WAEC Mathematics Past Questions and Answers PDF Download

2023 WAEC Mathematics Past Questions and Answers: Your Ultimate Guide to Success

Introduction

Are you preparing for the upcoming WAEC Mathematics exam and searching for reliable past questions and answers in PDF format? You’ve come to the right place! In this post, we provide a comprehensive collection of WAEC Mathematics past questions and answers to help you study effectively and ace your exams with confidence. This article includes free Reading and download links for the most recent and updated WAEC Mathematics past questions, answers, and exam tips.

Why WAEC Mathematics Past Questions are Important

WAEC (West African Examinations Council) is a crucial exam for students in West Africa, and Mathematics is one of the core subjects. Preparing with WAEC past questions can give you insights into the exam format, types of questions, and key topics to focus on. This boosts your confidence and enhances your chances of scoring higher marks.

Here are a few reasons why using past questions is beneficial:

• Familiarity with Question Patterns: Understand how questions are structured.
• Identify Key Topics: Focus on frequently asked topics.
• Time Management Practice: Practice working within the exam’s time limits.
• Test Your Knowledge: Gauge your readiness and identify areas for improvement.

Step-by-Step solutions to 2023 WAEC Mathematics Past Questions for Free

1. Evaluate, correct to three decimal places:

\[ 4.314 \times 0.000056 \div 0.0067 \]

A. 0.361

B. 0.036

C. 0.037

D. 0.004

2023 WAEC Mathematics past questions and answers

Q1. Evaluate, correct to three decimal places: \[ 4.314 \times 0.000056 \div 0.0067 \] Step 1: Multiply the two numbers first. \[ 4.314 \times 0.000056 = 0.000241584 \] Step 2: Now, divide the result by 0.0067. \[ 0.000241584 \div 0.0067 = 0.036 \] So, the correct answer is B. 0.036

2023 WAEC Mathematics past questions and answers

2. There are 30 students in a class. 15 study woodwork and 13 study metalwork. 6 study neither of the two subjects. How many students study woodwork but not metalwork?
A. 5

B. 9

C. 11

D. 13

2023 WAEC Mathematics past questions and answers

Q2. There are 30 students in a class. 15 study woodwork and 13 study metalwork. 6 study neither of the two subjects. How many students study woodwork but not metalwork?
Step 1: Find the number of students studying either subject.

Since 6 students study neither, the remaining number of students studying either woodwork or metalwork is:
\[ 30 – 6 = 24 \]
Step 2: Use the inclusion-exclusion principle. Let \( W \) be the set of students who study woodwork, and \( M \) be the set of students who study metalwork. We are given: \[ |W| = 15, \quad |M| = 13, \quad |W \cap M| = x \] The total number of students studying woodwork or metalwork is 24, so: \[ |W \cup M| = |W| + |M| – |W \cap M| = 24 \] \[ 15 + 13 – x = 24 \] \[ 28 – x = 24 \quad \Rightarrow \quad x = 4 \] Step 3: Find how many study only woodwork. The number of students studying only woodwork is: \[ |W| – |W \cap M| = 15 – 4 = 11 \] So, the correct answer is C. 11.

2023 WAEC Mathematics past questions and answers

3. solve:
\[ \frac{2^{5x}}{2^x} = \sqrt[5]{2^{10}} \]

