NECO Mathematics Questions and Answers 2020-Essay and Theory | Free PDF

NECO Mathematics Questions and Answers 2020-Essay and Theory | Free PDF.

NECO Mathematics Past Questions and Answers Free PDF Download…Are you currently writing NECO Examination or you about to? Here are the likely questions that you will encounter in NECO Mathematics Examination 2020.

NECO Mathematics Questions and Answers 2020
NECO Mathematics Questions and Answers 2020

The National Examinations Council (NECO) 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 Questions and Answers 2020

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 ש 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
=========================

Tips on How to Pass NECO Examination 2020

If you want to Past the NECO Examination this year, you will need to follow the instructions below.

  • Practice daily with NECO Past Questions and Answers
  • Write out your subjects, the dates and time.
  • Engage in general revision few days to the examination and dwell more on the subjects you find difficult.
  • Focus on only your first paper a week to the exam. This would make it sweet and you will have energy to read for other papers.
  • Don’t read any other subject when it is two days to a particular subject.
  • When you have two papers that are separated by only one day, read the second one first before going to the subject you have first.
  • Ensure to be in the NECO examination venue at least two hours to the examination so that you would be balanced.
  • Try to be up to date in cases of changes in the timetable.
  • Have a personal copy of the NECO timetable and syllabus.
  • Go through your timetable daily so that you don’t miss any paper.
  • Don’t go to the hall when you are not having paper that day.

Disclaimer: Please note these are not the actual questions and answers that you will find in the NECO Examination this year, but they are the likely questions. to get the actual NECO Past Questions and answers, please call 07063986527.

Download Actual NECO Past Questions and Answers.

If this article on NECO Mathematics Questions and Answers has been helpful Please share and Like us on Facebook@Examstuts

Follow us on twitter, instagram, LinkedIn, Pinterest

Be the first to comment

Leave a Reply

Your email address will not be published.


*