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NECO 2018 - MATHS ANSWERS

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NECO 2018 - MATHS ANSWERS


Monday 4th June 2018
OBJ – General Mathematics – 10:00am – 11:45am
Essay – General Mathematics – 12:00noon –2:30pm
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COMPLETED
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MATHS OBJ:
1-10: CDAAEABAEC
11-20: ACDDCDCDAC
21-30: CEADEDCABC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA
51-60: BCECCBBCEE
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(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

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(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

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(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

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(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


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(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

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(6a)
log2 = 0.3010
Log3 base 10 = 0.4771
(i) Log10 3.6 = Log10 36/10
= log10 36 - log10 base 10
= log10 (9x4) -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

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(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

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(8a)
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

(8ai)
Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y

(8aii)
When y=7
x=4+3(7)
x=4+21
x=25


(8b) 
3x/x+2 - 5x/3x - 1 + 1/3

Find the L. C. M
3(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)


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(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| = TanR
100/|BR| = Tan60°
|BR| = 100/tan60
|BR| = 100√3
|BR| = 100√3 * √3/√3
=100√3/3m OR 57.7m

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(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
integrating
6x²/2+3x/1

6
= {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] - [9]
= 126-9
= 117


(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

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(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

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COMPLETED
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