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# NECO Mathematics Obj and Essay Answer – 2018 June/July Expo

NECO 2018 MATHEMATICS QUESTIONS AND ANSWER.

Mr Spy – 4th, June 2018
==> MATHEMATICS OBJ:
1-10: CDAAEABAEC
11-20:ACDDCDCDAC
21-30: CEBDEDCBBC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA
51-60:BCEACBBCEE
————————————————————–

THEORY

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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

================================

2

(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

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(4)
Draw the diagram

(i) Arc length = Tita/360*2?r
= 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*?r²
=72/360*22/7*14/1*14/1
=1/5*22*2*14
=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

================================

6

================================

(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 <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
|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

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