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)

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Previous Year WAEC F/Maths Obj Answers.

1DCEBDBABAC.

11ABDDDCCDAC.

21CDBACAEBCB.

31CEAEEDEBBD

41CEBCDDCDDA

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

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

The 2023 Waec Further Mathematics Theory questions and answers are loading! 2023 Further Mathematics Essay answers Loading!! 2023 Waec Further Mathematics Theory Answers Loading!!! Kindly bookmark the website for the answers that will be released. Better still reload the site to check if the answers for the 2023 Waec Further Mathematics questions and answers have dropped.

Previous year F/Maths Theory Answers

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)

cosx=adj/hyp=12/13

then tanx=opp/adj=5/12

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 |

__WAEC Grading Percentage Scores__

- A1 Excellent 75% – 100%
- B2 Very good 70% – 74%
- B3 Good 65% – 69%
- C4 Credit 60% – 64%
- C5 Credit 55% – 59%
- C6 Credit 50% – 54%
- D7 Pass 45% – 49%
- E8 Pass 40% – 45%
- F9 Failure 0% – 44%

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