Mathematics (Core), 1985, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1985

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 47 Questions

#	Question	Ans
31.	Without using table, calculate the value of 1 + sec2 30o A. 2\(\frac{1}{3}\) B. \(\frac{2}{15}\) C. \(\frac{5}{3}\) D. 3\(\frac{1}{2}\) Show Content Detailed Solution 1 + sec2 30o = sec 30o = \(\frac{2}{\sqrt{3}}\) \(\frac{(2)^2}{3}\) = \(\frac{4}{3}\) 1 + sec2 30o = sec 30o = 1 + \(\frac{4}{3}\) = 2\(\frac{1}{3}\)	A
32.	What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3? A. \(\frac{7}{10}\) B. \(\frac{3}{5}\) C. \(\frac{4}{5}\) D. \(\frac{3}{10}\) Show Content Detailed Solution Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10) Prime: (2, 3, 5, 7) multiples of 3: (3, 6, 9) Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6) Probability = \(\frac{6}{10}\) = \(\frac{3}{5}\)	B
33.	Find the area of a regular hexagon inscribed in a circle of radius 8cm A. 16\(\sqrt{3}\) cm³ B. 96\(\sqrt{3}\) cm³ C. 192\(\sqrt{3}\) cm³ D. 16\(\sqrt{3}\) cm² E. 33cm² Show Content Detailed Solution Area of a regular hexagon = 8 x 8 x sin 60o = 32 x \(\frac{\sqrt{3}}{2}\) Area = 16\(\sqrt{3}\) x 6 = 96 \(\sqrt{3}\)cm2	B
34.	If cos \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90o. Calculate \(\frac{\cot(90 - \theta)}{sin^2\theta}\) A. \(\frac{4}{\sqrt{3}}\) B. \(4 \sqrt{3}\) C. \(\sqrt{\frac{3}{2}}\) D. \(\frac{1}{\sqrt{3}}\) E. \(\frac{2}{\sqrt{3}}\) Show Content Detailed Solution \(\frac{\cot (90 - \theta)}{sin^2\theta}\) \(\cot (90 - \theta) = \tan \theta\) \(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\) \(\tan \theta = \sqrt{3}\) \(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\) \(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{\frac{1}{4}}\) = \(4 \sqrt{3}\)	B
35.	A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively? A. 5kg B. 16kg C. 19kg D. 6kg Show Content Detailed Solution \(\frac{1\sqrt{3}}{(\frac{1}{2})^2}\) = \(\frac{4}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{4\sqrt{3}}{\sqrt{3}}\) m = 64kg, V = \(\frac{4\pi r^3}{3}\) = \(\frac{4\pi(4)^3}{3}\) = \(\frac{256\pi}{3}\) x 10^-6m³ density(P) = \(\frac{\text{Mass}}{\text{Volume}}\) = \(\frac{64}{\frac{256\pi}{3 \times 10^{-6}}}\) = \(\frac{64 \times 3 \times 10^{-6}}{256}\) = \(\frac{3}{4 \times 10^{-6}}\) m = PV = \(\frac{3}{4 \pi \times 10^{-6}}\) x \(\frac{4}{3}\) \(\p	A
36.	PRSQ is a trapezium of area 14cm² in which PQ\|\|RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm² A. 7.0 B. 6.0 C. 5.2 D. 5.0 Show Content Detailed Solution Area of trapezium = 14cm² Area of trapezium = \(\frac{1}{2}\)(a + b)h 14 = \(\frac{1}{2}\)(4 + 3)h 14 = \(\frac{7}{2}\)h h = \(\frac{14 \times 2}{7}\) = 4 Area of SQR = \(\frac{1}{2}\)(3 x 4) = \(\frac{12}{2}\) = 6.0	B
37.	A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red? A. \(\frac{2}{3}\) B. \(\frac{2}{15}\) C. \(\frac{1}{2}\) D. \(\frac{1}{3}\) Show Content Detailed Solution P(R¹) = \(\frac{6}{10}\) = \(\frac{2}{3}\) P(R¹ n R¹¹) = P(both red) \(\frac{3}{5}\) x \(\frac{5}{9}\) = \(\frac{1}{3}\)	D
38.	Two points X and Y both on latitude 60^oS have longitude147^oE and 153^oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude(Take 2\(\pi\)R = 4 x 10⁴km, where R is the radius of the earth) A. 16667km B. 28850km C. 8333km D. 2233km Show Content Detailed Solution Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r Longitude difference = 147 + 153 = 30N^oN distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60^o = \(\frac{300}{360}\) x 4 x 10⁴ x \(\frac{1}{2}\) = 1.667 x 104km(1667 km)	A
39.	In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Z^o at the centre of the circle, what is Z? A. 15 B. 27 C. 50 D. 52 E. 54 Show Content Detailed Solution Total students = 120 grade = 18 Z^o = \(\frac{18}{120}\) x \(\frac{360}{1}\) = 54^o	E
40.	If a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = 6x. Find a in its simplest form A. x² - 1 B. x² + 1 C. x² + 4 D. x² - 4 E. 1 Show Content Detailed Solution a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = a{\(\frac{(x + 1)(x + 2)- (x - 1)(x - 2)}{(x - 2)(x + 2)}\)} = 6 \(\frac{6x}{x^2 - 4}\) = 6x a = x² - 4	D

