Year :
2001
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 50 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Solve the equations A. (2, 3) and ( 3, 5) B. (2, 5) and (5, 2) C. (5, 2) and ( 5, 3) D. (5, 3) and (3, 5) Detailed Solutionm2 + n2 = 29 .......(1)m + n = 7 ............(2) From (2), m = 7 - n but m2 + n2 = 29, substituting; (7-n)2 + n2 = 29 49 - 14n + n2 + n2 = 29 => 2n2 -14n + 20 = 0 Thus n2 -7n + 10 = 0 Factorizing; (n-5)(n-2) = 0 n - 5 = 0, => n = 5 n - 2 = 0, => n = 2. When n = 5, m + n = 7, => m = 2, When n = 2, |
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| 12. |
An operation * is defined on the set of real numbers by a*b = a + b + 1. If the identity elements is -1, find the inverse of the element 2 under *. A. 4 B. zero C. -2 D. -4 Detailed SolutionBy definition a*b = a + b + 1.Let the inverse of the element 2 be x, Therefore 2*x = -1 i.e. 2 + x + 1 = -1 3 + x = -1 x = -1 - 3 x = -4 |
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| 13. |
The sixth term of an A.P is half of its twelfth term. The first term of the A.P is equal to A. zero B. half of the common difference C. double the common difference D. the common difference Detailed Solution1st statement: U6 = 1/2(U12)a + (n -1)d = 1/2[a + (n-1)d] a + 5d = a + 11d 2(a + 5d) = a + 11d 2a + 10d = a + 11d Solving, => a = d Hence the first term is equal to the common difference |
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| 14. |
Factorize 4x2 - 9y2 + 20x + 25 A. (2x -3y + 5)(2x - 3y - 5) B. (2x - 3y)(2x + 3y) C. (2x - 3y +5)(2x + 3y + 5) D. (2x + 5)(2x - 9y +5) Detailed SolutionGiven: 4x2 - 9y2 + 20x + 25Collect like terms: 4x2 + 20x + 25 - 9y2 (2x + 5)(2x + 5) - 9y2 (2x + 5)2 - (3y)2 (2x - 3y +5)(2x + 3y + 5) |
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| 15. |
A sector of a circle of radius 7.2cm which subtends an angle of 300° at the centre is used to form a cone. What is the radius of the base of the cone? A. 8cm B. 6cm C. 9cm D. 7cm Detailed Solution(r/L) = (θ/360°)Given θ = 300, and L = 7.2cm, => r = (300 x 7.2)/360 r = 6cm |
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| 16. |
A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30° A. 4√3cm B. 8cm C. 8√3cm D. 4cm Detailed SolutionHint: Make a sketch of an isosceles triangle with two of its sides and angles angles.: PQ (r) = PR (q) = 8cm : R° = Q° = 30° Sum of angles in a triangle = 180° P° + Q° + R° = 180° P° + 30° + 30° = 180° P° = 180° - 60° p° = 120°. : PQ = r, PR = q, QR = p Using sine rule: \(\frac{p}{sinP}\) = \(\frac{q}{sinq}\) \(\frac{p}{sin120°}\) = \(\frac{8}{sin30°}\) cross multiply p = \(\frac{8 X sin120°}{sin30°}\) p = \(\frac{8 X √3/2 }{1/2}\) |
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| 17. |
A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line. A. y = (x/10) + 5 B. y = x + 5 C. √3y = - x + 5√3 D. √3y = x + 5√3 Detailed SolutionCos 30 = 5/xx cos 30 = 5, => x = 5√3 Coordinates of P = -5, 3, 0 Coordinates of Q = 0, 5 Gradient of PQ = (y2 - y1) = 5/5√3 = 1/√3 Equation of PQ = y - y1 = m (x -x1) y - 0 = 1/√3 (x -(-5√3)) Thus: √3y = x + 5√3 |
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| 18. |
Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units. A. 100° B. 120° C. 30° D. 60° Detailed SolutionCos θ° = t2 + t2 -(√3t)2= 2t2 - 3t2 = -t2/2t2 = -1/2 Thus θ = cos-1 (-0.5) = 120° |
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| 19. |
Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3). A. 4 B. 6 C. 3 Detailed SolutionThe line joining (P, 4) and (6, -2).Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\) The line joining (2, P) and (-1, 3) Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\) For perpendicular lines, the product of their gradient = -1. \((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\) \(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\) \(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\) |
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| 20. |
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle. A. 6 B. 2 C. 3 D. 8 Detailed SolutionLet the ext. angle = xThus int. angle = 2x But sum of int + ext = 180 (angle of a straight line). 2x + x = 180 3x = 180 x = 180/3 = 60 Each ext angle = 360/n => 60 = 360/n n = 360/60 = 6 |
