# NECO 2018 Mathematics Questions Online [OBJ and THEORY Answers]

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**NECO 2018 Mathematics Questions Online [OBJ and THEORY Answers].**

* NECO Mathematics Questions 2018…* The National Examinations Council is an examination body in Nigeria that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and December/January respectively.

**NECO 2018 Mathematics Questions Online**

**NECO 2018 Mathematics Questions Online**

1a)

(2x+1)/(3-4x)=2/3

3(2x+1)=2(3-4x)

6x+3=6-8x

6x+8x=6-3

14x/14=3/14

x=3/14

1bi)

E=MV^2/2

2E/M =MV^2/M

V^2=2E/M

V=sqr2E/M

1bii)

Vsqr2E/M

Vsqr2*64/2

Vsqr64

V=8

2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

MATHEMATICS OBJ 100% VERIFIED :

1-10: ADADDDCEBE

11-20: CCDCAACECA

21-30: DCDAECAEED

31-40: EAADBEEADC

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

8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12

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

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**Tags**: 2018 NECO Mathematics Answers, NECO, NECO Mathematics Questions, NECO Mathematics Questions 2018

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MATHEMATICS OBJ ANSWERS.

1 ecadddceee 10

11 ccdcaaceca 20

21 dcdeeccded 30

31 eaadbeeadc 40

41 bbdccdcecb 50

51 cadbaecdec 60

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1(a)

Tabulate

x – 1 ,2 ,3 ,4

1 – 1 ,2 ,3 ,4

2 – 2 _ , 4 , 0 _ ,2 _

3- 3 , 0_ , 3, 0 _

4 – 4 _ , 2_ , 0_ , 4

1b)

I = PRT/100, p=N15000 R=10% and

I=3years

A = P+ I

where I = 15000*10*3/100=N4500

A=4500+15000 =N19500

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

2a)

using sine rule

b/sin20 = 6/sin30

bsin30 = 6sin120

b 6sin120/sin30

b = 6×0.2511/0.4540

b = 5.7063/0.4540

b = 12.57 ≠ 12.6cm

2bi)

the diagram is euivalent triangles.

where

|AX|/|BC| = |BY|/|AC| = |XY|/|YC|

XY = 9, BY = 7

YC = 18-7=11

9/11 = 7/|AC|

9|AC| = 77

|AC| = 77/9

|AC| = 8cm

2bii)

XY/AB = BY/AC

9/|AB| = 7/8.6

|AB| = 9×8.6/7

|AB| = 11cm

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

2a)

using sine rule

b/sin20 = 6/sin30

bsin30 = 6sin120

b 6sin120/sin30

b = 6×0.2511/0.4540

b = 5.7063/0.4540

b = 12.57 ≠ 12.6cm

2bi)

the diagram is euivalent triangles.

where

|AX|/|BC| = |BY|/|AC| = |XY|/|YC|

XY = 9, BY = 7

YC = 18-7=11

9/11 = 7/|AC|

9|AC| = 77

|AC| = 77/9

|AC| = 8cm

2bii)

XY/AB = BY/AC

9/|AB| = 7/8.6

|AB| = 9×8.6/7

|AB| = 11cm

3a)

let the son age be x

man = 5 x

son= x

4 yrs ago; the man age = 5 x – 4

the son age = x – 4

the product of their ages

(5 x – 4 )( x – 4 ) = 448

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

4 a)

volume of fuel = cross -sectional area of X

depth of fuel

rectangular

tank

30, 000litres = 7 . 5* 4 .2 * d m ^ 3

but ; 1000litres = 1 m ^ 3

therefore ; 30( M ^3 ) = 7 .5 *4 . 2* d ( M

^

3)

30= 31.5 d

==== d = 30/31. 5 = 0 .95( 2 d .p )

4 b)

to fill the tank /volume of fuel needed

= 7 .5 * 4. 2 *1 .2

= 37.8 m ^ 3

= 37, 800 litres

addition fuel = 37, 800- 30, 000

= 7 , 800 litres

therefore , 7 , 800 more litres would be

needed

5 a)

sector for building project =

48000 /144000*

360 = 120degree

sector for education = 32, 000 /144000*

360

=80degree

sector for saving = 19200 /144000* 360 =

48degree

sector for maintenance = 12000 /

144000*360 = 30degree

sector for miscellaneous = 7200/144000*

360 =18degree

sector for food items = 360 -( 120 + 80+

48+30+ 18)

