 WAEC Further Mathematics Questions and Answers for free to all Weac candidates In Ghana, Liberia, Nigeria, Sierra Leone, and The Gambia. On this page, all the Waec Further Mathematics questions and answers for 2023 and the most common questions and answers are released here.

Waec Candidates that applied for the West African Examination Council (WAEC) SSCE Examination will write their Waec Further Mathematics For general students. All details you need for you to be successful and pass this 2023 Waec Exam will also be given and make sure you read all through.

### 2023 Waec Further Mathematics Exam Papers

2023 Waec Further Mathematics Exam Papers Are

• Waec Further Mathematics Essay Questions
• Waec Further Mathematics Objectives Questions,

You are writing the 2 papers in only one day. In this post, the previous Year’s Waec questions and answers for Further Mathematics are released and the 2023 Waec Further Maths Exam Questions will also be released for those participating in the 2023 Waec examination.

### 2023 Waec Further Mathematics Questions and Answers Objective Questions (paper 1)

Previous Year WAEC F/Maths Obj Answers.

1DCEBDBABAC.
11ABDDDCCDAC.
21CDBACAEBCB.
31CEAEEDEBBD
41CEBCDDCDDA

==================================

### 2023 Waec Further Mathematics Questions and Answers THEORY (paper 2)

1a)
A=(2 -1)(1 3)
|A|=(2*3–1)
|A|=6+1
|A|=7
since the determinant of A is not 0 the A is not non singular
matrix
To find A^-1
A^-1=(3 1)(-1 2)
then 1/7(3 1)(-1 2)
A^-1=(3/7 1/7)(-1/7 2/7)
======================

2a)
3root2-3root5/5root2+2roo5
conjugate=5root2-2root5
=(3root2-3root5)(5root2-2root5) / (5root2+2root5)(5root2-2root5)
=(15*2-6root10-15root10-6*5) / (25*2-10root10+10root10-4*5)
=(30-31root10-30) / (50-20)
=-31root10/30
2b)
siny=3/5
tany=3/4
tan(x+y)=tanx+tany/1-tanxtany
=(5/12+3/4)/(1-(5/12)(3/4)
=(14/12)/(1-15/48)
=36/125+18/125+16/125
=70/125
=14/25
======================

3a)
Pr(A passed)=4/5,pr(A failed)=1/5
Pr(B passed)=3/5,Pr(B failed)=2/5
Pr(C passed)-2/5,Pr(C failed)=3/5
Pr(atleast one passed)
=4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5
=24/125+9/125+4/125
=47/125
3b)
Pr(atleast one failed)
=4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5
=101/125
======================

4)
(3+2y)^5 with (a+b)^5 shows
a=3,b=2y
Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5
and 1
(3+2y)^5
=(3)^5+5(3)^4(2y) + 10(3)^3(2y)^2 + 10(3)^2(2y)^3 + 5(3)(2y)^4 +
(2y)^5
=243+810y+1080y^2+720y^3+240y^4+32y^5
(3.04)^5=(3+0.04)^5
=3^5+5(3)^4(0.04) + 10(3)^2(0.04)^2 +10(3)^2(0.04)^3 + 5(3)(0.04)
^4 + (0.04)^5
=243 + 5*81(0.04) + 10*27(0.0016) + 10*9(0.000064) +
15(0.0000026) + 0.0000001
=243 +16.2 + 0.432 + 0.00576 + 0.000039 + 0.0000001
=259.63783
======================

