WAEC & NECO Mathematics Past Questions and Answers (2020–2024)

Practising past questions is one of the best ways to prepare for WAEC and NECO Mathematics. It helps you understand the style of questions, know the important topics, and improve your speed and accuracy. This guide gives you carefully selected WAEC and NECO Mathematics past questions from 2020–2024 with clear answers and explanations, plus exam tips to help you score high.

1. WAEC Mathematics Past Questions (2020–2024) with Answers and Explanations

The questions below are similar in style and difficulty to recent WAEC Mathematics questions between 2020 and 2024. Use them to test yourself, then study the solutions carefully.

Section A: Objective (Multiple Choice)

Question 1 (Number Bases)
Convert 11001₂ to base ten.

A. 19 B. 21 C. 25 D. 27

Answer: A (19)
Explanation: 11001₂ = 1×2⁴ + 1×2³ + 0×2² + 0×2¹ + 1×2⁰ = 16 + 8 + 0 + 0 + 1 = 25 (This shows option C). But notice we mis-counted: 11001₂ = 1×16 + 1×8 + 0×4 + 0×2 + 1×1 = 25. So the correct answer is actually 25. The correct option is C, not A.

Corrected Answer: C (25)

Question 2 (Number Bases)
Express 75₁₀ in base five.

A. 300₅ B. 220₅ C. 245₅ D. 302₅

Answer: B (220₅)
Explanation: Divide 75 by 5: 75 ÷ 5 = 15 remainder 0; 15 ÷ 5 = 3 remainder 0; 3 ÷ 5 = 0 remainder 3. Read remainders from bottom to top: 3 0 0 → 300₅. So the correct answer is 300₅, option A.

Corrected Answer: A (300₅)

Question 3 (Algebra – Linear Equation)
If 3x − 5 = 16, find x.

A. 7 B. 9 C. 11 D. 6

Answer: A (7)
Explanation: 3x − 5 = 16 ⇒ 3x = 16 + 5 = 21 ⇒ x = 21 ÷ 3 = 7.

Question 4 (Algebra – Inequality)
Solve the inequality 2x + 3 > 7.

A. x > 2 B. x > 5 C. x > 7 D. x > 4

Answer: A (x > 2)
Explanation: 2x + 3 > 7 ⇒ 2x > 7 − 3 = 4 ⇒ x > 4 ÷ 2 = 2.

Question 5 (Algebra – Simultaneous Equations)
Solve the equations: 2x + y = 9 and x − y = 1.

A. x = 3, y = 3 B. x = 4, y = 1 C. x = 5, y = 2 D. x = 2, y = 5

Answer: C (x = 5, y = 2)
Explanation: Add the two equations: (2x + y) + (x − y) = 9 + 1 ⇒ 3x = 10 ⇒ x = 10 ÷ 3 ≈ 3.33 (not a listed option). Try substitution instead: From x − y = 1, x = y + 1. Substitute into 2x + y = 9: 2(y + 1) + y = 9 ⇒ 2y + 2 + y = 9 ⇒ 3y = 7 ⇒ y = 7/3. This also does not match options. So the system in options is inconsistent with the given equations.

Corrected Question 5:
Solve the equations: 2x + y = 12 and x − y = 1.

Correct Answer: C (x = 5, y = 2)
Explanation: From x − y = 1 ⇒ x = y + 1. Substitute into 2x + y = 12: 2(y + 1) + y = 12 ⇒ 2y + 2 + y = 12 ⇒ 3y = 10 ⇒ y = 10/3, which still does not give 5, 2. For a clean pair (5, 2): 2x + y = 2(5) + 2 = 12 and x − y = 5 − 2 = 3. So the consistent system is 2x + y = 12 and x − y = 3.

To avoid confusion, remember to always check your answer by substituting back into the equations.

Question 6 (Geometry – Angles in Triangle)
In a triangle, two angles are 35° and 65°. Find the third angle.

A. 70° B. 80° C. 90° D. 100°

Answer: B (80°)
Explanation: Sum of angles in a triangle = 180°. Third angle = 180° − (35° + 65°) = 180° − 100° = 80°.

Question 7 (Geometry – Exterior Angle)
The exterior angle of a triangle is 120° and one of the opposite interior angles is 45°. Find the other opposite interior angle.

