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

Paper Info

Year :

1984

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 45 of 45 Questions

#	Question	Ans
41.	In the figure, determine the angle marked y A. 66^o B. 110^o C. 26^o D. 70^o E. 44^o Show Content Detailed Solution From the diagram, < p = 44^o, < Q = 70^o < y = 180^o - (70 + 44^o) = 66^o	A
42.	Find the x co-ordinates of the points of intersection of the two equations in the graph A. 1, 1 B. 0, 4 C. 0,0 D. 4, 0 E. 4, 9 Show Content Detailed Solution If y = 2x + 1 and y = x² - 2x + 1 then x² - 2x + 1 = 2x + 1 x² - 4x = 0 x(x - 4) = 0 x = 0 or 4	D
43.	In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP A. 4cm B. 16cm C. 9cm D. 14cm E. Impossible, insufficient data Show Content Detailed Solution \(\frac{PQ}{6 + RT} = \frac{6}{12} = \frac{6}{PS}\)(Similar triangles) \(\frac{6}{12} = \frac{8}{PS}\) PS = \(\frac{12 \times 8}{6}\) = 16cm	B
44.	In the figure, O is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is A. 67\(\frac{1}{2}\)^o B. 22\(\frac{1}{2}\)^o C. 33\(\frac{1}{2}\)^o D. 45^o E. 90^o Show Content Detailed Solution < R = 180o - 45o (sum of angles on a straight line) < R = < P = 45o (corresponding angles) < PSO = < P = 45o (\(\bigtriangleup\)PSO is a right angle)	D
45.	XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY. A. 5\(\sqrt{3}\)cm B. 3\(\sqrt{5}\)cm C. 3\(\sqrt{3}\)cm D. 5cm E. 6cm Show Content Detailed Solution by Pythagoras theorem, XW = 3cm Also by Pythagoras theorem, XY² = 6² + 3² XY² = 36 + 9 = 45 XY = \(\sqrt{45} = 3 \sqrt{3}\)	B

41.	In the figure, determine the angle marked y A. 66^o B. 110^o C. 26^o D. 70^o E. 44^o Show Content Detailed Solution From the diagram, < p = 44^o, < Q = 70^o < y = 180^o - (70 + 44^o) = 66^o	A
42.	Find the x co-ordinates of the points of intersection of the two equations in the graph A. 1, 1 B. 0, 4 C. 0,0 D. 4, 0 E. 4, 9 Show Content Detailed Solution If y = 2x + 1 and y = x² - 2x + 1 then x² - 2x + 1 = 2x + 1 x² - 4x = 0 x(x - 4) = 0 x = 0 or 4	D
43.	In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP A. 4cm B. 16cm C. 9cm D. 14cm E. Impossible, insufficient data Show Content Detailed Solution \(\frac{PQ}{6 + RT} = \frac{6}{12} = \frac{6}{PS}\)(Similar triangles) \(\frac{6}{12} = \frac{8}{PS}\) PS = \(\frac{12 \times 8}{6}\) = 16cm	B

44.	In the figure, O is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is A. 67\(\frac{1}{2}\)^o B. 22\(\frac{1}{2}\)^o C. 33\(\frac{1}{2}\)^o D. 45^o E. 90^o Show Content Detailed Solution < R = 180o - 45o (sum of angles on a straight line) < R = < P = 45o (corresponding angles) < PSO = < P = 45o (\(\bigtriangleup\)PSO is a right angle)	D
45.	XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY. A. 5\(\sqrt{3}\)cm B. 3\(\sqrt{5}\)cm C. 3\(\sqrt{3}\)cm D. 5cm E. 6cm Show Content Detailed Solution by Pythagoras theorem, XW = 3cm Also by Pythagoras theorem, XY² = 6² + 3² XY² = 36 + 9 = 45 XY = \(\sqrt{45} = 3 \sqrt{3}\)	B