Mathematics (Core)
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1992
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1992
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
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| 1. |
Find n if 34n = 100112 A. 5 B. 6 C. 7 D. 8
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Detailed SolutionTo find n if 34n = 100112, convert both sides to base 10= 3n + 4 = (1 x 24) + (0 x23) + (0 x 22) + (1 x 21) + 1 x 2o = 3n + 4 = 16 + 0 + 0 + 2 + 1 3n + 4 = 19 3n = 15 n = 5 |
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| 2. |
The radius of a circle is given as 5cm subject to an error of 0.1cm. What is the percentage error in the area of the circle? A. \(\frac{1}{25}\) B. \(\frac{1}{4}\) C. 4 D. 25
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Detailed Solution% error in Area = \(\frac{\pi(5.1)^2 - \pi(5)^2 \times 100%}{\pi(5)^2}\)= \(\frac{\pi 26.01 - 25 \times 100%}{\pi(25)}\) = \(\frac{1.01}{25}\) x 100% = 4.04% |
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| 3. |
Evaluate \(\log_{b} a^{n}\) if \(b = a^{\frac{1}{n}}\). A. n2 B. n C. \(\frac{1}{n}\) D. \(\frac{1}{n^2}\)
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Detailed SolutionLet \(\log_{b} a^{n} = x\)\(\therefore a^{n} = b^{x}\) \(a^{n} = (a^{\frac{1}{n}})^{x}\) \(a^{n} = a^{\frac{x}{n}} \implies n = \frac{x}{n}\) \(x = n^{2}\) |
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| 4. |
What is the value of x satisfying the equation \(\frac{4^{2x}}{4^{3x}}\) = 2? A. -2 B. -\(\frac{1}{2}\) C. \(\frac{1}{2}\) D. 2
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Detailed Solution\(\frac{4^{2x}}{4^{3x}}\) = 242x - 3x = 2 4-x = 2 (22)-x = 21 Equating coefficients: -2x = 1 x = -\(\frac{1}{2}\) |
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| 5. |
Simplify \(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\) A. 4.0 x 10-3 B. 5.0 x 10-2 C. 2.0 x 10-1 D. 5.0 x 10-3
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Detailed Solution\(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\) = \(\frac{1.25 \times 2}{6.25}\) x 104 - 1 - 5\(\frac{2.50}{6.25}\) x 10-2 = \(\frac{250}{625}\) x 10-2 0.4 x 10-2 = 4.0 x 10-3 |
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| 6. |
Simplify 5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) A. 17\(\sqrt{4}\) B. 4\(\sqrt{17}\) C. 17\(\sqrt{2}\) D. 12\(\sqrt{4}\)
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Detailed Solution5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) = 5(3\(\sqrt{2}\)) - 3(6\(\sqrt{2}\)) + 4(5\(\sqrt{2}\))15\(\sqrt{2}\) - 18\(\sqrt{2}\) + 20\(\sqrt{2}\) = 35\(\sqrt{2}\) - 18\(\sqrt{2}\) = 17\(\sqrt{2}\) |
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| 7. |
If x = 3 - \(\sqrt{3}\), find x2 + \(\frac{36}{x^2}\) A. 9 B. 18 C. 24 D. 27
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Detailed Solutionx = 3 - \(\sqrt{3}\)x2 = (3 - \(\sqrt{3}\))2 = 9 + 3 - 6\(\sqrt{34}\) = 12 - 6\(\sqrt{3}\) = 6(2 - \(\sqrt{3}\)) ∴ x2 + \(\frac{36}{x^2}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{36}{6(2 - \sqrt{3})}\) 6(2 - \(\sqrt{3}\)) + \(\frac{6}{2 - \sqrt{3}}\) = 6(- \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{4 - 3}\) 6(2 - \(\sqrt{3}\)) + 6(2 + \(\sqrt{3}\)) = 12 + 12 = 24 |
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| 8. |
If x = (all prime factors of 44) and y = (all prime factors of 60), the elements of X ∪ Y and X ∩ Y respectively are A. (2, 4, 3, 5, 11) and (4) B. (4, 3, 5, 11) and (3, 4) C. (2, 5, 11) and (2) D. (2, 3, 5, 11) and (2)
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Detailed Solutionx = (all prime factors of 44) and y = (all prime factors of 60)∴ x = (2, 11), y = (2, 3, 5) X ∪ Y = (2, 3, 5, 11), X ∩ Y = (2) |
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| 9. |
If U = (1, 2, 3, 6, 7, 8, 9, 10) is the universal set. E = (10, 4, 6, 8, 10) and F = {x: 1x\(^{2}\) = 2\(^{6}, x is odd}. Find (E ∩ F)', where ' means the complement of a set. A. (0) B. U C. (8) D. \(\phi\)
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Detailed SolutionU = (1, 2, 3, 6, 7, 8, 9, 10)E = (10, 4, 6, 8, 10) F = (x : x\(^2\) = 2\(^6\), x is odd) ∴ F = \(\phi\) Since x\(^2\) = 2\(^6\) = 64 x = \(\pm 8\) which is even ∴ E ∩ F = \(\phi\) Since there are no common elements |
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| 10. |
Factorize \(9p^2 - q^2 + 6qr - 9r^2\) A. (3p - 3q + r)(3p - q - 3r) B. (6p - 3q - 3r)(3p - q - 4r) C. (3p - q + 3r)(3p + q - 3r) D. (3q - p + 3r)(3q - p + 3r)
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Detailed Solution\(9p^{2} - q^{2} + 6qr - 9r^{2}\)= \(9p^{2} - (q^{2} - 6qr + 9r^{2})\) = \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\) = \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\) = \(9p^{2} - (q - 3r)^{2}\) = \((3p + (q - 3r))(3p - (q - 3r))\) = \((3p + q - 3r)(3p - q + 3r)\) |
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