2023 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

$$text{Since the letter L is repeated twice, the total number of permutations is given by } \frac{8!}{2!} = 20160. \text{ This accounts for the repeated letters.}$$

$$\text{Simplify by approximate multiplication: } 160.16 \approx 160, \ 20.2 \approx 20. \text{ Thus, } 160 \times 160 \times 20 = 512000.$$

$$\text{First compute } -4 * 2 = (-4)^2 \cdot 2 = 32, \text{ and } 7 * -1 = 7^2 \cdot -1 = -49. \text{ Then } 32 ^ -49 = 2(32) + (-49) = 64 – 49 = 15.$$

$$\text{Use substitution: let } u = 4x – 6x^2, \text{ then differentiate to find the integral, resulting in } -85.\text{ Verify the simplifications for consistency.}$$

$$\text{Percentage decrease is given by } \frac{\text{Decrease}}{\text{Original Population}} \times 100 = \frac{1230 – 1040}{1230} \times 100 = 15.45\%. $$

$$\text{Let the exterior angle be } x. \text{ Then, the interior angle is } 5x. \text{ Since their sum is } 180^{\circ}, \text{ solve } x + 5x = 180^{\circ} \text{ to get } x = 30^{\circ}. \text{ Number of sides } = \frac{360}{30} = 12. $$

$$\text{Given } A = \pi r^2, \text{ differentiate to get } \frac{dA}{dt} = 2\pi r \frac{dr}{dt}. \text{ Substitute } A = 2, \text{ so } r = \sqrt{\frac{2}{\pi}}. \text{ Solve for } \frac{dr}{dt} \text{ to get } 0.200 \text{ cm} \text{s}^{-1}. $$

$$\text{Cross-multiply: } y(4x + 5d) = 3x – 9c. \text{ Rearrange: } x = \frac{9c – 5dy}{4y – 3}. $$

$$\text{Expand: } 3x – 3 \leq 2x – 6. \text{ Simplify: } x \leq -3. $$

$$\text{Using tangent properties, } \angle PQR = 180^{\circ} – 2 \angle PSR. \text{ Substitute: } \angle PQR = 180^{\circ} – 72^{\circ} = 144^{\circ}. $$

$$\text{Speed = } \frac{225}{4.5} = 50 \text{ km/hr. Time for 150 km } = \frac{150}{50} = 3 \text{ hrs.}$$

$$\text{Perform binary division: } 1101001_2 \div 101_2 = 11101_2.$$

$$\text{Combine like terms: } a = 9, \ b = 5. \text{ Substitute and verify.}$$

$$\text{Using the distance formula: } \sqrt{(t + 3)^2 + (-5 + 14)^2} = 9. \text{ Solve: } t = -3.$$

$$\text{Interest rate: } R = \frac{200}{10} = 20\%. \text{ For quintupling: } T = \frac{400}{20} = 20 \text{ years.}$$

$$\text{Simplify: } \frac{(x – 2)(x + 6)}{x(x – 2)} = \frac{x + 6}{x}. \text{ Substitute: } \frac{2 + 6}{2} = 4.$$

$$\text{Use midpoint formula for grouped data: } \mu = \frac{(5.5 \times 29) + (7.5 \times 40) + (9.5 \times 38)}{107} = 7.7.$$

$$\text{Majority means at least 3 women. Use combinations: } \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1} = 60.$$

$$\tan^{-1}\left(\frac{6}{8}\right) = 36.87^{\circ}. \text{ Add } 180^{\circ}: \text{ bearing } = 217^{\circ}. $$

$$\text{Probability of no heads: } \left(\frac{1}{2}\right)^3 = \frac{1}{8}. \text{ So, } P(\text{at least one head}) = 1 – \frac{1}{8} = \frac{7}{8}. $$

$$\text{The slope of the given line is } -\frac{3}{2}. \text{ Perpendicular slope: } \frac{2}{3}. \text{ Equation: } y – 3 = \frac{2}{3}(x – 2). $$

$$\frac{dy}{dx} = 10x^{5/3} – 8x^{2/3}. \text{ Differentiate using the product and chain rules.}$$

$$\text{Find intersection points by solving: } 2x^2 + 10 = 4x + 16. \text{ Then integrate over the bounds.}$$

