NECO 2018 MATHEMATICS OBJ & THEORY ANSWER NOW AVAILABLE.

MATHEMATICS OBJ:
1-10: CDAAEABAEC
11-20:ACDDCDCDAC
21-30: CEBDEDCBBC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA
51-60:BCEACBBCEE

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========COMPLETED=======

(1a)
Log10(20x – 10) – log10(x+3) = log10^5
Log10(20x-10/x+3)= log10^5
20x – 10/x + 3 = 5
Cross multiplying
20x – 10 = 5(x + 3)
5(4x – 2) = 5(x + 3)
4x – 2 = X + 3
4x – x = 3+2
3x = 5
X = 5/3 OR 1 whole no 2/3

(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
Therefore actual amount paid on the article
=#4,000-#600
=#3,400
Actual amount paid on the article =#3,400

(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

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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

(10a)
Obtuse

(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