## 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 <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°

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 <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°

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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