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

Paper Info

Year :

1981

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 47 Questions

#	Question	Ans
31.	The area of a circular plate is one-sixteenth the surface area of a ball of a ball, If the area of the plate is given as P cm², then the radius of the ball is A. \(\frac{2P}{\pi}\) B. \(\frac{P}{\sqrt{\pi}}\) C. \(\frac{P}{\sqrt{2\pi}}\) D. 2\(\frac{P}{\pi}\) Show Content Detailed Solution Surface area of a sphere = 4\(\pi\)r² \(\frac{1}{16}\) of 4\(\pi\)r² = \(\frac{\pi r^2}{4}\) P = \(\frac{\pi r^2}{4}\) r = 2\(\frac{P}{\pi}\)	D
32.	Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is A. sin x B. cos²x C. 2 D. 3 Show Content Detailed Solution \(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\) \(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2.	C
33.	The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81 A. 12 B. 27 C. 9 D. 4 E. 36 Show Content Detailed Solution 1st term a = 3, 5th term = 9, sum of n = 81 nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 S_n = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n² + 9n = 324 3n² + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3	C
34.	Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) A. \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) B. \(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\) C. \(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\) D. \(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\) Show Content Detailed Solution T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\) = \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\) = \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\) T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)	A
35.	If b = a + cp and r = ab + \(\frac{1}{2}\)cp², express b² in terms of a, c, r. A. b² = aV + 2cr B. b² = ar + 2c²r C. b² = a² = \(\frac{1}{2}\) cr² D. b² = \(\frac{1}{2}\)ar² + c E. b² = 2cr - a² Show Content Detailed Solution b = a + cp....(i) r = ab + \(\frac{1}{2}\)cp².....(ii) expressing b² in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i) b - a = cp = \(\frac{b - a}{c}\) sub. for p in eqn.(ii) r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\) 2cr = 2ab + b² - 2ab + a² b² = 2cr - a²	E
36.	Multiply x² + x + 1 by x² - x + 1 A. x⁴ - x + x² B. x⁴ - x² + x² C. x⁴ + x² + 1 D. x⁴ + x² Show Content Detailed Solution (x² + x + 1)( x² - x + 1) = x²(x² + x + 1) - x(x² + x + 1) + (x³ + x + 1) = x⁴ - x³ + x² + x³ - x² - x + 1 = x⁴ + x² + 1	C
37.	A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used? A. 12.25kg B. 6.75kg C. 8.75kg D. 15.75kg E. 8.25kg Show Content Detailed Solution Sugar : flour = 2.5 : 4.5 Total = 7 sugar used = \(\frac{2.5}{7}\) x 24.5 = \(\frac{61.25}{7}\) = 8.75	C
38.	The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) greater by A. \(\frac{23}{28}\) B. \(\frac{24}{28}\) C. \(\frac{50}{56}\) D. \(\frac{27}{28}\) E. \(\frac{48}{56}\) Show Content Detailed Solution Difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) \(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\) The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\) = \(\frac{11}{7}\) \(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\) = \(\frac{50}{56}\)	C
39.	What is the size of an exterior angle of a regular pentagon? A. 36^o B. 60^o C. 72^o D. 120^o E. 360^o Show Content Detailed Solution Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o = 5 = (10 - 4) x 90o = 540o each interior angle = \(\frac{540}{5}\) = 108o = 180o let x represent ext. angle x + 108o = 180v x = 180o - 108o = 72o	C
40.	The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is A. x + 8x² - 5 = 0 B. 2x² - 5x + 8 = 0 C. 2x² - 8x + 5 = 0 D. x² + 8x - 5 = 0 Show Content Detailed Solution x2 - (sum of roots) x + (product of roots) = 0 x2 - \(\frac{5x}{2}\) + 4 = 0 2x2 - 5x + 8 = 0	B

31.	The area of a circular plate is one-sixteenth the surface area of a ball of a ball, If the area of the plate is given as P cm², then the radius of the ball is A. \(\frac{2P}{\pi}\) B. \(\frac{P}{\sqrt{\pi}}\) C. \(\frac{P}{\sqrt{2\pi}}\) D. 2\(\frac{P}{\pi}\) Show Content Detailed Solution Surface area of a sphere = 4\(\pi\)r² \(\frac{1}{16}\) of 4\(\pi\)r² = \(\frac{\pi r^2}{4}\) P = \(\frac{\pi r^2}{4}\) r = 2\(\frac{P}{\pi}\)	D
32.	Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is A. sin x B. cos²x C. 2 D. 3 Show Content Detailed Solution \(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\) \(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2.	C
33.	The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81 A. 12 B. 27 C. 9 D. 4 E. 36 Show Content Detailed Solution 1st term a = 3, 5th term = 9, sum of n = 81 nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 S_n = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n² + 9n = 324 3n² + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3	C
34.	Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) A. \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) B. \(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\) C. \(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\) D. \(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\) Show Content Detailed Solution T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\) = \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\) = \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\) T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)	A
35.	If b = a + cp and r = ab + \(\frac{1}{2}\)cp², express b² in terms of a, c, r. A. b² = aV + 2cr B. b² = ar + 2c²r C. b² = a² = \(\frac{1}{2}\) cr² D. b² = \(\frac{1}{2}\)ar² + c E. b² = 2cr - a² Show Content Detailed Solution b = a + cp....(i) r = ab + \(\frac{1}{2}\)cp².....(ii) expressing b² in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i) b - a = cp = \(\frac{b - a}{c}\) sub. for p in eqn.(ii) r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\) 2cr = 2ab + b² - 2ab + a² b² = 2cr - a²	E

36.	Multiply x² + x + 1 by x² - x + 1 A. x⁴ - x + x² B. x⁴ - x² + x² C. x⁴ + x² + 1 D. x⁴ + x² Show Content Detailed Solution (x² + x + 1)( x² - x + 1) = x²(x² + x + 1) - x(x² + x + 1) + (x³ + x + 1) = x⁴ - x³ + x² + x³ - x² - x + 1 = x⁴ + x² + 1	C
37.	A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used? A. 12.25kg B. 6.75kg C. 8.75kg D. 15.75kg E. 8.25kg Show Content Detailed Solution Sugar : flour = 2.5 : 4.5 Total = 7 sugar used = \(\frac{2.5}{7}\) x 24.5 = \(\frac{61.25}{7}\) = 8.75	C
38.	The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) greater by A. \(\frac{23}{28}\) B. \(\frac{24}{28}\) C. \(\frac{50}{56}\) D. \(\frac{27}{28}\) E. \(\frac{48}{56}\) Show Content Detailed Solution Difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) \(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\) The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\) = \(\frac{11}{7}\) \(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\) = \(\frac{50}{56}\)	C
39.	What is the size of an exterior angle of a regular pentagon? A. 36^o B. 60^o C. 72^o D. 120^o E. 360^o Show Content Detailed Solution Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o = 5 = (10 - 4) x 90o = 540o each interior angle = \(\frac{540}{5}\) = 108o = 180o let x represent ext. angle x + 108o = 180v x = 180o - 108o = 72o	C
40.	The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is A. x + 8x² - 5 = 0 B. 2x² - 5x + 8 = 0 C. 2x² - 8x + 5 = 0 D. x² + 8x - 5 = 0 Show Content Detailed Solution x2 - (sum of roots) x + (product of roots) = 0 x2 - \(\frac{5x}{2}\) + 4 = 0 2x2 - 5x + 8 = 0	B