Year :
1986
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y. A. (12, 6) B. (23, 17) C. (17, 11) D. (18, 12) Detailed Solutionx - y = 6.......(i)xy = 187.......(ii) From equation (i), x(6 + y) sub. for x in equation (ii) = y(6 + y) = 187 y2 + 6y = 187 y2 + 6y - 187 = 0 (y + 17)(y - 11) = 0 y = -17 or y = 11 y cannot be negative, y = 11 Sub. for y in equation(i) = x - 11 = 16 x = 6 + 11 = 17 ∴(x, y) = (17, 11) |
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| 12. |
In 1984, Ike was 24 yrs old and his father was 45 yrs old. In what year was Ike exactly half his father's age? A. 1981 B. 1979 C. 1982 D. 1978 Detailed SolutionLet the no. of years be y24 - y = \(\frac{1}{2}\)(45 - y) 45 - y = 2(24 - y) 45 - y = 48 - 2y 2y - y = 48 - 45 ∴ y = 3 The exact year = 1984 1984 - 3 = 1981 |
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| 13. |
Simplify \((\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}\) A. \(\frac{\sqrt{3}}{\sqrt{5}}\) B. \(\frac{2 \sqrt{3}}{7}\) C. -2 D. -1 Detailed Solution\((\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}\)\(\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}\) \(\frac{(\sqrt{5} - \sqrt{3}) - (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})}\) = \(\frac{\sqrt{5} - \sqrt{3} - \sqrt{5} - \sqrt{3}}{5 - \sqrt{15} + \sqrt{15} - 3}\) = \(\frac{-2\sqrt{3}}{2}\) = \(- \sqrt{3}\) \(\therefore (\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}} = - \sqrt{3} \times \frac{1}{\sqrt{3}}\) = \(-1\) |
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| 14. |
Find n if log\(_{2}\) 4 + log\(_{2}\) 7 - log\(_{2}\) n = 1 A. 10 B. 14 C. 27 D. 28 Detailed Solutionlog\(_2\) 4 + log\(_2\) 7 - log\(_2\) n = 1= log\(_2\) (4 x 7) - log\(_2\) n = 1 \(\therefore\) log\(_2\) 28 - log\(_2\) n = 1 = \(\frac{28}{n} = 2^1\) \(\frac{28}{n}\) = 2 2n = 28 ∴ n = 14 |
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| 15. |
Simplify \(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) A. \(\frac{1}{3}\) B. 1 C. 3 D. 9 Detailed Solution\(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) = \(\frac{(3^2)^{\frac{1}{3}} \times (3^3)^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\)= \(\frac{3^{\frac{2}{3}} \times 3^{\frac{2}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) = \(\frac{3^{\frac{5}{6}}}{3^{-\frac{5}{6}}}\) = 1 |
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| 16. |
If x varies directly as y3 and x = 2 when y = 1, find x when y = 5 A. 2 B. 10 C. 125 D. 250 Detailed Solutionx \(\alpha\) y3x = ky3 k = \(\frac {x}{y^3}\) when x = 2, y = 1 k = 2 Thus x = 2y3 - equation of variation = 2(5)3 = 250 |
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| 17. |
Factorize completely 8a + 125ax3 A. (2a + 5x2)(4 + 26ax) B. a(2 + 5x)(4 - 10x + 25x2) C. (2a + 5x)(4 - 10ax + 25x2) D. a(2 + 5x)(4 + 10ax + 25x2) Detailed Solution\(8a + 125ax^{3} = (2^{3})a + 5^{3} ax^{3}\)= \(a(2^3 + 5^3 x^3)\) ∴\(a[2^3 + (5x)^3]\) \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) ∴ \(a(2^3 + (5x)^3)\) = \(a(2 + 5x)(4 - 10x + 25x^2)\) |
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| 18. |
If y = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) find y when x = -2 A. -\(\frac{3}{5}\) B. \(\frac{3}{5}\) C. -\(\frac{7}{5}\) D. \(\frac{2}{5}\) Detailed Solutiony = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) when x = -2y = \(\frac{-2}{-5}\) + \(\frac{(-2)}{-2 + 4}\) = \(\frac{2}{5}\) + \(\frac{-2}{2}\) = \(\frac{4 -10}{10}\) = \(\frac{-6}{10}\) = -\(\frac{3}{5}\) |
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| 19. |
Find all real numbers x which satisfy the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4) A. x < 11 B. x < -1 C. x > 6 D. x > 11 Detailed Solution\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)= \(\frac{x + 1}{3}\) - 1 > \(\frac{x + 4}{5}\) \(\frac{x + 1}{3}\) - \(\frac{x + 4}{5}\) - 1 > 0 = \(\frac{5x + 5 - 3x - 12}{15}\) = 2x - 7 > 15 = 2x > 22 = x > 11 |
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| 20. |
Factorize \(x^2 + 2a + ax + 2x\) A. (x + 2a)(x + 1) B. (x + 2a)(x - 1) C. (x2 - 1)(x - a) D. (x + 2)(x + a) Detailed Solution\(x^{2} + 2a + ax + 2x\)\(x^{2} + 2a + ax + 2x\) \(x(x + 2) + a(x + 2)\) \((x + 2)(x + a)\) |
