Year :
1991
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
31 - 40 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
One angle of a rhombus is 60o. The shorter of the two diagonals is 8cm long. Find the length of the longer one. A. 8\(\sqrt{3}\) B. \(\frac{16}{\sqrt{3}}\) C. \(\sqrt{3}\) D. \(\frac{10}{\sqrt{3}}\) Detailed Solution\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)x = 4 tan60 = 4\(\sqrt{3}\) BD = 2x = 8\(\sqrt{3}\) |
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| 32. |
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x A. 118o B. 72o C. 62o D. 36o Detailed SolutionThe sum of the exterior angles of a regular polygon = 360°.\(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°\) \(5x + 50 = 360° \implies 5x = 360° - 50° = 310°\) \(x = \frac{310°}{5} = 62°\) |
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| 33. |
A flagstaff stands on the top of a vertical tower. A man standing 60 m away from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64o and 62o respectively. Find the length of the flagstaff. A. 60 (tan 62o - tan 64o) B. 60 (cot 64o - cot 62o) C. 60 (cot 62o - cot 64o) D. 60 (tan 64o - tan 62o) Detailed Solution\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\)BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o) |
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| 34. |
Simplify \(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\) A. tan x B. tanx secx C. \(\sec^2 x\) D. cosec2x Detailed Solution\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)= \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\) = \(1 + \tan^{2} x\) = \(1 + \frac{\sin^{2} x}{\cos^{2} x}\) = \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\) = \(\frac{1}{\cos^{2} x} = \sec^{2} x\) |
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| 35. |
If cos x = \(\sqrt{\frac{a}{b}}\) find cosec x A. \(\frac{b}{\sqrt{b - a}}\) B. \(\sqrt{\frac{b}{a}}\) C. \(\sqrt{\frac{b}{b - a}}\) D. \(\sqrt{\frac{b - a}{a}}\) Detailed Solutioncosx = \(\sqrt{\frac{a}{b}}\)y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras y2 = b - a ∴ y = b - a cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\) \(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\) = \(\sqrt{\frac{b}{b - a}}\) |
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| 36. |
From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X. A. 0.30o B. 045o C. 060o D. 090o Detailed Solutiontan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\)\(\theta\) = tan \(\frac{1}{3}\) = 60o Bearing of y from x = 060o |
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| 37. |
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm A. 56\(\pi\) cm2 B. 72\(\pi\) cm2 C. 96\(\pi\) cm2 D. 192\(\pi\) cm2 Detailed SolutionThe total surface area of a cylinder = \(2\pi r (r + h)\)= \(2 \pi (4) (4 + 5)\) = \(8 \pi (9) = 72 \pi cm^{2}\) |
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| 38. |
3% of a family's income is spent on electricity, 59% on food, 20% on transport, 11% on education and 7% on extended family. The angles subtended at the centre of the pie chart under education and food are respectively A. 76.8o and 25.2o B. 10.8o and 224.6o C. 112.4o and 72.0o D. 39.6o and 212.4o Detailed SolutionEducation = 11% = \(\frac{11}{100} \times 360° = 39.6°\)Food = 59% = \(\frac{59}{100} \times 360° = 212.4°\) |
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| 39. |
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result A. 6.7, 6 B. 7.6, 5 C. 5.7, 87 D. 34, 6 Detailed Solution\(Mean = \frac{\sum fx}{\sum f}\)= \(\frac{333}{50} = 6.66 \approxeq 6.7\) The median is the average of the 25th and 26th position = 6. |
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| 40. |
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result A. 96 B. 94 C. 92 D. 90 Detailed Solution% of boxes containing at least 5 defective bolts each= \(\frac{7 + 17 + 10 + 8 + 6}{50}\) = \(\frac{48}{50}\) x \(\frac{100}{1}\) = 96% |
| 31. |
One angle of a rhombus is 60o. The shorter of the two diagonals is 8cm long. Find the length of the longer one. A. 8\(\sqrt{3}\) B. \(\frac{16}{\sqrt{3}}\) C. \(\sqrt{3}\) D. \(\frac{10}{\sqrt{3}}\) Detailed Solution\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)x = 4 tan60 = 4\(\sqrt{3}\) BD = 2x = 8\(\sqrt{3}\) |
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| 32. |
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x A. 118o B. 72o C. 62o D. 36o Detailed SolutionThe sum of the exterior angles of a regular polygon = 360°.\(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°\) \(5x + 50 = 360° \implies 5x = 360° - 50° = 310°\) \(x = \frac{310°}{5} = 62°\) |
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| 33. |
A flagstaff stands on the top of a vertical tower. A man standing 60 m away from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64o and 62o respectively. Find the length of the flagstaff. A. 60 (tan 62o - tan 64o) B. 60 (cot 64o - cot 62o) C. 60 (cot 62o - cot 64o) D. 60 (tan 64o - tan 62o) Detailed Solution\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\)BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o) |
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| 34. |
Simplify \(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\) A. tan x B. tanx secx C. \(\sec^2 x\) D. cosec2x Detailed Solution\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)= \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\) = \(1 + \tan^{2} x\) = \(1 + \frac{\sin^{2} x}{\cos^{2} x}\) = \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\) = \(\frac{1}{\cos^{2} x} = \sec^{2} x\) |
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| 35. |
If cos x = \(\sqrt{\frac{a}{b}}\) find cosec x A. \(\frac{b}{\sqrt{b - a}}\) B. \(\sqrt{\frac{b}{a}}\) C. \(\sqrt{\frac{b}{b - a}}\) D. \(\sqrt{\frac{b - a}{a}}\) Detailed Solutioncosx = \(\sqrt{\frac{a}{b}}\)y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras y2 = b - a ∴ y = b - a cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\) \(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\) = \(\sqrt{\frac{b}{b - a}}\) |
| 36. |
From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X. A. 0.30o B. 045o C. 060o D. 090o Detailed Solutiontan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\)\(\theta\) = tan \(\frac{1}{3}\) = 60o Bearing of y from x = 060o |
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| 37. |
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm A. 56\(\pi\) cm2 B. 72\(\pi\) cm2 C. 96\(\pi\) cm2 D. 192\(\pi\) cm2 Detailed SolutionThe total surface area of a cylinder = \(2\pi r (r + h)\)= \(2 \pi (4) (4 + 5)\) = \(8 \pi (9) = 72 \pi cm^{2}\) |
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| 38. |
3% of a family's income is spent on electricity, 59% on food, 20% on transport, 11% on education and 7% on extended family. The angles subtended at the centre of the pie chart under education and food are respectively A. 76.8o and 25.2o B. 10.8o and 224.6o C. 112.4o and 72.0o D. 39.6o and 212.4o Detailed SolutionEducation = 11% = \(\frac{11}{100} \times 360° = 39.6°\)Food = 59% = \(\frac{59}{100} \times 360° = 212.4°\) |
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| 39. |
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result A. 6.7, 6 B. 7.6, 5 C. 5.7, 87 D. 34, 6 Detailed Solution\(Mean = \frac{\sum fx}{\sum f}\)= \(\frac{333}{50} = 6.66 \approxeq 6.7\) The median is the average of the 25th and 26th position = 6. |
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| 40. |
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result A. 96 B. 94 C. 92 D. 90 Detailed Solution% of boxes containing at least 5 defective bolts each= \(\frac{7 + 17 + 10 + 8 + 6}{50}\) = \(\frac{48}{50}\) x \(\frac{100}{1}\) = 96% |