Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1983
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1983
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. |
If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8 then (M, D) is A. (6, 5) B. (5, 8) C. (5, 7) D. (5, 5) E. (7, 5)
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Detailed Solutionfirst re-arrange the given data in the form 3, 5, 5, 7, 8, 9 Mean(x) = \(\frac{\sum x}{N}\)= \(\frac{42}{7}\) = 6, re-arrange thenumbers, 3, 5, 5| 7, 8, 9 median(D) = 5 (m, d) = (5, 5) |
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| 2. |
A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received N5,000 more than X. What is the total profit of the company for the year? A. N20, 000 B. N25,000 C. N30,000 D. N15,000 E. N45, 000
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Detailed SolutionTotal sharing ratio is 9X has 4, Y has 4 + 1 If 1 is N5000 Total profit = 5000 x 9 = N45,000 |
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| 3. |
Given a regular hexagon, calculate each interior angle of the hexagon A. 60o B. 30o C. 120o D. 45o E. 135o
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Detailed SolutionSum of interior angles of polygon = (2n - 4)rt < ssum of interior angles of an hexagon (2 x 6 - 4) x 90o = (12 - 4) x 90o = 8 x 90o = 720o each interior angle will have \(\frac{720^o}{6}\) = 120o |
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| 4. |
If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled? A. 4:1 increase B. 2:1 increase C. 1:1 no change D. 1:4 decrease E. 3:4 decrease
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Detailed Solution\(x \propto y^{3} z^{4}\)\(x = ky^{3} z^{4}\) If y is halved and z is doubled, we have \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\) |
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| 5. |
Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12 A. x = 5, y = 2 B. x = 2, y = 5 C. x = 5, y = -2 D. x = -2, y = -5 E. x = -5, y = -2
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Detailed Solution4x - 3 = 3x + y = x - y = 3.......(i)3x + y = 2y + 5x - 12.........(ii) eqn(ii) + eqn(i) 3x = 15 x = 5 substitute for x in equation (i) 5 - y = 3 y = 2 |
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| 6. |
In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS A. 2\(\sqrt{3}\)cm B. 4\(\sqrt{6}\)cm C. 2\(\sqrt{6}\)cm D. 8\(\sqrt{6}\)cm E. 8cm
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Detailed SolutionFrom the diagram, sin 60o = \(\frac{PR}{8}\)PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\) |
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| 7. |
Given that cos z = L , whrere z is an acute angle, find an expression for \(\frac{\cot z - \csc z}{\sec z + \tan z}\) A. \(\frac{1 - L}{1 + L}\) B. \(\frac{L^2 \sqrt{3}}{1 + L}\) C. \(\frac{1 + L^3}{L^2}\) D. \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
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Detailed SolutionGiven Cos z = L, z is an acute angle\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z = \(\frac{\text{cos z}}{\text{sin z}}\) cosec z = \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\) sec z = \(\frac{1}{\text{cos z}}\) tan z = \(\frac{\text{sin z}}{\text{cos z}}\) sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\) |
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| 8. |
If 0.0000152 x 0.042 = A x 108, where 1 \(\leq\) A < 10, find A and B A. A = 9, B = 6.38 B. A = 6.38, B = -9 C. A = -6.38, B = -9 D. A = -9, B = -6.38
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Detailed Solution0.0000152 x 0.042 = A x 1081 \(\leq\) A < 10, it means values of A includes 1 - 9 0.0000152 = 1.52 x 10-5 0.00042 = 4.2 x 10-4 1.52 x 4.2 = 6.384 10-5 x 10-4 = 10-5-4 = 10-9 = 6.38 x 10-9 A = 6.38, B = -9 |
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| 9. |
If (x + 2) and (x - 1) are factors of the expression \(Lx^{3} + 2kx^{2} + 24\), find the values of L and k. A. 1 = -6, k = -9 B. 1 = -2, k = 1 C. k = -1, 1 = -2 D. 1 = 0, k = 1 E. k = 0,1 = 6
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Detailed Solutionf(x) = Lx3 + 2kx2 + 24f(-2) = -8L + 8k = -24 4L - 4k = 12 f(1):L + 2k = -24 L - 4k = 3 3k = -27 k = -9 L = -6 |
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| 10. |
Make T the subject of the equation \(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\) A. T = \(\frac{3av}{1 - v}\) B. T = \(\frac{1 + v}{2a^2v^3}\) C. T = \(\frac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2}\) D. \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
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Detailed Solution\(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\) |
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