Year :
1989
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Solve the pair of equation for x and y respectively \(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\) A. -1, 2 B. 1, 2 C. 2, 1 D. 2, -1 Detailed Solution\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)Let \(x^{-1}\) = a and \(y^{-1}\)= b 2a - 3b = 4 .......(i) 4a + b = 1 .........(ii) (i) x 3 = 12a + 3b = 3........(iii) 2a - 3b = 4 ...........(i) (i) + (iii) = 14a = 7 ∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\) From (i), 3b = 2a - 4 3b = 1 - 4 3b = -3 ∴ b = -1 From substituting, \(2^{-1} = x^{-1}\) ∴ x = 2 \(y^{-1} = -1, y = -1\) |
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| 22. |
What value of Q will make the expression 4x2 + 5x + Q a complete square? A. \(\frac{25}{16}\) B. \(\frac{25}{64}\) C. \(\frac{5}{8}\) D. \(\frac{5}{4}\) Detailed Solution4x2 + 5x + QTo make a complete square, the coefficient of x2 must be 1 = x2 + \(\frac{5x}{4}\) + \(\frac{Q}{4}\) Then (half the coefficient of x2) should be added i.e. x2 + \(\frac{5x}{4}\) + \(\frac{25}{64}\) ∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\) Q = \(\frac{4 \times 25}{64}\) = \(\frac{25}{16}\) |
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| 23. |
Find the range of values of values of r which satisfies the following inequality, where a, b and c are positive \(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 A. r > \(\frac{abc}{bc + ac + ab}\) B. r < abc C. r > \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) D. \(\frac{1}{abc}\) Detailed Solution\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\) |
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| 24. |
Express \(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) as a single algebraic fraction A. \(\frac{-3}{(x + 1)(2 - x)}\) B. \(\frac{3}{(x + 1)(2 - x)}\) C. \(\frac{-1}{(x + 1)}\) D. \(\frac{1}{(x + 1)(x - 2)}\) Detailed Solution\(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) = \(\frac{x - 2 - x - 1}{(x + 1)(x - 2)}\)= \(\frac{-3}{(x + 1)(2 - x)}\) |
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| 25. |
Simplify \(\frac{x(x + 1)^{\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\) A. \(\frac{-1}{x + 1}\) B. \(\frac{1}{x + 1}\) C. \(\frac{1}{x}\) D. \(\frac{1}{x - 1}\) Detailed Solution\(\frac{x}{(x + 1)}\) - \(\frac{\sqrt{(x + 1)}{\sqrt{x + 1}}\)= \(\frac{x}{x + 1}\) - 1 \(\frac{x - x - 1}{x + 1}\) = \(\frac{-1}{x + 1}\) |
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| 26. |
The sum of the first two terms of a geometric progression is x and sum of the last terms is y. If there are n terms in all, then the common ratio is A. \(\frac{x}{y}\) B. \(\frac{y}{x}\) C. (\(\frac{x}{y}\))\(\frac{1}{n - 2}\) D. (\(\frac{y}{x}\))\(\frac{1}{n - 2}\) Detailed SolutionSum of nth term of a G.P = Sn = \(\frac{ar^n - 1}{r - 1}\)sum of the first two terms = \(\frac{ar^2 - 1}{r - 1}\) x = a(r + 1) sum of the last two terms = Sn - Sn - 2 = \(\frac{ar^n - 1}{r - 1}\) - \(\frac{(ar^{n - 1})}{r - 1}\) = \(\frac{a(r^n - 1 - r^{n - 2} + 1)}{r - 1}\) (r2 - 1) ∴ \(\frac{ar^{n - 2}(r + 1)(r - 1)}{1}\)= arn - 2(r + 1) = y = a(r + 1)r^n - 2 y = xrn - 2 = yrn - 2 |
