Year :
2002
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
The binary operation * is defined on the set of integers p and q by p*q = pq + p + q. Find 2 * (3 * 4). A. 59 B. 19 C. 67 D. 38 Detailed Solution\(p \ast q = pq + p + q\)\(2 \ast (3 \ast 4) \) \(3 \ast 4 = 12 + 3 + 4 = 19\) \(2 \ast 19 = 38 + 2 + 19 = 59\) |
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| 22. |
The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is A. 11/3 B. 10/3 C. 5/2 D. 3/2 Detailed SolutionThe series is geometric with common ratio \(\frac{1}{3}\).\(S_{\infty} = \frac{a}{1 - r}\) = \(\frac{1}{1 - \frac{1}{3}} \) = \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\) |
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| 23. |
If x varies directly as √n and x = 9 when n = 9, find x when n = (17/9) A. 4 B. 27 C. √3 D. √17 Detailed Solution\(x \propto \sqrt{n}\)\(x = k \sqrt{n}\) \(9 = k \sqrt{9} \implies 9 = 3k\) \(k = 3\) \(x = 3 \sqrt{n}\) When n = 17/9, \(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\) |
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| 24. |
A chord of a circle subtends an angle of 120° degrees at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector. A. 4π cm2 B. 32 π cm2 C. 16 π cm2 D. 8 π cm2 Detailed SolutionDiameter = 4\(\sqrt{3}\) cmradius = 2\(\sqrt{3}\) cm Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240°\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\) |
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| 25. |
If tan θ = 4/3, calculate sin\(^2\) θ - cos\(^2\) θ. A. 16/25 B. 24/25 C. 7/25 D. 9/25 Detailed Solution\(\tan \theta = \frac{opposite}{adjacent} = \frac{4}{3}\)Hyp\(^2\) = 4\(^2\) + 3\(^2\) Hyp = 5. \(\sin \theta = \frac{4}{5}; \cos \theta = \frac{3}{5}\) \(\sin^{2} \theta - \cos^{2} \theta = \frac{16}{25} - \frac{9}{25}\) = \(\frac{7}{25}\) |
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| 26. |
Find the equation of the set of points which are equidistant from the parallel lines x = 1 and x = 7 A. y = 3 B. x = 3 C. x = 4 D. y = 4 Detailed SolutionThe line equidistant from x = 1 and x = 7 is\(x = \frac{1 + 7}{2} \implies x = 4\) |
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| 27. |
A bucket is 12 cm in diameter at the top, 8 cm in diameter at the bottom and 4 cm deep. Calculate its volume. A. 304π/3 cm3 B. 144π cm3 C. 128π cm3 D. 72π cm3 Detailed SolutionVolume of a frustrum with top of radius R and bottom r and height h = \(\frac{1}{3} \pi (R^{2} + Rr + r^{2})\)V = \(\frac{1}{3} \pi \times 4 \times (6^2 + 6 \times 4 + 4^2)\) = \(\frac{304}{3} \pi cm^{3}\) |
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| 28. |
The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have? A. 12 B. 20 C. 40 D. 10 Detailed SolutionThe formula for the sum of the interior angles of a regular polygon = (2n - 4) x 90°Given: Sum = 20 right angles (2n - 4) x 90° = 20 \times 90° ⇒ 2n - 4 = 20 2n = 24; n = 12. |
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| 29. |
Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8 A. (2, 0) B. (1, -2) C. (-1, -2) D. (1, 2) Detailed Solution2y = 4x - 8 \(\implies\) y = 2x - 4.When x = 0, y = -4. When y = 0, x = 2. The midpoint between (0, -4) and (2, 0) = \((\frac{0 + 2}{2}, \frac{-4 + 0}{2})\) = \((1, -2)\) |
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| 30. |
A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45°. if the distance between the hunter and the tree is 10.4 m, find the height of the tree. A. 9.0 m B. 0 m C. 8.8 m D. 10.4 m Detailed Solution \(\tan 45 = \frac{x}{10.4}\) \(x = 10.4m\) |
