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In this post are verified regularly repeated General Maths Objective and Theory questions and answers for 2018 May/June SSCE exam that holds on 18/4/2018. Get them in your finger tips and you are sure of a high grade in this exams. 2018 WAEC General Maths Questions And Answers: Objectives & Theory.

### 2018 WAEC General Maths Questions And Answers: Objectives & Theory

Table of Contents

#### WAEC FOR ALL

The West African Examinations Council (WAEC) is the West African countries Examination Body. It conducts the West African Senior School Certificate Examination for in-house school students and General Certificate Examination for private candidates in West African Countries. These are entry qualification examinations for higher education institutions in these countries.

As you know, no University, Polytechnic, or college of education in these countries 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, 2018 WAEC General Maths Questions And Answers: Objectives & Theory, our goal here is to help you achieve this your life dream. Listed here are Maths, Objective and Theory questions and answers relevant to 18/4/2018 Maths Subject examination.

### 2018 WAEC General Maths Questions And Answers: Objectives & Theory

**THE OBJECTIVES**

Verified MATHS OBJ:

1-10: BCABCADDAA

11-20: DDEACDBACA

21-30: ABAACECADB

31-40: ACAEBEBBCD

41-50: EDEDDEDDCA

51-60: CADCEDDEAB

#### THEORY SESSION;

(keys)

/ means DIVISION

* means multiplication

tita u should know dat.

rais means raise to power

^ means raise to power

sqr rut/ squar rut means square root

—-(1) means equation 1

yr means year

f(x-xbar) means x MINUS x BAR as in statistics which means that MINUS WILL BE ON TOP OF x.

, (Comma) means NEXT LINE, (/quote)

(1) n(AUBUC)=n(A)+n(B)+n(C)-n(AnB)-n(BnC)+n(AnBnC)

55=21+24+23-6-8-5+2x+x

55=49+3x

3x=6

x=2

(1i)

21-6-5-x = 21-11-2

physics only=8

(1ii) 6+5+8= 19

(2a) Using almighty formular

x= -b+sqrt(b^2- 4ac)/2a

=5+sqrt(25-4(2)(3))/2(2)

=(5+1)/4 or (5-1)/4

= 1 or (1)1/2

(2b) Volume of cylinder= pieR^2h

352=22/7*4*4*h

English

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2018 WAEC Time Table

Physics

Further Maths

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2018 WAEC Time Table

### 2018 WAEC General Maths Questions And Answers: Objectives & Theory

(3a) RD^2=(37)^2- (12)^2

= 1369-144

RD^2=1225

RD=sqrt(1225)

=35cm

Area= 35*12=420cm^2

(3b) ((2p+q)/(p-3q))^1/2=h

sqrt(h)=(2p+q)/(p-3q)

Psqrt(h)- 3qsqrt(h)=2p+q

Psqrt(h)-2P= q+3qsqrt(h)

P(sqrt(h)-2)= q+3qsqrt(h)

P=q+3qsrt(h)/sqrt(h)-2

(4a) (n-2)180=

xn+x=180

x(n+1)

180n-360=xn+x

180n-360=x(n+1)

x=180n-360/n+1

(4b) (x+5)/3-(2x-1)/5=5/6

30((x+5)/3) – 30((2x-1)/5)= 30(5/6)

10x + 50-12x+6=25

56-2x=25

-2x=-31

x=15.5

(5a) perimeter of sector= (tita/360)2pieR+2r

=2(9)+120/360*2*22/7*9

=18+18.86

=36.9cm

(5b) 150/h=tan30

h=150/tan30

h=259.8 approx 260

(6a) Snth= n/2(2a+11d)

-192=24a+132d

-96=12a+66d

-32=4a+22d

-16=2a+11d —-equ(1)

-252=18(2a+17d)/2

-504=18(2a+17d)

-28=2a+17d——-equ(2)

(-16=2a+11d)

-(-28=2a+17d)

12=-6d

d=-2

(6b) from equ(1)

-16=29+11(-2)

-16=29-22

2a=6

a=3

(6c) Tnth= a+(n-1)d

=3+(15-1)(-2)

T15th= -25

T4th=a+(4-1)d

=3+3(-2)

=-3

Product of 15th and 4th= -3*-25=75

(7ai) STB= 180-75-40

(sum of angles on a straight line)

STB= 65

SRP= 180-65(opposite angles are complementary)

=115degree

(7aii) SQP= 115(angles substended from the same circumference are equal)

(7aiii)

SPQ=180-115-12 (sum of angle in a triangle)

=53degree

(7b) C=R+KT

2800=R+K(6)

2800= R+6k ———equ(1)

– 3600= R+10K——equ(2)

-800 = -4k

k=200

From eqn(1)

2800=R+200(6)

2800=1200R

R=2(1/3)

at 4hours,

C= 2(1/3) + 200(4)

= 800 +2(1/3)

=802(1/3)

=#802.3

( 8 ) Distance from Airport to the North= Time*Speed

= 500*1(1/2)= 750km

2nd distance= xkm

From cosine rule, a^2=b^2+c^2-2bc CosA

x^2=750^2+400^2-2(750)(400)cos127

=562500+160000-(-361089.01)

x^2=1083589.01

x=sqrt(1083589.01)= 1,040.96km

Bearing from the airport

from sine rule, a/sinA=b/sinB

1040.96/sin127=750/sinb

sinb=0.5754

b=35.13

d= 150-127-35.13(sum of anglesin a triangle)

d= 17.89 approx = 17.9

Bearing from the airport=180+17.9=197.9

### 2018 WAEC General Maths Questions And Answers: Objectives & Theory

(9a) a^2= b^2+c^2(pythagora’s theorem)

h^2=40^2+16^2

=1600+256

=1856

h=sqrt(1856)= 43.1cm

(9b) Total surface area

Area of one of the four triangular faces=40*16=640cm^2

for four triangles= 4*640=2560cm^2

Area of square base=32*32=1024cm^2

Total surface area=1024+2560=3584cm^2

(12) TABULATE THE TABLE

MARK%: 51-60, 61-60, 71-80, 81-90, 91-100

BOUNDARIES: 50.5-60.5, 60.5-70.5, 70.5-80.5,80.5-90.5, 90.5-100.5

CLASS MARK(X): 55.5,65.5,75.5,85.5,95.5

FREQUENCY: 11, 23,39,17,10

(FX): 610.5,1506.5,2944.5, 1453.5, 955

/X-XBAR/: 19.2, 9.2, 0.8, 10.8, 20.8

/X-XBAR/^2: 368.64, 84.64, 0.64, 116.64, 432.64

F/X-XBAR/: 211.2, 211.6, 31.2, 183.6,208

(i) Mean= Efx/fx

= (610.5+1506.5+2944.5+1453.5+955)/100

=7470/100

=74.7 approx =7

(ii) Mean Deviation= sqrt(Ef/x-xbar/)/Efx

sqrt(211.2+ 211.6+31.2+183.6+20 /100

845.6/100=8.456 approx = 8

(iii) Standard Deviation=sqrt(Ef/x-xbar/^2)/Efx

sqrt(12336/100)=sqrt(123.36)

=1

### 2018 WAEC General Maths Questions And Answers: Objectives & Theory

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