# NECO Mathematics Questions 2020 See OBJ and Theory Answers Here

NECO Mathematics Questions 2020 See OBJ and Theory Answers Here.

Mathematics NECO Questions: In this article, I will be showing you past Mathematics objective and theory random repeated questions for free. You will also understand how NECO Mathematics questions are set and many other examination guides. Stay focus and read through.

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 Mathematics Question and Answers

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 :
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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

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