Year :
1978
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
A man runs a distance of 9km at a constant speed for the first 4 km and then 2 km\h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4 km is A. 6 km/h B. 8 km/h C. 9 km\h D. 11 km\h E. 13 km\h Detailed SolutionLet the average speed for the first 4 km = x km/h.Hence, the last 5 km, speed = (x + 2) km/h Recall: \(Time = \frac{Distance}{Speed}\) Total time = 1 hour. \(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\) \(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\) \(9x + 8 = x^{2} + 2x\) \(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\) \(x^{2} - 8x + x - 8 = 0\) \(x(x - 8) + 1(x - 8) = 0\) \(x = -1; x = 8\) Since speed cannot be negative, x = 8km/h. |
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| 22. |
A pyramid is constructed on a cuboid. The figure has A. twelve faces B. thirteen vertices C. fourteen edges D. fifteen edges E. sixteen edges Detailed SolutionThe pyramid has 8 edges in itself while the cuboid has 12 edges. When merging the two shapes together, the edge of the base of the pyramid becomes same as the edges of the top of the cuboid.Hence, the new structure will have (12 + 8) - 4 = 16 edges. |
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| 23. |
In a geometric progression, the first term is 153 and the sixth term is \(\frac{17}{27}\). The sum of the first four terms is A. \(\frac{860}{3}\) B. \(\frac{680}{3}\) C. \(\frac{608}{3}\) D. \(\frac{680}{3}\) Detailed Solutiona = 153 - 1st term, 6th term = \(\frac{17}{27}\)nth term = arn Sn = a(1 - rn) where r < 1 6th term = 153\(\frac{1 - 0.4^4}{1 - 0.4}\) = \(\frac{680}{3}\) |
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| 24. |
An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is A. 351 B. 531 C. 135 D. 153 Detailed Solution1st term a = 11, 4th term = 32nth term = a + (n - 1)d 4th term = 11 = (4 - 1)d = 11 + 3d = 32 3d = 21 d = 7 sn = n(2a + (n - 1)d) sn = \(\frac{9}{2}\)(2 \times 11) + (9 - 1)7 \(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78 = 351 |
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| 25. |
A triangle has angles 30o, 15o and 135o. The side opposite to the angle 30o is length 6cm. The side opposite to the angle 135o is equal to A. 12cm B. 6cm C. 6\(\sqrt{2}\)cm D. 12\(\sqrt{2}\)cm Detailed Solution\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\)\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\) x = \(\frac{6 \times \sin 45}{\sin 30}\) = \(6 \sqrt{2}\)cm |
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| 26. |
A regular hexagon is constructed inside a circle of diameter 12cm. The area of the hexagon is A. 36\(\pi\)cm2 B. 54\(\sqrt{3}\)cm2 C. \(\sqrt{3}\)cm2 D. \(\frac{1}{x - 1}\) Detailed SolutionSum of interior angle of hexagon = [2(6) - 4]90o= 720o sum of central angle = 360o Each central angle = \(\frac{360}{6}\) = 60o Area of Hexagon = \(\frac{1}{2}\) x 6 x 6 sin 60o \(\frac{36 \times 6\sqrt{3}}{2 \times 2}\) = \(54 \sqrt{3}\)cm2 |
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| 27. |
In a soccer competition in one season, a club had scored the following goals: 2, 0, 3, 3, 2, 1, 4, 0, 0, 5, 1, 0, 2, 2, 1, 3, 1, 4, 1 and 1. The mean, median and mode are respectively A. 1, 1.8 and 1.5 B. 1.8, 1.5 and 1 C. 1.8, 1 and 1.5 D. 1.51, 1 and 1.8 E. 1.5, 1.8 and 1 Detailed SolutionBy re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5.Mean = \(\frac{36}{20}\) = 1.8 Median = \(\frac{1 + 2}{2}\) = \(\frac{3}{2}\) = 1.5 Mode = 1 = 1.8, 1.5 and 1 |
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| 28. |
If sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3, then the angle \(\theta\) is equal to A. 30o B. 45o C. 60o D. 90o E. 105o Detailed Solutionsec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\) 1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 2 tan\(^2\) \(\theta\) = 2 tan\(^2\) \(\theta\) = 1 tan\(\theta\) = √1 where √1 = 1 tan\(\theta\) = 1 And tan 45° = 1 ∴ \(\theta\) = 45° |
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| 29. |
A hollow right prism of equilateral triangular base of side 4cm is filled with water up to a certain height. If a sphere of radius \(\frac{1}{2}\)cm is immersed in the water, then the rise of water is A. 1cm B. \(\sqrt{\frac{3\pi}{24}}\) C. \(\frac{\pi}{24\sqrt{3}}\) D. 24\(\sqrt{3}\) Detailed SolutionThe rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\)Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) = \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\) = \(\frac{\pi}{24\sqrt{3}}\) |
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| 30. |
The set of value of x and y which satisfies the equations x2 - y - 1 = 0 and y - 2x + 2 = 0 is A. 1, 0 B. 1, 1 C. 2, 2 D. 0, 2 E. 1, 2 Detailed Solutionx2 - y - 1 = 0.......(i)y - 2x + 2 = 0......(ii) By re-arranging eqn. (ii) y = 2x - 2........(iii) Subst. eqn. (iii) in eqn (i) x2 - (2x - 2) - 1 = 0 x2 - 2x + 1 = 0 = (x - 1) = 0 When x - 1 = 0 x = 1 Sub. for x = 1 in eqn. (iii) y = 2 - 2 = 0 x = 1, y = 0 |
