Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1994
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
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
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
Get full scholarship paid tuition from 5 countries with free education in 2022 for international students
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
| # | 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
Show Content
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. |
|
| 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
Show Content
Detailed SolutionLet the sum of the 12 numbers be x and the 13th number be y.\(\frac{x}{12} = 3 \implies x = 36\) \(\frac{36 + y}{13} = 5 \implies 36 + y = 65\) \(y = 65 - 36 = 29\) |
|
| 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. \(\frac{84}{13}\)
Show Content
Detailed Solution\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\)\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\) = \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\) \(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\) = \(\frac{28}{39}\) |
|
| 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
Show Content
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 |
|
| 5. |
Evaluate \(\frac{log_5 (0.04)}{log_3 18 - log_3 2}\) A. 1 B. -1 C. \(\frac{2}{3}\) D. -\(\frac{2}{3}\)
Show Content
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 |
|
| 6. |
Without using table, solve the equation 8x-2 = \(\frac{2}{25}\) A. 4 B. 6 C. 8 D. 10
Show Content
Detailed Solution8x-2 = \(\frac{2}{25}\)= 200x-2 = 2 = 100x-2 = 1 x-2 = \(\frac{1}{100}\) x-2 = 10-2 x = 10 |
|
| 7. |
Simplify \(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\) A. 5√3 B. 6√3 C. 8√3 D. 18√3
Show Content
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 |
|
| 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
Show Content
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 |
|
| 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
Show Content
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') = \(\phi\) |
|
| 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. \(\frac{m - 2u}{m + 5u}\)
Show Content
Detailed Solution\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\)= \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\) = \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\) = \(\frac{3}{5}\) |
Preview displays only 10 out of the 48 Questions