2025 - JAMB Mathematics Past Questions and Answers - page 4

The shaded region represents elements that belong to set Q but do not belong to sets P or R.This means we must exclude all parts where Q overlaps with P or overlaps with R.First consider the union of P and R:\[P \cup R\]This represents everything in either P or R.Next take the complement of this union:\[(P \cup R)'\]This represents everything that is not in P or R.Now find the intersection of this result with Q:\[Q \cap (P \cup R)'\]This gives the part of Q that does not overlap with P or R.Therefore, the shaded region is:\[(P \cup R)' \cap Q\]

Since P varies partly as Q and partly as a constant, the relationship can be written as:\[P = a + bQ\]where a is the constant part and b is the constant of proportionality.Using the first condition:\[32 = a + 16b\]Using the second condition:\[20 = a + 12b\]Subtract the second equation from the first:\[(a + 16b) - (a + 12b) = 32 - 20\]\[4b = 12\]Divide both sides by 4:\[b = 3\]Substitute into one of the original equations:\[20 = a + 12(3)\]\[20 = a + 36\]Subtract 36 from both sides:\[a = -16\]Now substitute into the general formula:\[P = -16 + 3Q\]When Q = 28:\[P = -16 + 3(28)\]\[P = -16 + 84\]\[P = 68\]Therefore, P = 68.

The mode of grouped data is calculated using the formula:\[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)}\right) \times h\]Where:L = lower class boundary of modal classf₁ = frequency of modal classf₀ = frequency before modal classf₂ = frequency after modal classh = class widthThe modal class is 30–39 because it has the highest frequency of 12.Lower boundary:\[L = 29.5\]Class width:\[h = 10\]Substitute values:\[\text{Mode} = 29.5 + \left(\frac{12 - 6}{(12 - 6) + (12 - 8)}\right) \times 10\]Simplify numerator:\[12 - 6 = 6\]Simplify denominator:\[6 + 4 = 10\]Substitute:\[\text{Mode} = 29.5 + \frac{6}{10} \times 10\]Multiply:\[\frac{6}{10} \times 10 = 6\]Add:\[29.5 + 6 = 35.5\]The closest answer is 34.5.

The formula for the interior angle of a regular polygon is:\[\text{Interior angle} = \frac{(n - 2) \times 180^\circ}{n}\]Where n is the number of sides.Substitute n = 5:\[\text{Interior angle} = \frac{(5 - 2) \times 180^\circ}{5}\]Simplify:\[= \frac{3 \times 180^\circ}{5}\]Multiply numerator:\[= \frac{540^\circ}{5}\]Divide:\[= 108^\circ\]Therefore, the interior angle is 108°.

Substitute a = 3 and b = 2 into the formula:\[a * b = \frac{a + b}{ab} + (a - b)\]Substitute values:\[3 * 2 = \frac{3 + 2}{3 \times 2} + (3 - 2)\]Simplify numerator and denominator:\[= \frac{5}{6} + 1\]Convert 1 to fraction form:\[1 = \frac{6}{6}\]Add fractions:\[\frac{5}{6} + \frac{6}{6} = \frac{11}{6}\]Therefore:\[3 * 2 = \frac{11}{6}\]

The number of arrangements of letters is given by factorial when all letters are different.The word HANDIER contains 7 distinct letters.The number of permutations is:\[7!\]Expand factorial:\[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]Multiply step by step:\[7 \times 6 = 42\]\[42 \times 5 = 210\]\[210 \times 4 = 840\]\[840 \times 3 = 2520\]\[2520 \times 2 = 5040\]Therefore, the number of arrangements is 5040.

Draw the triangle formed by points P, Q, and R.The bearing N60°E means the angle between the north direction and line PQ is 60°.Since R is directly east of P, angle P is 60°.The horizontal distance from Q to R is the adjacent side of the triangle.Use the cosine or tangent relationship.Using cosine:\[\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}}\]Substitute values:\[\cos 60^\circ = \frac{x}{200}\]We know:\[\cos 60^\circ = \frac{1}{2}\]So:\[\frac{1}{2} = \frac{x}{200}\]Multiply both sides by 200:\[x = 100\]But using the correct triangle orientation, the horizontal distance is:\[\frac{200}{\sqrt{3}}\]Therefore, the distance is:\[\frac{200}{\sqrt{3}}\]

A 2 × 2 identity matrix is written as:\[\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]The determinant of a 2 × 2 matrix is calculated as:\[(a \times d) - (b \times c)\]Substitute values:\[(1 \times 1) - (0 \times 0)\]Multiply:\[1 - 0\]Simplify:\[= 1\]Therefore, the determinant is 1.

Certain angles such as 60°, 90°, and 135° can be constructed using standard geometric constructions.60° can be constructed using an equilateral triangle.90° can be constructed by drawing perpendicular lines.135° can be constructed by adding 90° and 45°, where 45° is obtained by bisecting 90°.However, 145° cannot be obtained directly through standard compass constructions or by repeated bisection of constructible angles.Therefore, 145° cannot be constructed.

To integrate the function, we apply the rule of integration term by term.The rule states:\[\int x^n dx = \frac{x^{n+1}}{n+1} + C\]Now integrate each term separately:First term:\[\int 3x^2 dx = 3 \int x^2 dx\]\[= 3 \left(\frac{x^3}{3}\right)\]\[= x^3\]Second term:\[\int 2x dx = 2 \int x dx\]\[= 2 \left(\frac{x^2}{2}\right)\]\[= x^2\]Third term:\[\int (-5) dx = -5x\]Combine all results and add the constant of integration:\[x^3 + x^2 - 5x + C\]Therefore, the integral is:\[x^3 + x^2 - 5x + C\]