11a)
5x^2-2x+3=0
Divide both sides by 5
x^2-2/5x+3/5=0–eq1
Compare eq1 above with
x^2-(alpha+Beta)x+alphabeta=o
alpha+beta=2/5
alphabeta=3/5
11ai)
alpha^3+beta^3 = (alpha+beta) (alpha+beta)^2 – 3alphabeta
=2/5(2/5)^2-3(3/5)
=2/5(4/25-9/5)
=2/5(-41/25)
=-82/125
11aii)
Sum of alpha +1/beta and beta+1/alpha
=(alphabeta+1)/beta + (alpha beta + 1)/alpha
=alphabeta^2+alpha^2beta+beta/alphabeta
=alphabeta(alpha+beta) + (alpha+beta)/alphabeta
product root
(alpha +1/beta)(beta+1/alpha)
=alpha(beta+1/alpha)=1/beta(beta+1/alpha)
=alphabeta+1+1+1q/alphabeta
sum=alphabeta(alpha+beta) + (alpha+beta)/alphabeta
=3/5(2/5)+(2/5)
=6/25+2/5
=(6+10)/25
=16/25
Product alphabeta+2+1/alphabeta
=3/5+2+1/(3/5)
=3/5+2/1+5/3
=64/15.
Recall
x^2-(alpha+beta)+alphabeta=0
x^2-16/25x+64/15=0
11bi)
f(x)=x^3+2x-kx-6
x-2=0
x=+2
f(2)=2^3+2(2)^2-k(2)-6=0
8+8-2k-6=0
16-6-2k0
10-2k=0
2k=10
k=10/2
k=5
11bii)
(x^2+4x+3)
(x-2)rootx^3+2x^2-5x-6
=x^3-2x^2
=4x^2-5x
=4x^2-8x
=3x-6
=3x-6
=0
=>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)
11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3
======================

10a)
T4=ar^4=162—(1)
T8=ar^7=4374—(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6—(1)
T4=ar^3=54—(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
======================.

10a)
T4=ar^4=162—(1)
T8=ar^7=4374—(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6—(1)
T4=ar^3=54—(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
======================

15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309
(15b)
Pr(at most 2)=Pr(0)+Pr(1)+Pr(2)
n=1000,Pr=0.05 Pr(2)=0.95
Pr(x)=e^-landa(landa^x)
Pr(0.05)<0.1 and landa=1.000*0.05=50
e^-50(50^0)/o!+e^-50(50^1)/1!+e^-50(50^2)/2!
1/e^50+50/e^50+250.2e^50
=1/e^50[1+50+25/2]
======================

(12a)
w=weight of the plank
moment=force*perpendicular distance
clockwise moment=Anticlockwise
Make A the reference point
Pw=2.5m,
AM=(2.5-0.8)m
=1.7m g=10m/s
AB=3.2m,PA=0.8m
Convert 30kg to weight
=mg=30*10=300N
therefore 300*0.8=w*1.7
w=240/1.7=141.18N
Distance of its C.G from P=2.5m

(12b)
For R1=A, R2=B
R1=R2=2R1
Let Q be the reference point
300x*141.18*2.5=2R1(4.8+1)
300x+352.95=11.6R1
300x-11.6R1=-352.95—(1)
Let P in the reference point
2R1(0.8+4)=141.15*2.5
9.6R1=352.95
R1=352.95/9.6
=36.77N

Before you leave this page kindly make sure you understand and know how WASSCE grades your WAEC Subject. The reason why we are doing this, is not to frighten or scare you but to put you in check and see reasons why you need to be serious with your Further Mathematics studies for the 2023 Further Mathematics Exam.

### Waec Grading System For all Waec Candidates

Do you know that the West African Examination Council (WAEC) Board has published the Waec grading system of results? Kindly check below to see the meaning of the Waec Grading Result.

The table below shows candidates’ positioning of the Waec grading of results ranging from A1, B2, B3, C4, C5, C6, D7, E8, F9 are the complete list of Waec Grading Result. All Waec candidates must fall into one of the Waec Grading Systems.

 GRADE NUMERIC VALUE INTERPRETATION A1 1 EXCELLENT B2 2 VERY GOOD B3 3 GOOD C4 4 CREDIT C5 5 CREDIT C6 6 CREDIT D7 7 PASS E8 8 PASS F9 9 FAIL

1. A1 Excellent 75% – 100%
2. B2 Very good 70% – 74%
3. B3 Good 65%  69%
4. C4 Credit 60% – 64%
5. C5 Credit 55% – 59%
6. C6 Credit 50% – 54%
7. D7 Pass 45% – 49%
8. E8 Pass 40% – 45%
9. F9 Failure 0% – 44%

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