## 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.

###### Similarly on NECO Mathematics Question and Answers

## 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 ×© 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

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## 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

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.

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## 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

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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## 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

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( 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.

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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]

## MATHEMATICS OBJ 100% VERIFIED :

Presently 1-10: ADADDDCEBE

Furthermore 11-20: CCDCAACECA

Finally 21-30: DCDAECAEED

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

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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

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

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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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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

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## Finally NECO One Mathematics Questions

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