**NECO 2018 Mathematics Obj And Essay Answer – June/July Expo**

**Mathematics OBJ:**

1CDAAEABAEC

11AEDDCDCDCC

21CEBDEDCBBC

31CBEEECBDCC

41DBCBCDDBCA

51BCBDCDCCEC

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CLICK HERE FOR no.1 THE IMAGE

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*=600*100

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

CLICK HERE FOR no.2 THE IMAGE

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

CLICK HERE FOR no.4 THE IMAGE

4i)

length of Arc of the sector

Titter= 72degree, r = 14cm

L= titter / 360 x 2 pie r

==> L= 72/360 x 2 x 22/7 x 14

=44352/2520

=17.6cm

4ii) perimeter of the sector

Perimeter = titter/360 x 2 pie r + 2r

=17.6 +(2×14)

=17.6+28

=45.6cm

4iii) Area of the sector

Area = Titter/360 x pie r^2

=72/360 x 22/7 x (14)^2

=72 x 22 x 196/2520

Area= 310464/2520

=123.2cm^2

CLICK HERE FOR no.5 THE IMAGE

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

CLICK HERE FOR no.6 THE IMAGE

6a)

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

CLICK HERE FOR no.7ai THE IMAGE

CLICK HERE FOR no.7aiii THE IMAGE

CLICK HERE FOR no.7b THE IMAGE

CLICK HERE FOR no.7bii THE IMAGE

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

CLICK HERE FOR no.8 THE IMAGE

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

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)

CLICK HERE FOR no.10a THE IMAGE

10a)

Obtuse 105 + reflexReflex =255°

Now 2w = reflex2w =255°

W = 255/2 =127.5°

Also 2x = obtuse2x = 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°

CLICK HERE FOR no.10b THE IMAGE

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

CLICK HERE FOR no.11 THE IMAGE

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

12bi)

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

12bii)

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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NECO 2018 Mathematics Obj And Essay Answer – June/July Expo
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