# WAEC 2023 Mathematics OBJ Answers

By | April 1, 2023

WAEC 2018 Mathematics OBJ Answers, here are MathsObjective and Theory questions and answers relevant to 2018  Maths Subject examination. 2023 WAEC General Maths ..Verified MATHS OBJ: 2023,

How to Pass Mathematics Examination in WAEC 2023

## How to Pass Mathematics Examination in WAEC

The following are tips to pass your mathematics examination in the West African Examination Council (WAEC);

1. Develop the Passion for Maths
2. Know the Exam Format
3. Get WAEC Syllabus
4. Practice Daily
5. Get the Past Questions
6. Join a Group
8. Seek for Help
9. Attend WAEC Tutorials
10. Have a Shortcut
11. Avoid Social Media Distraction

## How to Study Mathematics for WAEC

Here are ten (10) ways to study mathematics. They are:

• Avoid distraction
• Solve problems
• Know different maths formulas
• Study more
• Join study groups
• Rest and eat well
• Have a mentor

### 2023 WAEC General Maths Questions And Answers: Objectives & Theory THE OBJECTIVES
Verified MATHS OBJ:
11-20: DDEACDBACA
31-40: ACAEBEBBCD
41-50: EDEDDEDDCA

#### THEORY SESSION;

—-(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
h=7

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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
Category: WAEC 