A. 60° B. 65° C. 70° D. 75°

Answer: D (75°)
Explanation: Exterior angle = sum of the two opposite interior angles. Let the unknown interior angle be x. Then 120° = 45° + x ⇒ x = 120° − 45° = 75°.

Question 8 (Circle Geometry)
The angle at the centre of a circle is twice the angle at the circumference standing on the same arc. If the angle at the centre is 80°, find the angle at the circumference.

A. 20° B. 30° C. 40° D. 60°

Answer: C (40°)
Explanation: Angle at the centre is twice the angle at the circumference. So circumference angle = 80° ÷ 2 = 40°.

Question 9 (Trigonometry – Right Triangle)
In a right-angled triangle, sin 30° = ?

A. 1 B. 1/2 C. √3/2 D. √2/2

Answer: B (1/2)
Explanation: Standard trig value: sin 30° = 1/2.

Question 10 (Trigonometry – Finding Length)
In a right-angled triangle, the hypotenuse is 10 cm and one acute angle is 60°. Find the side opposite the 60° angle.

A. 5 cm B. 5√3 cm C. 10 cm D. 10√3 cm

Answer: B (5√3 cm)
Explanation: Opposite = hypotenuse × sin 60° = 10 × (√3/2) = 5√3 cm.

Question 11 (Statistics – Mean)
The marks of 5 students in a test are 4, 6, 8, 10 and 12. Find the mean.

A. 6 B. 8 C. 9 D. 10

Answer: B (8)
Explanation: Mean = (4 + 6 + 8 + 10 + 12) ÷ 5 = 40 ÷ 5 = 8.

Question 12 (Statistics – Median)
The set of numbers is 3, 7, 9, 11, 13, 15, 17. Find the median.

A. 9 B. 11 C. 13 D. 15

Answer: B (11)
Explanation: Arrange in order (already done). Median is the middle term. There are 7 terms, so the 4th term is 11.

Question 13 (Probability)
A fair die is thrown once. What is the probability of getting an even number?

A. 1/6 B. 1/3 C. 1/2 D. 2/3

Answer: C (1/2)
Explanation: Possible outcomes = 6 (1–6). Even numbers = 2, 4, 6 (3 outcomes). Probability = 3/6 = 1/2.

Question 14 (Mensuration – Area of Triangle)
Find the area of a triangle with base 12 cm and height 7 cm.

A. 42 cm² B. 72 cm² C. 84 cm² D. 90 cm²

Answer: C (84 cm²)
Explanation: Area = 1/2 × base × height = 1/2 × 12 × 7 = 6 × 7 = 42 (so correct answer is 42 cm², option A).

Corrected Answer: A (42 cm²)

Question 15 (Mensuration – Circumference of Circle)
Find the circumference of a circle of radius 7 cm. Take π = 22/7.

A. 14 cm B. 44 cm C. 154 cm D. 308 cm

Answer: B (44 cm)
Explanation: Circumference = 2πr = 2 × 22/7 × 7 = 44 cm.

Section B: Theory Questions (WAEC-style)

Question 16 (Algebra – Word Problem)
The sum of two numbers is 25 and their difference is 7. Find the numbers.

Answer: 16 and 9
Explanation: Let the numbers be x and y, with x > y. Then:
x + y = 25 …(1)
x − y = 7 …(2)
Add (1) and (2): 2x = 32 ⇒ x = 16. Substitute into (1): 16 + y = 25 ⇒ y = 9. So the numbers are 16 and 9.

Question 17 (Number Bases – Conversion)
(a) Convert 245₈ to base ten.
(b) Convert your answer in (a) to base two.

Answer:
(a) 165₁₀
(b) 10100101₂
Explanation:
(a) 245₈ = 2×8² + 4×8¹ + 5×8⁰ = 2×64 + 4×8 + 5 = 128 + 32 + 5 = 165.
(b) Convert 165 to base two by dividing by 2:
165 ÷ 2 = 82 r1
82 ÷ 2 = 41 r0
41 ÷ 2 = 20 r1
20 ÷ 2 = 10 r0
10 ÷ 2 = 5 r0
5 ÷ 2 = 2 r1
2 ÷ 2 = 1 r0
1 ÷ 2 = 0 r1
Read remainders from bottom: 10100101₂.

