Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1982
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1982
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. |
Rationalize the expression \(\frac{1}{\sqrt{2} + \sqrt{5}}\) A. \(\frac{1}{3}\)(\(\sqrt{5} - \sqrt{2}\) B. \(\frac{\sqrt{2}}{3}\) + \(\frac{\sqrt{5}}{5}\) C. \(\sqrt{2} - \sqrt{5}\) D. 5(\(\sqrt{2} - \sqrt{5}\) E. \(\frac{1}{3}\)(\(\sqrt{2} - \sqrt{5}\)
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Detailed Solution\(\frac{1}{\sqrt{2} + \sqrt{5}}\)\(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\) = \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\) = \(\frac{\sqrt{2} - \sqrt{5}}{-3}\) = \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\) |
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| 2. |
Simplify 3 - 2 \(\div\) \(\frac{4}{5}\) + \(\frac{1}{2}\) A. 1\(\frac{3}{4}\) B. -1 C. 1\(\frac{3}{10}\) D. 1 E. 1\(\frac{9}{10}\)
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Detailed Solution3 - 2 \(\div\) (\(\frac{4}{5}\)) + \(\frac{1}{2}\)3 - (2 x \(\frac{5}{4}\)) + \(\frac{1}{2}\) = 3 - \(\frac{10}{4}\) + \(\frac{1}{2}\) = 3 - \(\frac{5}{2}\) + \(\frac{1}{2}\) = \(\frac{6 - 5 + 1}{2}\) = \(\frac{2}{2}\) = 1 |
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| 3. |
If N560.70 is shared in the ratio 7 : 2 : 1, what is the smallest share? A. N392.49 B. N56.70 C. N113.40 D. N112.14 E. N56.07
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Detailed Solution7 + 2 + 1 = 10\(\frac{1}{10}\) x 560.70 = N56.07 |
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| 4. |
Seven years ago, the age of a father was three times that of his son, but in six years time the age of the son will be half that of his father, representing the present ages of the father and son by x and y, respectively, the two equations relating x and y are A. 3y - x = 0; 2y - x = 0 B. 3y - x = 14; x - 2y = 6 C. 3y - x =7; x - 2y = 6 D. 3y - x = 14; y - 2x = 6 E. x + 3y = 7; x = 2y = 12
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Detailed Solution7 years ago, Father(x - 7) years old, Son (y - 7) yearsx - 7 = 3(y - 7) x - 7 = 3y - 21 3y - x = -7 + 21 = 14 3y - x = 14 ... (1) In six years time, x + 6 = 2(y + 6) x + 6 = 2y + 12 2y + 12 = x + 6 12 - 6 = x - 2y 6 = x - 2y ... (2) |
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| 5. |
The factors of 6x - 5 - x2 are A. -(x + 3)(x + 2) B. (x + 5)(x + 1) C. (x - 5)(1 - x) D. (x + 1)(x + 5)
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Detailed Solution6x - 5 - x2 = (-1)(-x2 - 5 + 6x)= x2 - 6x + 5 = (x - 5)(x - 1) -(x - 1) = 1 - x = (x - 5)(1 - x) |
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| 6. |
The solution of the quadratic equation bx2 + cx + a = 0 is given by A. x = b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) B. x = c \(\pm\) \(\frac{\sqrt{b^2 - 4ab}}{2b}\) C. x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\) D. x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2b}\)
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Detailed Solutionbx2 + cx + a = 0a = b; b = c; c = a x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\) |
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| 7. |
The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves A. y = x3 and y = 3x2 + x - 28 B. y = x3 + 3x2 + 4x + 4 and the line y = \(\frac{28}{x}\) C. y = x3 + 3x2 + 4x and y D. y = x2 + 3x + 4 and y = \(\frac{28}{x}\) E. y = x2 + 3x + 4 and line y = 28x
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Detailed SolutionThe graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curvesy = x2 + 3x + 4 and y = \(\frac{28}{x}\)`. |
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| 8. |
Write the equation 2 log2x - x log2(1 + y) = 3 in a form not involving logarithms A. 2x(1 + y) = 3 B. 2x - x(1 + y) = 8 C. x2 = 8(1 + y)x D. x2 - x(1 + y) = 8 E. x2 - (1 + y)2 = 8
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Detailed Solution2log2 x - x log2 (1 + y) = 3log2 \(\frac{x^2}{(1 + y)^x}\) = 3 = \(\frac{x^2}{(1 + y)^x}\) = 23 = 8 = x2 = 8(1 + y)x |
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| 9. |
Find \(\alpha\) and \(\beta\) such that x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) A. \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\) B. \(\alpha\)= 1, \(\beta\) = -\(\frac{5}{7}\) C. \(\alpha\)= \(\frac{3}{5}\), \(\beta\) = -6 D. \(\alpha\)= 1, \(\beta\) = -\(\frac{3}{5}\)
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Detailed Solutionx\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\)x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x y\(\frac{1}{7}\) = x\(\alpha\) = x\(\frac{3}{8}\) + \(\frac{5}{8}\) + y\(\frac{6}{7}\) + \(\frac{1}{7}\) = x\(\alpha\)y\(\beta\) x1y\(\frac{-5}{7}\) = x\(\alpha\)y\(\beta\) \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\) |
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| 10. |
Which of the following lines is not parallel to the line 3y + 2x + 7 = 0? A. 3y + 2x - 7 = 0 B. 9y + 6x + 17 = 0 C. 24y + 16x + 19 = 0 D. 3y - 2x + 7 = 0 E. 15y + 10x - 13 = 0
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Detailed SolutionTwo lines are said to be parallel if the slope of the two lines are equal.The equation : \(3y + 2x + 7 = 0\) \(3y = -2x - 7\) \(y = \frac{-2}{3} x - \frac{7}{3}\) \(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\) All the options have the same slope except \(3y - 2x + 7 = 0\). |
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