2025 - JAMB Mathematics Past Questions and Answers - page 2

The total number of students is given as 12.The frequencies are expressed as:\[x + 1 + 2x + 2 + 3\]Step 1: Combine like terms.\[x + 2x = 3x\]\[1 + 2 + 3 = 6\]So total becomes:\[3x + 6\]Step 2: Set equal to total number of students.\[3x + 6 = 12\]Step 3: Solve for x.Subtract 6 from both sides:\[3x = 6\]Divide both sides by 3:\[x = 2\]Step 4: Substitute x into the frequencies.54 kg → x = 2 students56 kg → 1 student58 kg → 2x = 4 students60 kg → 2 students62 kg → 3 studentsStep 5: Identify highest frequency.The largest frequency is 4, corresponding to weight 58 kg.Final answer:58 kg

Step 1: Express the number using powers of 10.\[0.16 = 16 \times 10^{-2}\]Step 2: Apply the fourth root.\[\sqrt[4]{0.16} = (16 \times 10^{-2})^{\frac{1}{4}}\]Step 3: Apply index law:\[(ab)^n = a^n b^n\]\[= 16^{\frac{1}{4}} \times 10^{-\frac{2}{4}}\]Step 4: Simplify each part.\[16^{\frac{1}{4}} = 2\]\[-\frac{2}{4} = -\frac{1}{2}\]So result becomes:\[2 \times 10^{-\frac{1}{2}}\]Final answer:\[2 \times 10^{-\frac{1}{2}}\]

Step 1: Simplify each term using index laws.First term:\[125 = 5^3\]\[125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}}\]Multiply powers:\[= 5^{-1}\]\[= \frac{1}{5}\]Second term:\[49 = 7^2\]\[49^{-\frac{1}{2}} = (7^2)^{-\frac{1}{2}}\]\[= 7^{-1}\]\[= \frac{1}{7}\]Third term:\[10^0 = 1\]Step 2: Multiply all results.\[\frac{1}{5} \times \frac{1}{7} \times 1\]\[= \frac{1}{35}\]Final answer:\[\frac{1}{35}\]

Step 1: Find total number of fruits.\[6 + 11 + 13 = 30\]Step 2: Find favorable outcomes.Grapes or bananas:\[6 + 11 = 17\]Step 3: Apply probability formula.\[\text{Probability} = \frac{\text{favorable outcomes}}{\text{total outcomes}}\]\[= \frac{17}{30}\]Final answer:\[\frac{17}{30}\]

Step 1: Count total letters.S, C, H, O, O, L → total = 6 lettersStep 2: Identify repeated letters.Letter O appears twice.Step 3: Use permutation formula with repetition.\[\frac{n!}{r!}\]\[= \frac{6!}{2!}\]Step 4: Expand factorials.\[6! = 720\]\[2! = 2\]\[= \frac{720}{2}\]\[= 360\]Final answer:360

Step 1: Express 64 as a power of 2.\[64 = 2^6\]Step 2: Express using base \(\frac{1}{2}\).\[\frac{1}{2} = 2^{-1}\]\[64 = \left(\frac{1}{2}\right)^{-6}\]Step 3: Equate exponents since bases are equal.\[t - 1 = -6\]Step 4: Solve for t.Add 1 to both sides:\[t = -5\]Final answer:-5

Step 1: Convert mixed numbers to improper fractions.\[2\frac{1}{4} = \frac{9}{4}\]\[1\frac{7}{11} = \frac{18}{11}\]\[3\frac{2}{3} = \frac{11}{3}\]Step 2: Simplify division.\[\frac{3}{4} \div \frac{9}{4}\]Multiply by reciprocal:\[\frac{3}{4} \times \frac{4}{9}\]\[= \frac{12}{36}\]\[= \frac{1}{3}\]Step 3: Simplify subtraction.\[\frac{11}{3} - \frac{15}{6}\]Convert to common denominator:\[\frac{22}{6} - \frac{15}{6}\]\[= \frac{7}{6}\]Step 4: Multiply all parts.\[\frac{1}{3} \times \frac{18}{11} \times \frac{7}{6}\]\[= \frac{126}{198}\]Step 5: Simplify fraction.\[= \frac{7}{11}\]Final answer:\[\frac{7}{11}\]

Step 1: Divide 54 by 4.\[54 \div 4 = 13 \text{ remainder } 2\]Step 2: Divide 13 by 4.\[13 \div 4 = 3 \text{ remainder } 1\]Step 3: Divide 3 by 4.\[3 \div 4 = 0 \text{ remainder } 3\]Step 4: Write remainders from bottom to top.\[312_4\]Step 5: Express to three decimal places.\[312.000_4\]Final answer:\[312.000_4\]

Given operation:\[a*b = ab + a + b\]Step 1: Substitute values.\[a = 1, b = 3\]\[1*3 = (1 \times 3) + 1 + 3\]Step 2: Multiply first.\[1 \times 3 = 3\]Step 3: Add remaining values.\[3 + 1 + 3\]\[= 7\]Final answer:7

To find the determinant of the matrix, we expand along the first row.The determinant of a \(3 \times 3\) matrix\[\begin{pmatrix}a & b & c \\d & e & f \\g & h & i\end{pmatrix}\]is calculated as:\[a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}\]Substitute the values from the matrix:\[\text{det}(A) = (-2)\begin{vmatrix}2 & 1 \\ 4 & 2\end{vmatrix} - 3\begin{vmatrix}p & 1 \\ 1 & 2\end{vmatrix} + 1\begin{vmatrix}p & 2 \\ 1 & 4\end{vmatrix}\]Now evaluate each \(2 \times 2\) determinant using:\[\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad - bc\]First determinant:\[(2 \times 2) - (1 \times 4) = 4 - 4 = 0\]Second determinant:\[(p \times 2) - (1 \times 1) = 2p - 1\]Third determinant:\[(p \times 4) - (2 \times 1) = 4p - 2\]Substitute these results:\[\text{det}(A) = (-2)(0) - 3(2p - 1) + (4p - 2)\]Simplify step by step:\[= 0 - 6p + 3 + 4p - 2\]\[= -2p + 1\]We are given that the determinant equals -5, so:\[-2p + 1 = -5\]Subtract 1 from both sides:\[-2p = -6\]Divide both sides by -2:\[p = 3\]Therefore, the value of p is 3.