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
2023 WAEC Mathematics past questions and answers
14. Make 𝑥 the subject of the relation: \( y = \frac{ax^3 – b}{3z} \)\[ \boxed{A. \; x = \sqrt[3]{\frac{3yzb}{a}}} \]
\[ \boxed{B. \; x = \sqrt[3]{\frac{3yz + b}{a}}} \]
\[ \boxed{C. \; x = \sqrt[3]{\frac{3yz – b}{a}}} \]
\[ \boxed{D. \; x = \sqrt[3]{\frac{3y^3 + b}{a}}} \]
2023 WAEC Mathematics past questions and answers
To make \( x \) the subject of the equation \( y = \frac{ax^3 – b}{3z} \), follow these steps:Step 1: Multiply both sides by \( 3z \) to eliminate the denominator.
\[ y \times 3z = ax^3 – b \] \[ 3yz = ax^3 – b \] Step 2: Add \( b \) to both sides. \[ 3yz + b = ax^3 \] Step 3: Divide both sides by \( a \) to isolate \( x^3 \). \[ x^3 = \frac{3yz + b}{a} \] Step 4: Take the cube root of both sides to solve for \( x \). \[ x = \sqrt[3]{\frac{3yz + b}{a}} \] Thus, the correct expression for \( x \) is: \[ x = \sqrt[3]{\frac{3yz + b}{a}} \] The correct answer is: \[ \boxed{B. \; x = \sqrt[3]{\frac{3yz + b}{a}}} \]
2023 WAEC Mathematics past questions and answers
15.The price of a shoe was decreased by 22%. If the new price was $27.30, what was the original price?– A. $62.30
– B. $42.30
– C. $72.00
– D. $35.00
2023 WAEC Mathematics past questions and answers
Solution: Let the original price be \( P \). 1. If the price was decreased by 22%, then 78% of the original price is the new price. Mathematically: \[ 78\% \times P = 27.30 \] This can be written as: \[ 0.78P = 27.30 \] 2. Solve for \( P \): \[ P = \frac{27.30}{0.78} \] \[ P = 35 \] Thus, the original price of the shoe is $35. The correct answer is: \[ \boxed{D. \; 35} \]2023 WAEC Mathematics past questions and answers
16.The radius and height of a solid cylinder are 8 cm and 14 cm, respectively. Find, correct to two decimal places, the total surface area. (Take \(\pi = \frac{22}{7}\))– A. 1,106.28 cm²
– B. 1,016.29 cm²
– C. 1,106.29 cm²
– D. 1,206.27 cm²
2023 WAEC Mathematics past questions and answers
Solution: The formula for the total surface area of a cylinder is: \[ \text{Total Surface Area} = 2\pi r(r + h) \] where \( r \) is the radius and \( h \) is the height. 1. **Substitute the values** \( r = 8 \) cm and \( h = 14 \) cm: \[ \text{Total Surface Area} = 2 \times \frac{22}{7} \times 8 \times (8 + 14) \] \[ = 2 \times \frac{22}{7} \times 8 \times 22 \] 2. Simplify the expression: \[ = \frac{44}{7} \times 8 \times 22 \] \[ = \frac{44}{7} \times 176 = 1106.28 \, \text{cm}^2 \] Thus, the total surface area is approximately \( 1106.28 \, \text{cm}^2 \). The correct answer is: \[ \boxed{A. \; 1106.28 \, \text{cm}^2} \]2023 WAEC Mathematics past questions and answers
17.In the diagram, \( O \) is the centre of the circle \( NST \). \( \overline{NT} = \overline{ST} \) and \( \angle LNTS = 36^\circ \). Find the measure of the angle marked \( t \).– A. \( 36^\circ \)
– B. \( 72^\circ \)
– C. \( 108^\circ \)
– D. \( 54^\circ \)
2023 WAEC Mathematics past questions and answers
Solution: Since \( \overline{NT} = \overline{ST} \), triangle \( NST \) is an **isosceles triangle. The base angles of an isosceles triangle are equal. 1. **Let the base angles be \( t \). The sum of angles in any triangle is \( 180^\circ \). Therefore: \[ 36^\circ + t + t = 180^\circ \] \[ 36^\circ + 2t = 180^\circ \] \[ 2t = 180^\circ – 36^\circ \] \[ 2t = 144^\circ \] \[ t = \frac{144^\circ}{2} = 72^\circ \] Thus, the measure of angle \( t \) is \( 72^\circ \). The correct answer is: \[ \boxed{B. \; 72^\circ} \]2023 WAEC Mathematics past questions and answers
18. A sphere has radius 3 cm. Find, in terms of \(\pi\), its volume.– A. \( 27\pi \, \text{cm}^3 \)
– B. \( 108\pi \, \text{cm}^3 \)
– C. \( 36\pi \, \text{cm}^3 \)
– D. \( 30\pi \, \text{cm}^3 \)
2023 WAEC Mathematics past questions and answers
Solution: The formula for the volume of a sphere is: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius. 1. Substitute the radius \( r = 3 \) cm: \[ V = \frac{4}{3} \pi (3)^3 \] \[ V = \frac{4}{3} \pi \times 27 \] \[ V = 36\pi \, \text{cm}^3 \] Thus, the volume of the sphere is \( 36\pi \, \text{cm}^3 \). The correct answer is: \[ \boxed{C. \; 36\pi \, \text{cm}^3} \]2023 WAEC Mathematics past questions and answers
