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

NECO 2018 MATHEMATICS QUESTIONS AND ANSWER.

1-10: CDAAEABAEC

11-20:ACDDCDCDAC

21-30: CEBDEDCBBC

31-40: CBEECCBDCC

41-50: DDCBCDDBBA

51-60:BCEACBBCEE

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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

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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

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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*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

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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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6

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(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

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(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

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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

(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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