2025 WAEC Mathematics likely questions and answers

2025 WAEC Mathematics likely questions and answers

INTRODUCTION

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2025 WAEC Mathematics likely questions and answers

QUESTION

1. A man invests a sum of money at 4

A. ₦7,640.00 B.₦6,250.00 C. ₦6,160.00 D. ₦5,833.33

ANSWER

Simple Interest Calculation Given: Rate of interest $$r = 4%$$ Time $$t = 3$$ years Principal + Interest (Final Amount) = ₦7,000 Step 1: Use the Simple Interest Formula $$A = P + SI$$ Where: $$SI = \frac{P \times r \times t}{100}$$ Substituting the given values: $$SI = \frac{P \times 4 \times 3}{100} = \frac{12P}{100} = 0.12P$$ Thus, the total amount: $$A = P + 0.12P = 1.12P$$ $$7,000 = 1.12P$$ Step 2: Solve for $P$ $$P = \frac{7,000}{1.12} = 6,250$$ Final Answer: B. ₦6,250.00

QUESTION

2. 12 men complete a job in 9 days. How many men, working at the same rate, are needed to complete the job in 6 days?

A. 24 B. 18 C. 12 D. 9 E. 8

ANSWER

$$\text{12 men complete the job in 9 days.}$$ Find how many men $x$ will complete it in 6 days. Step 1: Use Work Formula $$\text{Men} \times \text{Days} = \text{Constant Work}$$ $$12 \times 9 = x \times 6$$ Step 2: Solve for $x$ $$x = \frac{12 \times 9}{6} = \frac{108}{6} = 18$$ Answer: B. 18 men

QUESTION

The probability that Amaka will pass an examination is $\frac{3}{7}$, and that Bala will pass is $\frac{4}{9}$. Find the probability that both will pass the examination. Options: A. $\frac{2}{21}$ B. $\frac{4}{21}$ C. $\frac{5}{21}$ D. $\frac{9}{21}$

QUESTION

(a) Using completing the square method, solve, correct to 2 decimal places, the equation: $$3y^2 – 5y + 2 = 0$$ (b) Given that: $$M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, \quad N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}, \quad$$ $$\text{and} \quad MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$$ Find the matrix $N.$ —

ANSWER

(a) Given: $\quad 3y^2 – 5y + 2 = 0$

Step 1: Divide all terms by 3:

$\quad y^2 – \frac{5}{3}y + \frac{2}{3} = 0$

Step 2: Move constant to the right:

$\quad y^2 – \frac{5}{3}y = -\frac{2}{3}$

Step 3: Complete the square:

$\quad \left(\frac{5}{6}\right)^2 = \frac{25}{36}$ $$y^2 – \frac{5}{3}y + \frac{25}{36} = -\frac{2}{3} + \frac{25}{36} = \frac{1}{36}$$ $$\left(y – \frac{5}{6}\right)^2 = \frac{1}{36} y – \frac{5}{6} = \pm \frac{1}{6}$$ $$y = \frac{5}{6} \pm \frac{1}{6} \Rightarrow y = 1 \text{ or } \frac{2}{3}$$ $$\boxed{y = 1.00 \quad \text{or} \quad 0.67}$$

ANSWER

(b) $$M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, \quad$$ $$N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}, \quad$$ $$MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$$

$$MN = \begin{pmatrix} 1\cdot m + 2\cdot n & 1\cdot x + 2\cdot y \\ 4\cdot m + 3\cdot n & 4\cdot x + 3\cdot y \end{pmatrix}$$

$$= \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix}$$

$$\begin{cases} m + 2n = 2 \\ 4m + 3n = 3 \\ x + 2y = 1 \\ 4x + 3y = 4 \end{cases}$$ $$m = 2 – 2n \\ $$

$$4(2 – 2n) + 3n = 3$$ $$\Rightarrow 8 – 8n + 3n = 3$$

$$\Rightarrow -5n = -5$$ $$\Rightarrow n = 1 \\ $$

$$m = 2 – 2(1) = 0$$ $$x = 1 – 2y \\ $$

$$4(1 – 2y) + 3y = 4$$ $$\Rightarrow 4 – 8y + 3y = 4$$

$$\Rightarrow -5y = 0$$ $$\Rightarrow y = 0 \\ $$

$$x = 1 – 2(0) = 1$$ $$\boxed{ N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} }$$

QUESTION

Question If $T = \{\text{prime numbers}$ $M = \{\text{odd numbers}$ are subsets of the universal set $\mu = \{x \mid 0 < x \leq 10,\, x \in \mathbb{Z}$ find: $T' \cap M'$ Options: (a) 4, 6, 8, 10 (b) 1, 4, 5, 8, 10 (c) 1, 2, 4, 6, 8, 10 (d) 1, 2, 3, 5, 7, 8, 9

QUESTION

The table below shows the distribution of ages of a number of children in a school: $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Age }(x) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Frequency }(f) & 2 & 6 & 5 & x & 6 & 9 & 8 & 5 \\ \hline \end{array}$$

If the mean of the distribution is 7, find:

(a) The value of $x$

(b) The standard deviation of their ages

Answers

Given Table: $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Age }(x) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Frequency }(f) & 2 & 6 & 5 & x & 6 & 9 & 8 & 5 \\ \hline \end{array}$$

Mean = 7 Part (a)

(a) Find the value of $x$

Mean formula: $$\bar{x} = \frac{\sum fx}{\sum f}$$ Step 1: Total frequency $$\sum f = 2 + 6 + 5 + x + 6 + 9 + 8 + 5$$ $$= 41 + x$$

Step 2: Total $fx$ $$\sum fx = (3 \cdot 2) + (4 \cdot 6) + (5 \cdot 5)$$ $$+ (6 \cdot x) + (7 \cdot 6)$$ $$+ (8 \cdot 9) + (9 \cdot 8) + (10 \cdot 5)$$ $$= 6 + 24 + 25 + 6x + 42 + 72 + 72 + 50$$ $$= 291 + 6x$$

Step 3: Plug into mean formula $$7 = \frac{291 + 6x}{41 + x}$$

Step 4: Cross-multiply and simplify $$7(41 + x) = 291 + 6x$$ $$287 + 7x = 291 + 6x$$ $$7x – 6x = 291 – 287$$ $$x = 4$$

$x = 4$

Part (b)

(b) Find the Standard Deviation

\text{Updated table:} $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline f & 2 & 6 & 5 & 4 & 6 & 9 & 8 & 5 \\ \hline \end{array}$$

Mean $\bar{x} = 7$

Standard deviation formula: $$\sigma = \sqrt{\frac{\sum f(x – \bar{x})^2}{\sum f}}$$

Compute each row: $$\begin{array}{|c|c|c|c|c|} \hline x & f & x – 7 & (x – 7)^2 & f(x – 7)^2 \\ \hline 3 & 2 & -4 & 16 & 32 \\ 4 & 6 & -3 & 9 & 54 \\ 5 & 5 & -2 & 4 & 20 \\ 6 & 4 & -1 & 1 & 4 \\ 7 & 6 & 0 & 0 & 0 \\ 8 & 9 & 1 & 1 & 9 \\ 9 & 8 & 2 & 4 & 32 \\ 10 & 5 & 3 & 9 & 45 \\ \hline \sum & 45 & & & 196 \\ \hline \end{array}$$

Standard deviation: $$\sigma = \sqrt{\frac{196}{45}} \approx \sqrt{4.36} \approx 2.09$$ $\sigma \approx 2.09$

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