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# Neco 2023/2024 ssce Mathematics Syllabus/Brochure Questions & Answers

NECO 2023 MATHEMATICS ANSWERS  MATHEMATICS OBJ: 1-10 CDAAEABAEC 11-20 AEDDCDCDCC 21-30 CEBDEDCBBC 31-40 CBEEECBDCC 41-50 DBCBCDDBCA 51-60 BCBDCDCCEC (1a) Log10(20x-10) – log10(x+3)= log10^5 Log10(20x-10)/x+3) =log10^5 20x-10/x+3 = 5 Cross multiply 20x-10 = 5(x+3) 5(4x-2) = 5(x+3) 4x-x = 3+2 3x= 5 X=3/5 (1b) Let actual amount be #x 15% of #x = #600 15x/100 =600 x =(100/15) * 600 x = 100 * 40 x = 4,000 Actual 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 – √2 Multiply both sides by (k+√2)(k-√2) √2(k-√2) = k+√2 √2k-√2 = k+√2 √2k-k = 2+√2 K(√2 -1) = 2+√2 K = 2+√2/√2-1 K = -(2+√2)/1-√2 Rationalizing K = -(2+√2) * 1+√2/1-√2 K = -(2+√2)(1+√2)/1 – 2 K = (2+√2)(1+√2) K = 2+2√2 + √2+2 K = 4+3√2 (3) V = Mg√1 – r² Square both sides V² = m²g²(1-r²) V²/m²g² = 1-r² r² = 1 – v²/m²g² r = √1-(v/mg)² If v = 15, m = 20, and g = 10 r = √1 – (15/20*10)² r = √1 – (0.075)² r= √(1.075)(0.925) r = √0.994375 r = 0.9972 (4ai) length of Arc of the sector Titter= 72?, r = 14cm L= titter / 360 x 2 pie r ==> L= 72/360 x 2 x 22/7 x 14 =44352/2520 = 17.6cm (ii) perimeter of the sector Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm (iii) Area of the sector Area = Titter/360 x pie r? = 72/360 x 22/7 x (14)? = 72 x 22 x 196/2520 Area= 310464/2520 = 123.2cm? (5a) Mode = mass with highest frequency = 35kg Median is the 18th mass = 40kg. (5b) IN A TABULAR FORM Under Masses(x kg) 30,35,40,45,50,55 Under frequency(f) 5,9,7,6,4,4 Ef = 35 Under X-A -10, -5, 0, 5, 10, 15 Under F(X-A) -50, -45, 0, 30, 40, 60 Ef(X – A) = 35 Mean = A + (Ef(X – A)/Ef) = 40 + 35/35 = 40 + 1 = 41kg (6) (a) log2 = 0.3010 Log3 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 = 0.5562 (6aii) Log10 0.9 = log10 9/10 = log10 9-log10 10 = 2log10 3 – 1 = 2(0.4771)-1 = -0.0458 = 1.9542 (6b) (3√5 – 4√5)(3√5-4√5)/(3√5+4√5)(3√5-4√5) = 45 – 60 + 80 = 60 45-60+60-80 = 5/35 = 1/7 (7ai) T3=>a+2d=6(eqi) T7=>a+6d=30(eqii) Eqii minus eqi gives 6d-2d=30-6 4d=24 d=24/4 d=6 Common difference=6 (7aii) Putting d=6 into eqi a+2(6)=6 a+12=6 a=6-12 a=-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 eqi ar^5/ar²=243/16 divided by 9/2 r³=243/16*2/9 r³=27/8 r³=3³/2³ r=3/2 Putting this into eqi a(3/2)²=9/2 a(9/4)=9/2 a=9/2*4/9 a=4/2=2 (7bii) Common ratio r=3/2 as above (8) x=a+by(eqi) when y=5 and x=19 19=a+5b(eqii) when y=10 and x=34 34=a+10b(eqiii) solving eqii and eqiii a+10b=34 a+5b=19 =>5b=15 b=15/5=3 putting b=3 in eqii 19=a+5(3) 19=a+15 a=19-15 a=4 (8i) Putting a=4 and b=3 in eqi x=4+3y This is the relationship between xand y (8ii) When y=7 x=4+3(7) x=4+21 x=25 (8b) 3x/x+2 – 5x/3x-1 + 1/5 Find the LCM 3(3x-1)(3x)-3(x+2)(5x)+(+2)3x-1/(x+2)(3x-1)(3) 27x² = 9x – 15x²-30x + 3x²-x+6x-2/3(x+2)(3x-1) Collect the lime terms 15x² – 34 – 2/3(x+2)(3x-1) (10a) Obtuse <BOD + Reflex<BOD = 360degrees (angle at a point) 105 + reflex<BOD = 360degrees Reflex <BOD= 360 – 105 =255° Now 2w = reflex<BOD(angle at centre = twice angle at circumference) 2w =255° W = 255/2 =127.5° Also 2x = obtuse<BOD(angle at centre = twice angle at circumference) 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| = TanR 100/|BR| = Tan60° |BR| = 100/tan60 |BR| = 100√3 |BR| = 100√3 * √3/√3 =100√3/3m OR 57.7m 11a) x+y/2 =11 x+y= 11*2 x+y= 22 —(1) x-y= 4 —-(11) x+y = 22—-(1) – x-y= 4—-(11) ____________ 2y = 18 y= 18/2 y=9 Substitute y=9 in equ 1 x+9=22 x=22-9 x=13 x=13, y=9 x+y= 13+9= 22 Sum of the two number (11b) (6x + 3) dx (6x + 3)dx (6x +3)^6 – (6x + 3)^1 (6 x + 3)^5 (7776x^5 + 243) 38,880x/6 + 243 6480 x^6 + 243x 9(720x^6 + 27x) (11c) y = x² + 5x – 3 (x = 2) y = 2² + 5(2) – 3 y = 4 + 10 – 3 y = 14 – 3 y = 11 Gradient of the curve = 11 (12a) Pr of Abu to pass = 3/7 Pr of Abu to fail = 1 – 3/7 = 7-3/7 = 4/7 Pr of kuranku to pass = 5/9 Pr of kuranku to fail = 1 – 5/9 = 9 – 5/9 = 4/9 Pr of musa to pass = 12/13 Pr 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 is Prof(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