A. \( \frac{1}{3} \)

B. \( \frac{1}{5} \)

C. \( \frac{1}{2} \)

D. \( \frac{1}{4} \)

2023 WAEC Mathematics past questions and answers

We are given the equation: \[ \frac{2^{5x}}{2^x} = \sqrt[5]{2^{10}} \] Step 1: Simplify the left-hand side Using the exponent rule \(\frac{a^m}{a^n} = a^{m-n}\), we can simplify the left-hand side: \[ \frac{2^{5x}}{2^x} = 2^{5x – x} = 2^{4x} \] So, the equation becomes: \[ 2^{4x} = \sqrt[5]{2^{10}} \] Step 2: Simplify the right-hand side The fifth root of \(2^{10}\) can be written as: \[ \sqrt[5]{2^{10}} = 2^{\frac{10}{5}} = 2^2 = 4 \] So, the equation now becomes: \[ 2^{4x} = 4 \] Step 3: Express \(4\) as a power of \(2\) We know that \(4 = 2^2\), so the equation becomes: \[ 2^{4x} = 2^2 \] Step 4: Set the exponents equal Since the bases are the same, we can set the exponents equal: \[ 4x = 2 \] Step 5: Solve for \(x\) Now, divide both sides by 4: \[ x = \frac{2}{4} = \frac{1}{2} \] Final Answer: The solution is \(x = \frac{1}{2}\).

2023 WAEC Mathematics past questions and answers

4. solve: \( 1 + \sqrt[3]{x – 3} = 4 \)
A. 6

B. 12

C. 30

D. 66

2023 WAEC Mathematics past questions and answers

We are given the equation: \[ 1 + \sqrt[3]{x – 3} = 4 \] Step 1: Isolate the cube root term First, subtract 1 from both sides to isolate the cube root: \[ \sqrt[3]{x – 3} = 4 – 1 \] \[ \sqrt[3]{x – 3} = 3 \] Step 2: Cube both sides Now, cube both sides of the equation to eliminate the cube root: \[ (\sqrt[3]{x – 3})^3 = 3^3 \] \[ x – 3 = 27 \] Step 3: Solve for \(x\) Now, add 3 to both sides: \[ x = 27 + 3 \] \[ x = 30 \] Final Answer: The solution is \(x = 30\).

2023 WAEC Mathematics past questions and answers

5. Express 413\(_{10}\) in base 5.
A. 1131\(_5\)

B. 1311\(_5\)

C. 2311\(_5\)

D. 2132\(_5\)

2023 WAEC Mathematics past questions and answers

Q5.Express 413\(_{10}\) in base 5.
Step 1: Divide 413 by 5 and record the remainder.
\[ 413 \div 5 = 82 \, \text{remainder} \, 3 \] Step 2: Divide the quotient 82 by 5. \[ 82 \div 5 = 16 \, \text{remainder} \, 2 \] Step 3: Divide the quotient 16 by 5. \[ 16 \div 5 = 3 \, \text{remainder} \, 1 \] Step 4: Finally, divide the quotient 3 by 5. \[ 3 \div 5 = 0 \, \text{remainder} \, 3 \] Now, read the remainders from bottom to top: \[ 413_{10} = 1313_5 \] So, the correct answer is B. 1311\(_5\).

2023 WAEC Mathematics past questions and answers

6. Solve: \( \log_3 x + \log_3 (x – 8) = 2 \)
A. 6

B. 7

C. 8

D. 9

2023 WAEC Mathematics past questions and answers

We are given the equation: \[ \log_3{x} + \log_3{(x – 8)} = 2 \] Step 1: Use the logarithm product rule Using the logarithmic property \(\log_b{A} + \log_b{B} = \log_b{(A \cdot B)}\), we can combine the two logarithms: \[ \log_3{[x(x – 8)]} = 2 \] This simplifies to: \[ \log_3{(x^2 – 8x)} = 2 \] Step 2: Convert the logarithmic equation to an exponential equation Recall that \(\log_b{A} = C\) is equivalent to \(A = b^C\). Using this property, we can rewrite the equation as: \[ x^2 – 8x = 3^2 \] \[ x^2 – 8x = 9 \] Step 3: Rearrange the equation Subtract 9 from both sides to set the equation to zero: \[ x^2 – 8x – 9 = 0 \] Step 4: Solve the quadratic equation We will solve the quadratic equation \(x^2 – 8x – 9 = 0\) using the quadratic formula: \[ x = \frac{-(-8) \pm \sqrt{(-8)^2 – 4(1)(-9)}}{2(1)} \] \[ x = \frac{8 \pm \sqrt{64 + 36}}{2} \] \[ x = \frac{8 \pm \sqrt{100}}{2} \] \[ x = \frac{8 \pm 10}{2} \] Step 5: Find the solutions Now we solve for \(x\): \[ x = \frac{8 + 10}{2} = \frac{18}{2} = 9 \] \[ x = \frac{8 – 10}{2} = \frac{-2}{2} = -1 \] Step 6: Check for extraneous solutions Since \(\log_3(x)\) is undefined for non-positive values of \(x\), we must reject \(x = -1\). Thus, the only valid solution is: \[ x = 9 \] Final Answer: The solution is \(x = 9\).

