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WAEC GCE Mathematics questions and answers with details explanations WAEC GCE Mathematics questions and answers Are you preparing for this upcoming WAEC GCE Mathematics examination, looking for the right place to get real WAEC GCE Mathematics Questions and Answers, macoach.com.ng is the best place or onlinle to boost your success. Mathematics remains a core subject, and performing well in it is essential for gaining admission into higher institutions.In this post, we have compiled updated WAEC GCE Mathematics Questions and Answers based on previous years’ exams, likely questions with detailed solutions. this post will help you understand the exam pattern and practice effectively. InstructorMA COACH Last updatedtoday Total Viewers 1k+ ClassWAEC GCE Why Use WAEC GCE Mathematics Questions and Answers? Understand Exam Format – Going through our complied WAEC GCE Mathematics Questions and Answers helps you know how WAEC GCE sets its Questions and understand Exam formats. Boost Confidence – Practicing with real exam-style questions prepares you academically to pass. Identify Weak Areas – The answers guide you in knowing which topics need more study time. Save Study Time – Focusing on likely exam questions with details solutions helps you prepare smart, not hard. WAEC GCE Mathematics Questions and Answers Question Answers Video Answers Simplify $10 dfrac{2}{5} - 6 dfrac{2}{3} + 3$ A $6 dfrac{1}{5}$ B $6 dfrac{11}{15}$ C $7 dfrac{1}{5}$ D $7 dfrac{1}{15}$ We change mixed numbers to improper fractions $$ frac{52}{5} - frac{20}{3} + 3 $$ LCM of the denominators 5 and 3 is 15 $$ = frac{156}{15} - frac{100}{15} + frac{45}{15} $$ $$ = frac{101}{15} = 6 dfrac{11}{15} ; (B) $$ Please, Check back later Loading.. Simplify $10 dfrac{2}{5} - 6 dfrac{2}{3} + 3$ A $6 dfrac{1}{5}$ B $6 dfrac{11}{15}$ C $7 dfrac{1}{5}$ D $7 dfrac{1}{15}$ We change mixed numbers to improper fractions $$ frac{52}{5} - frac{20}{3} + 3 $$ LCM of the denominators 5 and 3 is 15 $$ = frac{156}{15} - frac{100}{15} + frac{45}{15} $$ $$ = frac{101}{15} = 6 dfrac{11}{15} ; (B) $$ Please, Check back later Loading.. Question Answers Video Answers Simplify: $$ 6 dfrac{3}{4} ; div ; Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ Options: A. $6 dfrac{3}{4}$ B. $4$ C. $dfrac{1}{4}$ D. $dfrac{4}{27}$ We are asked to simplify: $$ 6 dfrac{3}{4} ; div ; Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ Step 1: Change all mixed numbers to improper fractions $$ 6 dfrac{3}{4} = frac{27}{4} $$ $$ 21 dfrac{1}{3} = frac{64}{3} $$ $$ 2 dfrac{1}{4} = frac{9}{4} $$ Step 2: Work inside the bracket We have: $$ Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ First, multiply: $$ frac{64}{3} times frac{9}{4} = frac{64 times 9}{3 times 4} $$ Simplify step-by-step: $64 ÷ 4 = 16$ So we get: $frac{16 times 9}{3}$ $9 ÷ 3 = 3$ So: $16 × 3 = 48$ Therefore: $$ frac{64}{3} times frac{9}{4} = 48 $$ Now subtract 21: $$ 48 - 21 = 27 $$ So the bracket simplifies to $27$. Step 3: Now divide We now have: $$ frac{27}{4} ; div ; 27 $$ Dividing by 27 means multiplying by $frac{1}{27}$: $$ frac{27}{4} × frac{1}{27} = frac{27}{108} $$ Simplify $frac{27}{108}$: $27 ÷ 27 = 1$ $108 ÷ 27 = 4$ $$ frac{27}{108} = frac{1}{4} $$ ✅ Final Answer: $$ boxed{ dfrac{1}{4} } $$ Correct Option: C. $dfrac{1}{4}$ Please, Check back later Loading.. Simplify: $$ 6 dfrac{3}{4} ; div ; Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ Options: A. $6 dfrac{3}{4}$ B. $4$ C. $dfrac{1}{4}$ D. $dfrac{4}{27}$ We are asked to simplify: $$ 6 dfrac{3}{4} ; div ; Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ Step 1: Change all mixed numbers to improper fractions $$ 6 dfrac{3}{4} = frac{27}{4} $$ $$ 21 dfrac{1}{3} = frac{64}{3} $$ $$ 2 dfrac{1}{4} = frac{9}{4} $$ Step 2: Work inside the bracket We have: $$ Big( 21 dfrac{1}{3} times 2 dfrac{1}{4} - 21 Big) $$ First, multiply: $$ frac{64}{3} times frac{9}{4} = frac{64 times 9}{3 times 4} $$ Simplify step-by-step: $64 ÷ 4 = 16$ So we get: $frac{16 times 9}{3}$ $9 ÷ 3 = 3$ So: $16 × 3 = 48$ Therefore: $$ frac{64}{3} times frac{9}{4} = 48 $$ Now subtract 21: $$ 48 - 21 = 27 $$ So the bracket simplifies to $27$. Step 3: Now divide We now have: $$ frac{27}{4} ; div ; 27 $$ Dividing by 27 means multiplying by $frac{1}{27}$: $$ frac{27}{4} × frac{1}{27} = frac{27}{108} $$ Simplify $frac{27}{108}$: $27 ÷ 27 = 1$ $108 ÷ 27 = 4$ $$ frac{27}{108} = frac{1}{4} $$ ✅ Final Answer: $$ boxed{ dfrac{1}{4} } $$ Correct Option: C. $dfrac{1}{4}$ Please, Check back later Loading.. Question Answers Video Answers A sales girl gets a commission of $8%$ of the value of the things she sells. We need to find her commission for selling: $3$ keyboards at $N 25,000.00$ each, and $5$ calculators at $N 600.00$ each. Step 1. Find the total value of the keyboards Price of one keyboard = $N 25,000.00$ Number of keyboards = $3$ $$ text{Total value of keyboards} = 3 × 25,000 = N 75,000.00 $$ Step 2. Find the total value of the calculators Price of one calculator = $N 600.00$ Number of calculators = $5$ $$ text{Total value of calculators} = 5 × 600 = N 3,000.00 $$ Step 3. Find the total value of goods sold $$ text{Total sales} = N 75,000.00 + N 3,000.00 = N 78,000.00 $$ Step 4. Calculate commission (8%) $$ text{Commission} = 8% times 78,000 $$ $$ = frac{8}{100} × 78,000 $$ $$ = 0.08 × 78,000 = N 6,240.00 $$ ✅ Final Answer: $$ boxed{ N 6,240.00 } $$ Correct Option: D. $N 6,240.00$ Please, Check back later Loading.. A sales girl gets a commission of $8%$ of the value of the things she sells. We need to find her commission for selling: $3$ keyboards at $N 25,000.00$ each, and $5$ calculators at $N 600.00$ each. Step 1. Find the total value of the keyboards Price of one keyboard = $N 25,000.00$ Number of keyboards = $3$ $$ text{Total value of keyboards} = 3 × 25,000 = N 75,000.00 $$ Step 2. Find the total value of the calculators Price of one calculator = $N 600.00$ Number of calculators = $5$ $$ text{Total value of calculators} = 5 × 600 = N 3,000.00 $$ Step 3. Find the total value of goods sold $$ text{Total sales} = N 75,000.00 + N 3,000.00 = N 78,000.00 $$ Step 4. Calculate commission (8%) $$ text{Commission} = 8% times 78,000 $$ $$ = frac{8}{100} × 78,000 $$ $$ = 0.08 × 78,000 = N 6,240.00 $$ ✅ Final Answer: $$ boxed{ N 6,240.00 } $$ Correct Option: D. $N 6,240.00$ Please, Check back later Loading.. Question Answers Video Answers For musical shows, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the: (i) value of x; (ii) ratio of the number of children to the number adults; (iii) average amount paid per person (iv) percentage profit it the organizers spent D720 on the show. Let number of children = $x$. Then number of adults = $x+60$. Total collected: adults pay D5 each, children D2 each, so $$ 5(x+60)+2x=1280. $$ (i) Find $x$ Work the equation out digit-by-digit: $5(x+60)+2x=1280$ $5x+300+2x=1280$ $7x+300=1280$ $7x=1280-300=980$ $x=dfrac{980}{7}=140.$ So $x=140$ (children). Adults $=x+60=140+60=200.$ (ii) Ratio of children to adults Children : Adults $=140:200$. Simplify by dividing both terms by 20: $$ 140div20 : 200div20 = 7:10. $$ So the ratio is $7:10$. (iii) Average amount paid per person Total people $=140+200=340.$ Total collected $=D1280.$ Average $=dfrac{1280}{340}$. Reduce fraction stepwise: Divide numerator and denominator by 20: $dfrac{1280div20}{340div20}=dfrac{64}{17}.