Mathematics (Core)
Paper 1 | Objectives | 51 Questions
JAMB Exam
Year: 1993
Level: SHS
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Type: Question Paper
Answers provided
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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
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| # | Question | Ans |
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| 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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