WAEC GCE MATHEMATICS QUESTIONS AND ANSWERS

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.

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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.
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Save Study Time – Focusing on likely exam questions with details solutions helps you prepare smart, not hard.

WAEC GCE Mathematics Questions and 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 $$ = \frac{156}{15} - \frac{100}{15} + \frac{45}{15} $$ $$ = \frac{101}{15} = 6 \dfrac{11}{15} \; (B) $$

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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. $\dfrac{1}{4}$

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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. $N 6,240.00$

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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:

$$ 140\div20 : 200\div20 = 7:10. $$

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: $\dfrac{1280\div20}{340\div20}=\dfrac{64}{17}.$

As a decimal: $\dfrac{64}{17}\approx 3.76470588\ldots$

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 $=\dfrac{7}{9}\times100\% \approx 77.78\%$.

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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.

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