Year :
1989
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\) A. (\(\frac{3}{q}\))3 B. (\(\frac{q}{3}\))\(\frac{1}{3}\) C. (\(\frac{q}{3}\))3 D. (\(\frac{3}{q}\))\(\frac{1}{3}\) Detailed Solution\(\log_{3} p + 3\log_{3} q = 3\)\(\log_{3} p + \log_{3} q^{3} = 3\) \(\implies \log_{3} (pq^{3}) = 3\) \(pq^{3} = 3^{3} = 27\) \(\therefore p = \frac{27}{q^{3}}\) = \((\frac{3}{q})^{3}\) |
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| 12. |
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ? A. -1 and 0 B. -1 and 1 C. 1 and 3 D. 0 and 1 Detailed Solution\(9^{y} - 4 \times 3^{y} + 3 = 0\)\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\) \((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\) Let \(3^{y}\) be r. Then, \(r^{2} - 4r + 3 = 0\) Solving the equation, \(r^{2} - 3r - r + 3 = 0\) \(r(r - 3) - 1(r - 3) = 0\) \((r - 3)(r - 1) = 0\) \(\therefore \text{r = 3 or 1}\) Recall, \(3^{y} = r\) \(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\) \(\implies \text{y = 1 or 0}\) |
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| 13. |
Make R the subject of the fomula S = \(\sqrt{\frac{2R + T}{2RT}}\) A. R = \(\frac{T}{(TS^2 + 1)}\) B. R = \(\frac{T}{2(TS^2 - 2)}\) C. R = \(\frac{T}{2(TS^2 + 1)}\) D. R = \(\frac{R}{2(TS^2 + 1)}\) Detailed SolutionS = \(\sqrt{\frac{2R + T}{2RT}}\)Squaring both sides, \(S^{2} = \frac{2R + T}{2RT}\) \(S^{2} (2RT) = 2R + T\) \(2S^{2} RT - 2R = T\) \(R = \frac{T}{2TS^{2} - 2}\) = \(\frac{T}{2(TS^{2} - 1)} |
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| 14. |
The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is N74.00 when the number of is 20 and N96.00 when the number is 30, find the cost when there are 15 students A. N68.50 B. N63.00 C. N60.00 D. N52.00 Detailed SolutionC = a + kS. If C = 74, S = 20C = 96, S = 30, C= ? S = 15 74 = a + 20k......(1) 96 = a + 30k......(2) subtract equation (1) from (2) 96 = a + 30k - 74 = a + 20k -------------- 22 = 10k k = 2.2 find a 74 = a + 44 a = 30 C = 30 + 2.2S when S = 15, C = 30 + 2.2 x 15 = 30 + 33 = N63 |
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| 15. |
If \(f(x) = 2x^2 - 5x + 3\), find f(x + 1). A. 2x2 - x B. 2x2 - x + 10 C. 4x2 + 3x + 2 D. 4x2 + 3x + 12 Detailed Solution\(f(x) = 2x^2 - 5x + 3\)\(f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3\) = \(2(x^2 + 2x + 1) - 5x - 5 + 3\) = \(2x^2 + 4x + 2 - 5x - 2\) = \(2x^2 - x\) |
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| 16. |
Solve for a positive number x such that \(2^{(x^3 - x^2 - 2x)} = 1\) A. 4 B. 3 C. 2 D. 1 Detailed Solution\(2^{(x^3 - x^2 - 2x)} = 1\)\(x^3 - x^2 - 2x = 0\) \(x(x^2 - x - 2) = 0\) \(x^2 - x - 2 = 0\) \((x + 1)(x - 2) = 0\) x = 2 is the positive answer. |
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| 17. |
Simplify \(\frac{324 - 4x^2}{2x + 18}\) A. 2(x - 9) B. 2(9 + x) C. 81 - x2 D. -2(x - 9) Detailed Solution\(\frac{324 - 4x^2}{2x + 18}\) = \(\frac{18^2 - (2x)^2}{2x + 18}\)= \(\frac{(18 - 2x)(18 + 2x)}{(2x + 18)}\) 18 - 2x = 2(9 - x) or -2(x - 9) |
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| 18. |
Factorize completely \(y^3 -4xy + xy^3 - 4y\) A. (y + xy)(y + 2)(y - 2) B. (y - xy)(y - 2) C. y(1 + x)(y + 2)(y -2) D. y(1 - x)(y + 2)(y - 2) Detailed Solution\(y^3 -4xy + xy^3 - 4y \)= \(y^3(1 + x) - 4y(1 + x) \((y^3 - 4y)(1 + x) = (y^3(1 + x) - 4y(1 + x))\) ∴ = \(y(1 + x)(y + 2)(y - 2)\) |
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| 19. |
Factorize 4a2 - 12ab - C2 + 9b2 A. 4a(a - 3b) + (3b - c)2 B. (2a + 3b - c)(2a + 3b + c) C. (2a - 3b + c)(2a - 3b - c) D. 4a(a - 3b) + (3b + c) Detailed Solution4a2 - 12ab - C2 + 9b2rearranges: (4a2 - 12ab + 9b2) - c2 (2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2 = (2a - 3b + c)(2a - 3b - c) |
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| 20. |
What are K and L respectively if \(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2) A. -12, \(\frac{9}{2}\) B. -6, 9 C. 6, 9 D. 12, \(\frac{9}{2}\) Detailed Solution\(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2)\(\frac{1}{2}\)(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2 \(\frac{9}{2}\)y2 - 12xy) = kxy + Ly2 k = -12 ∴ L = \(\frac{9}{2}\) |