$$\text{Convert } 40_2 \text{ to decimal: } 4 \cdot 2^1 + 0 \cdot 2^0 = 4. \text{ Solve: } 4y = 10 \Rightarrow y = 2.$$

$$\text{Combine logs: } \log((y + 8)(y – 8)) = \log(3^2 \cdot 5^2). \text{ Simplify: } y^2 – 64 = 225 \Rightarrow y = \pm17.$$

$$\text{Use inclusion-exclusion: } n(A \cup B) = n(A) + n(B) – n(A \cap B). \text{ Solve: } 500 = 350 + 400 – x \Rightarrow x = 250.$$

$$\text{Difference of squares: } (2x – y)(2x + y)(4x^2 + y^2). \text{ Check expansion for correctness.}$$

$$\text{Find } A – B = \{1, 3, 5\}, B – A = \{8\}. \text{ Union: } \{1, 3, 5, 8\}. $$

$$\text{If } n \text{ is the number of white balls, } \frac{n}{8} = \frac{1}{2} \Rightarrow n = 4. \text{ Total probability: } \frac{8}{12}, \frac{4}{12}. $$

$$\text{Total frequency } = 107. \text{ Median is the 54th value in cumulative frequency: } 7-8. \text{ Median midpoint: } 7.5. $$

$$\text{Use inverse operations: Solve } A = \begin{bmatrix} 2 & -1/2 \\ 1 & -1/2 \end{bmatrix}. $$

$$\text{Find cumulative frequency and interpret the values to determine the range for scores above 25%.}$$

$$\text{Substitute } x = -1 \text{ in the polynomial: } -2(-1)^3 + 6(-1)^2 + 17(-1) – 21 = -30. $$

$$\text{A binary operation is commutative if } a*b = b*a \text{ for all } a, b \in A. $$

$$\text{Simplify: } x^2 – x – 6 \leq 0. \text{ Factorize: } (x – 3)(x + 2) \leq 0. \text{ Solution: } -2 \leq x \leq 3.$$

$$\text{Common difference: } 8 – 3 = 5. \text{ General term: } a_n = 5n – 2. $$

$$\text{The locus is the pair of angle bisectors formed by the intersecting lines.}$$

$$\text{Let the third number be } x. \text{ Ratios: } 1.35x \text{ and } 1.8x. \text{ Simplify: } \frac{1.35x}{1.8x} = \frac{3}{4}. $$

$$\text{Using trigonometry, } h = 24 + 24\sqrt{3} – 24. \text{ The result is } 24\sqrt{3}. $$

$$\text{Using Pythagoras or the given formula, the length } AB \text{ is } 36.4 \text{ cm. Ensure all inputs are accurate.}$$

$$\text{Cost price: } \frac{230}{1.15} = 200. \text{ Loss: } 200 – 180 = 20. \text{ Loss percentage: } \frac{20}{200} \times 100 = 10\%. $$

$$\text{Use the arc formula: } l = \theta \cdot r \text{ (in radians). Substitute given values for accuracy.}$$

$$\text{Max area: } 38.5 \cdot 52.5 = 2021.25. \text{ Min area: } 37.5 \cdot 51.5 = 1931.25. \text{ Range: } 1931.25 \leq A < 2021.25.$$

$$\text{Let total items be } T. \frac{1}{9}T + \frac{5}{8}(\frac{8}{9}T) + 81 = T. \text{ Solve: Brand B exceeds Brand C by 243.}$$

$$\text{Sides: } 40, 60, 80. \text{ Use Heron’s formula: } A = \sqrt{s(s-a)(s-b)(s-c)}. \text{ Area } \approx 1162 \text{ cm}^2.$$

$$\text{CI: } A = P(1 + r)^t. \text{ Substitute: } 15700(1 + 0.08)^2 = 18312.48. \text{ CI } = 18312.48 – 15700 = 2612.48.$$

$$\text{Volume: } V = \pi r^2 h. \text{ Substitute given values: } V \approx 9856 \text{ cm}^3.$$

$$\text{Common difference: } d = \frac{12 – 6}{5 – 3} = 3. \text{ First term: } a = 6 – 2d = 0. \text{ Sum: } S_n = \frac{n}{2}(2a + (n-1)d). \text{ Substitute: } S_{12} = 198.$$