2023 WAEC Mathematics past questions and answers

7. Mr. Manu is 4 times as old as his son, Adu. 7 years ago, the sum of their ages was 76 years. How old is Adu?
A. 12 years

B. 15 years

C. 18 years

D. 22 years

2023 WAEC Mathematics past questions and answers

Q7. Mr. Manu is 4 times as old as his son, Adu. 7 years ago, the sum of their ages was 76 years. How old is Adu?
Let Adu’s age be \( x \). Then, Mr. Manu’s age is \( 4x \).

Step 1: Write the equation for 7 years ago.
\[ (x – 7) + (4x – 7) = 76 \] Step 2: Simplify and solve for \( x \). \[ x – 7 + 4x – 7 = 76 \] \[ 5x – 14 = 76 \] \[ 5x = 90 \quad \Rightarrow \quad x = 18 \] So, Adu is 18 years old, and the correct answer is C. 18 years.

2023 WAEC Mathematics past questions and answers

8. Factorize completely: \( x^2 – (y + z)^2 \)
A. \( (x – y + z)(x – y – z) \)

B. \( (x + y + z)(x – y – z) \)

C. \( (x + y + z)(x + y – z) \)

D. \( (x – y – z)(x – y – z) \)

2023 WAEC Mathematics past questions and answers

Q8.Factorize completely: \[ x^2 – (y + z)^2 \] This is a difference of squares. The difference of squares formula is: \[ a^2 – b^2 = (a + b)(a – b) \] Step 1: Apply the formula. Here, \( a = x \) and \( b = y + z \), so: \[ x^2 – (y + z)^2 = (x + (y + z))(x – (y + z)) \] \[ = (x + y + z)(x – y – z) \] So, the correct answer is B. \( (x + y + z)(x – y – z) \).

2023 WAEC Mathematics past questions and answers

9. Find the roots of the quadratic equation: \[ 3m^2 – 2m – 65 = 0 \]
A. \( \left( \frac{13}{3}, 5 \right) \)

B. \( \left( -\frac{13}{3}, -5 \right) \)

C. \( \left( -\frac{13}{3}, 5 \right) \)

D. \( \left( \frac{13}{3}, -5 \right) \)

2023 WAEC Mathematics past questions and answers

Q9.Find the roots of the quadratic equation: \[ 3m^2 – 2m – 65 = 0 \] Step 1: Use the quadratic formula. The quadratic formula is: \[ m = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \] Here, \( a = 3 \), \( b = -2 \), and \( c = -65 \). Step 2: Calculate the discriminant. \[ b^2 – 4ac = (-2)^2 – 4(3)(-65) = 4 + 780 = 784 \] Step 3: Find the roots. \[ m = \frac{2 \pm \sqrt{784}}{2(3)} = \frac{2 \pm 28}{6} \] Thus, the two solutions are: \[ m = \frac{2 + 28}{6} = \frac{30}{6} = 5 \quad \text{and} \quad m = \frac{2 – 28}{6} = \frac{-26}{6} = -\frac{13}{3} \] So, the correct answer is C. \( \left( -\frac{13}{3}, 5 \right) \).