31.	Without using table, calculate the value of 1 + sec2 30o A. 2\(\frac{1}{3}\) B. \(\frac{2}{15}\) C. \(\frac{5}{3}\) D. 3\(\frac{1}{2}\) Show Content Detailed Solution 1 + sec2 30o = sec 30o = \(\frac{2}{\sqrt{3}}\) \(\frac{(2)^2}{3}\) = \(\frac{4}{3}\) 1 + sec2 30o = sec 30o = 1 + \(\frac{4}{3}\) = 2\(\frac{1}{3}\)	A
32.	What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3? A. \(\frac{7}{10}\) B. \(\frac{3}{5}\) C. \(\frac{4}{5}\) D. \(\frac{3}{10}\) Show Content Detailed Solution Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10) Prime: (2, 3, 5, 7) multiples of 3: (3, 6, 9) Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6) Probability = \(\frac{6}{10}\) = \(\frac{3}{5}\)	B
33.	Find the area of a regular hexagon inscribed in a circle of radius 8cm A. 16\(\sqrt{3}\) cm³ B. 96\(\sqrt{3}\) cm³ C. 192\(\sqrt{3}\) cm³ D. 16\(\sqrt{3}\) cm² E. 33cm² Show Content Detailed Solution Area of a regular hexagon = 8 x 8 x sin 60o = 32 x \(\frac{\sqrt{3}}{2}\) Area = 16\(\sqrt{3}\) x 6 = 96 \(\sqrt{3}\)cm2	B
34.	If cos \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90o. Calculate \(\frac{\cot(90 - \theta)}{sin^2\theta}\) A. \(\frac{4}{\sqrt{3}}\) B. \(4 \sqrt{3}\) C. \(\sqrt{\frac{3}{2}}\) D. \(\frac{1}{\sqrt{3}}\) E. \(\frac{2}{\sqrt{3}}\) Show Content Detailed Solution \(\frac{\cot (90 - \theta)}{sin^2\theta}\) \(\cot (90 - \theta) = \tan \theta\) \(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\) \(\tan \theta = \sqrt{3}\) \(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\) \(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{\frac{1}{4}}\) = \(4 \sqrt{3}\)	B
35.	A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively? A. 5kg B. 16kg C. 19kg D. 6kg Show Content Detailed Solution \(\frac{1\sqrt{3}}{(\frac{1}{2})^2}\) = \(\frac{4}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{4\sqrt{3}}{\sqrt{3}}\) m = 64kg, V = \(\frac{4\pi r^3}{3}\) = \(\frac{4\pi(4)^3}{3}\) = \(\frac{256\pi}{3}\) x 10^-6m³ density(P) = \(\frac{\text{Mass}}{\text{Volume}}\) = \(\frac{64}{\frac{256\pi}{3 \times 10^{-6}}}\) = \(\frac{64 \times 3 \times 10^{-6}}{256}\) = \(\frac{3}{4 \times 10^{-6}}\) m = PV = \(\frac{3}{4 \pi \times 10^{-6}}\) x \(\frac{4}{3}\) \(\p	A

36.	PRSQ is a trapezium of area 14cm² in which PQ\|\|RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm² A. 7.0 B. 6.0 C. 5.2 D. 5.0 Show Content Detailed Solution Area of trapezium = 14cm² Area of trapezium = \(\frac{1}{2}\)(a + b)h 14 = \(\frac{1}{2}\)(4 + 3)h 14 = \(\frac{7}{2}\)h h = \(\frac{14 \times 2}{7}\) = 4 Area of SQR = \(\frac{1}{2}\)(3 x 4) = \(\frac{12}{2}\) = 6.0	B
37.	A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red? A. \(\frac{2}{3}\) B. \(\frac{2}{15}\) C. \(\frac{1}{2}\) D. \(\frac{1}{3}\) Show Content Detailed Solution P(R¹) = \(\frac{6}{10}\) = \(\frac{2}{3}\) P(R¹ n R¹¹) = P(both red) \(\frac{3}{5}\) x \(\frac{5}{9}\) = \(\frac{1}{3}\)	D
38.	Two points X and Y both on latitude 60^oS have longitude147^oE and 153^oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude(Take 2\(\pi\)R = 4 x 10⁴km, where R is the radius of the earth) A. 16667km B. 28850km C. 8333km D. 2233km Show Content Detailed Solution Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r Longitude difference = 147 + 153 = 30N^oN distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60^o = \(\frac{300}{360}\) x 4 x 10⁴ x \(\frac{1}{2}\) = 1.667 x 104km(1667 km)	A
39.	In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Z^o at the centre of the circle, what is Z? A. 15 B. 27 C. 50 D. 52 E. 54 Show Content Detailed Solution Total students = 120 grade = 18 Z^o = \(\frac{18}{120}\) x \(\frac{360}{1}\) = 54^o	E
40.	If a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = 6x. Find a in its simplest form A. x² - 1 B. x² + 1 C. x² + 4 D. x² - 4 E. 1 Show Content Detailed Solution a{\(\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}\)} = a{\(\frac{(x + 1)(x + 2)- (x - 1)(x - 2)}{(x - 2)(x + 2)}\)} = 6 \(\frac{6x}{x^2 - 4}\) = 6x a = x² - 4	D