Mathematics (Core)
Paper 1 | Objectives | 48 Questions
JAMB Exam
Year: 1983
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
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
11 hot tips for highest score in IELTS 2022 [ get band 8 or 9] Reading, Listening, Writing and speaking
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
8 Tips To Be More Productive in Test Revision for good grades and higher performance in exams
| # | Question | Ans |
|---|---|---|
| 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)
Show Content
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) |
|
| 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
Show Content
Detailed SolutionTotal sharing ratio is 9X has 4, Y has 4 + 1 If 1 is N5000 Total profit = 5000 x 9 = N45,000 |
|
| 3. |
Given a regular hexagon, calculate each interior angle of the hexagon A. 60o B. 30o C. 120o D. 45o E. 135o
Show Content
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 |
|
| 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
Show Content
Detailed Solution\(x \propto y^{3} z^{4}\)\(x = ky^{3} z^{4}\) If y is halved and z is doubled, we have \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\) |
|
| 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
Show Content
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 |
|
| 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
Show Content
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\(\sqrt{6}\) |
|
| 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. \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
Show Content
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 = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\) |
|
| 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
Show Content
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 |
|
| 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
Show Content
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 |
|
| 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. \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
Show Content
Detailed Solution\(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\) |
Preview displays only 10 out of the 48 Questions