$ As a decimal: $dfrac{64}{17}approx 3.76470588ldots$ So the average paid per person is $dfrac{64}{17}$ D, or about D3.76 (to 2 d.p.). (iv) Percentage profit if cost = D720 Profit $=$ Revenue $-$ Cost $=1280-720=560.$ Percentage profit $=dfrac{text{Profit}}{text{Cost}}times100=dfrac{560}{720}times100.$ Simplify fraction: $dfrac{560}{720}=dfrac{56}{72}=dfrac{14}{18}=dfrac{7}{9}.$ So percentage $=dfrac{7}{9}times100=dfrac{700}{9}%approx77.777%$. Rounded to two decimal places: 77.78%. Final answers (i) $x=140$ children. (ii) Children : Adults $=7:10$. (iii) Average paid per person $=dfrac{64}{17}$ D $approx$ D3.76. (iv) Percentage profit $=dfrac{7}{9}times100% approx 77.78%$. Please, Check back later Loading.. For musical shows, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the: (i) value of x; (ii) ratio of the number of children to the number adults; (iii) average amount paid per person (iv) percentage profit it the organizers spent D720 on the show. Let number of children = $x$. Then number of adults = $x+60$. Total collected: adults pay D5 each, children D2 each, so $$ 5(x+60)+2x=1280. $$ (i) Find $x$ Work the equation out digit-by-digit: $5(x+60)+2x=1280$ $5x+300+2x=1280$ $7x+300=1280$ $7x=1280-300=980$ $x=dfrac{980}{7}=140.$ So $x=140$ (children). Adults $=x+60=140+60=200.$ (ii) Ratio of children to adults Children : Adults $=140:200$. Simplify by dividing both terms by 20: $$ 140div20 : 200div20 = 7:10. $$ So the ratio is $7:10$. (iii) Average amount paid per person Total people $=140+200=340.$ Total collected $=D1280.$ Average $=dfrac{1280}{340}$. Reduce fraction stepwise: Divide numerator and denominator by 20: $dfrac{1280div20}{340div20}=dfrac{64}{17}.$ As a decimal: $dfrac{64}{17}approx 3.76470588ldots$ So the average paid per person is $dfrac{64}{17}$ D, or about D3.76 (to 2 d.p.). (iv) Percentage profit if cost = D720 Profit $=$ Revenue $-$ Cost $=1280-720=560.$ Percentage profit $=dfrac{text{Profit}}{text{Cost}}times100=dfrac{560}{720}times100.$ Simplify fraction: $dfrac{560}{720}=dfrac{56}{72}=dfrac{14}{18}=dfrac{7}{9}.$ So percentage $=dfrac{7}{9}times100=dfrac{700}{9}%approx77.777%$. Rounded to two decimal places: 77.78%. Final answers (i) $x=140$ children. (ii) Children : Adults $=7:10$. (iii) Average paid per person $=dfrac{64}{17}$ D $approx$ D3.76. (iv) Percentage profit $=dfrac{7}{9}times100% approx 77.78%$. Please, Check back later Loading.. Question Answers Video Answers A man bought 250 electric lanterns in the United States of America (USA) for $5,000. He sold them in Lagos at N1,920 each. At the time of his journey back to USA, the exchange rate was N 96 to the dollar. (i) If by selling the lanterns at N 1,920 each, he made a profit of 20%. Calculate the exchange rate at the time he bought the lanterns (ii) How much in dollar will the sales amount to at the time of his journey back to USA? Given: he bought 250 lanterns for $$$5,000$$. He sold them at $text{N}1{,}920$ each. So total sales in naira $=250times1{,}920=text{N}480{,}000.$ We are told selling at N1,920 each produced a 20% profit (presumably measured on his cost). (i) Exchange rate at time he bought the lanterns If selling produced 20% profit, then selling price $= $ cost $times 1.20$. So cost (in naira) $= dfrac{text{Total sales}}{1.20}=dfrac{480{,}000}{1.20}=text{N}400{,}000.