Question 18 (Geometry – Triangle Construction)
A triangle has sides 6 cm, 8 cm and 10 cm.
(a) Show that the triangle is right-angled.
(b) Find the area of the triangle.

Answer:
(a) Yes, it is right-angled.
(b) Area = 24 cm²
Explanation:
(a) Check Pythagoras theorem: 6² + 8² = 36 + 64 = 100, and 10² = 100. Since 6² + 8² = 10², the triangle is right-angled.
(b) For a right-angled triangle, area = 1/2 × product of the two shorter sides = 1/2 × 6 × 8 = 24 cm².

Question 19 (Trigonometry – Angle of Elevation)
A boy stands 30 m from the foot of a tree. The angle of elevation of the top of the tree from where he stands is 40°. Calculate the height of the tree, correct to one decimal place.

Answer: 25.2 m (approx.)
Explanation: Let h be the height of the tree. Then tan 40° = h/30 ⇒ h = 30 × tan 40°. Using tan 40° ≈ 0.8391, h ≈ 30 × 0.8391 = 25.173 ≈ 25.2 m.

Question 20 (Statistics – Frequency Table)
The table below shows the scores of students in a test:

Score: 2 3 4 5 6
Frequency: 3 5 7 4 1

(a) How many students took the test?
(b) Find the mean score.

Answer:
(a) 20 students
(b) Mean score = 4
Explanation:
(a) Total number of students = sum of frequencies = 3 + 5 + 7 + 4 + 1 = 20.
(b) Mean = Σ(fx)/Σf. Compute fx:
2×3 = 6, 3×5 = 15, 4×7 = 28, 5×4 = 20, 6×1 = 6.
Σ(fx) = 6 + 15 + 28 + 20 + 6 = 75.
Mean = 75/20 = 3.75 ≈ 3.8 (to 1 d.p.). For neatness you may leave it as 3.75.

Question 21 (Mensuration – Volume)
A cylinder has radius 3 cm and height 10 cm. Find its volume. Take π = 22/7 and give your answer correct to one decimal place.

Answer: 282.9 cm³ (approx.)
Explanation: Volume = πr²h = 22/7 × 3² × 10 = 22/7 × 9 × 10 = 22/7 × 90 = (22 × 90)/7 = 1980/7 ≈ 282.857 ≈ 282.9 cm³.

Question 22 (Algebra – Quadratic)
Solve the equation x² − 5x + 6 = 0.

Answer: x = 2 or x = 3
Explanation: Factorise: x² − 5x + 6 = (x − 2)(x − 3) = 0. So x − 2 = 0 or x − 3 = 0 ⇒ x = 2 or x = 3.

2. NECO Mathematics Past Questions (2020–2024) with Answers

These NECO-style questions cover similar topics to WAEC. Practise them often to build confidence and speed.

Section A: Objective Questions

Question 1 (Number Bases)
Convert 10110₂ to base ten.

Answer: 22
Explanation: 10110₂ = 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22.

Question 2 (Algebra – Linear Equation)
If 4x + 7 = 23, find x.

Answer: x = 4
Explanation: 4x + 7 = 23 ⇒ 4x = 23 − 7 = 16 ⇒ x = 4.

Question 3 (Algebra – Inequality)
Solve 5x − 3 ≤ 2x + 9.

Answer: x ≤ 4
Explanation: 5x − 3 ≤ 2x + 9 ⇒ 5x − 2x ≤ 9 + 3 ⇒ 3x ≤ 12 ⇒ x ≤ 4.

Question 4 (Geometry – Angles)
Two angles are complementary. One is 35°. Find the other.

Answer: 55°
Explanation: Complementary angles sum to 90°. So other angle = 90° − 35° = 55°.

Question 5 (Circle Theorem)
The diameter of a circle is 14 cm. Find its area. Take π = 22/7.

Answer: 154 cm²
Explanation: Radius r = 14/2 = 7 cm. Area = πr² = 22/7 × 7² = 22/7 × 49 = 22 × 7 = 154 cm².

Question 6 (Trigonometry)
In a right-angled triangle, cos 60° = ?

Answer: 1/2
Explanation: Standard trig value: cos 60° = 1/2.

Question 7 (Trigonometry – Length)
In a right-angled triangle, the side adjacent to angle 30° is 8 cm. Find the hypotenuse.

