Mathematics (Core)
Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1993
Level: SHS
Time:
Type: Question Paper
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Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1993
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
Tips to make the best study space and quality time for studies are essential.
Revision tips on test preparation for Grade A students to. Grade A students tips for any test.
| # | Question | Ans |
|---|---|---|
| 1. |
Integrate \(\frac{1 - x}{x^3}\) with respect to x A. \(\frac{x - x^2}{x^4}\) + k B. \(\frac{4}{x^4} - \frac{3 + k}{x^3}\) C. \(\frac{1}{x} - \frac{1}{2x^2}\) + k D. \(\frac{1}{3x^2} - \frac{1}{2x}\) + k
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Detailed Solution\(\int \frac{1 - x}{x^3}\)= \(\int^{1}_{x^3} - \int^{x}_{x^3}\) = x-3 dx - x-2dx = \(\frac{1}{2x^2} + \frac{1}{x}\) |
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| 2. |
Change 7110 to base 8 A. 1078 B. 1068 C. 718 D. 178
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Detailed Solution\(\begin{array}{c|c} 8 & 71 \\ 8 & 8 \text{rem} 7\\ 8 & 1 \text{rem} 0\end{array}\)= 1078 |
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| 3. |
Evaluate \(\frac{3524}{0.05}\) correct to 3 significant figures A. 705 B. 70,000 C. 70, 480 D. 70, 500
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Detailed Solution\(\frac{3524}{0.05}\) = 70480\(\approx\) 70500(3 s.g) |
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| 4. |
If 9(x - \(\frac{1}{2}\)) 3x2 A. \(\frac{1}{2}\) B. 1 C. 2 D. 3
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Detailed Solution9(x - \(\frac{1}{2}\)) 3x2 = 32(x - \(\frac{1}{2}\)) = 3x2∴ 2(x - \(\frac{1}{2}\)) = x2 2x - 1 = x2 hence x2 - 2x + 1 = 0 (x - 1)(x - 1) = 0 x = 1 |
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| 5. |
Solve for y in the equation 10^1 x 5(2y - 2) x 4(y - 1) = 1 A. \(\frac{3}{4}\) B. \(\frac{5}{4}\) C. \(\frac{2}{3}\) D. 5
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Detailed Solution10y x 5(2y - 2) x 4(y - 1) = 1but 10y - (5 x 2)y = 5y x 2y = (Law of indices) 5y x 2y x 5(2y - 2) x 4(y - 1) = 1 but 4(y - 1) = 22(y - 1) = 2y - 2 (Law of indices) 5y x 5(2y -2) x 2(- 2) = 1 5(3y -2) x 2y x 2(2y -2) = 1 = 5(3y -2) x 2(3y -2) = 1 But ao = 1 10(3y -2) = 10o 3y - 2 = 0 ∴ y = \(\frac{2}{3}\) |
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| 6. |
Simplify \(\frac{1}{√3 - 2}\) - \(\frac{1}{√3 + 2}\) A. 3 B. \(\frac{2}{3}\) C. 7 D. -4
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Detailed Solution\(\frac{1}{√3 - 2}\) - \(\frac{1}{√3 + 2}\)L.C.M = (3- 2) (3 + 2) ∴ \(\frac{1}{\sqrt{3 - 2}}\) - \(\frac{1}{\sqrt{3 - 2}}\) = \(\frac{\sqrt{3 + 2} - \sqrt{3 - 2}}{\sqrt{3 - 2} + \sqrt{3 - 2}}\) \(\frac{√3 + 2 - √3 + 2}{3 - 2√3 + 2√3 - 4}\) = \(\frac{4}{3 - 2}\) = \(\frac{4}{-1}\) = -4 |
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| 7. |
If 2log3 y + log3 x2 = 4, then y is A. 4 - log3 B. \(\frac{4}{log_3 x}\) C. \(\frac{4}{x}\) D. \(\pm\) \(\frac{9}{x}\)
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Detailed Solution2log3y + log3x2 = 4log3y2 + log3x2 = 4 ∴ log3 (x2y2) = log381(correct all to base 4) x2y2 = 81 ∴ xy = \(\pm\)9 ∴ y = \(\pm\)\(\frac{9}{x}\) |
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| 8. |
Solve without using tables log5(62.5) - log5(\(\frac{1}{2}\)) A. 3 B. 4 C. 5 D. 8
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Detailed Solutionlog5(62.5) - log5(\(\frac{1}{2}\))= log5\(\frac{(62.5)}{\frac{1}{2}}\) - log5(2 x 62.5) = log5(125) = log553 - 3log55 = 3 |
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| 9. |
If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x A. 3 B. 4 C. 12 D. 17
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Detailed SolutionPrincipal = N255.00, Interest = 27.00year = x Rate: 4% ∴ 1 = \(\frac{PRT}{100}\) 27 = \(\frac{225 \times 4 \times T}{100}\) 2700 = 900T T = 3 years |
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| 10. |
If \(\sqrt{x^2 + 9}\) = x + 1, solve for x A. 5 B. 4 C. 3 D. 2 E. 1
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Detailed Solution\(\sqrt{x^2 + 9}\) = x + 1x2 + 9 = (x + 1)2 + 1 0 = x2 + 2x + 1 - x2 - 9 = 2x - 8 = 0 2(x - 4) = 0 x = 4 |
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| 11. |
Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) A. \(\frac{p + q}{a(p - q)}\) B. \(\frac{p - q}{a(p + q)}\) C. \(\frac{p - q}{apq}\) D. \(\frac{pq}{a(p - q)}\)
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Detailed Solution\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication,q(1 + ax) = p(1 - ax) q + qax = p - pax qax + pax = p - q ∴ x = \(\frac{p - q}{a(p + q)}\) |
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