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| 27. |
If -8, m, n, 19 are in arithmetic progression, find (m, n) A. 1, 10 B. 2, 10 C. 3, 13 D. 4, 16 Detailed Solution-8, m, n, 19 = m + 8= 19 - n m + n = 11 i.e. 1, 10 |
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| 28. |
A regular polygon of (2k + 1) sides has 140° as the size of each interior angle. Find k A. 4 B. 4\(\frac{1}{2}\) C. 8 D. 8\(\frac{1}{2}\) Detailed SolutionA regular has all sides and all angles equal. If each interior angle is 140° each exterior angle must be180° - 140° = 40° The number of sides must be \(\frac{360^o}{40^o}\) = 9 sides hence 2k + 1 = 9 2k = 9 - 1 8 = 2k k = \(\frac{8}{2}\) = 4 |
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| 29. |
PQRS is a rhombus. If PR\(^2\) + QS\(^2\) = kPQ\(^2\), determine k. A. 1 B. 2 C. 3 D. 4 Detailed SolutionPR\(^2\) + QS\(^2\) = kPQ\(^2\)SQ\(^2\) = SR\(^2\) + RQ\(^2\) PR\(^2\) + SQ\(^2\) = PQ\(^2\) + SR\(^2\) + 2RQ\(^2\) = 2PQ\(^2\) + 2RQ\(^2\) = 4PQ\(^2\) ∴ K = 4 |
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| 30. |
In XYZ, y = z = 30° and XZ = 3cm. Find YZ A. \(\frac{\sqrt{3}}{2}\) cm B. 3\(\frac{\sqrt{3}}{2}\) cm C. 3\(\sqrt{3}\) cm D. 2\(\sqrt{3}\) cm Detailed SolutionY \(\to\) 30°X \(\to\) 120° Z \(\to\) 30° \(\frac{3}{\text{sin 30}}\) = \(\frac{YZ}{\text{sin 120}}\) YZ = \(\frac{3\times \sin 120}{\sin 30}\) \(3 \times \frac{\sqrt{3}}{2} \times 2 = 3\sqrt{3}\) |
| 21. |
Solve the pair of equation for x and y respectively \(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\) A. -1, 2 B. 1, 2 C. 2, 1 D. 2, -1 Detailed Solution\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)Let \(x^{-1}\) = a and \(y^{-1}\)= b 2a - 3b = 4 .......(i) 4a + b = 1 .........(ii) (i) x 3 = 12a + 3b = 3........(iii) 2a - 3b = 4 ...........(i) (i) + (iii) = 14a = 7 ∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\) From (i), 3b = 2a - 4 3b = 1 - 4 3b = -3 ∴ b = -1 From substituting, \(2^{-1} = x^{-1}\) ∴ x = 2 \(y^{-1} = -1, y = -1\) |
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| 22. |
What value of Q will make the expression 4x2 + 5x + Q a complete square? A. \(\frac{25}{16}\) B. \(\frac{25}{64}\) C. \(\frac{5}{8}\) D. \(\frac{5}{4}\) Detailed Solution4x2 + 5x + QTo make a complete square, the coefficient of x2 must be 1 = x2 + \(\frac{5x}{4}\) + \(\frac{Q}{4}\) Then (half the coefficient of x2) should be added i.e. x2 + \(\frac{5x}{4}\) + \(\frac{25}{64}\) ∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\) Q = \(\frac{4 \times 25}{64}\) = \(\frac{25}{16}\) |
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| 23. |
Find the range of values of values of r which satisfies the following inequality, where a, b and c are positive \(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 A. r > \(\frac{abc}{bc + ac + ab}\) B. r < abc C. r > \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) D. \(\frac{1}{abc}\) Detailed Solution\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\) |
