2025 - JAMB Mathematics Past Questions and Answers - page 5

Let the total salary be represented by 1.Fraction spent on shirts:\[\frac{1}{5}\]Fraction remaining after buying shirts:\[1 - \frac{1}{5} = \frac{4}{5}\]Fraction spent on transport from the remainder:\[\frac{1}{3} \times \frac{4}{5} = \frac{4}{15}\]Fraction remaining after transport:\[\frac{4}{5} - \frac{4}{15}\]Convert to common denominator:\[\frac{4}{5} = \frac{12}{15}\]Subtract:\[\frac{12}{15} - \frac{4}{15} = \frac{8}{15}\]Therefore, the fraction left is:\[\frac{8}{15}\]

Start with the inequality:\[2x + 3 > 5x + 8\]Subtract 2x from both sides:\[3 > 3x + 8\]Subtract 8 from both sides:\[-5 > 3x\]Divide both sides by 3:\[-\frac{5}{3} > x\]Rewrite in standard form:\[x < -\frac{5}{3}\]Therefore, the solution is:\[x < -\frac{5}{3}\]

To convert a number to base 5, repeatedly divide by 5 and record the remainders.First division:\[137 \div 5 = 27 \text{ remainder } 2\]Second division:\[27 \div 5 = 5 \text{ remainder } 2\]Third division:\[5 \div 5 = 1 \text{ remainder } 0\]Fourth division:\[1 \div 5 = 0 \text{ remainder } 1\]Now write the remainders from bottom to top:\[1022\]Therefore:\[137 = 1022_5\]

Use the identity:\[\sin^2 A + \cos^2 A = 1\]Substitute the value of cosine:\[\sin^2 A + \left(\frac{12}{13}\right)^2 = 1\]Calculate square:\[\sin^2 A + \frac{144}{169} = 1\]Subtract:\[\sin^2 A = \frac{25}{169}\]Take square root:\[\sin A = \frac{5}{13}\]Use definition of tangent:\[\tan A = \frac{\sin A}{\cos A}\]Substitute values:\[\tan A = \frac{5/13}{12/13}\]Simplify:\[\tan A = \frac{5}{12}\]Therefore:\[\tan A = \frac{5}{12}\]

First express the number in powers of 10.\[0.00001 = 10^{-5}\]Now apply logarithm rule:\[\log_{10}(10^{-5}) = -5\]Apply the negative sign outside:\[-(-5) = 5\]Therefore, the simplified value is:\[5\]

Mean = Total ÷ NumberTotal weight of 15 bars:\[15 \times 1000 = 15000\]Total weight of remaining 14 bars:\[14 \times 995 = 13930\]Weight of heaviest bar:\[15000 - 13930 = 1070\]Therefore, the weight is:\[1070 \text{ kg}\]

Use the chain rule:\[\frac{d}{dx}[\sin u] = \cos u \times \frac{du}{dx}\]Let:\[u = 4x\]Differentiate u:\[\frac{du}{dx} = 4\]Apply chain rule:\[\frac{dy}{dx} = \cos(4x) \times 4\]Simplify:\[\frac{dy}{dx} = 4\cos 4x\]Therefore, the derivative is:\[4\cos 4x\]

Direct variation means:\[A = kb^2\]Substitute given values:\[4 = k(1)^2\]\[k = 4\]Now find A when b = 2:\[A = 4(2)^2\]\[A = 4 \times 4\]\[A = 16\]Therefore:\[A = 16\]

The modal class is the class interval with the highest frequency.From the table, the class interval 21–30 has the highest frequency.To find the upper class boundary, add 0.5 to the upper class limit:\[30 + 0.5 = 30.5\]Therefore, the upper class boundary is:\[30.5\]

To evaluate the limit, substitute x = 2 directly into the expression:\[\frac{2^3 - 2(2^2) + 6(2) - 12}{2 - 2}\]Simplify numerator step by step:\[2^3 = 8\]\[2(2^2) = 8\]\[6(2) = 12\]Substitute values:\[\frac{8 - 8 + 12 - 12}{0}\]This gives:\[\frac{0}{0}\]This is an indeterminate form, so factor the numerator.Group terms:\[(x^3 - 2x^2) + (6x - 12)\]Factor each group:\[x^2(x - 2) + 6(x - 2)\]Factor out common term (x − 2):\[(x - 2)(x^2 + 6)\]Substitute back into original expression:\[\frac{(x - 2)(x^2 + 6)}{x - 2}\]Cancel (x − 2):\[x^2 + 6\]Now substitute x = 2:\[2^2 + 6 = 4 + 6 = 10\]Therefore, the limit is:\[10\]