| 11. |
The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y. A. (12, 6) B. (23, 17) C. (17, 11) D. (18, 12) Detailed Solutionx - y = 6.......(i)xy = 187.......(ii) From equation (i), x(6 + y) sub. for x in equation (ii) = y(6 + y) = 187 y2 + 6y = 187 y2 + 6y - 187 = 0 (y + 17)(y - 11) = 0 y = -17 or y = 11 y cannot be negative, y = 11 Sub. for y in equation(i) = x - 11 = 16 x = 6 + 11 = 17 ∴(x, y) = (17, 11) |
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| 12. |
In 1984, Ike was 24 yrs old and his father was 45 yrs old. In what year was Ike exactly half his father's age? A. 1981 B. 1979 C. 1982 D. 1978 Detailed SolutionLet the no. of years be y24 - y = \(\frac{1}{2}\)(45 - y) 45 - y = 2(24 - y) 45 - y = 48 - 2y 2y - y = 48 - 45 ∴ y = 3 The exact year = 1984 1984 - 3 = 1981 |
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| 13. |
Simplify \((\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}\) A. \(\frac{\sqrt{3}}{\sqrt{5}}\) B. \(\frac{2 \sqrt{3}}{7}\) C. -2 D. -1 Detailed Solution\((\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}\)\(\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}\) \(\frac{(\sqrt{5} - \sqrt{3}) - (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})}\) = \(\frac{\sqrt{5} - \sqrt{3} - \sqrt{5} - \sqrt{3}}{5 - \sqrt{15} + \sqrt{15} - 3}\) = \(\frac{-2\sqrt{3}}{2}\) = \(- \sqrt{3}\) \(\therefore (\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}} = - \sqrt{3} \times \frac{1}{\sqrt{3}}\) = \(-1\) |
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| 14. |
Find n if log\(_{2}\) 4 + log\(_{2}\) 7 - log\(_{2}\) n = 1 A. 10 B. 14 C. 27 D. 28 Detailed Solutionlog\(_2\) 4 + log\(_2\) 7 - log\(_2\) n = 1= log\(_2\) (4 x 7) - log\(_2\) n = 1 \(\therefore\) log\(_2\) 28 - log\(_2\) n = 1 = \(\frac{28}{n} = 2^1\) \(\frac{28}{n}\) = 2 2n = 28 ∴ n = 14 |
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| 15. |
Simplify \(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) A. \(\frac{1}{3}\) B. 1 C. 3 D. 9 Detailed Solution\(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) = \(\frac{(3^2)^{\frac{1}{3}} \times (3^3)^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\)= \(\frac{3^{\frac{2}{3}} \times 3^{\frac{2}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}}\) = \(\frac{3^{\frac{5}{6}}}{3^{-\frac{5}{6}}}\) = 1 |
| 16. |
If x varies directly as y3 and x = 2 when y = 1, find x when y = 5 A. 2 B. 10 C. 125 D. 250 Detailed Solutionx \(\alpha\) y3x = ky3 k = \(\frac {x}{y^3}\) when x = 2, y = 1 k = 2 Thus x = 2y3 - equation of variation = 2(5)3 = 250 |
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| 17. |
Factorize completely 8a + 125ax3 A. (2a + 5x2)(4 + 26ax) B. a(2 + 5x)(4 - 10x + 25x2) C. (2a + 5x)(4 - 10ax + 25x2) D. a(2 + 5x)(4 + 10ax + 25x2) Detailed Solution\(8a + 125ax^{3} = (2^{3})a + 5^{3} ax^{3}\)= \(a(2^3 + 5^3 x^3)\) ∴\(a[2^3 + (5x)^3]\) \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) ∴ \(a(2^3 + (5x)^3)\) = \(a(2 + 5x)(4 - 10x + 25x^2)\) |
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| 18. |
If y = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) find y when x = -2 A. -\(\frac{3}{5}\) B. \(\frac{3}{5}\) C. -\(\frac{7}{5}\) D. \(\frac{2}{5}\) Detailed Solutiony = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) when x = -2y = \(\frac{-2}{-5}\) + \(\frac{(-2)}{-2 + 4}\) = \(\frac{2}{5}\) + \(\frac{-2}{2}\) = \(\frac{4 -10}{10}\) = \(\frac{-6}{10}\) = -\(\frac{3}{5}\) |
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| 19. |
Find all real numbers x which satisfy the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4) A. x < 11 B. x < -1 C. x > 6 D. x > 11 Detailed Solution\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)= \(\frac{x + 1}{3}\) - 1 > \(\frac{x + 4}{5}\) \(\frac{x + 1}{3}\) - \(\frac{x + 4}{5}\) - 1 > 0 = \(\frac{5x + 5 - 3x - 12}{15}\) = 2x - 7 > 15 = 2x > 22 = x > 11 |
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| 20. |
Factorize \(x^2 + 2a + ax + 2x\) A. (x + 2a)(x + 1) B. (x + 2a)(x - 1) C. (x2 - 1)(x - a) D. (x + 2)(x + a) Detailed Solution\(x^{2} + 2a + ax + 2x\)\(x^{2} + 2a + ax + 2x\) \(x(x + 2) + a(x + 2)\) \((x + 2)(x + a)\) |