Mathematics (Core), 1983, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1983

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 48 Questions

#	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 Solution first 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)	D
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 Solution Total sharing ratio is 9 X has 4, Y has 4 + 1 If 1 is N5000 Total profit = 5000 x 9 = N45,000	E
3.	Given a regular hexagon, calculate each interior angle of the hexagon A. 60^o B. 30^o C. 120^o D. 45^o E. 135^o Show Content Detailed Solution Sum of interior angles of polygon = (2n - 4)rt < s sum of interior angles of an hexagon (2 x 6 - 4) x 90^o = (12 - 4) x 90^o = 8 x 90^o = 720^o each interior angle will have \(\frac{720^o}{6}\) = 120^o	C
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 \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\)	B
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 Solution 4x - 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	A
6.	In a figure, PQR = 60^o, PRS = 90^o, RPS = 45^o, 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 Solution From the diagram, sin 60^o = \(\frac{PR}{8}\) PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45^o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45^o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\)	B
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 Solution Given 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 = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)	D
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 Solution 0.0000152 x 0.042 = A x 10⁸ 1 \(\leq\) 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	B
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 Solution f(x) = Lx3 + 2kx2 + 24 f(-2) = -8L + 8k = -24 4L - 4k = 12 f(1):L + 2k = -24 L - 4k = 3 3k = -27 k = -9 L = -6	A
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 \(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\) \(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)	D

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 Solution first 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)	D
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 Solution Total sharing ratio is 9 X has 4, Y has 4 + 1 If 1 is N5000 Total profit = 5000 x 9 = N45,000	E
3.	Given a regular hexagon, calculate each interior angle of the hexagon A. 60^o B. 30^o C. 120^o D. 45^o E. 135^o Show Content Detailed Solution Sum of interior angles of polygon = (2n - 4)rt < s sum of interior angles of an hexagon (2 x 6 - 4) x 90^o = (12 - 4) x 90^o = 8 x 90^o = 720^o each interior angle will have \(\frac{720^o}{6}\) = 120^o	C
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 \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\)	B
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 Solution 4x - 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	A

6.	In a figure, PQR = 60^o, PRS = 90^o, RPS = 45^o, 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 Solution From the diagram, sin 60^o = \(\frac{PR}{8}\) PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45^o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45^o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\)	B
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 Solution Given 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 = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)	D
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 Solution 0.0000152 x 0.042 = A x 10⁸ 1 \(\leq\) 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	B
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 Solution f(x) = Lx3 + 2kx2 + 24 f(-2) = -8L + 8k = -24 4L - 4k = 12 f(1):L + 2k = -24 L - 4k = 3 3k = -27 k = -9 L = -6	A
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 \(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\) \(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)	D