Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1980
Level: SHS
Time:
Type: Question Paper
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# | Question | Ans |
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1. |
Find, correct to three significant figures, the value of \(\sqrt{41830}\). A. 205 B. 647 C. 2050 D. 6470 E. 64.7
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Detailed Solution\(\sqrt{41830} = 204.5238\)\(\approxeq 205\) (to three significant figures) |
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2. |
Write down the number 0.0052048 correct to three significant figures A. 0.005 B. 0.0052 C. 0.00520 D. 5.2048 E. 5204
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Detailed Solution0.0052048 = 0.00520 |
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3. |
Evaluate \((2^{0} + 4^{-\frac{1}{2}})^{2}\) A. \(\frac{1}{4}\) B. \(\frac{5}{4}\) C. \(\frac{9}{4}\) D. 4 E. 9
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Detailed Solution\((2^{0} + 4^{-\frac{1}{2}})^{2}\)= \((1 + (\frac{1}{4})^{\frac{1}{2}})^{2}\) = \((1 + \frac{1}{2})^{2}\) = \((\frac{3}{2})^{2} = \frac{9}{4}\) |
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4. |
Which of the following is NOT a factor of 12\(^{4}\) - 5\(^{4}\)? A. 7 B. 13 C. 17 D. 49 E. 169
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Detailed Solution124 - 54 = (122)2 - (52)2= (122 + 52) x (122 - 52) (144 + 25)(144 - 25) = (169)(119) Factors of above 169, 17, 13, 7 |
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5. |
Rationalize the denominator of the given expression \(\frac{\sqrt{1 + a} - \sqrt{a}}{1 + a + \sqrt{a}}\) A. 1 + 2a - 2\(\sqrt{a(1 + a)}\) B. \(\sqrt{1(1 + a)}\) C. 2a - 2\(\sqrt{a(1 + a)}\) D. 1 + 2a - 2\(\sqrt{a + b}\)
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Detailed Solution\(\frac{\sqrt{1 + a} - \sqrt{a}}{1 + a + \sqrt{a}}\) = \(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} + \sqrt{a}}\) x \(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} - \sqrt{a}}\)= \(\frac{\sqrt{1 + a + a}}{1 + a - a}\) = 2a + a(1 + a) = 1 + 2a - 2\(\sqrt{a(1 + a)}\) |
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6. |
When a dealer sells a bicycle for N81 he makes a profit of 80%. What did he pay for the bicycle? A. N73 B. N74.52 C. N75 D. N87.48 E. N75.52
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Detailed SolutionProfit = 8%\(\therefore S.P = N81 = (100 + 8)% = 108%\) \(108% = N81\) \(100% = \frac{81}{108} \times 100\) = \(\frac{8100}{108} = N75\) |
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7. |
The median of the set of numbers 4, 9, 4, 13, 7, 14, 10, 17 is A. 13 B. 7 C. \(\frac{19}{2}\) D. \(\frac{39}{4}\) E. 10
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Detailed SolutionRearranging in increasing order 4, 4, 7, 9, 10, 13, 14, 17, themedian = \(\frac{(9 + 10)}{2}\) = \(\frac{19}{2}\) |
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8. |
List all integer values of x satisfying the inequality -1 < 2x - 5 \(\leq\) 5 A. <2, 3, 4, 5 B. 3, 4, 5 C. 2, 3, 4 D. 3, 4
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Detailed Solution\(-1 < 2x - 5 \leq 5\)\(\implies -1 + 5 < 2x - 5 + 5 \leq 5 + 5\) \(4 < 2x \leq 10\) \(\implies 2 < x \leq 5 \) = 3, 4, 5. |
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9. |
Find the roots of the equation 10x2 - 13x - 3 = 0 A. x = \(\frac{3}{5}\) or -\(\frac{1}{2}\) B. x = \(\frac{3}{10}\) or -1 C. x = \(\frac{3}{10}\) or 1 D. x = \(\frac{1}{5}\) or \(\frac{-3}{2}\) E. x = -\(\frac{1}{5}\) or \(\frac{3}{2}\)
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Detailed Solution10x2 - 13x - 3 = 0 = 10x2 - 15x + 2x - 3 = 05x(2x - 3) + 2x - 3 = 0 = (5x + 1)(2x - 3) = 0 5x + 1 = 0 or 2x - 3 = 0 x = -\(\frac{1}{5}\) or \(\frac{3}{2}\) |
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10. |
A solid cylinder of radius 3cm has a total surface area of 36\(\pi\)cm2. Find its height A. 2cm B. 3cm C. 4cm D. 5cm E. 6cm
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Detailed SolutionArea 2\(\pi\)r2 + 2 \(\pi\)r(r + h)= 2\(\pi\)r(r + h) 36\(\pi\) = 6\(\pi\)(r + h) 36\(\pi\) = 6\(\pi\)(3 + h) 36\(\pi\) = 18\(\pi\) + 6\(\pi\)h 36\(\pi\) - 18\(\pi\) = 6\(\pi\)h Divide both side by 6\(\pi\) h = 3cm |
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