WAEC 2018 Mathematics OBJ Answers, here are Maths, Objective 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);
- Develop the Passion for Maths
- Know the Exam Format
- Get WAEC Syllabus
- Practice Daily
- Get the Past Questions
- Join a Group
- Test your Knowledge
- Seek for Help
- Attend WAEC Tutorials
- Have a Shortcut
- Avoid Social Media Distraction
Develop the Passion for Maths
How to Study Mathematics for WAEC
Here are ten (10) ways to study mathematics. They are:
- Avoid distraction
- Revise your notes
- Solve problems
- Ask questions
- Know different maths formulas
- Study more
- Join study groups
- Rest and eat well
- Compete with your mates
- Have a mentor
2023 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;
—-(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 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