= 360- 296

= 64degree

5 b)

amount spent =144000- [48, 000 + 32000+

19200 + 12000

+7200]

= 144000-118400

= N 25600

6)

/————————— —

/ 41.02 × √0.7124

/ ———— ———— — —

√42.87 × 0.207 × 0.0404

| No | Log |

| 41.02 | 1.6130 | 1.6130

| 0.7124 | T. 8527 ÷ 2 | T.9263/1.5393

| 42.77 | 1.6321 |

| 0.207 | T.3160 |

| 0.0404 | 2. 6064 / |

| | T.5544 | T.5544/1.9849

Antilog = 9658 ≈ 96.58

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

7 a)

3^ 2 n + 1 – 4 ( 3 ^ n + 1 )+ 9 = 0

3^ 2 -3 – 4 ( 3^ n – 3) + 9 = 0

(3 ^ n )^2- 3 – 4 (3 ^ n -3 )+ 9 = 0

let 3^ n = p

p ^ 2 -3 – 4 (p -3 ) + 9 = 0

3p ^ 2 /3 – 12 p /3 + 9 / 3 = 0

p ^ 2 – 4 p + 3 = 0

p ^ 2 – 3 p – p + 3 = 0

p ^ 2 p ( p – 3 ) – 1 (p – 3) = 0

(p – 1 ) (p – 3) = 0

p -1 = 0 or p – 3 = 0

p = 1 or 3

Recall 3 ^ n = p

when p = 1

3^ n = 3^ 0

n = 0

when p = 3

3^ n = 3^ 1

n = 1

7 b )

log (x ^ 2 + 4 ) = 2 + logx – log ^ 20

log (x ^ 2 + 4 ) = log ^ 100 = log ^ x – log

^

20

(x ^ 2 + 4 ) = log ( xx )

x ^ 2 + 4 = 5 x

x ^ 2 -5 x + 4 = 0

x ^ 2 -4 x – x + 4 = 0

x (x -4 ) – 1 ( x -4 ) = 0

(x -1 ) (x -4 ) = 0

x -1 = 0 or x -4 = 0

x = 1 or 4

8 a)

| AD | ^ 2 = 13^2 -5 ^ 2

| AD | ^ 2 = 169- 25

| AD | ^ 2 = 144

AD =sqr 144

AD =12CM

| AD | = 12- r

r ^2 = ( 12-r )^2 – 5 ^ 2

r ^2 = ( 12-r ) ( 12- r ) + 25

r ^2 = 144 -24r + 25

r ^2 = 169 -24r

r ^2 + 24r -169 = 0

r ^2 + 24r = 169

r ^2 + 24r + 14^2 = 169 + 14^2

( r + 14)^2= 169 + 196

( r + 14)^2= 365

( r + 14=sqr 365

r + 14=19. 105

r = 19.105 -14

r = 5. 105

r = 5. 1 cm

8 aii )

circumfrenece of a circle = 2 pie R

C = 2 x22/ 2* ( 5 .1 )^2

C = 1144.44/ 7

C = 163 .4914cm

C = 163 .5 cm

8 b)

y2 -y1 / x2 -x1 = y- y1 /x- x1

6 -2 /2 – ( -1 )=y- 2/ x- ( -1 )

4 /2 + 1 = y- 2 /x+ 1

4 /3 = y-2 / x+ 1

3 ( y-2 )=4 ( x+ 1 )

3 y-6 = 4 x+ 4

3 y-4 x= 4 + 6

3 y-4 x= 10

y= 4 x/3 + 10/ 3

9 a)

let Xy represent the two digit number

x-y =5 — — ( i )

3 xy – ( 10x + y)=14

3 xy – 10x – y = 14 – — -( ii )

from eqn ( i )

x= 5 +y

3 y( 5 +y ) -10( 5 + y) -y= 14

15y+ 3 y^2 – 50 – 10y – y =14

3 y^2 + 4 y – 50 = 14

3 y^2 + 4 y – 50 – 14 = 0

3 y^2 + 4 y – 64 = 0

3 y^2 + 12y + 16y – 64 = 0

( 3 y^2 – 12y ) ( + 16y – 64)=0

by

( y-4 ) + 16( y- 4)= 0

( y-4 )=0

ii )

( 3 y+ 16) ( y-4 )=0

3 y+ 16=0 or y- 4 =0

3 y= -16 or y= 4

y= -16/ 3 or y= 4

when y =4!

x= 5 +y

x= 5 +4

x= 9

the no is 94

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