Mathematics (Core)
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1988
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1988
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Norway Kvaroy Arctic Women in Aquaculture Postgraduate Scholarship Program 2022 for African Women
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The goal of mocks tests is to create a bench-marking tool to help students assess their performances.
| # | Question | Ans |
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| 1. |
Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X. A. 4≤X≤5 B. 5≤X≤8 C. 5≤X≤10 D. 8≤X≤10
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Detailed SolutionNumber of oranges = X; Costing N5X.Number of mangoes = 2X; Costing N8X. \(\therefore 65 \leq 5X + 8X \leq 130\) \(5 \leq X \leq 10\) |
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| 2. |
If (1P03)4 = 11510 find P A. o B. 1 C. 2 D. 3
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Detailed Solution1 x 43 + P x 42 + 0 x 4 + 3 = 1151016p + 67 = 115 p = \(\frac{48}{16}\) = 3 |
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| 3. |
If \(\frac{1}{p}\) = \(\frac{a^2 + 2ab + b^2}{a - b}\) and \(\frac{1}{q}\) = \(\frac{a + b}{a^2 - 2ab + b^2}\) Find \(\frac{p}{q}\) A. \(\frac{a + b}{a - b}\) B. \(\frac{1}{a^2 - b^2}\) C. \(\frac{a - b}{a + b}\) D. a2 - b2
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Detailed Solution\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\) \(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\) \(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\) \(\therefore p = \frac{a - b}{(a + b)^{2}}\) \(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\) = \(\frac{1}{(a + b)(a - b)}\) = \(\frac{1}{a^{2} - b^{2}}\) |
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| 4. |
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3 A. \(\frac{1}{3}\) B. \(\frac{2}{3}\) C. \(\frac{8}{27}\) D. \(\frac{4}{9}\)
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Detailed Solution\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)When y = 8, x = 1 \(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\) \(k = 2\) \(\therefore x = \frac{2}{\sqrt[3]{y}}\) When x = 3, \(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\) \(y = (\frac{2}{3})^{3} = \frac{8}{27}\) |
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| 5. |
If a = -3, b = 2, c = 4, evaluate \(\frac{a^3 - b^3 - c^{\frac{1}{2}}}{b - a - c}\) A. 37 B. \(\frac{-37}{5}\) C. \(\frac{37}{5}\) D. -37
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Detailed Solution\(\frac{a^{3} - b^{3} - c^{\frac{1}{2}}}{b - a - c} = \frac{(-3)^{3} - (2)^{3} - 4^{\frac{1}{2}}}{2 - (-3) - 4}\)= \(\frac{-37}{1} = -37\) |
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| 6. |
If (g(y)) = \(\frac{y - 3}{11}\) + \(\frac{11}{y^2 - 9}\). what is g(y + 3)? A. \(\frac{y}{11} + \frac{11}{y(y + 6)}\) B. \(\frac{y}{11} + \frac{11}{y(y + 3)}\) C. \(\frac{y + 30}{11} + \frac{11}{y(y + 3)}\) D. \(\frac{y + 3}{11} + \frac{11}{y(y - 6)}\)
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Detailed Solution\(g(y) = \frac{y - 3}{11} + \frac{11}{y^{2} - 9}\)\(\therefore g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^{2} - 9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9 - 9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\) = \(\frac{y}{11} + \frac{11}{y(y + 6)}\) |
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| 7. |
Factorize completely \((x^2 + x)^2 - (2x + 2)^2\) A. (x + 1)(x + 2)(x - 2) B. (x + 1) 2(x + 2) (x - 2) C. (x + 1)2 (x + 2)2 D. (x + 1)2 (x + 2)(x - 2)
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Detailed Solution\((x^{2} + x)^{2} - (2x + 2)^{2}\)= \((x^{2} + x + 2x + 2)(x^{2} + x - (2x + 2))\) = \((x^{2} + 3x + 2)(x^{2} - x - 2)\) = \(((x + 1)(x + 2))((x + 1)(x - 2))\) = \((x + 1)^{2} (x + 2)(x - 2)\) |
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| 8. |
Simplify \(\frac{x - y}{x^{\frac{1}{3}} - x^{\frac{1}{3}}}\) A. x2 + xy + y2 B. x\(\frac{2}{3}\) + x \(\frac{1}{3}\) + y\(\frac{2}{3}\) C. x\(\frac{2}{3}\) - x\(\frac{1}{3}\)y\(\frac{2}{3}\) D. y\(\frac{2}{3}\) |
B |
| 9. |
What is the solution of the equation x2 - x - 1 + 0? A. x = 1.6 and x = -0.6 B. x = -1.6 and x = 0.6 C. x = 1.6 and x = 0.6 D. x = -1.6 and x = -0.6
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Detailed Solution\(x^{2} - x - 1 = 0\)Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) a = 1, b = -1, c = -1. \(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\) \(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\) \(x = \frac{1 + 2.24}{2} ; x = \frac{1 - 2.24}{2}\) \(x = \frac{3.24}{2}; x = \frac{-1.24}{2}\) \(x = 1.62 ; x = -0.61 \) \(x \approxeq 1.6; -0.6\) |
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| 10. |
For what values of x is the curve y = \(\frac{x^2 + 3}{x + 4}\) decreasing? A. -3 < x \(\leq\) 0 B. -3 \(\geq\) x < 0 C. 0 < x < 3 D. 0 \(\leq\) x \(\leq\) 3 |
D |
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