2023 WAEC Mathematics past questions and answers

10. \( M \) varies jointly as the square of \( n \) and the square root of \( q \). If \( M = 24 \) when \( n = 2 \) and \( q = 4 \), find \( M \) when \( n = 5 \) and \( q = 9 \).
A. 288

B. 400

C. 300

D. 225

2023 WAEC Mathematics past questions and answers

Q10.** \( M \) varies jointly as the square of \( n \) and the square root of \( q \). If \( M = 24 \) when \( n = 2 \) and \( q = 4 \), find \( M \) when \( n = 5 \) and \( q = 9 \).
step 1: Set up the equation for joint variation.
\[ M = k n^2 \sqrt{q} \] We are given \( M = 24 \), \( n = 2 \), and \( q = 4 \), so: \[ 24 = k (2^2) \sqrt{4} \] \[ 24 = k \times 4 \times 2 = \[ 24 = 8k \] Step 2: Solve for \( k \). \[ k = \frac{24}{8} = 3 \] Step 3: Find \( M \) when \( n = 5 \) and \( q = 9 \). Now substitute \( n = 5 \), \( q = 9 \), and \( k = 3 \) into the equation: \[ M = 3 \times (5^2) \times \sqrt{9} \] \[ M = 3 \times 25 \times 3 = 3 \times 75 = 225 \] So, the correct answer is D. 225.

2023 WAEC Mathematics past questions and answers

11. If \( m : n = \frac{2}{3} : 1 \frac{1}{3} \) and \( n : q = 1 \frac{1}{2} : 1 \frac{3}{4} \), find \( q : m \).
A. 16 : 35

B. 35 : 16

C. 18 : 35

D. 35 : 18

2023 WAEC Mathematics past questions and answers

Q11.** If \( m : n = \frac{2}{3} : 1 \frac{1}{3} \) and \( n : q = 1 \frac{1}{2} : 1 \frac{3}{4} \), find \( q : m \).
Step 1: Express the ratios as fractions.

First, convert the mixed numbers to improper fractions:
\[ m : n = \frac{2}{3} : \frac{4}{3} \] This simplifies to: \[ m : n = \frac{2}{3} \div \frac{4}{3} = \frac{2}{3} \times \frac{3}{4} = \frac{2}{4} = \frac{1}{2} \] Next, express \( n : q \) as: \[ n : q = \frac{3}{2} : \frac{7}{4} \] This simplifies to: \[ n : q = \frac{3}{2} \div \frac{7}{4} = \frac{3}{2} \times \frac{4}{7} = \frac{12}{14} = \frac{6}{7} \] Step 2: Combine the two ratios. We know \( m : n = \frac{1}{2} \) and \( n : q = \frac{6}{7} \). To find \( q : m \), multiply the reciprocal of both ratios: \[ q : m = \frac{7}{6} \times 2 = \frac{14}{6} = \frac{7}{3} \] So, the correct answer is B. 35 : 16.

2023 WAEC Mathematics past questions and answers

12. One-third the sum of two numbers is 12. Twice their difference is 12. Find the numbers.
A. 21 and 15

B. 20 and 16

C. 22 and 14

D. 23 and 13

2023 WAEC Mathematics past questions and answers

Q12. One-third the sum of two numbers is 12. Twice their difference is 12. Find the numbers.
Let the two numbers be \( x \) and \( y \).
Step 1: Set up the equations based on the problem. We are told that one-third the sum of the two numbers is 12, so: \[ \frac{x + y}{3} = 12 \quad \Rightarrow \quad x + y = 36 \] We are also told that twice their difference is 12, so: \[ 2(x – y) = 12 \quad \Rightarrow \quad x – y = 6 \] Step 2: Solve the system of equations. Now we have: \[ x + y = 36 \quad \text{and} \quad x – y = 6 \] Add the two equations together: \[ (x + y) + (x – y) = 36 + 6 \] \[ 2x = 42 \quad \Rightarrow \quad x = 21 \] Substitute \( x = 21 \) into \( x + y = 36 \): \[ 21 + y = 36 \quad \Rightarrow \quad y = 15 \] So, the two numbers are 21 and 15, and the correct answer is A. 21 and 15.

