NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives

NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives 

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In this post are researched  regularly repeated NECO Mathematics  Questions – Objectives and Theory questions and answers for 2018 June/July SSCE exam.  Get them in your finger tips and you are sure of a high grade in this exams. NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers.

NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives 

WHAT NECO STANDS FOR

The National Examinations Council – NECO is  a Nigerian Government owned examination body  that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and December/January respectively. These are entry qualification examinations for higher education institutions in Nigeria.
NECO was created  in April 1999 with a mandated to take over the responsibilities of the National Board of Education Measurement (NBEM). 
NECO therefore conducts Senior Secondary Certificate Examination (internal and external), Junior Secondary Certificate Examination (JSCE) for Federal Unity College and other Federal secondary schools and  Private secondary schools, and National Common Entrance Examination for primary 6 pupils interested in admission into Federal Unity Colleges.
The Senior Secondary Certificate Examinations (internal and external) are entry qualification examinations for higher education institutions in Nigeria.

NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives 

NECO RESULTS & HIGHER SCHOOL ADMISSION

As you know, no University, Polytechnic, or college of education in Nigeria would admit you into any of their courses without a good grade in any of these examinations. That is why over three million candidates register and take these subject exams annually.
Consequently, in NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers, our goal is to help you achieve this your life dream. Listed here are Mathmatics, Objective and Theory questions and answers relevant to 2018 June/July Mathematics Subject examination.
NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives

NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives 

VERIFIED NECO 2018 MATHEMATICS QUESTIONS AND ANSWERS

Paper III: Objective – General Mathematics
 Paper II: Essay – General Mathematics 
1-10 deacdacbbe 
11-20 cbaeebdacc 
21-30 deedacbbce 
31-40 abdcebaeec 
41-50 caadedbaea 
51-60 edeacbacdd 
1a)
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 equivalent 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
3)
let the son age be x
man=5x
son=x
4yrs ago; the man age = 5x – 4
the son age = x – 4
the product of their ages
(5x – 4)(x – 4) =448
4a)
volume of fuel = cross-sectional area of X depth of fuel rectangular tank
30,000litres = 7.5*4.2*d m^3
but; 1000litres =1m^3
therefore;30(M^3) = 7.5*4.2*d(M^3)
30=31.5d
====> d = 30/31.5 = 0.95(2d.p)
4b)
to fill the tank/volume of fuel needed = 7.5*4.2*1.2 = 37.8m^3 = 37,800
litres addition fuel = 37,800-30,000 = 7,800
litres therefore, 7,800
more litres would be needed 
================ 
5a)
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
5b)
amount spent=144000- [48,000 32000 19200 12000 7200] =144000-118400 =N25600 
===============
7a)
3²ⁿ ¹ — 4(3ⁿ ¹) 9 = 0
3²ⁿ × 3 — 4(3ⁿ× 3¹) 9 = 0
(3ⁿ)² × 3 — 4(30ⁿ× ) 9 = 0
Let 3ⁿ= x
3x² — 4 × 3 × x 9 = 0
3x² — 12x 9 = 0
Divide all by 3
3x²/3 — 12x/3 9/3 = 0
x² — 4x 3 = 0
x² — 3x — x 3 = 0
x(x—3) -1(x—3) = 0
(x—3)(x—3) = 0
x—3 = 0 or x—1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3ⁿ= 3°
n = 1 or n= 0
7b)
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 = 5x
x^2-5x 4 = 0
x^2-4x – 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)
|BC|² = |BD| + |CD|²
13² = BD² + 5²
169 = BD² + 25
√BD² = √144
BD= 12m
Both OB = OD and OB + OD
= 2OB/2 = 12cm/2
OB = 6cm
8b) Circumference = 2πr
r = 6cm, π = 22/7
= 22 × 2 × 6/7cm
= 264/7cm
= 37.7cm to 1 decimal places
=================
9a) Let the digits be y
10(5 + y)+y =3y(5 +y) —14
50 + 10y + y = 3y(5+y) —14
50 + 11y = 15y + 3y² — 14
3y² + 15y — 11y — 50 — 14 = 0
3y² + 4y — 64 = 0
(3y² — 12y) + (16y — 64) = 0
3y(y — 4)(3y + 16) = 0
y —4 = 0 or 3y + 16 = 0
y = 4 or —16/3
9b)
3—2x/4 + 2x—3″3
= 3(3—2x) + 4(2x—3)/12
= 9—6x + 8x—12/12
= 2x—3/12
10a)
y=(2x^2 + 3)^5
let U=2x^2 + 3
Y=u^5
du/dx = 4x
dy/du = 5u^4
dy/du = (2x^2 + 3)^4
dy/dx = du/dx dy/du
dy/dx = 4x.5(2x^2 + 3)^4
dy/dx = 20x(2x^2 + 3)^4
10b)
y=3x^2 + 2x +5
dy/dx =6x + 2
dy/dx =6(3) +2
dy/dx =18+2
dy/dx =20
10c)
R-W=Wv^2/gx
Wv^2=gx(R-W)
Wv^2=gRx-Wgx
Wv^2+Wgx=gRx
W(v^2 + gx) =gRx
W=gRx/V^2 + gx
R=2, g=10, x=3/2, V=3
W= 10*2*3/2/3^2 + 10*3/5
W=30/9+15
W=30/24
W=5/4
Good luck.


NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives 

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