Mathematics (Core), 1984, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1984

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 45 Questions

#	Question	Ans
31.	A right circular cone has a base radius r cm and a vertical angle 2y^o. The height of the cone is A. r tan y^o cm B. r sin y^o cm C. r cot y^o cm D. r cos y^o cm E. r cosec y^o cm Show Content Detailed Solution \(\frac{r}{h}\) = tan y^o h = \(\frac{r}{tan y^o}\) = r cot y^o	C
32.	Two fair dice are rolled. What is the probability that both show up the same number of points. A. \(\frac{1}{36}\) B. \(\frac{7}{36}\) C. \(\frac{1}{2}\) D. \(\frac{1}{3}\) E. \(\frac{1}{6}\) Show Content Detailed Solution A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face Prob. of both showing the same number of points = \(\frac{6}{36}\) = \(\frac{1}{6}\)	E
33.	The larger value of y for which (y - 1)² = 4y - 7 is A. 2 B. 4 C. 6 D. 7 E. 8 Show Content Detailed Solution (y - 1)² = 4y - 7 y² - 2xy + 1 = 4y - 7 y² - 6y + 8 = 0 (y - 4)(y - 2) y = 4 or 2 = 4	B
34.	If sin \(\theta\) = \(\frac{x}{y}\) and 0^o < 90^o then find \(\frac{1}{tan\theta}\) A. \(\frac{x}{\sqrt{y^2 - x^2}}\) B. \(\frac{y}{x}\) C. \(\frac{\sqrt{y^2 + x^2}}{y^2 - x^2}\) D. \(\frac{y^2 - x^2}{x}\) Show Content Detailed Solution \(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\) sin\(\theta\) = \(\frac{x}{y}\) cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)	D
35.	Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below \(\begin{array}{c\|c} \text{Class boundary in cm} & \text{frequency} & \\\hline 3.35 - 3.45 & 3\\ 3.45 - 3.55 & 6\\ 3.55 - 3.65 & 7\\ 3.65 - 3.75 & 4\end{array}\) What is the mean diameter of the copper spheres? A. 3.40cm B. 3.58cm C. 3.56cm D. 3.62cm Show Content Detailed Solution \(\begin{array}{c\|c} \text{x(mid point)} & f & fx\\ \hline 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0\\3.6 & 7 & 25.2\\3.7 & 4 & 14.8\end{array}\) \(\sum f\) = 20 \(\sum fx\) = 71.2 mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{71.2}{20}\) = 3.56	C
36.	What is the volume of this regular three dimensional figure? A. 160cm² B. 48cm² C. 120cm² D. 40cm² Show Content Detailed Solution Volume of the three dimensional figures = v = A x h A = \(\frac{1}{2}\) x 4 x 3 = 6cm² V = 6 x 8 = 48cm²	B
37.	Using \(\bigtriangleup\)XYZ in the figure, find XYZ A. 29^o B. 31^o C. 31^o 20' D. 31^o 18' Show Content Detailed Solution \(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\) sin 120^o = sin 60^o 5 sin y = 3 sin 60^o sin y = \(\frac{3 \sin 60^o}{5}\) \(\frac{3 \times 0.866}{5}\) = \(\frac{2.598}{5}\) y = sin^-1 0.5196 = 30^o 18'	D
38.	Find the area of the shaded portion of the semicircular figure. A. \(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\) B. \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\) C. \(\frac{1}{2}r^2 \pi\) D. \(\frac{1}{8}r^2 \sqrt{3}\) E. \(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\) Show Content Detailed Solution A_sector = \(\frac{60}{360} \times \pi r^2\) = \(\frac{1}{6} \pi r^2\) A_{\(\bigtriangleup\)} = \(\frac{1}{2}r^2 \sin 60^o\) \(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\) A_{\(\text{shaded portion}\)} = A_sector - A_{\(\bigtriangleup\)} = (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\) = \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\) = \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)	B
39.	In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS A. 60^o B. 90^o C. 120^o D. 30^o E. 80^o Show Content Detailed Solution From the diagram, PQRSTW is a regular hexagon. Hexagon is a six sided polygon. Sum of interior angles of polygon = (2n - 4)90^o = [2 x 6 - 4] x 90 = 8 x 90 = 720^o each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)	A
40.	In the figure, QRS is a line, PSQ = 35o, SPR = 30o and O is the centre of the circle. Find OQP. A. 35^o B. 30^o C. 130^o D. 25^o E. 65^o Show Content Detailed Solution < PSQ = 35o, < SPR = 30o O is the centre of the circle, PRQ = 65o < OQP = 90 - 65 = 25o	D