2023 WAEC Mathematics past questions and answers

13. Find the quadratic equation whose roots are \( \frac{2}{3} \) and \( \frac{3}{4} \).
A. \( y^2 + y – 6 = 0 \)

B. \( 12y^2 – y – 6 = 0 \)

C. \( 12x^2 + y – 6 = 0 \)

D. \( 12x^2 + y + 6 = 0 \)

2023 WAEC Mathematics past questions and answers

Q13.Find the quadratic equation whose roots are \( \frac{2}{3} \) and \( \frac{3}{4} \).
Step 1: Use the sum and product of roots formula.

For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( -\frac{b}{a} \) and the product of the roots is \( \frac{c}{a} \).
The sum of the roots is: \[ \frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12} \] The product of the roots is: \[ \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2} \] Step 2: Form the quadratic equation. The quadratic equation with these roots is: \[ 12x^2 – 17x + 6 = 0 \] So, the correct answer is B. \( 12x^2 – 17x + 6 = 0 \).

CONTINUE READING : QUESTION 1 TO QUESTION 13 HERE 2023 WAEC Mathematics Past Questions and Answers

CONTINUE READING : QUESTION 14 TO QUESTION 22 HERE 2023 WAEC Mathematics Past Questions and Answers

CONTINUE READING : QUESTION 23 TO QUESTION 31 HERE 2023 WAEC Mathematics Past Questions and Answers

CONTINUE READING : QUESTION 32 TO QUESTION 39 HERE 2023 WAEC Mathematics Past Questions and Answers

CONTINUE READING : QUESTION 40 TO QUESTION 44 HERE 2023 WAEC Mathematics Past Questions and Answers

CONTINUE READING : QUESTION 45 TO QUESTION 50 HERE 2023 WAEC Mathematics Past Questions and Answers

Topics Covered in WAEC Mathematics Past Questions

The WAEC Mathematics exam typically covers a wide range of topics from both Junior and Senior Secondary levels. Below is a list of essential topics covered in the past questions:

• Algebra
• Geometry and Trigonometry
• Probability
• Statistics
• Number and Numeration
• Sets and Logic
• Mensuration
• Coordinate Geometry

Each of these topics has numerous subtopics that are frequently featured in the exam. By working through the past questions, you can ensure that you’re fully prepared for any surprises in the exam.

Exam Tips for Scoring High in WAEC Mathematics

Here are some tips to help you maximize your performance in the WAEC Mathematics exam:

1. Start Early: Begin your revision months before the exam to avoid last-minute pressure.
2. Practice Regularly: Solve as many past questions as possible to enhance your problem-solving speed.
3. Focus on Weak Areas: Identify the topics you struggle with and focus on improving them.
4. Time Management: Always practice with a timer to simulate exam conditions.
5. Use a Calculator Wisely: Make sure you’re familiar with your calculator’s functions before the exam day.
6. Study the Marking Scheme: Understanding how marks are allocated can help you plan your answers better.

Frequently Asked Questions (FAQ)

1. Are WAEC Mathematics past questions helpful?
Yes, past questions are extremely useful for understanding the exam pattern and important topics. Many examiners often repeat questions or stick to certain question formats.

2. How can I download WAEC Mathematics past questions for free?
You can download the WAEC Mathematics past questions and answers by clicking the provided download link in this article.

3. Are the answers included in the PDFs?
Yes, all the PDFs come with both questions and answers to help you verify your solutions.

4. What is the best way to prepare for WAEC Mathematics?
The best approach is to practice past questions consistently, focus on areas where you need improvement, and review mathematical concepts regularly.

ALSO READ: WAEC Mathematics Past Questions and Answers PDF Download (1988 – 2024)

Conclusion

Using WAEC Mathematics past questions and answers is one of the most effective ways to prepare for the exam. The free PDF download links provided in this post will give you access to numerous past questions that will help you build confidence and improve your performance. Don’t wait—start preparing now!

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