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Mathematics (Core)
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1988
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper has been added to Cart!
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
Apply tips for stress-free eating before and during exams and what to avoid for good grades.
Good jobs available to people without a college degree, how to get good job without a college degree?
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
| # | 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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