Paper 1  Objectives  45 Questions
JAMB Exam
Year: 2004
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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#  Question  Ans 

1. 
An arc of a circle of length 22 cm subtends an angle of 3x° at the center of the circle. Find the value of x if the diameter of the circle is 14 cm A. 60^{o} B. 120^{o} C. 180^{o} D. 30^{o}
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Detailed Solution\( ARC\hspace{1mm}length = (\frac{\theta}{360})\times 2\pi r\\22=\frac{3x}{360}\times \left(2 \times(\frac{22}{7})\times(\frac{7}{1})\right)\\3x = 180\\x = \frac{180}{3}\\x = 60^{\circ}\) 

2. 
Find the value of α^{2} + β^{2} if α + β = 2 and the distance between points (1, α) and (β, 1)is 3 units A. 14 B. 3 C. 5 D. 11
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Detailed Solution\(PQ = \sqrt{(β  1)^{2} + (1  α)^{2}}\\3 =\sqrt{(β^{2} 2β^{2} + 1 + 1  2α + α^{2})}\\ 3 = \sqrt{(α^{2} + β^{2}  2α + 2β + 2)}\\ 3 = \sqrt{(α^{2} + β^{2}  2(α + β) + 2)}\\ 3 = \sqrt{(α^{2} + β^{2}  2 * 2 + 2)}\\ 3 = \sqrt{(α^{2} + β^{2}  2)}\\ 9 = (α^{2} + β^{2}  2)\\ α^{2} + β^{2} = 9 + 2\\ α^{2} + β^{2} = 11\) 

3. 
The sum of the interior angles of a pentagon is 6x + 6y. Find y in the terms of x A. y = 90  x B. y = 150  x C. y = 60  x D. y = 120 x
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Detailed Solution6x + 6y = (n  2) 1806x + 6y= (5  2) 180 6(x + y) = 3 * 180 x + y = (3 * 180)/6 x + y = 90^{o} y = 90  x 

4. 
In the diagram above, PQ = 4 cm and TS = 6 cm. If the area of parallelogram PQTU is 32 cm\(^2\), find the area of the trapezium PQRU A. 60 cm^{2} B. 72 cm^{2} C. 48 cm^{2} D. 24 cm^{2}
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Detailed SolutionArea of parallelogram PQTU = base * height32 = 4 * h w = 32/4 w = 8 ∴ = Area of Trapezium PQRU = (1/2)(4 + 14) *8 = 1/2 * (18 * 8) = 72 cm\(^2\) 

5. 
Find the midpoint of the line joining P(3, 5) and Q(5, 3). A. (1, 1) B. (2, 2) C. (4, 4) D. (4, 4)
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Detailed Solution\(Mid point = \frac{(x_1 + x_2)}{2} ; \frac{(y_1 + y_2)}{2}\\= \frac{(3 + 5)}{2} ; \frac{(5  3)}{2}\\ = \frac{2}{2} ; \frac{2}{2}\\ = (1, 1)\) 

6. 
Determine the locus of a point inside a square PQRS which is eqidistant from PQ and QR A. The diagonal QS B. the perpendicular bisector of PQ C. The diagonal PR D. side SR
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Detailed SolutionThe diagonal QS bisects the angle formed by PQ and QR∴ [A] 

7. 
Find the value of x in the figure above A. 15√6 B. 20√6 C. 3√6 D. 5√6
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Detailed Solution\(\frac{X}{sin45}=\frac{15}{sin60}\\X=\frac{15sin45}{sin60}\\X=\frac{15\times(\frac{1}{\sqrt{2}})}{\sqrt{\frac{3}{2}}}\\X=\frac{15}{\sqrt{2}}\times\frac{2}{\sqrt{3}}\\X=\frac{30}{\sqrt{6}}\\X=\frac{30}{\sqrt{6}}\times\frac{\sqrt{6}}{\sqrt{6}}=\frac{30\sqrt{6}}{6}=5\sqrt{6}\) 

8. 
P, R and S lie on a circle center as shown above while Q lies outside the circle. Find ∠PSO A. 45^{o} B. 55^{o} C. 35^{o} D. 40^{o}
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Detailed Solutionx^{o} = 35 + 29 (Exterior angle = sum of two interior opposite angles)x = 55^{o} (∠ at the center twice ∠ at circumference) y = 110^{o} ∠PSO = ∠SPO (base ∠S of 1sc Δ b/c PO = SO) ∴ ∠PSO = (180  110)/2 = 35^{o} 

9. 
The locus of a point which is 5 cm from the line LM is a A. line distance 10 cm from LM and parallel to LM B. pair of line on opposite sides of LM and parallel to it, each distance 5 cm from LM C. line parallel to LM and 5 cm from LM D. pair of parallel lines on one side of LM and parallel to LM
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Detailed SolutionThe locus of a point which is 5cm from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM. 
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