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| 24. |
Express \(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) as a single algebraic fraction A. \(\frac{-3}{(x + 1)(2 - x)}\) B. \(\frac{3}{(x + 1)(2 - x)}\) C. \(\frac{-1}{(x + 1)}\) D. \(\frac{1}{(x + 1)(x - 2)}\) Detailed Solution\(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) = \(\frac{x - 2 - x - 1}{(x + 1)(x - 2)}\)= \(\frac{-3}{(x + 1)(2 - x)}\) |
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| 25. |
Simplify \(\frac{x(x + 1)^{\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\) A. \(\frac{-1}{x + 1}\) B. \(\frac{1}{x + 1}\) C. \(\frac{1}{x}\) D. \(\frac{1}{x - 1}\) Detailed Solution\(\frac{x}{(x + 1)}\) - \(\frac{\sqrt{(x + 1)}{\sqrt{x + 1}}\)= \(\frac{x}{x + 1}\) - 1 \(\frac{x - x - 1}{x + 1}\) = \(\frac{-1}{x + 1}\) |
| 26. |
The sum of the first two terms of a geometric progression is x and sum of the last terms is y. If there are n terms in all, then the common ratio is A. \(\frac{x}{y}\) B. \(\frac{y}{x}\) C. (\(\frac{x}{y}\))\(\frac{1}{n - 2}\) D. (\(\frac{y}{x}\))\(\frac{1}{n - 2}\) Detailed SolutionSum of nth term of a G.P = Sn = \(\frac{ar^n - 1}{r - 1}\)sum of the first two terms = \(\frac{ar^2 - 1}{r - 1}\) x = a(r + 1) sum of the last two terms = Sn - Sn - 2 = \(\frac{ar^n - 1}{r - 1}\) - \(\frac{(ar^{n - 1})}{r - 1}\) = \(\frac{a(r^n - 1 - r^{n - 2} + 1)}{r - 1}\) (r2 - 1) ∴ \(\frac{ar^{n - 2}(r + 1)(r - 1)}{1}\)= arn - 2(r + 1) = y = a(r + 1)r^n - 2 y = xrn - 2 = yrn - 2 |
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| 27. |
If -8, m, n, 19 are in arithmetic progression, find (m, n) A. 1, 10 B. 2, 10 C. 3, 13 D. 4, 16 Detailed Solution-8, m, n, 19 = m + 8= 19 - n m + n = 11 i.e. 1, 10 |
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| 28. |
A regular polygon of (2k + 1) sides has 140° as the size of each interior angle. Find k A. 4 B. 4\(\frac{1}{2}\) C. 8 D. 8\(\frac{1}{2}\) Detailed SolutionA regular has all sides and all angles equal. If each interior angle is 140° each exterior angle must be180° - 140° = 40° The number of sides must be \(\frac{360^o}{40^o}\) = 9 sides hence 2k + 1 = 9 2k = 9 - 1 8 = 2k k = \(\frac{8}{2}\) = 4 |
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| 29. |
PQRS is a rhombus. If PR\(^2\) + QS\(^2\) = kPQ\(^2\), determine k. A. 1 B. 2 C. 3 D. 4 Detailed SolutionPR\(^2\) + QS\(^2\) = kPQ\(^2\)SQ\(^2\) = SR\(^2\) + RQ\(^2\) PR\(^2\) + SQ\(^2\) = PQ\(^2\) + SR\(^2\) + 2RQ\(^2\) = 2PQ\(^2\) + 2RQ\(^2\) = 4PQ\(^2\) ∴ K = 4 |
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| 30. |
In XYZ, y = z = 30° and XZ = 3cm. Find YZ A. \(\frac{\sqrt{3}}{2}\) cm B. 3\(\frac{\sqrt{3}}{2}\) cm C. 3\(\sqrt{3}\) cm D. 2\(\sqrt{3}\) cm Detailed SolutionY \(\to\) 30°X \(\to\) 120° Z \(\to\) 30° \(\frac{3}{\text{sin 30}}\) = \(\frac{YZ}{\text{sin 120}}\) YZ = \(\frac{3\times \sin 120}{\sin 30}\) \(3 \times \frac{\sqrt{3}}{2} \times 2 = 3\sqrt{3}\) |