19. Arrange the following numbers in ascending order of magnitude: \( 110_{\text{two}} \), \( 31_{\text{eight}} \), \( 42_{\text{five}} \)– A. \( 110_{\text{two}} \), \( 31_{\text{eight}} \), \( 42_{\text{five}} \)
– B. \( 110_{\text{two}} \), \( 42_{\text{five}} \), \( 31_{\text{eight}} \)
– C. \( 31_{\text{eight}} \), \( 42_{\text{five}} \), \( 110_{\text{two}} \)
– D. \( 42_{\text{five}} \), \( 31_{\text{eight}} \), \( 110_{\text{two}} \)
2023 WAEC Mathematics past questions and answers
Solution: We need to convert each of these numbers to base 10 to compare them. 1. Convert \( 110_{\text{two}} \) to base 10: \[ 110_{\text{two}} = 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 4 + 2 + 0 = 6 \] 2. Convert \( 31_{\text{eight}} \) to base 10: \[ 31_{\text{eight}} = 3 \times 8^1 + 1 \times 8^0 = 24 + 1 = 25 \] 3. Convert \( 42_{\text{five}} \) to base 10: \[ 42_{\text{five}} = 4 \times 5^1 + 2 \times 5^0 = 20 + 2 = 22 \] So, in base 10, the numbers are 6, 25, and 22. Arranging them in ascending order gives: \[ 6, 22, 25 \] The correct answer is: \[ \boxed{B. \; 110_{\text{two}}, 42_{\text{five}}, 31_{\text{eight}}} \]2023 WAEC Mathematics past questions and answers
20.A notebook of length 15 cm was measured by a student as 16.8 cm. Calculate, correct to two decimal places, the percentage error in the measurement.– A. 12.00%
– B. 11.71%
– C. 11.00%
– D. 10.71%
2023 WAEC Mathematics past questions and answers
Solution: The formula for percentage error is: \[ \text{Percentage Error} = \frac{\text{Measured Value} – \text{True Value}}{\text{True Value}} \times 100 \] 1. Substitute the measured and true values: \[ \text{Percentage Error} = \frac{16.8 – 15}{15} \times 100 \] \[ = \frac{1.8}{15} \times 100 \] \[ = 0.12 \times 100 = 12\% \] Thus, the percentage error is 12%. The correct answer is: \[ \boxed{A. \; 12.00\%} \]2023 WAEC Mathematics past questions and answers
21.** Find the value of \( m \) in the diagram.– A. \( 140^\circ \)
– B. \( 130^\circ \)
– C. \( 50^\circ \)
– D. \( 40^\circ \)
2023 WAEC Mathematics past questions and answers
2023 WAEC Mathematics past questions and answers
22. A line \( L \), passing through the point \( (6, -13) \), is parallel to the line which passes through \( (7, 4) \) and \( (-3, 9) \). Find the equation of the line \( L \).– A. \( y = -\frac{1}{2}x – 10 \)
– B. \( y = -\frac{1}{2}x + 10 \)
– C. \( y = -\frac{1}{2}x – 10 \)
– D. \( y = \frac{1}{2}x + 10 \)
2023 WAEC Mathematics past questions and answers
22. A line \( L \), passing through the point \( (6, -13) \), is parallel to the line which passes through \( (7, 4) \) and \( (-3, 9) \). Find the equation of the line \( L \). Solution: To solve this, we need to find the equation of a line parallel to another line, and that passes through a given point. The key here is that parallel lines have the same slope. 1. Find the slope of the line passing through \( (7, 4) \) and \( (-3, 9) \): The slope formula is: \[ m = \frac{y_2 – y_1}{x_2 – x_1} \] Substituting the points \( (7, 4) \) and \( (-3, 9) \): \[ m = \frac{9 – 4}{-3 – 7} = \frac{5}{-10} = -\frac{1}{2} \] So, the slope of the line is \( -\frac{1}{2} \). 2. Use the point-slope form to find the equation of the new line: The point-slope form is: \[ y – y_1 = m(x – x_1) \] We know the slope \( m = -\frac{1}{2} \) and the point \( (6, -13) \). Substituting these values: \[ y + 13 = -\frac{1}{2}(x – 6) \] 3. Simplify the equation: \[ y + 13 = -\frac{1}{2}x + 3 \] \[ y = -\frac{1}{2}x + 3 – 13 \] \[ y = -\frac{1}{2}x – 10 \] Thus, the equation of the line is: \[ \boxed{A. \; y = -\frac{1}{2}x – 10} \]CONTINUE READING : QUESTION 1 TO QUESTION 13 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
CONTINUE READING : QUESTION 14 TO QUESTION 22 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:
- Start Early: Begin your revision months before the exam to avoid last-minute pressure.
- Practice Regularly: Solve as many past questions as possible to enhance your problem-solving speed.
- Focus on Weak Areas: Identify the topics you struggle with and focus on improving them.
- Time Management: Always practice with a timer to simulate exam conditions.
- Use a Calculator Wisely: Make sure you’re familiar with your calculator’s functions before the exam day.
- 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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