$ That $text{N}400{,}000$ was equal to $$$5,000$$ (his purchase in dollars). Therefore the exchange rate then was $$ text{Naira per dollar}=frac{400{,}000}{5{,}000}=80. $$ Answer (i): $$boxed{text{N}80 = $1}$$. (ii) How much (in dollars) will the sales amount to at the time of his journey back (rate N96/$1)? Total sales in naira $=text{N}480{,}000$. Convert at N96 per dollar: $$ text{Dollar proceeds}=frac{480{,}000}{96}=$5{,}000. $$ Answer (ii): $$boxed{$5{,}000}$$. Please, Check back later Loading.. A man bought 250 electric lanterns in the United States of America (USA) for $5,000. He sold them in Lagos at N1,920 each. At the time of his journey back to USA, the exchange rate was N 96 to the dollar. (i) If by selling the lanterns at N 1,920 each, he made a profit of 20%. Calculate the exchange rate at the time he bought the lanterns (ii) How much in dollar will the sales amount to at the time of his journey back to USA? Given: he bought 250 lanterns for $$$5,000$$. He sold them at $text{N}1{,}920$ each. So total sales in naira $=250times1{,}920=text{N}480{,}000.$ We are told selling at N1,920 each produced a 20% profit (presumably measured on his cost). (i) Exchange rate at time he bought the lanterns If selling produced 20% profit, then selling price $= $ cost $times 1.20$. So cost (in naira) $= dfrac{text{Total sales}}{1.20}=dfrac{480{,}000}{1.20}=text{N}400{,}000.$ That $text{N}400{,}000$ was equal to $$$5,000$$ (his purchase in dollars). Therefore the exchange rate then was $$ text{Naira per dollar}=frac{400{,}000}{5{,}000}=80. $$ Answer (i): $$boxed{text{N}80 = $1}$$. (ii) How much (in dollars) will the sales amount to at the time of his journey back (rate N96/$1)? Total sales in naira $=text{N}480{,}000$. Convert at N96 per dollar: $$ text{Dollar proceeds}=frac{480{,}000}{96}=$5{,}000. $$ Answer (ii): $$boxed{$5{,}000}$$. Please, Check back later Loading.. WAEC GCE Mathematics Questions and Answers PDF [dflip id="748"][/dflip] Tips for Using WAEC GCE Mathematics Questions and Answers Effectively Practice Daily – Dedicate at least one hour daily to solving past questions. Time Yourself – Simulate exam conditions to improve your speed and accuracy. Understand, Don’t Cram – Go through the solutions carefully to understand how each answer is derived. Focus on Frequently Asked Topics – WAEC often repeats certain question patterns. Seek Help When Stuck – Discuss challenging problems with teachers, friends, or online study groups. How to Download WAEC GCE Mathematics Questions and Answers To download WAEC GCE Mathematics Questions and Answers, a fee of ₦500 is required. We have it in PDF format. To download WAEC GCE Mathematics Questions and Answers, please click on download button below OR contact the instructor to download it using the number or Email. Download Now OR Phone / WhatsApp Number +2349061221656, +2348062853040 Email admin@macoach.com.ng Share to: Trending Updates Latest postLesson NoteWAEC GCE Questions and AnswersWAEC Questions and Answers BackDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackNursery School Lesson NoteDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackWAEC GCE Mathematics Questions and AnswersWAEC GCE MATHEMATICS QUESTIONS AND ANSWERS1 CommentWAEC GCE Mathematics questions and answers with details explanations WAEC GCE Mathematics questions and answers Are you preparing for this...Read MoreDownload Nursery 2 Lesson Note For All SubjectsNo CommentsDownload Nursery 2 Lesson Note For All Subjects Download Nursery 2 Lesson Note For All Subjects Teaching Nursery 2 can...Read MoreDownload Nursery 1 Lesson Note For All SubjectsNo CommentsDownload Nursery 1 Lesson Note For All Subjects Download Nursery 1 Lesson Note For All Subjects Looking for a simple,...Read MorePOST UTMENo CommentsRead MoreMaster Roman Numerals in Minutes! | JSS1 Mathematics Made Easy | 1st term jss1 scheme of workNo CommentsWelcome to our JSS1 Mathematics class! In this video, we will help you master Roman numerals in just a few...Read More OUR PAGES All PagesLesson NoteWAEC GCE Questions and AnswersWAEC Questions and Answers BackDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackNursery School Lesson NoteDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackWAEC GCE Mathematics Questions and AnswersDownload Senior Secondary School Lesson Note For All Subjects and ClassesRead MoreDownload Junior Secondary School Lesson Note For All Subjects and ClassesRead MoreDownload Primary school Lesson Note For All Subjects and ClassesRead MoreDownload Nursery School Lesson Note For All Subjects and ClassesNursery School Lesson Note For All Subjects and Classes Nursery school Lesson Note For All Subjects All Subjects 1K+ Dowloaded...Read MoreLesson Note for all schoolsLesson Note For All Classes Nursery | Primary | Junior Secondary | Senior Secondary Lesson Note For All Classes Save...Read MoreBest Skills To Learn NigeriaRead MoreMathematics Textbooks Exercises SolutionRead MoreExam’s past Questions and AnswersRead MoreMathematics LessonRead MoreLoad MoreEnd of Content. More than 2 results are available in the PRO version (This notice is only visible to admin users) Popular Posts All PostsLesson NoteWAEC GCE Questions and AnswersWAEC Questions and Answers BackDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackNursery School Lesson NoteDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackWAEC GCE Mathematics Questions and AnswersWAEC GCE MATHEMATICS QUESTIONS AND ANSWERSDownload Nursery 2 Lesson Note For All SubjectsDownload Nursery 1 Lesson Note For All SubjectsPOST UTMEMaster Roman Numerals in Minutes! | JSS1 Mathematics Made Easy | 1st term jss1 scheme of work Blog Category Lesson Note (0) Uncategorized (2) WAEC GCE Questions and Answers (0) Get in Touch Phone Number +2349061221656, +2348062853040 Email admin@macoach.com.ng Address 2nd Floor former joybell nursery & primary school opposite lotogbe junction ondo city, ondo state Business Hours Monday — Friday 8am – 11pm Saturday — 8am – 10pm Sunday — Closed Tags download Nursery 1 lesson note (1) download Nursery 2 lesson note (1) Download Nursery 2 Lesson Note for All Subjects (1) early childhood education (2) free Nursery 1 PDF (1) free Nursery 2 PDF (1) GCE Mathematics Answers (1) lesson notes for Nursery 1 (1) lesson notes for Nursery 2 (1) Mathematics WAEC Practice (1) Nursery 1 curriculum Nigeria (1) Nursery 1 education guide (1) Nursery 1 lesson note (1) Nursery 1 lesson plan (1) Nursery 2 curriculum Nigeria (1) Nursery 2 lesson note (1) Nursery 2 lesson plan (1) WAEC GCE 2025 (1) WAEC GCE Exam (1) WAEC GCE Mathematics (1) WAEC GCE Study Guide (1) WAEC Mathematics Preparation (1) WAEC Mathematics Questions (1) WAEC Maths Solutions (1) WAEC Past Questions (1) Newsletter Signup [wpforms id="426"] Join Us Facebook Youtube Instagram Linkedin Twitter About Us MACOACH.com.ng –Your No.1 Online Academy for Daily Mathematics Lessons, Textbook Solutions, and Exam Preparation in Nigeria. +2348062853040, +2349061221656 admin@macoach.com.ng 2nd Floor former joybell nursery & primary school opposite lotogbe junction ondo city, ondo state Departments Mathematics Lesson Lesson Note Exam's past Questions Maths Textbooks Exercises Solution Learn Skills Quick Links About Us Terms and Conditions Privacy Policy Contact Us Declaimer Refund Policy Recent news All PostLesson NoteWAEC GCE Questions and AnswersWAEC Questions and Answers BackDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackNursery School Lesson NoteDownload Nursery 1 Lesson Note For All SubjectsDownload Nursery 2 Lesson Note for All Subjects BackWAEC GCE Mathematics Questions and AnswersWAEC GCE MATHEMATICS QUESTIONS AND ANSWERSSeptember 9, 2025Download Nursery 2 Lesson Note For All SubjectsAugust 29, 2025 © 2025 Created with MA COACH

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