| 21. |
The binary operation * is defined on the set of integers p and q by p*q = pq + p + q. Find 2 * (3 * 4). A. 59 B. 19 C. 67 D. 38 Detailed Solution\(p \ast q = pq + p + q\)\(2 \ast (3 \ast 4) \) \(3 \ast 4 = 12 + 3 + 4 = 19\) \(2 \ast 19 = 38 + 2 + 19 = 59\) |
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| 22. |
The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is A. 11/3 B. 10/3 C. 5/2 D. 3/2 Detailed SolutionThe series is geometric with common ratio \(\frac{1}{3}\).\(S_{\infty} = \frac{a}{1 - r}\) = \(\frac{1}{1 - \frac{1}{3}} \) = \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\) |
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| 23. |
If x varies directly as √n and x = 9 when n = 9, find x when n = (17/9) A. 4 B. 27 C. √3 D. √17 Detailed Solution\(x \propto \sqrt{n}\)\(x = k \sqrt{n}\) \(9 = k \sqrt{9} \implies 9 = 3k\) \(k = 3\) \(x = 3 \sqrt{n}\) When n = 17/9, \(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\) |
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| 24. |
A chord of a circle subtends an angle of 120° degrees at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector. A. 4π cm2 B. 32 π cm2 C. 16 π cm2 D. 8 π cm2 Detailed SolutionDiameter = 4\(\sqrt{3}\) cmradius = 2\(\sqrt{3}\) cm Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240°\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\) |
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| 25. |
If tan θ = 4/3, calculate sin\(^2\) θ - cos\(^2\) θ. A. 16/25 B. 24/25 C. 7/25 D. 9/25 Detailed Solution\(\tan \theta = \frac{opposite}{adjacent} = \frac{4}{3}\)Hyp\(^2\) = 4\(^2\) + 3\(^2\) Hyp = 5. \(\sin \theta = \frac{4}{5}; \cos \theta = \frac{3}{5}\) \(\sin^{2} \theta - \cos^{2} \theta = \frac{16}{25} - \frac{9}{25}\) = \(\frac{7}{25}\) |
| 26. |
Find the equation of the set of points which are equidistant from the parallel lines x = 1 and x = 7 A. y = 3 B. x = 3 C. x = 4 D. y = 4 Detailed SolutionThe line equidistant from x = 1 and x = 7 is\(x = \frac{1 + 7}{2} \implies x = 4\) |
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| 27. |
A bucket is 12 cm in diameter at the top, 8 cm in diameter at the bottom and 4 cm deep. Calculate its volume. A. 304π/3 cm3 B. 144π cm3 C. 128π cm3 D. 72π cm3 Detailed SolutionVolume of a frustrum with top of radius R and bottom r and height h = \(\frac{1}{3} \pi (R^{2} + Rr + r^{2})\)V = \(\frac{1}{3} \pi \times 4 \times (6^2 + 6 \times 4 + 4^2)\) = \(\frac{304}{3} \pi cm^{3}\) |
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| 28. |
The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have? A. 12 B. 20 C. 40 D. 10 Detailed SolutionThe formula for the sum of the interior angles of a regular polygon = (2n - 4) x 90°Given: Sum = 20 right angles (2n - 4) x 90° = 20 \times 90° ⇒ 2n - 4 = 20 2n = 24; n = 12. |
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| 29. |
Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8 A. (2, 0) B. (1, -2) C. (-1, -2) D. (1, 2) Detailed Solution2y = 4x - 8 \(\implies\) y = 2x - 4.When x = 0, y = -4. When y = 0, x = 2. The midpoint between (0, -4) and (2, 0) = \((\frac{0 + 2}{2}, \frac{-4 + 0}{2})\) = \((1, -2)\) |
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| 30. |
A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45°. if the distance between the hunter and the tree is 10.4 m, find the height of the tree. A. 9.0 m B. 0 m C. 8.8 m D. 10.4 m Detailed Solution\(\tan 45 = \frac{x}{10.4}\) \(x = 10.4m\) |