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NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers – related to these are NECO past questions 2016,2016/2017 NECO mathematics answers,2015 NCO past questions and answers, NECO mathematics past questions and answers, NECO past questions 2016 mathematics and NECO mathematics question and answer 2015

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.

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.

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.

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.