Mathematics (Core)
Paper 1 | Objectives | 44 Questions
JAMB Exam
Year: 1990
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
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
Scholarship in Norway universities are open for application in 2022 also for developing countries.
8 Tips To Be More Productive in Test Revision for good grades and higher performance in exams
These are the best study techniques and methods that get higher grades in any school tests or exams.
| # | 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. \(\frac{10}{21}\)
Show Content
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 |
|
| 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)
Show Content
Detailed Solutiona2bx + ab2x; a2b - b2abx(a + b); b(a2 - b2) b(a + b)(a + b) ∴ H.C.F. = (a + b) |
|
| 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
Show Content
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) |
|
| 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%
Show Content
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% |
|
| 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. \(\frac{13}{16}\)
Show Content
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 = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\) |
|
| 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
Show Content
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 |
|
| 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
Show Content
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 = \(\frac{12 000}{\frac{2}{9}}\) |
|
| 8. |
Simplify \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) A. 9r2 B. 12\(\sqrt{3r}\) C. 13r D. \(\sqrt{13r}\)
Show Content
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 |
|
| 9. |
Simplify \(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) A. 4\(\sqrt{3}\) B. \(\frac{4}{\sqrt{3}}\) C. 3\(\sqrt{3}\) D. \(\frac{\sqrt{3}}{4}\)
Show Content
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\(\sqrt{3}\) |
Preview displays only 9 out of the 44 Questions