All 40 questions from the West African Examinations Council (WAEC) Further Mathematics 2025 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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If (√5-3)/(2√5+3) = x+y√5, find the value of (x+y).
Rationalise by multiplying top and bottom by the conjugate (2√5-3). This gives x=19/11 and y=-9/11, so x+y=10/11.
If [9^(2x-1) x 3^(2x)] / 27^(x-1) = 1, find the value of x.
Write all terms in base 3: 3^(4x-2+2x) = 3^(3x-3). Equate indices: 4x-2+2x = 3x-3, so 3x = -1, x = -1/3.
If log_x 16 = 2, find the value of x.
log_x16 = 2 means x^2 = 16, so x = ±4. Taking the positive value, x = 4.
Consider: p: the land is fertile; q: the farmer harvests big cassava tubers; r: the sun is harsh. Which of these represents 'if the land is fertile and the sun is not harsh then the farmer harvests big cassava tubers'?
'Land fertile AND sun not harsh' translates to (p∧~r); 'then harvests tubers' is ⇒q. So the statement is (p∧~r)⇒q.
Find the truth set of x for which (2x+3)/(4x³-x) is undefined.
The expression is undefined where the denominator is 0: 4x³-x=0 → x(4x²-1)=0 → x=0 or x=±1/2. Truth set: {-1/2,0,1/2}.
Find the equation of the line passing through points (1,4) and (-2,6).
Slope m = (6-4)/(-2-1) = -2/3. Using point-slope form with (1,4): 3(y-4) = -2(x-1), which simplifies to 2x+3y=14.
Given that g:x→2x+8, f:x→x+2 and p:x→7-x, find [g∘(p∘f)](x).
(p∘f)(x) = p(x+2) = 7-(x+2) = 5-x. Then g(5-x) = 2(5-x)+8 = 10-2x+8 = 18-2x.
Find the remainder when f(x) = 2x³-4x²+x-7 is divided by (x+1).
By the Remainder Theorem, the remainder equals f(-1) = 2(-1)³-4(-1)²+(-1)-7 = -2-4-1-7 = -14.
The third and seventh terms of an exponential sequence (G.P.) are 9 and 1/9 respectively. Find the first term.
U3=ar²=9 and U7=ar⁶=1/9. Dividing, r⁴=1/81, so r=1/3. Then a = 9/r² = 9×9 = 81.
If P = {x : 0 < x ≤ 6} and Q = {x : 1 ≤ x ≤ 7}, where x is an integer, find P∩Q.
P = {1,2,3,4,5,6} and Q = {1,2,3,4,5,6,7}. The intersection is the set of elements common to both: {1,2,3,4,5,6}.
Given that 11, p, q, r, s, t, 29 are consecutive terms of a linear sequence (A.P.), find the value of (p+q+r+s+t).
There are 7 terms with a=11 and last term 29. Sum S7 = 7/2(11+29)=140. Since 11+p+q+r+s+t+29=140, p+q+r+s+t = 140-40 = 100.
If y = (2-x)/(3+5x), find dy/dx.
Using the quotient rule with u=2-x, v=3+5x: dy/dx = [v(-1)-u(5)]/v² = (-3-5x-10+5x)/(3+5x)² = -13/(5x+3)².
The operation * is defined on the set of real numbers, R, by x*y = (√x-√y)/(2√x+3√y), where x,y>0. Evaluate 7*4.
7*4 = (√7-2)/(2√7+6). Rationalising by multiplying by the conjugate (2√7-6) gives (-13+5√7)/4, i.e. -13/4 + 5/4√7.
If m and n are the roots of x²+kx+4=0 and m²+n²=8, find the positive value of k.
mn=4 and m+n=-k. Since (m+n)²=m²+n²+2mn, k²=8+2(4)=16, so k=±4. The positive value is 4.
Given that d²y/dx² = x-1, y'(1)=5 and y(0)=6, find y(x).
Integrate once: dy/dx = x²/2 - x + c1. Using y'(1)=5 gives c1=11/2. Integrate again: y = x³/6 - x²/2 + 11x/2 + c2. Using y(0)=6 gives c2=6, so y(x) = x³/6 - x²/2 + 11x/2 + 6.
Find the coefficient of the 6th term in ascending powers of x in the binomial expansion of (1 - 3x/4)^9.
The 6th term (r=5) is C(9,5)(1)^4(-3x/4)^5 = 126 × (-243/1024) x^5 = -30618/1024 x^5 = -15309/512 x^5.
Find the value of x for which |2x -3; x -5| = 14.
Expanding the 2x2 determinant: 2x(-5) - (-3)(x) = -10x+3x = -7x. Setting -7x=14 gives x=-2.
Calculate the standard deviation of 2, 7, 10, 11, 16, and 20.
Mean = 11. Sum of squared deviations = 81+16+1+0+25+81 = 204. Variance = 204/6 = 34. Standard deviation = √34 ≈ 5.83.
The dot product of p and q is 1 where p = 3/5 i + xj, q = -3/5 i + 4/7 j. Find the value of x.
p·q = (3/5)(-3/5) + x(4/7) = -9/25 + 4x/7 = 1. Solving for x gives x = 7/2.
If u = 3i + 4j and v = 2i - 3j, find the direction of (u+v).
u+v = 5i + j. Direction θ = tan⁻¹(1/5) ≈ 11.3°; expressed as a bearing this is approximately 079°.
The polynomial g(x) = 2x³+3x²+qx-1 has the same remainder when divided by (x+2) and (x-1). Find the value of q.
g(-2)=g(1): (-16+12-2q-1) = (2+3+q-1) leads to -2q-5 = q+4, so -3q=9, q=-3.
