Mathematics (Core), 1981, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1981

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 47 of 47 Questions

#	Question	Ans
41.	Solve the given equation \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\) A. 27 B. 9 C. \(\frac{1}{27}\) D. 18 E. 81 Show Content Detailed Solution \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\) Let \(\log_{3} x = a\). \(a^{2} - 6a + 9 = 0\) \(a^{2} - 3a - 3a + 9 = 0\) \(a(a - 3) - 3(a - 3) = 0\) \((a - 3)(a - 3) = 0\) \(\implies a = 3 (twice)\) \(\log_{3} x = 3 \implies x = 3^{3} = 27\)	A
42.	What is the length of an arc of a circle that substends 2\(\frac{1}{2}\) radians at the centre when the raduis of the circle = \(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\) then A. p< 0 B. p\(\geq\) 0 C. p \(\leq\) 0 D. p < 1 E. p > 0 Show Content Detailed Solution \(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\) = \(\frac{k^2 + (k + 1)^2}{k(k + 10}\) = \(\frac{2k^2 + 2k + 1}{k(k + 1}\) let k = \(\frac{1}{2}\) p = \(\frac{10}{3}\) p > 0	E
43.	In the figure, the chords XY and ZW are produced to meet at T such that YT = WT, ZYW = 40o and YTW = 30o. What is YXW? A. 50^o B. 75^o C. 35^o D. 115^o E. 30^o Show Content Detailed Solution < WYT = \(\frac{180 - 30}{2}\) = \(\frac{150}{2}\) = 75o(base angles of isosceles D) YXW = 75 - 40 = 35o(Exterior angle is equal to sum of interior angles)	C
44.	In the figure, find the area of XYZW A. 60cm² B. 54cm² C. 36cm² D. 54\(\sqrt{2}\)cm² E. 27\(\sqrt{2}\)cm² Show Content Detailed Solution Average of trapezium XYZW = \(\frac{1}{2}\)(a + b)h from rt < YDZ, h = 8 sin 30o area = \(\frac{1}{2}\)(10 + 8)4 = 8 x \(\frac{1}{2}\) = 4 \(\frac{1}{2}\)(18) x 4 = 36cm2	C
45.	In the figure, find x in terms of a, b and c. A. a + b + c B. 180^o - (a + b + c) C. a - b - c D. a + b E. a + c Show Content Detailed Solution 180 - x + a + b + c = 180(sum of interior angle in triangle) a + b + c = x.	A
46.	In the diagram, XZ is the diameter of a circle's radius \(\frac{5}{2}\). If XY is 4cm, then the area of the triangle XYZ is A. 12cm² B. 28\(\frac{1}{2}\)cm² C. 16cm² D. 10cm² E. 6cm² Show Content Detailed Solution Area of the triangle XYZ = \(\frac{1}{2}\)bh = \(\frac{1}{2}\)ZY x XY = \(\frac{1}{2}\) x 3 x 4 = 6cm2.	E
47.	In the figure, WU//YZ, WY//YZ = 12cm, VZ = 6cm, XU = 8cm. Determine the length of WU. A. 1cm``3cm B. 6cm C. 2cm D. 4cm Show Content Detailed Solution From similar triangle, \(\frac{x}{6}\) = \(\frac{8}{12}\) 12x = 48 x = \(\frac{48}{12}\) = 4	D

41.	Solve the given equation \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\) A. 27 B. 9 C. \(\frac{1}{27}\) D. 18 E. 81 Show Content Detailed Solution \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\) Let \(\log_{3} x = a\). \(a^{2} - 6a + 9 = 0\) \(a^{2} - 3a - 3a + 9 = 0\) \(a(a - 3) - 3(a - 3) = 0\) \((a - 3)(a - 3) = 0\) \(\implies a = 3 (twice)\) \(\log_{3} x = 3 \implies x = 3^{3} = 27\)	A
42.	What is the length of an arc of a circle that substends 2\(\frac{1}{2}\) radians at the centre when the raduis of the circle = \(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\) then A. p< 0 B. p\(\geq\) 0 C. p \(\leq\) 0 D. p < 1 E. p > 0 Show Content Detailed Solution \(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\) = \(\frac{k^2 + (k + 1)^2}{k(k + 10}\) = \(\frac{2k^2 + 2k + 1}{k(k + 1}\) let k = \(\frac{1}{2}\) p = \(\frac{10}{3}\) p > 0	E
43.	In the figure, the chords XY and ZW are produced to meet at T such that YT = WT, ZYW = 40o and YTW = 30o. What is YXW? A. 50^o B. 75^o C. 35^o D. 115^o E. 30^o Show Content Detailed Solution < WYT = \(\frac{180 - 30}{2}\) = \(\frac{150}{2}\) = 75o(base angles of isosceles D) YXW = 75 - 40 = 35o(Exterior angle is equal to sum of interior angles)	C
44.	In the figure, find the area of XYZW A. 60cm² B. 54cm² C. 36cm² D. 54\(\sqrt{2}\)cm² E. 27\(\sqrt{2}\)cm² Show Content Detailed Solution Average of trapezium XYZW = \(\frac{1}{2}\)(a + b)h from rt < YDZ, h = 8 sin 30o area = \(\frac{1}{2}\)(10 + 8)4 = 8 x \(\frac{1}{2}\) = 4 \(\frac{1}{2}\)(18) x 4 = 36cm2	C

45.	In the figure, find x in terms of a, b and c. A. a + b + c B. 180^o - (a + b + c) C. a - b - c D. a + b E. a + c Show Content Detailed Solution 180 - x + a + b + c = 180(sum of interior angle in triangle) a + b + c = x.	A
46.	In the diagram, XZ is the diameter of a circle's radius \(\frac{5}{2}\). If XY is 4cm, then the area of the triangle XYZ is A. 12cm² B. 28\(\frac{1}{2}\)cm² C. 16cm² D. 10cm² E. 6cm² Show Content Detailed Solution Area of the triangle XYZ = \(\frac{1}{2}\)bh = \(\frac{1}{2}\)ZY x XY = \(\frac{1}{2}\) x 3 x 4 = 6cm2.	E
47.	In the figure, WU//YZ, WY//YZ = 12cm, VZ = 6cm, XU = 8cm. Determine the length of WU. A. 1cm``3cm B. 6cm C. 2cm D. 4cm Show Content Detailed Solution From similar triangle, \(\frac{x}{6}\) = \(\frac{8}{12}\) 12x = 48 x = \(\frac{48}{12}\) = 4	D