WAEC Further Mathematics Objective and Theory Questions and Answers in 2024

The WAEC further maths examination may seem like a very difficult one for certain students. That is understandable. However, we have provided WAEC Further Mathematics Objective and Theory Questions for you to study with.

The West African Examinations Council (WAEC) is an examination board that administers the West African Senior School Certificate Examination for University and Jamb admission exams in West African countries.

What WAEC is All About

Every year, more than three million candidates register for WAEC-coordinated exams.

The 2024 WAEC Further Mathematics exam will comprise Papers 2 & 1 Essay and Objective. That means the examination will last for three hours (3hrs) only.

There will be two papers to complete: paper 1 and paper 2.

PAPER 1: This paper will contain forty multiple-choice objective questions that will cover the entire course.

Candidates will have one hour to answer all questions for 40 points. The following areas of the curriculum will generate questions:

30 questions on pure mathematics

4 questions about statistics and probability

6 questions on vectors and mechanics

PAPER 2: There will be two portions, Sections A and B, that must be completed in two hours for 100 points.

Section A will include eight obligatory questions of an elementary kind worth 48 points. The following is how the questions will be distributed:

We will post samples of the WAEC Further Mathematics questions for candidates who will take part in the examination for practice.

WAEC Further Mathematics Objective Answers

Today’s WAEC Further Math OBJ Answers are below:

1-10. CBBDCAADCB

11-20. DCBDCBCDCC

21-30. ABDCDABBCB

31-40. CDCADBCAAD

41-50. ABDCBACDDC

Theory Questions and Answers

Keep the following in mind:

(1) pr(age)=4/5

pr(fully)=3/4

pr(must)=2/3

pr(age not admitted)=1-4/5

=1/5

pr(fully not admitted)=1-3/4

=1/4

pr(must not be admitted)=1-2/3

=1/3

Therefore pr(none admitted)=1/5*1/4*1/3

=1/60

(2) Tabulate

Marks| 1-10, 11-20, 21-30, 31-40, 41-50,51-60, 61-70, 71-80, 81-90, 91-100

F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6

C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,40.5-505, 50.5-605, 60.5-705, 70.5-805,80.5-905, 90.5-1005

C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146,146+77=243, 243+115=358, 358+101=459,459+64=523, 523+21=544, 544+6=550

(3) Given: f(x)={(4x-x^2)dx

f(x)=2x^2 – x^3/3 + K

f(3)=2(3)^2 – (3)^2/3 + K =21

18 – 9 + K=11

9+K=21

K=21-9

K=12

Therefore, f(x)= -x^3 + 2x^2 + 12

(4) 1/1-cos tita + 1/1+cos tita

=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita) = 2/1+cos tita – cos tita – cos^2 tita = 2/1-cos^2 tita

Recall that :

Cos^2 tita + sin^2 tita = 1

.:. Cos^2 tita = 1-sin^2 tita

.:. 1/1-cos^2 tita + 1/1+cos tita = 2/1-(1-sin^2 tita)

(5) At stationary points,

dy/dx=0.

y=x^0(x-3)

Let u=x^2,v=x-3.

du/dx=2x dv/dx=1.

dy/dx= Udv/dx + Vdu/dx

dy/dx=x^2(1)+(x-3)(2x)

.:. dy/dx=x^2+2x^2-6x

dy/dx=3x^2-6x

At a stationary point,

dy/dx=0..

.:.3x^2-6x=0

Equation of line=> 3x^2-6x=0

(6ai) P:F=4:1 =4x+1x=100

5x=100

x=100/5 x=20

pass=20*4=80%

fail= 20*1=20%

p(pass)=80/100=0.8

p(fail)=20/100=0.2

n=7

(6aii) P(between 3 and 6 failed)

P=0.2

q=0.8

P(36)

P(x=3) + P (x=4)+p(x=5)+P(x=6)

p(x=3) 7^C3 (0.2)^3 (0.8)^4

=0.114688

p(x=4)=7^C4 (0.2)^4 (0.8)^3

0.028672

P(x=5)=7^C5 (0.2)^5 (0.8)^2

=0.0043008

P(x=6)=7^C6 (0.2)^6 (0.8)^1

=0.0003584

p(36)

=0.114688+0.028672+0.0043008 +0.0003584 =0.1480192 =0.15(2d.p)

DISCLAIMER! These are not real WAEC Further Mathematics questions but are likely repeated questions over the years to help the candidate understand the nature of their examinations. Ensure to note every question provided on this page.