31.	A right circular cone has a base radius r cm and a vertical angle 2y^o. The height of the cone is A. r tan y^o cm B. r sin y^o cm C. r cot y^o cm D. r cos y^o cm E. r cosec y^o cm Show Content Detailed Solution \(\frac{r}{h}\) = tan y^o h = \(\frac{r}{tan y^o}\) = r cot y^o	C
32.	Two fair dice are rolled. What is the probability that both show up the same number of points. A. \(\frac{1}{36}\) B. \(\frac{7}{36}\) C. \(\frac{1}{2}\) D. \(\frac{1}{3}\) E. \(\frac{1}{6}\) Show Content Detailed Solution A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face Prob. of both showing the same number of points = \(\frac{6}{36}\) = \(\frac{1}{6}\)	E
33.	The larger value of y for which (y - 1)² = 4y - 7 is A. 2 B. 4 C. 6 D. 7 E. 8 Show Content Detailed Solution (y - 1)² = 4y - 7 y² - 2xy + 1 = 4y - 7 y² - 6y + 8 = 0 (y - 4)(y - 2) y = 4 or 2 = 4	B
34.	If sin \(\theta\) = \(\frac{x}{y}\) and 0^o < 90^o then find \(\frac{1}{tan\theta}\) A. \(\frac{x}{\sqrt{y^2 - x^2}}\) B. \(\frac{y}{x}\) C. \(\frac{\sqrt{y^2 + x^2}}{y^2 - x^2}\) D. \(\frac{y^2 - x^2}{x}\) Show Content Detailed Solution \(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\) sin\(\theta\) = \(\frac{x}{y}\) cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)	D
35.	Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below \(\begin{array}{c\|c} \text{Class boundary in cm} & \text{frequency} & \\\hline 3.35 - 3.45 & 3\\ 3.45 - 3.55 & 6\\ 3.55 - 3.65 & 7\\ 3.65 - 3.75 & 4\end{array}\) What is the mean diameter of the copper spheres? A. 3.40cm B. 3.58cm C. 3.56cm D. 3.62cm Show Content Detailed Solution \(\begin{array}{c\|c} \text{x(mid point)} & f & fx\\ \hline 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0\\3.6 & 7 & 25.2\\3.7 & 4 & 14.8\end{array}\) \(\sum f\) = 20 \(\sum fx\) = 71.2 mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{71.2}{20}\) = 3.56	C

36.	What is the volume of this regular three dimensional figure? A. 160cm² B. 48cm² C. 120cm² D. 40cm² Show Content Detailed Solution Volume of the three dimensional figures = v = A x h A = \(\frac{1}{2}\) x 4 x 3 = 6cm² V = 6 x 8 = 48cm²	B
37.	Using \(\bigtriangleup\)XYZ in the figure, find XYZ A. 29^o B. 31^o C. 31^o 20' D. 31^o 18' Show Content Detailed Solution \(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\) sin 120^o = sin 60^o 5 sin y = 3 sin 60^o sin y = \(\frac{3 \sin 60^o}{5}\) \(\frac{3 \times 0.866}{5}\) = \(\frac{2.598}{5}\) y = sin^-1 0.5196 = 30^o 18'	D
38.	Find the area of the shaded portion of the semicircular figure. A. \(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\) B. \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\) C. \(\frac{1}{2}r^2 \pi\) D. \(\frac{1}{8}r^2 \sqrt{3}\) E. \(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\) Show Content Detailed Solution A_sector = \(\frac{60}{360} \times \pi r^2\) = \(\frac{1}{6} \pi r^2\) A_{\(\bigtriangleup\)} = \(\frac{1}{2}r^2 \sin 60^o\) \(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\) A_{\(\text{shaded portion}\)} = A_sector - A_{\(\bigtriangleup\)} = (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\) = \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\) = \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)	B
39.	In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS A. 60^o B. 90^o C. 120^o D. 30^o E. 80^o Show Content Detailed Solution From the diagram, PQRSTW is a regular hexagon. Hexagon is a six sided polygon. Sum of interior angles of polygon = (2n - 4)90^o = [2 x 6 - 4] x 90 = 8 x 90 = 720^o each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)	A
40.	In the figure, QRS is a line, PSQ = 35o, SPR = 30o and O is the centre of the circle. Find OQP. A. 35^o B. 30^o C. 130^o D. 25^o E. 65^o Show Content Detailed Solution < PSQ = 35o, < SPR = 30o O is the centre of the circle, PRQ = 65o < OQP = 90 - 65 = 25o	D