2025 WAEC Mathematics likely questions and answers

2025 WAEC Mathematics likely questions and answers

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2025 WAEC Mathematics likely questions and answers

QUESTION

1. A man invests a sum of money at 4% per annum simple interest. After 3 years, the principal amounts to ₦7,000.00. Find the sum invested.

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

ANSWER

Simple Interest Calculation Given: Rate of interest $$r=4\mathrm{\%}$$ 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: $$M=\left(\begin{array}{cc}1& 2\\ 4& 3\end{array}\right),N=\left(\begin{array}{cc}m& x\\ n& y\end{array}\right),$$ $$\text{and}MN=\left(\begin{array}{cc}2& 1\\ 3& 4\end{array}\right)$$ Find the matrix $N.$ —

ANSWER

(a) Given: $3y^2–5y+2=0$

Step 1: Divide all terms by 3:

$y^2–\frac{5}{3}y+\frac{2}{3}=0$

Step 2: Move constant to the right:

$y^2–\frac{5}{3}y=-\frac{2}{3}$

Step 3: Complete the square:

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

ANSWER

(b) $$M=\left(\begin{array}{cc}1& 2\\ 4& 3\end{array}\right),$$ $$N=\left(\begin{array}{cc}m& x\\ n& y\end{array}\right),$$ $$MN=\left(\begin{array}{cc}2& 1\\ 3& 4\end{array}\right)$$

$$MN=\left(\begin{array}{cc}1\cdot m+2\cdot n& 1\cdot x+2\cdot y\\ 4\cdot m+3\cdot n& 4\cdot x+3\cdot y\end{array}\right)$$

$$=\left(\begin{array}{cc}m+2n& x+2y\\ 4m+3n& 4x+3y\end{array}\right)$$

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

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

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

$$m=2–2(1)=0$$ $$x=1–2y$$

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

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

$$x=1–2(0)=1$$ $$\boxed{{\displaystyle N=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)}}$$

QUESTION

Question If $T=\{\text{prime numbers}$ $M=\{\text{odd numbers}$ are subsets of the universal set $\mu =\{x\mid 0<x\le 10,x\in \mathbb{Z}$ find: $T^{\prime }\cap M^{\prime }$ 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: $$\begin{array}{|ccccccccc|}\hline \text{Age }(x)& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{Frequency }(f)& 2& 6& 5& x& 6& 9& 8& 5\\ \hline\end{array}$$

If the mean of the distribution is 7, find:

(a) The value of $x$

(b) The standard deviation of their ages

Answers

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

Mean = 7 Part (a)

(a) Find the value of $x$

Mean formula: $$\overline{x}=\frac{\sum fx}{\sum f}$$ 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:} $$\begin{array}{|ccccccccc|}\hline x& 3& 4& 5& 6& 7& 8& 9& 10\\ f& 2& 6& 5& 4& 6& 9& 8& 5\\ \hline\end{array}$$

Mean $\overline{x}=7$

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

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

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