Mathematics (Core)
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1992
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
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
Tips to make the best study space and quality time for studies are essential.
Middle East and North Africa MENA Scholarship Program (MSP) initiative provides scholarships
12 Most Effective Ways to Pass Any Exams Without Studying Hard even when you don't have enough time.
| # | Question | Ans |
|---|---|---|
| 1. |
Find n if 34n = 100112 A. 5 B. 6 C. 7 D. 8
Show Content
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 |
|
| 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
Show Content
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% |
|
| 3. |
Evaluate \(\log_{b} a^{n}\) if \(b = a^{\frac{1}{n}}\). A. n2 B. n C. \(\frac{1}{n}\) D. \(\frac{1}{n^2}\)
Show Content
Detailed SolutionLet \(\log_{b} a^{n} = x\)\(\therefore a^{n} = b^{x}\) \(a^{n} = (a^{\frac{1}{n}})^{x}\) \(a^{n} = a^{\frac{x}{n}} \implies n = \frac{x}{n}\) \(x = n^{2}\) |
|
| 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
Show Content
Detailed Solution\(\frac{4^{2x}}{4^{3x}}\) = 242x - 3x = 2 4-x = 2 (22)-x = 21 Equating coefficients: -2x = 1 x = -\(\frac{1}{2}\) |
|
| 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
Show Content
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 |
|
| 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\(\sqrt{4}\)
Show Content
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\(\sqrt{2}\) |
|
| 7. |
If x = 3 - \(\sqrt{3}\), find x2 + \(\frac{36}{x^2}\) A. 9 B. 18 C. 24 D. 27
Show Content
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 |
|
| 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)
Show Content
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) |
|
| 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. \(\phi\)
Show Content
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 |
|
| 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)
Show Content
Detailed Solution\(9p^{2} - q^{2} + 6qr - 9r^{2}\)= \(9p^{2} - (q^{2} - 6qr + 9r^{2})\) = \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\) = \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\) = \(9p^{2} - (q - 3r)^{2}\) = \((3p + (q - 3r))(3p - (q - 3r))\) = \((3p + q - 3r)(3p - q + 3r)\) |
Preview displays only 10 out of the 49 Questions