| 11. |
Solve the equations A. (2, 3) and ( 3, 5) B. (2, 5) and (5, 2) C. (5, 2) and ( 5, 3) D. (5, 3) and (3, 5) Detailed Solutionm2 + n2 = 29 .......(1)m + n = 7 ............(2) From (2), m = 7 - n but m2 + n2 = 29, substituting; (7-n)2 + n2 = 29 49 - 14n + n2 + n2 = 29 => 2n2 -14n + 20 = 0 Thus n2 -7n + 10 = 0 Factorizing; (n-5)(n-2) = 0 n - 5 = 0, => n = 5 n - 2 = 0, => n = 2. When n = 5, m + n = 7, => m = 2, When n = 2, |
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| 12. |
An operation * is defined on the set of real numbers by a*b = a + b + 1. If the identity elements is -1, find the inverse of the element 2 under *. A. 4 B. zero C. -2 D. -4 Detailed SolutionBy definition a*b = a + b + 1.Let the inverse of the element 2 be x, Therefore 2*x = -1 i.e. 2 + x + 1 = -1 3 + x = -1 x = -1 - 3 x = -4 |
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| 13. |
The sixth term of an A.P is half of its twelfth term. The first term of the A.P is equal to A. zero B. half of the common difference C. double the common difference D. the common difference Detailed Solution1st statement: U6 = 1/2(U12)a + (n -1)d = 1/2[a + (n-1)d] a + 5d = a + 11d 2(a + 5d) = a + 11d 2a + 10d = a + 11d Solving, => a = d Hence the first term is equal to the common difference |
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| 14. |
Factorize 4x2 - 9y2 + 20x + 25 A. (2x -3y + 5)(2x - 3y - 5) B. (2x - 3y)(2x + 3y) C. (2x - 3y +5)(2x + 3y + 5) D. (2x + 5)(2x - 9y +5) Detailed SolutionGiven: 4x2 - 9y2 + 20x + 25Collect like terms: 4x2 + 20x + 25 - 9y2 (2x + 5)(2x + 5) - 9y2 (2x + 5)2 - (3y)2 (2x - 3y +5)(2x + 3y + 5) |
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| 15. |
A sector of a circle of radius 7.2cm which subtends an angle of 300° at the centre is used to form a cone. What is the radius of the base of the cone? A. 8cm B. 6cm C. 9cm D. 7cm Detailed Solution(r/L) = (θ/360°)Given θ = 300, and L = 7.2cm, => r = (300 x 7.2)/360 r = 6cm |
| 16. |
A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30° A. 4√3cm B. 8cm C. 8√3cm D. 4cm Detailed SolutionHint: Make a sketch of an isosceles triangle with two of its sides and angles angles.: PQ (r) = PR (q) = 8cm : R° = Q° = 30° Sum of angles in a triangle = 180° P° + Q° + R° = 180° P° + 30° + 30° = 180° P° = 180° - 60° p° = 120°. : PQ = r, PR = q, QR = p Using sine rule: \(\frac{p}{sinP}\) = \(\frac{q}{sinq}\) \(\frac{p}{sin120°}\) = \(\frac{8}{sin30°}\) cross multiply p = \(\frac{8 X sin120°}{sin30°}\) p = \(\frac{8 X √3/2 }{1/2}\) |
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| 17. |
A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line. A. y = (x/10) + 5 B. y = x + 5 C. √3y = - x + 5√3 D. √3y = x + 5√3 Detailed SolutionCos 30 = 5/xx cos 30 = 5, => x = 5√3 Coordinates of P = -5, 3, 0 Coordinates of Q = 0, 5 Gradient of PQ = (y2 - y1) = 5/5√3 = 1/√3 Equation of PQ = y - y1 = m (x -x1) y - 0 = 1/√3 (x -(-5√3)) Thus: √3y = x + 5√3 |
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| 18. |
Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units. A. 100° B. 120° C. 30° D. 60° Detailed SolutionCos θ° = t2 + t2 -(√3t)2= 2t2 - 3t2 = -t2/2t2 = -1/2 Thus θ = cos-1 (-0.5) = 120° |
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| 19. |
Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3). A. 4 B. 6 C. 3 Detailed SolutionThe line joining (P, 4) and (6, -2).Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\) The line joining (2, P) and (-1, 3) Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\) For perpendicular lines, the product of their gradient = -1. \((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\) \(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\) \(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\) |
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| 20. |
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle. A. 6 B. 2 C. 3 D. 8 Detailed SolutionLet the ext. angle = xThus int. angle = 2x But sum of int + ext = 180 (angle of a straight line). 2x + x = 180 3x = 180 x = 180/3 = 60 Each ext angle = 360/n => 60 = 360/n n = 360/60 = 6 |