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

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

##
**Mathematics OBJ:**

1CDAAEABAEC

11AEDDCDCDCC

21CEBDEDCBBC

31CBEEECBDCC

41DBCBCDDBCA

51BCBDCDCCEC

ALWAYS SUBSCRIBE IF YOU WANT YOUR ANSWERS BEFORE THE EXAM

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

**Mathematics OBJ:**

1CDAAEABAEC

11AEDDCDCDCC

21CEBDEDCBBC

31CBEEECBDCC

41DBCBCDDBCA

51BCBDCDCCEC

1CDAAEABAEC

11AEDDCDCDCC

21CEBDEDCBBC

31CBEEECBDCC

41DBCBCDDBCA

51BCBDCDCCEC

ALWAYS SUBSCRIBE IF YOU WANT YOUR ANSWERS BEFORE THE EXAM

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

Now 2w = reflex

W = 255/2 =127.5°

Also 2x = obtuse

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

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

*042tvseries.com send all exam expo answer earlier than others*============================

*NECO ssce 2018 Mathematics expo answers, Ssce 2018 Mathematics objective and essay Expo answers now available, nov/dec Mathematics theory Answers for 2018, Mathematics obj and theory Questions NECO Ssce Expo Answers Here, Free wassce Mathematics objective, essay & theory correct Expo 2018 answers Runz*

*NECO*

*Mathematics, NECO Ssce 2018 Free Mathematics essay, obj and theory Questions and Answers Expo for*

*NECO*

*Mathematics Paper 1 and Paper 2 objective and essay Questions and Answers Expo Now Posted, Free Mathematics objective ans for internal ssce, see*

*NECO*

*ssce expo Mathematics obj and theory, essay questions for free, loadedexpo correct wassce Economics ans, sure*

*NECO*

*ssce Mathematics objective answers runz runs expo free real and correct NECO ssce 2018 nov/dec Mathematics free expo runz runs ans answers online for free see free answers online, 2018 Verified*

*NECO*

*Free Ssce Mathematics Obj and theory Answers have been posted REAL NECO SSCE Mathematics QUESTIONS & ANSWERS Direct To Phone Number AS Text message, RE: 2018 WASSCE Mathematics (Expo) NECO 2018 certified runs Mathematics questions and answers*

*NECO*

*SSCE November/December 2018 Mathematics THEORY / ESSAY / OBJ QUESTION AND EXPO I need*

*NECO*

*ssce free expo site, how can I get WAEC Mathematics answer for free,*

*NECO*

*2018 Mathematics essay, Objective And Theory Question and Answer Now Posted, Real 2018/2019 nov/dec wassce*

*NECO*

*Mathematics objective, essay and theory 100% correct expo questions and answers runz chokes runs, verified 2018 NECO Mathematics theory, essay and obj*

*NECO*

*Mathematics expo website,*

*NECO*

*ssce Mathematics live cheats, sure November/December*

*NECO*

*SSCE 2018 Mathematics original objective, theory and essay questions and answer now available here, 100%*

*NECO*

*ssce Mathematics expo, 2018 wassce Mathematics obj, essay and theory runz now posted, correct*

*NECO*

*ssce Mathematics objective, essay (theory) answer now ready, free 2018 wassce Mathematics obj, theory and essay best expo site Neco ssce Mathematics questions and answers, password for certified WAEC ssce Mathematics essay theory obj expo, how to get/ where to get wassce Mathematics theory, essay and objective expo answer direct to my phone as text message, trusted*

*NECO*

*ssce Mathematics questions and answer, i need NECO Ssce nov/dec Mathematics essay theory objective obj Expo 2018 answers on whatsapp, gidifans Mathematics answer, solutionclass Mathematics answer, 042tvseries Mathematics answer, codedexam*

*NECO*

*Mathematics answer, wapbaze Mathematics answer, wapextra Mathematics expo, Martinlibrary Mathematics expo, examgists*

*NECO*

*Mathematics expo,*

*NECO*

*Mathematics objectives answer, naija*

*NECO*

*Mathematics expo free answer*

Real and Confirmed WAEC Maths Questions and Expo Answer – May/June 2018

Verified WAEC 2018 May/June Mathematics OBJ and Essay Answer and Solution to the questions.

**GOODLUCK!!!**