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

Filed in Exam, NECO News on June 1, 2018

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

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
area =?
© 2 = 360 / 12 =30 °
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 :
11-20: CCDCAACECA
21-30: DCDAECAEED

==================================
8a)
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
=========================

CSN Team

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9. jj says:

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 | = 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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