Mathematics (Core)
Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 2003
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 45 Questions
JAMB Exam
Year: 2003
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. |
Evaluate \(log_{\sqrt{2}}4+log_{\frac{1}{2}}16-log_{4}32\) A. -5.5 B. -2.5 C. 2.5 D. 5.5
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Detailed SolutionNote that if we set \(log_{\sqrt{2}}4 = x_{1}, \hspace{1mm}solving\hspace{1mm}gives\hspace{1mm}x_{1}=4\\if\hspace{1mm}log_{\frac{1}{2}}16 = x_{2}, \Rightarrow x_{2} = -4\\Also \hspace{1mm}for\hspace{1mm}log_{4}32 = x_{3} = 2.5\\Combining\hspace{1mm}results,\hspace{1mm}x_{1}+x_{2}+x_{3} = 4+(-4)-2.5=-2.5\)Note that the solution has been separated to simplify solving difficulty due to different bases. |
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| 2. |
Simplify 213\(_4\) x 23\(_4\) A. 103114 B. 103214 C. 122314 D. 132114
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Detailed SolutionYou can either multiply directly in base 4 or you can decide to convert to base 10 and do the multiplication and re-convert. |
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| 3. |
In a class of 40 students, 32 offer mathematics, 24 offer Physics, and 4 offer neither Mathematics nor Physics. How many offer both Mat A. 20 B. 16 C. 8 D. 4
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Detailed SolutionUsing a venn diagram, let x = number who offer both Maths and Physics. So that (32-x) offer Maths and another (24-x) offer Physics.(32-x) + (24-x) + (x) + (4 who offer neither) = 40 => 60 - x = 40 => -x = -20 Therefore x = 20. |
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| 4. |
Find (\(\frac{1}{0.06} \div \frac{1}{0.042}\))-1 A. 1.43 B. 1.53 C. 3.14 D. 4.42
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Detailed Solution\((\frac{1}{0.06} \div \frac{1}{0.042})^{-1}\)= \((\frac{100}{6} \div \frac{1000}{42})^{-1}\) = \((\frac{100}{6} \times \frac{42}{1000})^{-1}\) = \((\frac{7}{10})^{-1}\) = \(\frac{10}{7}\) = 1.4285 \(\approxeq\) 1.43. |
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| 5. |
A woman buys 270 oranges for N1800.00 and sells at 5 for N40.00. What is her profit? A. N2,160.00 B. N1, 620.00 C. N630.00 D. N360.00
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Detailed SolutionC.P = N1800S.P = (270/5) x 40 = N2160.00 Profit = S.P - C.P = N2160 - N1800 = N360.00 |
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| 6. |
Simplify 1 - (7/3 x 5/4) + 3/5 A. -16/15 B. -79/60 C. -37/15 D. -151/60
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Detailed Solution\(1 - (\frac{7}{3} \times \frac{5}{4}) + \frac{3}{5}\)= \(1 - \frac{35}{12} + \frac{3}{5}\) = \(\frac{60 - 175 + 36}{60}\) = \(\frac{-79}{60}\) |
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| 7. |
Simplify (√98 -√50)/√32 A. 3 B. 1 C. 1/2 D. 1/4
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Detailed Solution(√98 - √50)/√32 = √(49x2) - √(25x2)=> (7√2 - 5√2)/4√2 = 1/2 |
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| 8. |
A cinema hall contains a certain number of people. If 22\(\frac{1}{2}\)% are children, 47\(\frac{1}{2}\)% are men and 84 are women, find the number of men in the hall. A. 63 B. 84 C. 113 D. 133
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Detailed Solution45/2% + 95/2% = 70%100% - 70% = 30% (i.e women) => 30% = 84 1% = 84/30 95/2% = (84/30 x 95/2) = 133. |
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| 9. |
If \(\frac{9^{2x-1}}{27^{x+1}} = 1\), find the value of x. A. 8 B. 5 C. 3 D. 2
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Detailed Solution\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\)\(\implies 9^{2x - 1} = 27^{x + 1}\) \((3^{2})^{2x - 1} = (3^{3})^{x + 1}\) \(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\) \(4x - 3x = 3 + 2 \implies x = 5\) |
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