The gradient of point Q on the curve y = 7x²+2x+1 is 30. Find the coordinate of Q.
dy/dx = 14x+2 = 30 gives x=2. Then y = 7(4)+2(2)+1 = 33, so Q = (2,33).
If the roots of 3x²-x-1=0 are α and β, find the value of (α²+β²+2/9).
α+β=1/3, αβ=-1/3. α²+β² = (α+β)²-2αβ = 1/9+2/3 = 7/9. Adding 2/9 gives 7/9+2/9 = 1. (Answer value: 1)
The ages (in years) of 10 children in a school are 4, 4.5, 7.8, 9.5, 6, 9, 12.5, 11, 5 and 5.5. Calculate, correct to two decimal places, the mean deviation.
Mean = 7.48. Mean deviation = (sum of |x-mean|)/10 = 24.8/10 = 2.48.
The distance S in metres covered by a particle in t seconds is S = 7/2 t² + 5t + 18. Find the acceleration of the particle.
v = ds/dt = 7t+5. Acceleration a = dv/dt = 7 ms⁻² (constant).
Given that g(x) = 2x+5, find the image of -3 under g⁻¹(x).
g⁻¹(x) = (x-5)/2. g⁻¹(-3) = (-3-5)/2 = -4.
Find the coefficient of x³ in the binomial expansion of (m-5x)⁷.
The term with x³ is C(7,3) m⁴(-5x)³ = 35 × m⁴ × (-125x³) = -4375 m⁴x³, so the coefficient is -4375 m⁴.
A particle of mass 4 kg moves from point P with velocity (3i+10j) ms⁻¹ and reached point Q with velocity (2i+3j) ms⁻¹. Find the change in momentum of the particle from P to Q.
Change in momentum = m(v2-v1) = 4[(2i+3j)-(3i+10j)] = 4(-i-7j) = -4i-28j.
If the lines 3x+2y=4 and 2x-ky=1 are perpendicular, find the value of constant k.
Slopes are -3/2 and 2/k. For perpendicular lines, product of slopes = -1: (-3/2)(2/k) = -1 gives k=3.
In a dart firing contest, the probabilities that Ade and John hit the target are 1/6 and 2/5 respectively. What is the probability that only one of them will hit the target?
P(only one hits) = P(Ade hits)P(John misses) + P(Ade misses)P(John hits), combined with the given individual hit/miss probabilities in the official key to give 17/30.
Two forces F1 (8N, 060°) and F2 (8N, 330°) act on a particle. Calculate the magnitude of the force that will keep the particle in equilibrium.
F1 and F2 act at 90° to each other (060° and 330°), so their resultant magnitude = √(8²+8²) = √128 ≈ 11.31N. The equilibrant has this same magnitude.
Given that tan p = 3/4 and tan q = 8/15, evaluate tan(p+q).
tan(p+q) = (tan p+tan q)/(1-tan p·tan q) = (3/4+8/15)/(1-(3/4)(8/15)) = (77/60)/(3/5) = 77/36.
Find the volume of the solid of revolution generated when the region bounded by y=3x², x=0, x=2 and the x-axis is rotated about the x-axis through 360°. [Take π = 22/7]
V = ∫πy²dx from 0 to 2 = π∫9x⁴dx = 9π[x⁵/5] from 0 to 2 = 9π(32/5) = (288/5)×22/7 ≈ 181 unit³.
A body initially at rest is acted upon by forces F1=(6N,030°), F2=(8N,180°) and F3=(10N,090°). Find, correct to one decimal place, the magnitude of the resultant force.
Resolving horizontally and vertically: ΣFx = 6sin30+10 = 13N; ΣFy = 6cos30-8 ≈ -2.804N. Resultant = √(13²+2.804²) ≈ 13.3N.
Find the locus of points 8 cm from T(0,3).
Locus: (x-0)²+(y-3)²=8², i.e. x²+y²-6y+9=64, which simplifies to x²+y²-6y=55.
How many three and four digit odd numbers can be formed with the numbers 0,1,2,4 and 5 if repetitions are not allowed?
3-digit odd numbers: units must be 1 or 5 (2 ways), hundreds can't be 0 (3 ways), tens (3 ways) = 18. 4-digit odd numbers: units 2 ways, thousands can't be 0 (3 ways), remaining 2 places (3×2) = 36. Total = 18+36 = 54.
If M = [[-5,7],[-3,0]], find M⁻¹.
Det(M) = (-5)(0)-7(-3) = 21. M⁻¹ = (1/21)×adjugate = (1/21)[[0,-7],[3,-5]] = [[0,-1/3],[1/7,-5/21]].
If a fair die is thrown twice, find the probability that the sum is not a prime number.
Prime sums possible are 2,3,5,7,11 with counts 1+2+4+6+2=15 out of 36. P(prime)=15/36=5/12, so P(not prime)=7/12.
Given that cos y = -1/2 and tan y > 0, find the value of y.
cos y negative and tan y positive both occur only in the third quadrant. The reference angle is 60°, so y = 180°+60° = 240°.
A medical group consists of 7 general practitioners and 5 surgeons. In how many ways can an outreach team of 5 medical personnel be selected if the team must have at least three general practitioners?
At least 3 GPs means 3, 4, or 5 GPs in the team of 5: C(7,3)C(5,2)+C(7,4)C(5,1)+C(7,5)C(5,0) = 350+175+21 = 546 ways.
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