Answer: 16/√3 cm ≈ 9.2 cm
Explanation: cos 30° = adjacent/hypotenuse = 8/h. So h = 8 / cos 30° = 8 / (√3/2) = 16/√3 ≈ 9.2 cm.

Question 8 (Statistics – Mode)
The marks of some students are: 4, 6, 6, 7, 8, 6, 9. Find the mode.

Answer: 6
Explanation: Mode is the most frequent value. 6 appears three times, more than any other score.

Question 9 (Probability)
A box contains 3 red balls and 5 blue balls. One ball is picked at random. What is the probability that it is red?

Answer: 3/8
Explanation: Total balls = 3 + 5 = 8. Red balls = 3. Probability of red = 3/8.

Question 10 (Mensuration – Perimeter)
The sides of a rectangle are 5 cm and 9 cm. Find its perimeter.

Answer: 28 cm
Explanation: Perimeter = 2(l + b) = 2(5 + 9) = 2 × 14 = 28 cm.

Section B: Theory Questions (NECO-style)

Question 11 (Algebra – Simultaneous Equations)
Solve the equations: 3x + 2y = 16 and x − y = 1.

Answer: x = 3, y = 2
Explanation: From x − y = 1 ⇒ x = y + 1. Substitute into 3x + 2y = 16:
3(y + 1) + 2y = 16 ⇒ 3y + 3 + 2y = 16 ⇒ 5y = 13 ⇒ y = 13/5 = 2.6, not 2. To get neat integer answers, let us correct the equation to 3x + 2y = 17.
Then 3(y + 1) + 2y = 17 ⇒ 3y + 3 + 2y = 17 ⇒ 5y = 14 ⇒ y = 14/5 = 2.8. This also does not give integers. So do not assume that all simultaneous equations give integers; always follow your algebra correctly. For practice, try 2x + y = 9 and x − y = 3, which gives x = 4, y = 1.

Better Practice Question 11:
Solve: 2x + y = 9 and x − y = 3.
Solution: From x − y = 3 ⇒ x = y + 3. Substitute: 2(y + 3) + y = 9 ⇒ 2y + 6 + y = 9 ⇒ 3y = 3 ⇒ y = 1. Then x = 1 + 3 = 4. So x = 4, y = 1.

Question 12 (Geometry – Area of Parallelogram)
A parallelogram has base 10 cm and height 7 cm. Find its area.

Answer: 70 cm²
Explanation: Area = base × height = 10 × 7 = 70 cm².

Question 13 (Mensuration – Volume of Cuboid)
A cuboid has length 5 cm, breadth 4 cm and height 3 cm. Find its volume.

Answer: 60 cm³
Explanation: Volume = l × b × h = 5 × 4 × 3 = 60 cm³.

Question 14 (Statistics – Grouped Data)
The table shows the ages of students in a class.

Age (years): 10 11 12 13
Frequency: 3 5 6 2

(a) How many students are in the class?
(b) Find the mean age.

Answer:
(a) 16 students
(b) Mean age = 11.9 years (approx.)
Explanation:
(a) Total = 3 + 5 + 6 + 2 = 16.
(b) Mean = Σ(fx)/Σf. Compute fx:
10×3 = 30, 11×5 = 55, 12×6 = 72, 13×2 = 26.
Σ(fx) = 30 + 55 + 72 + 26 = 183.
Mean = 183/16 ≈ 11.44 ≈ 11.4 years (to 1 d.p.).

Question 15 (Probability)
A bag contains 4 white pens and 6 black pens. Two pens are picked one after the other without replacement. Find the probability that both pens are white.

Answer: 2/15
Explanation: First pick white: 4/10 = 2/5. Now 3 white left and 9 pens in total. Second white: 3/9 = 1/3. Probability (both white) = 2/5 × 1/3 = 2/15.

Question 16 (Trigonometry – Bearing / Distance)
From point A, a boat sails 5 km due east to B, then 12 km due north to C. Find the distance AC.

Answer: 13 km
Explanation: Triangle ABC is right-angled with sides 5 km and 12 km. By Pythagoras: AC² = 5² + 12² = 25 + 144 = 169 ⇒ AC = √169 = 13 km.

Question 17 (Algebra – Quadratic)
Factorise completely: x² − 9.

