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Saturday, 22nd November, 2025
Mathematics (Essay) – 08:30 hrs – 11:00 hrs.
Mathematics (Objective) – 13:00 hrs – 14:30 hrs

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Waec GCE 2025 Mathematics Questions And Answers

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5) 3 < x < 8 and 8 < y < 15

The mean of the numbers is 7, so we can set up the equation:
(2 + 3 + x + 8 + y + 15) / 6 = 7

Combine like terms:
(28 + x + y) / 6 = 7

Multiply both sides by 6:
28 + x + y = 42

Subtract 28 from both sides:
x + y = 14

Since x < 8 and y > 8, let’s try to find a combination that works:
x = 6, y = 8 doesn’t work (y > 8)
x = 5, y = 9 works!

So, x = 5 and y = 9.

The product of x and y is:
xy = 5 × 9 = 45

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Waec GCE 2025 Mathematics Questions And Answers

(7a)
The volume of the cylinder is:
V = πr²h
= 3.14 × (5)² × 10
= 3.14 × 25 × 10
= 785 m³

The volume of the sphere is:
V = (4/3)πr³

Since the volumes are equal:
785 = (4/3) × 3.14 × r³
785 = 4.187 × r³

Divide both sides by 4.187:
r³ = 187.5
r = ∛187.5
r ≈ 5.72

The radius of the sphere is approximately 5.72 m.

(7bi)
The longitude difference between A and B is:
Δλ = distance / (R × cos(latitude))
= 1500 / (6400 × cos(60°))
= 1500 / (6400 × 0.5)
= 1500 / 3200
= 0.46875 radians

Convert radians to degrees:
Δλ = 0.46875 × (180/π)
= 0.46875 × (180/3.14)
≈ 26.86°

The longitude of B is:
λ_B = λ_A + Δλ
= 20°E + 26.86°E
≈ 46.86°E

(7bii)
The latitude difference between A and C is:
Δφ = 70° – 60°
= 10°

The longitude of A and C is the same (20°E), so Δλ = 0.

The great circle distance is:
d = R × √(Δφ² + (cos(φ_A) × Δλ)²)
= R × Δφ (since Δλ = 0)
= 6400 × (10 × π / 180)
= 6400 × (10 × 3.14 / 180)
≈ 1116.44 km

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Waec GCE 2025 Mathematics Questions And Answers

(8a)
The perimeter of a rhombus is 52 cm, so each side is:
52 / 4 = 13 cm

Let’s call the diagonals d1 and d2. We know d1 = 24 cm.

The diagonals of a rhombus bisect each other at right angles. Using the Pythagorean theorem:
(13)² = (d1/2)² + (d2/2)²
169 = (24/2)² + (d2/2)²
169 = 12² + (d2/2)²
169 = 144 + (d2/2)²
(d2/2)² = 25
d2/2 = 5
d2 = 10

The area of a rhombus is:
Area = (d1 × d2) / 2
= (24 × 10) / 2
= 120

The area of the rhombus is 120 cm².

(8b)
Let the integers be x and x + 2 (since they’re consecutive odd integers).

The sum of their squares is 290:
x² + (x + 2)² = 290
x² + x² + 4x + 4 = 290
2x² + 4x – 286 = 0
x² + 2x – 143 = 0

Factoring the quadratic equation:
(x + 13)(x – 11) = 0

x = -13 (not possible since x is positive)
x = 11

The integers are 11 and 13.

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Waec GCE 2025 Mathematics Questions And Answers

10ai)
Let’s define the events:
X: Team X qualifies
Y: Team Y qualifies

P(X) = 3/2 is not possible (probabilities can’t be greater than 1). Assuming it’s 3/4 and 3/4 for X and Y respectively doesn’t match the question, let’s assume P(X) = 2/3 and P(Y) = 3/4.

P(only one team qualifies) = P(X and not Y) + P(Y and not X)
= P(X) × P(not Y) + P(Y) × P(not X)
= (2/3) × (1/4) + (3/4) × (1/3)
= 2/12 + 3/12
= 5/12

(10aii)
P(at least one team qualifies) = 1 – P(neither team qualifies)
= 1 – P(not X and not Y)
= 1 – (1/3) × (1/4)
= 1 – 1/12
= 11/12

(10b)
First, find the mean:
Mean = (11 + 15 + 8 + 10 + 14 + 12) / 6
= 70 / 6
= 11.67

Next, find the deviations from the mean:
|11 – 11.67| = 0.67
|15 – 11.67| = 3.33
|8 – 11.67| = 3.67
|10 – 11.67| = 1.67
|14 – 11.67| = 2.33
|12 – 11.67| = 0.33

Mean deviation = (0.67 + 3.33 + 3.67 + 1.67 + 2.33 + 0.33) / 6
= 12 / 6
= 2