Paper 1  Objectives  48 Questions
JAMB Exam
Year: 1994
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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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. 

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.\(\frac{x}{12} = 3 \implies x = 36\) \(\frac{36 + y}{13} = 5 \implies 36 + y = 65\) \(y = 65  36 = 29\) 

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\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} 1 + \frac{3}{4})]\)\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10}  \frac{10}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{1}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{2 + 15}{20})]\) = \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\) \(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\) = \(\frac{28}{39}\) 

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 

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 log_{5} 0.04 = x 5x = 0.04 x = \(\frac{4}{100}\) = 5^{2} Let log_{3} 9 = z 3^{2} = 3^{2} z = 3 

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 

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 

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 

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') = \(\phi\) 

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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