NECO One Mathematics Questions 2017 | NECO Mathematics OBJ & THEORY Answers.

Finally NECO One Mathematics Questions  All in one Mathematics NECO Questions 2017 | NECO Mathematics OBJ & THEORY Answers. As a matter of fact in this article, We will be unleashing 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. All we need you to do is to stay focus and don’t rush your self while reading.

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NECO One Mathematics Questions

NECO Mathematics Questions 2017NECO Mathematics Questions 2017

Furthermore 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.

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NECO One Mathematics Questions

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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
Finally V=8

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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
Finally A =130 – 95 cm^2

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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
Finally x = 7.

Finally NECO One Mathematics Questions

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
Finally = 877 km

NECO One Mathematics Questions

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
Finally = 9460 / 19, 460 = 0.486

NECO One Mathematics Questions

4 aii )
Pr one type of fruit is picked
( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )
= 6162 / 19, 460 + 3422 / 19 , 460
Finally = 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
Finally X ≤ 11 / 7 ===> X ≤ 1 4 / 9
=============================

NECO One Mathematics Questions

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)
Finally 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
As a matter of fact finally this is it = N 35 ,122 . 50
============================

NECO One Mathematics Questions

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
As a matter of fact finally this is it =2 X ^2 – X – 6

NECO One Mathematics Questions

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
As a matter of fact finally this is it = 522 . 67 cm3

NECO One Mathematics Questions

 

NECO One Mathematics Questions 2017

7 a )
Furthermore 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
Finally d ~ 6130 km .

ii ) distance along gent circle
Furthermore 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
Finally D = 6144 .03
As a matter of fact finally this is it = D =~ 6146 km .

Finally NECO One Mathematics Questions

7 b )
Length of sector tita/ 360 × 2 pie r
Furthermore L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1
L= 120 ×44 ×42 / 2520
L= 221760 / 2520
L= 88cm
L= 2pie r
Furthermore Where r is the radius of circumference
88 =2 × 22 / 7× r
88 ×7 = 44 r
R =88 ×7 /44
R =616 / 44
Meanwhile R =14 cm.

Finally NECO One Mathematics Questions

Curved surface area
= pie rc
A =22 / 7 ×14 × 42 / 1
A =22 ×14 ×4 ^2 /7
A =12936 /7
As a matter of fact A =1848 cm^ 2.

Above all Finally Meanwhile
Actually First, Second, Third Moreover
Afterward First and foremost Next
All things considered For this reason No doubt
Accordingly From here on Of course

NECO One Mathematics Questions

8 a )
Furthermore X =60 / t — – – – – – – – – > ( i )
Y = 180 / t – – – – – — – – – > ( ii )
T 1=60 / X
T 2=100 /Y
T 1+T 2= 5
Meanwhile 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
Finally Substract ( i ) from ( ii )
8 p = 8
P = 8 / 8 ÷ P = 1
Furthermore 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
Finally 4 /3 = 1/ Y ÷ Y = 3/ 4

 

NECO One Mathematics Questions

8 b )
2001 – – – – – — – – 25 , 700
2002 – – – – – — – – 15 / 100 X 25, 700 + 25,
700 = 29 , 555
Similarly Amount of tax in 2002
= 29 , 555 * 12 .5 /100
= N 3694 .375
Finally = N 3690

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All things considered For this reason No doubt
Accordingly From here on Of course

Finally NECO One Mathematics Questions

8 c )
Presently Log 25
Log 16 25 / 100
Log 16 2/ 4
Log 4 6 – 1
– 1/ 2 Log 4^4
– 1/ 2

9 ai)
Presently 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
Furthermore Substact ( ii ) from ( i )
12 k / 12 = 312 /12
K = 26
Finally Substract K into ( i )
16 k + C = 384
C = 384 – 416
C = – 32
W= k +C / t 2
As a matter of fact W= 26- 32 / t 2

Above all Finally Meanwhile
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Accordingly From here on Of course

( 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
Finally and above all = 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

Above all Finally Meanwhile
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All things considered For this reason No doubt
Accordingly From here on Of course

 

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

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Above all 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
Above all r=5.1cm
8aii)
circumfrenece of a circle=2pie R
C=2×22/2*(5.1)^2
C=1144.44/7
C=163.4914cm
Finally  C=163.5cm

Above all.  Finally Meanwhile
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Afterward First and foremost Next
All things considered For this reason No doubt
Accordingly From here on Of course

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
Finally =y=4x/3+10/3
======================

Above all Finally Meanwhile
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All things considered For this reason No doubt
Accordingly From here on Of course, Above all Finally Meanwhile
Actually First, Second, Third Moreover
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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

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All things considered For this reason No doubt
Accordingly From here on Of course

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
Finally =2x-2/12
=========================

Finally NECO One Mathematics Questions

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