Paper 1  Objectives  48 Questions
JAMB Exam
Year: 1983
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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#  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)
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Detailed Solutionfirst rearrange the given data in the form 3, 5, 5, 7, 8, 9 Mean(x) = \(\frac{\sum x}{N}\)= \(\frac{42}{7}\) = 6, rearrange 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
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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 

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}
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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 90^{o} = (12  4) x 90^{o} = 8 x 90^{o} = 720^{o} each interior angle will have \(\frac{720^o}{6}\) = 120^{o} 

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 \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\) 

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 

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
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Detailed SolutionFrom 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\(\sqrt{6}\) 

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 = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\) 

8. 
If 0.0000152 x 0.042 = A x 10^{8}, 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 10^{8}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 

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 

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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