Mathematics (Core), 1990, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1990

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 44 of 44 Questions

#	Question	Ans
41.	In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY A. 8cm B. 6cm C. 4cm D. 3cm	B
42.	In the figure, YXZ = 30^o, XYZ = 105^o and XY = 8cm. Calculate YZ A. 16\(\sqrt{2}\)cm B. 8\(\sqrt{2}\)cm C. 4\(\sqrt{2}\)cm D. 22cm Show Content Detailed Solution yzx + 105^o + 30^o = 180^o yzx = 180 - 155 = 45^o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm	C
43.	In the figure, PQR is a semicircle. Calculate the area of the shaded region A. 125^{\(\frac{2}{7}\)}² B. 149^{\(\frac{2}{7}\)}cm² C. 234^{\(\frac{1}{7}\)}cm² D. 267^{\(\frac{1}{2}\)}cm²	A
44.	Find the curved surface area of the frustrum in the figure A. 16\(\pi \sqrt{10}\)cm² B. 20\(\pi \sqrt{10}\)cm² C. 24\(\pi \sqrt{10}\)cm² D. 36\(\pi \sqrt{10}\)cm² Show Content Detailed Solution \(\frac{x}{4} = \frac{6 + x}{6}\) 6x = 4(6 + x) = 24 + 4x x = 12cm CSA = \(\pi RL - \pi rl\) = \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\) = 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\) = 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\) = 20\(\pi \sqrt{10}\)cm2	B

41.	In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY A. 8cm B. 6cm C. 4cm D. 3cm	B
42.	In the figure, YXZ = 30^o, XYZ = 105^o and XY = 8cm. Calculate YZ A. 16\(\sqrt{2}\)cm B. 8\(\sqrt{2}\)cm C. 4\(\sqrt{2}\)cm D. 22cm Show Content Detailed Solution yzx + 105^o + 30^o = 180^o yzx = 180 - 155 = 45^o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm	C

43.	In the figure, PQR is a semicircle. Calculate the area of the shaded region A. 125^{\(\frac{2}{7}\)}² B. 149^{\(\frac{2}{7}\)}cm² C. 234^{\(\frac{1}{7}\)}cm² D. 267^{\(\frac{1}{2}\)}cm²	A
44.	Find the curved surface area of the frustrum in the figure A. 16\(\pi \sqrt{10}\)cm² B. 20\(\pi \sqrt{10}\)cm² C. 24\(\pi \sqrt{10}\)cm² D. 36\(\pi \sqrt{10}\)cm² Show Content Detailed Solution \(\frac{x}{4} = \frac{6 + x}{6}\) 6x = 4(6 + x) = 24 + 4x x = 12cm CSA = \(\pi RL - \pi rl\) = \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\) = 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\) = 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\) = 20\(\pi \sqrt{10}\)cm2	B