| 11. |
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\) A. (\(\frac{3}{q}\))3 B. (\(\frac{q}{3}\))\(\frac{1}{3}\) C. (\(\frac{q}{3}\))3 D. (\(\frac{3}{q}\))\(\frac{1}{3}\) Detailed Solution\(\log_{3} p + 3\log_{3} q = 3\)\(\log_{3} p + \log_{3} q^{3} = 3\) \(\implies \log_{3} (pq^{3}) = 3\) \(pq^{3} = 3^{3} = 27\) \(\therefore p = \frac{27}{q^{3}}\) = \((\frac{3}{q})^{3}\) |
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| 12. |
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ? A. -1 and 0 B. -1 and 1 C. 1 and 3 D. 0 and 1 Detailed Solution\(9^{y} - 4 \times 3^{y} + 3 = 0\)\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\) \((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\) Let \(3^{y}\) be r. Then, \(r^{2} - 4r + 3 = 0\) Solving the equation, \(r^{2} - 3r - r + 3 = 0\) \(r(r - 3) - 1(r - 3) = 0\) \((r - 3)(r - 1) = 0\) \(\therefore \text{r = 3 or 1}\) Recall, \(3^{y} = r\) \(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\) \(\implies \text{y = 1 or 0}\) |
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| 13. |
Make R the subject of the fomula S = \(\sqrt{\frac{2R + T}{2RT}}\) A. R = \(\frac{T}{(TS^2 + 1)}\) B. R = \(\frac{T}{2(TS^2 - 2)}\) C. R = \(\frac{T}{2(TS^2 + 1)}\) D. R = \(\frac{R}{2(TS^2 + 1)}\) Detailed SolutionS = \(\sqrt{\frac{2R + T}{2RT}}\)Squaring both sides, \(S^{2} = \frac{2R + T}{2RT}\) \(S^{2} (2RT) = 2R + T\) \(2S^{2} RT - 2R = T\) \(R = \frac{T}{2TS^{2} - 2}\) = \(\frac{T}{2(TS^{2} - 1)} |
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| 14. |
The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is N74.00 when the number of is 20 and N96.00 when the number is 30, find the cost when there are 15 students A. N68.50 B. N63.00 C. N60.00 D. N52.00 Detailed SolutionC = a + kS. If C = 74, S = 20C = 96, S = 30, C= ? S = 15 74 = a + 20k......(1) 96 = a + 30k......(2) subtract equation (1) from (2) 96 = a + 30k - 74 = a + 20k -------------- 22 = 10k k = 2.2 find a 74 = a + 44 a = 30 C = 30 + 2.2S when S = 15, C = 30 + 2.2 x 15 = 30 + 33 = N63 |
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| 15. |
If \(f(x) = 2x^2 - 5x + 3\), find f(x + 1). A. 2x2 - x B. 2x2 - x + 10 C. 4x2 + 3x + 2 D. 4x2 + 3x + 12 Detailed Solution\(f(x) = 2x^2 - 5x + 3\)\(f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3\) = \(2(x^2 + 2x + 1) - 5x - 5 + 3\) = \(2x^2 + 4x + 2 - 5x - 2\) = \(2x^2 - x\) |
| 16. |
Solve for a positive number x such that \(2^{(x^3 - x^2 - 2x)} = 1\) A. 4 B. 3 C. 2 D. 1 Detailed Solution\(2^{(x^3 - x^2 - 2x)} = 1\)\(x^3 - x^2 - 2x = 0\) \(x(x^2 - x - 2) = 0\) \(x^2 - x - 2 = 0\) \((x + 1)(x - 2) = 0\) x = 2 is the positive answer. |
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| 17. |
Simplify \(\frac{324 - 4x^2}{2x + 18}\) A. 2(x - 9) B. 2(9 + x) C. 81 - x2 D. -2(x - 9) Detailed Solution\(\frac{324 - 4x^2}{2x + 18}\) = \(\frac{18^2 - (2x)^2}{2x + 18}\)= \(\frac{(18 - 2x)(18 + 2x)}{(2x + 18)}\) 18 - 2x = 2(9 - x) or -2(x - 9) |
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| 18. |
Factorize completely \(y^3 -4xy + xy^3 - 4y\) A. (y + xy)(y + 2)(y - 2) B. (y - xy)(y - 2) C. y(1 + x)(y + 2)(y -2) D. y(1 - x)(y + 2)(y - 2) Detailed Solution\(y^3 -4xy + xy^3 - 4y \)= \(y^3(1 + x) - 4y(1 + x) \((y^3 - 4y)(1 + x) = (y^3(1 + x) - 4y(1 + x))\) ∴ = \(y(1 + x)(y + 2)(y - 2)\) |
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| 19. |
Factorize 4a2 - 12ab - C2 + 9b2 A. 4a(a - 3b) + (3b - c)2 B. (2a + 3b - c)(2a + 3b + c) C. (2a - 3b + c)(2a - 3b - c) D. 4a(a - 3b) + (3b + c) Detailed Solution4a2 - 12ab - C2 + 9b2rearranges: (4a2 - 12ab + 9b2) - c2 (2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2 = (2a - 3b + c)(2a - 3b - c) |
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| 20. |
What are K and L respectively if \(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2) A. -12, \(\frac{9}{2}\) B. -6, 9 C. 6, 9 D. 12, \(\frac{9}{2}\) Detailed Solution\(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2)\(\frac{1}{2}\)(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2 \(\frac{9}{2}\)y2 - 12xy) = kxy + Ly2 k = -12 ∴ L = \(\frac{9}{2}\) |