(x + 3)4x - 2 = X + 34x - x = 3+23x = 5X = 5/3 OR 1 whole no 2/3(1b)Let actual amount be #X15% of #x = #60015x/100 = 600X = (100/15)*600X = 100*40X = 4,000Actual amount = #4,000 ===============================(2a)(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)^1/4(2b) √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-√2Rationalizing K = -(2+√2) * 1+√2/1-√2K = -(2+√2)(1+√2)/1 - 2K = (2+√2)(1+√2)K = 2+2√2 + √2+2K = 4+3√2 ==================================================(3)V = Mg√1 - r²Square both sides V² = 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 = 0.9972 =============================================(4)Draw the diagram (i) Arc length = Tita/360*2πr= 72/360*2*22/7*14=1/5*44*2=88/5=17.6cm(ii) Perimeter of Sector = arc length +2r=17.6+2(14)=17.6+28=45.6cm(iii) Area of sector = Tita/360*πr²=72/360*22/7*14/1*14/1=1/5*22*2*14=616/5=123.2cm2 =======================================(5a) 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= 41kg ============================================(7ai)T3=>a+2d=6(eqi)T7=>a+6d=30(eqii)Eqii minus eqi gives 6d-2d=30-64d=24d=24/4d=6Common difference=6(7aii)Putting d=6 into eqi a+2(6)=6a+12=6a=6-12a=-6(7aiii)10th term T10=a+9d=-6+9(6)=-6+54=48(7bi)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=2(7bii)Common ratio r=3/2 as above =============================================(8a)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=4(8ai)Putting a=4 and b=3 in eqix=4+3yThis is the relationship between xand y(8aii)When y=7x=4+3(7)x=4+21x=25(8b) 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 terms 15x²-34x-2/3(x+2)(3x-1) ==================================(10a) Obtuse 105 + reflex Reflex =255°Now 2w = reflex 2w =255°W = 255/2 =127.5°Also 2x = obtuse 2x = 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°(10b) Draw the diagram Opp/adj = TanR |TB|/|BR| = TanR100/|BR| = Tan60°|BR| = 100/tan60|BR| = 100√3|BR| = 100√3 * √3/√3=100√3/3m OR 57.7m ========================================(11a)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 number(11b) (6x + 3) dxintegrating6x²/2+3x/16= {3x^2+3x}1= 6{3x^2+3x}1= [3(6)²+3(6)] - [3(¹)²+³(1)]= [3(36)+18] - [3+4]= [108+18] - = 126-9= 117(11c)y = x² + 5x - 3 (x = 2)y = 2² + 5(2) - 3y = 4 + 10 - 3y = 14 - 3y = 11Gradient of the curve = 11 ==================================================(12a)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 = 1/13 Pr 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 = 32/117(12b) 10Red + 8green + 7blue = 25(i) 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 = 103/200(ii) 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 ===============================COMPLETED===============================

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