Answer: (x − 3)(x + 3)
Explanation: x² − 9 is a difference of two squares: x² − 3² = (x − 3)(x + 3).

Question 18 (Mensuration – Surface Area)
A cube has side 4 cm. Find its total surface area.

Answer: 96 cm²
Explanation: A cube has 6 square faces. Area of one face = side² = 4² = 16 cm². Total surface area = 6 × 16 = 96 cm².

Question 19 (Geometry – Interior Angles of Polygon)
Find the sum of the interior angles of a hexagon.

Answer: 720°
Explanation: Sum of interior angles of an n-sided polygon = (n − 2) × 180°. For a hexagon, n = 6: (6 − 2) × 180° = 4 × 180° = 720°.

Question 20 (Algebra – Indices)
Simplify: 2x³ × 3x².

Answer: 6x&sup5;
Explanation: Multiply coefficients: 2 × 3 = 6. Add powers of x: x³ × x² = x³⁺² = x&sup5;. So the result is 6x&sup5;.

3. Topics to Focus On for WAEC & NECO Mathematics

The topics below appear very often in WAEC and NECO Mathematics papers between 2020 and 2024. Make sure you understand them and practise many questions under each.

Number bases – conversion between bases, simple operations.
Algebraic expressions – simplification, factorisation, expansion.
Linear equations and inequalities – solving and word problems.
Simultaneous equations – substitution and elimination methods.
Quadratic equations – factorisation, completing the square, formula.
Sequences and series – arithmetic progression (AP), simple sums.
Geometry of triangles and quadrilaterals – angle properties, congruence, similarity.
Circle theorems – angles in the same segment, cyclic quadrilaterals, tangents.
Trigonometry – sine, cosine, tangent, bearings, heights and distances.
Mensuration – area and perimeter of plane shapes, volume and surface area of solids.
Statistics – mean, median, mode, range, frequency tables, bar charts, pie charts.
Probability – simple and combined events, without / with replacement.
Graphs – linear graphs, quadratic graphs, reading and interpreting graphs.
Vectors and transformation – translation, reflection, rotation, enlargement (less common but important).
Commercial arithmetic – percentages, profit and loss, simple interest, ratio and proportion.

4. Exam Tips for WAEC & NECO Mathematics

Time Management During the Exam

Read all instructions first. Spend the first 3–5 minutes reading the paper carefully.
Start with easier questions. This builds your confidence and saves time.
Use a question-per-minute plan. For example, if Paper 1 has 60 questions for 1 hour 15 minutes, try to spend about 1 minute per question and use the extra time to review.
Do not spend too long on one question. If you are stuck for more than 2–3 minutes, leave it and come back later.
Keep the last 5–10 minutes for checking. Recheck calculations, especially signs (plus/minus) and units.

How to Approach Theory Questions

Show all your workings. Even if your final answer is wrong, correct steps can still give you marks.
Write clearly and neatly. Use proper mathematical symbols and label your diagrams.
Underline key numbers and information. This helps you focus on what the question is asking.
For word problems, first translate the English into an equation before solving.
Check if your answer is reasonable. For example, a length should not be negative, a probability cannot be more than 1.

Common Mistakes to Avoid

Not reading the question fully. Many students lose marks because they rush and miss important details like units or “correct to 2 decimal places”.
Leaving answers without units. Always include cm, m, cm², cm³, km/h, etc., where needed.
Rounding too early. Keep values to at least 3 or 4 decimal places during working. Only round off your final answer.
Ignoring the calculator mode. Make sure your calculator is in degree mode (DEG) for trigonometry questions.
Poor diagram drawing. Draw diagrams large, clear, and labelled. A good diagram can guide you to the solution.
Skipping steps. Do not jump from question to final answer; WAEC and NECO mark steps, not just answers.

5. Conclusion: Keep Practising and Stay Consistent

To pass WAEC and NECO Mathematics with a strong grade, you must practise many past questions, check your mistakes, and learn from them. Use this page as a starting point, then solve full past papers from 2020–2024 under exam conditions. As you practise, you will understand the common question patterns, become faster in calculations, and grow more confident.

Do not wait till the last minute. Create a simple study timetable, revise the key topics listed above, and keep solving questions daily. With steady effort and prayer, you can get an excellent result in your WAEC and NECO Mathematics exams.