WAEC Futher Mathematics (Essay & Obj) Questions and Answers 2026/2027

WAEC Futher Mathematics (Essay & Obj) Questions and Answers 2026/2027 :WAEC Further Mathematics exam (WASSCE) evaluates advanced mathematical concepts across two papers. Paper 1 is a 1.5-hour objective test with 40 questions, while Paper 2 is a 2.5-hour essay section. Both sections span Pure Mathematics, Statistics, Probability, Vectors, and Mechanics.

WAEC Futher Mathematics (Essay and Obj) Questions and Answers 2026/2027

FURTHER MATHS
01-10: CABDCADACC
11-20: ABCDDDBDDC
21-30: DBCCBCDDCC
31-40: AACABBDBAC

COMPLETED

(6a)
From x:y = 1:2 and y:z = 1:3, combining gives x:y:z = 1:2:6
Total parts = 1 + 2 + 6 = 9
Number of z marbles = 6/9 × 27 = 18

(6b) Number of y marbles = 2/9 × 27 = 6
P(first y) = 6/27, P(second y) = 5/26
P(both y) = 6/27 × 5/26 = 30/702 = 5/117

(7)
m·n = |m||n|cosθ
m·n = (2)(−3) + (p)(4) = −6 + 4p
|m| = √(4 + p²), |n| = √(9 + 16) = √13
So: (−6 + 4p) / (√(4 + p²) · √13) = 6/(5√13)
−6 + 4p = 6√(4 + p²)/5
5(−6 + 4p) = 6√(4 + p²)
(−30 + 20p)² = 36(4 + p²)
900 − 1200p + 400p² = 144 + 36p²
364p² − 1200p + 756 = 0
91p² − 300p + 189 = 0
p = (300 ± √(90000 − 68796))/182 = (300 ± √21204)/182 = (300 ± 145.6)/182
p = 2.45 or p = 0.849

(8)
Mass = 12 kg, θ = 30°, μ = 2/3, g = 10
Normal reaction: N = 12g cos30° = 120 × (√3/2) = 60√3 N
Friction force: F = μN = (2/3)(60√3) = 40√3 N
Weight component along plane = 12 × 10 × sin30° = 60 N

(a) Moving down the plane: P acts down, friction acts up
At point of moving down: P + 60 = 40√3
P = 40√3 − 60 = 69.28 − 60 = 9.28 N

(b) Moving up the plane: P acts up, friction acts down
P = 60 + 40√3 = 60 + 69.28 = 129.28 N

(9)
Total: 7 mid-fielders, 5 attackers, 8 referees. Committee of 6, each discipline represented.

(a) Exactly 1 attacker, 1 mid-fielder → remaining 4 from referees:
C(5,1) × C(7,1) × C(8,4) = 5 × 7 × 70 = 2450

(b) Exactly 2 referees → remaining 4 split between attackers and mid-fielders (at least 1 each):
C(8,2) × [C(5,1)×C(7,3) + C(5,2)×C(7,2) + C(5,3)×C(7,1)]
= 28 × [5×35 + 10×21 + 10×7]
= 28 × [175 + 210 + 70]
= 28 × 455 = 12740

(c) At least 2 attackers. Cases: 2, 3, 4, or 5 attackers (with at least 1 each from other groups):
C(5,2)×C(7,1)×C(8,3) + C(5,2)×C(7,2)×C(8,2) + C(5,2)×C(7,3)×C(8,1) + C(5,3)×C(7,1)×C(8,2) + C(5,3)×C(7,2)×C(8,1) + C(5,4)×C(7,1)×C(8,1) + C(5,5)×C(7,1)×C(8,0)…
= 10×7×56 + 10×21×28 + 10×35×8 + 10×7×28 + 10×21×8 + 5×7×8 + 1×7×1
= 3920 + 5880 + 2800 + 1960 + 1680 + 280 + 7 = 16527

(10b)
p log₃x = log₈₁x
log₈₁x = log₃x / log₃81 = log₃x / 4
So p log₃x = (1/4) log₃x → p = 1/4

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