NECO 2018 MATHEMATICS ANSWER Verified Math Obj:
1-10: CDAAEABAEC
11-20:ACDDCDCDAC
21-30: CEBDEDCBBC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA 51-60:BCEDCBBCEE =========================== \(1a)
Log 10\(20-10)-log10\(+3)=log105
\(20-10/+3)=log10 =5
20-10/+3=5
5\(+3)=20-10
5+15=20-10 15+10=20-5
25=15
*=25/15
*=5/3=1 2/3 \(1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
15%=#600 15/100 =600
15=600100
15=60000
*=60000/15
*=#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 – \(15/2010)²
r = √1 – \(0.075)²
r= √\(1.075)\(0.925) r = √0.994375
r = 0.9972
=========== \(4)
Draw the diagram \(i) Arc length = Tita/3602πr
= 72/360222/714
=1/5442
=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/360πr²
=72/36022/714/114/1
=1/522214
=616/5
=123.2cm2 ==========
\(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 ============== \(8a)
x=a+by\(eq i)
when y=5 and x=19
19=a+5b\(eq ii)
when y=10 and x=34
34=a+10b\(eq iii) solving eq ii and eq iii
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 \(i) Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y \(ii) When y=7
x=4+3\(7)
x=4+21
x=25 ==============
\(10a)
Obtuse 105 + reflex 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= 112
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