Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1994
Level: SHS
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Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1994
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
Solve for x if \(25^{x} + 3(5^{x}) = 4\) A. 1 or -4 B. 0 C. 1 D. -4 or 0
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Detailed Solution\(25^{x} + 3(5^{x}) = 4\)Let \(5^{x}\) = y. \((5^{2})^{x} + 3(5^{x}) - 4 = 0\) \(y^{2} + 3y - 4 = 0\) \(y^{2} - y + 4y - 4 = 0\) \(y(y - 1) + 4(y - 1) = 0\) \((y + 4)(y - 1) = 0\) \(y = -4 ; y = 1\) y = -4 is not possible. y = 1 \(\implies\) x = 0. |
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| 2. |
The mean of twelve positive numbers is 3. When another number is added, the mean becomes 5. Find the thirteenth number A. 29 B. 26 C. 25 D. 24
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Detailed SolutionLet the sum of the 12 numbers be x and the 13th number be y.\(\frac{x}{12} = 3 \implies x = 36\) \(\frac{36 + y}{13} = 5 \implies 36 + y = 65\) \(y = 65 - 36 = 29\) |
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| 3. |
Evaluate \(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\) A. \(\frac{28}{39}\) B. \(\frac{13}{39}\) C. \(\frac{39}{28}\) D. \(\frac{84}{13}\)
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Detailed Solution\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\)\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\) = \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\) \(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\) = \(\frac{28}{39}\) |
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| 4. |
Evaluate \(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\) A. 0.013 B. 0.014 C. 0.14 D. 0.13
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Detailed Solution\(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\)= \(\frac{36}{420} \times \frac{54}{90} \times \frac{63}{240}\) = \(\frac{6}{70} \times \frac{18}{30} \times \frac{21}{80}\) = \(\frac{27}{2000}\) = 0.0135 \(\approx\) = 0.014 |
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| 5. |
Evaluate \(\frac{log_5 (0.04)}{log_3 18 - log_3 2}\) A. 1 B. -1 C. \(\frac{2}{3}\) D. -\(\frac{2}{3}\)
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Detailed Solution\(\frac{log_5 0.04}{log_3 18 - log_3 2}\)= \(\frac{log_5 0.04}{log_3(\frac{18}{2})}\) = \(\frac{log_5 0.04}{log_3 9}\) = \(\frac{-2}{2}\) = -1 Let log5 0.04 = x 5x = 0.04 x = \(\frac{4}{100}\) = 5-2 Let log3 9 = z 32 = 32 z = 3 |
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| 6. |
Without using table, solve the equation 8x-2 = \(\frac{2}{25}\) A. 4 B. 6 C. 8 D. 10
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Detailed Solution8x-2 = \(\frac{2}{25}\)= 200x-2 = 2 = 100x-2 = 1 x-2 = \(\frac{1}{100}\) x-2 = 10-2 x = 10 |
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| 7. |
Simplify \(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\) A. 5√3 B. 6√3 C. 8√3 D. 18√3
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Detailed Solution\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\)Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\) = (√16 x √3) + (√25 x √3) - \(\frac{9}{\sqrt{3}}\) =4√3 + 5√3 - \(\frac{9}{\sqrt{3}}\) Rationalize \(\to\) 9√3 = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{9\sqrt{3}}{\sqrt{9}}\) - \(\frac{9\sqrt{3}}{\sqrt{3}}\) = 3√3 |
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| 8. |
Given that \(\sqrt{2} = 1.414\), find without using tables, the value of \(\frac{1}{\sqrt{2}}\) A. 0.141 B. 0.301 C. 0.667 D. 0.707
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Detailed Solution\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\)= \(\frac{\sqrt{2}}{2}\) = \(\frac{1.414}{2}\) = 0.707 |
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| 9. |
Given that for sets A and B, in a universal set E, A \(\subseteq\) B then A \(\cap\)(A \(\cap\) B)' is A. A B. \(\phi\) C. B D. E
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Detailed SolutionA \(\subset\) B means A is contained in B i.e. A is a subset of B(A \(\cap\) B)' = A'A(A \(\cap\) B)' = A \(\cap\) A' The intersection of complement of a set P and P' has no element i.e. n(A \(\cap\) A') = \(\phi\) |
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| 10. |
Simplify \(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\) A. \(\frac{3}{5}\) B. \(\frac{2}{5}\) C. \(\frac{2m - u}{5m + u}\) D. \(\frac{m - 2u}{m + 5u}\)
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Detailed Solution\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\)= \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\) = \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\) = \(\frac{3}{5}\) |
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