| 21. |
A man runs a distance of 9km at a constant speed for the first 4 km and then 2 km\h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4 km is A. 6 km/h B. 8 km/h C. 9 km\h D. 11 km\h E. 13 km\h Detailed SolutionLet the average speed for the first 4 km = x km/h.Hence, the last 5 km, speed = (x + 2) km/h Recall: \(Time = \frac{Distance}{Speed}\) Total time = 1 hour. \(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\) \(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\) \(9x + 8 = x^{2} + 2x\) \(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\) \(x^{2} - 8x + x - 8 = 0\) \(x(x - 8) + 1(x - 8) = 0\) \(x = -1; x = 8\) Since speed cannot be negative, x = 8km/h. |
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| 22. |
A pyramid is constructed on a cuboid. The figure has A. twelve faces B. thirteen vertices C. fourteen edges D. fifteen edges E. sixteen edges Detailed SolutionThe pyramid has 8 edges in itself while the cuboid has 12 edges. When merging the two shapes together, the edge of the base of the pyramid becomes same as the edges of the top of the cuboid.Hence, the new structure will have (12 + 8) - 4 = 16 edges. |
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| 23. |
In a geometric progression, the first term is 153 and the sixth term is \(\frac{17}{27}\). The sum of the first four terms is A. \(\frac{860}{3}\) B. \(\frac{680}{3}\) C. \(\frac{608}{3}\) D. \(\frac{680}{3}\) Detailed Solutiona = 153 - 1st term, 6th term = \(\frac{17}{27}\)nth term = arn Sn = a(1 - rn) where r < 1 6th term = 153\(\frac{1 - 0.4^4}{1 - 0.4}\) = \(\frac{680}{3}\) |
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| 24. |
An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is A. 351 B. 531 C. 135 D. 153 Detailed Solution1st term a = 11, 4th term = 32nth term = a + (n - 1)d 4th term = 11 = (4 - 1)d = 11 + 3d = 32 3d = 21 d = 7 sn = n(2a + (n - 1)d) sn = \(\frac{9}{2}\)(2 \times 11) + (9 - 1)7 \(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78 = 351 |
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| 25. |
A triangle has angles 30o, 15o and 135o. The side opposite to the angle 30o is length 6cm. The side opposite to the angle 135o is equal to A. 12cm B. 6cm C. 6\(\sqrt{2}\)cm D. 12\(\sqrt{2}\)cm Detailed Solution\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\)\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\) x = \(\frac{6 \times \sin 45}{\sin 30}\) = \(6 \sqrt{2}\)cm |
| 26. |
A regular hexagon is constructed inside a circle of diameter 12cm. The area of the hexagon is A. 36\(\pi\)cm2 B. 54\(\sqrt{3}\)cm2 C. \(\sqrt{3}\)cm2 D. \(\frac{1}{x - 1}\) Detailed SolutionSum of interior angle of hexagon = [2(6) - 4]90o= 720o sum of central angle = 360o Each central angle = \(\frac{360}{6}\) = 60o Area of Hexagon = \(\frac{1}{2}\) x 6 x 6 sin 60o \(\frac{36 \times 6\sqrt{3}}{2 \times 2}\) = \(54 \sqrt{3}\)cm2 |
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| 27. |
In a soccer competition in one season, a club had scored the following goals: 2, 0, 3, 3, 2, 1, 4, 0, 0, 5, 1, 0, 2, 2, 1, 3, 1, 4, 1 and 1. The mean, median and mode are respectively A. 1, 1.8 and 1.5 B. 1.8, 1.5 and 1 C. 1.8, 1 and 1.5 D. 1.51, 1 and 1.8 E. 1.5, 1.8 and 1 Detailed SolutionBy re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5.Mean = \(\frac{36}{20}\) = 1.8 Median = \(\frac{1 + 2}{2}\) = \(\frac{3}{2}\) = 1.5 Mode = 1 = 1.8, 1.5 and 1 |
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| 28. |
If sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3, then the angle \(\theta\) is equal to A. 30o B. 45o C. 60o D. 90o E. 105o Detailed Solutionsec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\) 1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 2 tan\(^2\) \(\theta\) = 2 tan\(^2\) \(\theta\) = 1 tan\(\theta\) = √1 where √1 = 1 tan\(\theta\) = 1 And tan 45° = 1 ∴ \(\theta\) = 45° |
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| 29. |
A hollow right prism of equilateral triangular base of side 4cm is filled with water up to a certain height. If a sphere of radius \(\frac{1}{2}\)cm is immersed in the water, then the rise of water is A. 1cm B. \(\sqrt{\frac{3\pi}{24}}\) C. \(\frac{\pi}{24\sqrt{3}}\) D. 24\(\sqrt{3}\) Detailed SolutionThe rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\)Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) = \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\) = \(\frac{\pi}{24\sqrt{3}}\) |
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| 30. |
The set of value of x and y which satisfies the equations x2 - y - 1 = 0 and y - 2x + 2 = 0 is A. 1, 0 B. 1, 1 C. 2, 2 D. 0, 2 E. 1, 2 Detailed Solutionx2 - y - 1 = 0.......(i)y - 2x + 2 = 0......(ii) By re-arranging eqn. (ii) y = 2x - 2........(iii) Subst. eqn. (iii) in eqn (i) x2 - (2x - 2) - 1 = 0 x2 - 2x + 1 = 0 = (x - 1) = 0 When x - 1 = 0 x = 1 Sub. for x = 1 in eqn. (iii) y = 2 - 2 = 0 x = 1, y = 0 |