Mathematics (Core)
Paper 1 | Objectives | 44 Questions
JAMB Exam
Year: 1990
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 44 Questions
JAMB Exam
Year: 1990
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. |
Simplify \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) A. -7\(\frac{7}{8}\) B. \(\frac{-2}{7}\) C. \(\frac{-10}{21}\) D. \(\frac{10}{21}\)
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Detailed Solution\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\)\(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\f |
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| 2. |
The H.C.F. of a2bx + ab2x and a2b - b2 is A. b B. a + b C. b(a \(\div\) b) D. abx(a2 - b2)
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Detailed Solutiona2bx + ab2x; a2b - b2abx(a + b); b(a2 - b2) b(a + b)(a + b) ∴ H.C.F. = (a + b) |
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| 3. |
Correct 241.34(3 x 10-\(^3\))\(^2\) to 4 significant figures A. 0.0014 B. 0.001448 C. 0.0022 D. 0.002172
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Detailed Solutionfirst work out the expression and then correct the answer to 4 s.f = 241.34..............(A)(3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f) |
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| 4. |
At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50? A. \(\frac{1}{2}\)% B. 2\(\frac{1}{2}\)% C. 1.5% D. 25%
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Detailed SolutionInterest I = \(\frac{PRT}{100}\)∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5% |
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| 5. |
Three children shared a basket of mangoes in such a way that the first child took \(\frac{1}{4}\) of the mangoes and the second \(\frac{3}{4}\) of the remainder. What fraction of the mangoes did the third child take? A. \(\frac{3}{16}\) B. \(\frac{7}{16}\) C. \(\frac{9}{16}\) D. \(\frac{13}{16}\)
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Detailed SolutionYou can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\) |
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| 6. |
Simplify and express in standard form \(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) A. 8.8 x 10-1 B. 8.8 x 10-2 C. 8.8 x 10-3 D. 8.8 x 103
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Detailed Solution\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\)Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) 88 x 10-4 = 88 x 10-1 x 10-4 = 8.8 x 10-3 |
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| 7. |
Three brothers in a business deal share the profit at the end of a contact. The first received \(\frac{1}{3}\) of the profit and the second \(\frac{2}{3}\) of the remainder. If the third received the remaining N12000.00 how much profit did they share? A. N60 000.00 B. N54 000.00 C. N48 000.00 D. N42 000.00
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Detailed Solutionuse "T" to represent the total profit. The first receives \(\frac{1}{3}\) Tremaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) |
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| 8. |
Simplify \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) A. 9r2 B. 12\(\sqrt{3r}\) C. 13r D. \(\sqrt{13r}\)
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Detailed Solution\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\)Simplifying from the innermost radical and progressing outwards we have the given expression \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\) = \(\sqrt{160r^2 + \sqrt{81r^4}}\) \(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\) = 13r |
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| 9. |
Simplify \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) A. 4\(\sqrt{3}\) B. \(\frac{4}{\sqrt{3}}\) C. 3\(\sqrt{3}\) D. \(\frac{\sqrt{3}}{4}\)
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Detailed Solution\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\)= \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\) |
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