Mathematics (Core), 1982, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1982

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 48 Questions

#	Question	Ans
1.	Rationalize the expression \(\frac{1}{\sqrt{2} + \sqrt{5}}\) A. \(\frac{1}{3}\)(\(\sqrt{5} - \sqrt{2}\) B. \(\frac{\sqrt{2}}{3}\) + \(\frac{\sqrt{5}}{5}\) C. \(\sqrt{2} - \sqrt{5}\) D. 5(\(\sqrt{2} - \sqrt{5}\) E. \(\frac{1}{3}\)(\(\sqrt{2} - \sqrt{5}\) Show Content Detailed Solution \(\frac{1}{\sqrt{2} + \sqrt{5}}\) \(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\) = \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\) = \(\frac{\sqrt{2} - \sqrt{5}}{-3}\) = \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\)	A
2.	Simplify 3 - 2 \(\div\) \(\frac{4}{5}\) + \(\frac{1}{2}\) A. 1\(\frac{3}{4}\) B. -1 C. 1\(\frac{3}{10}\) D. 1 E. 1\(\frac{9}{10}\) Show Content Detailed Solution 3 - 2 \(\div\) (\(\frac{4}{5}\)) + \(\frac{1}{2}\) 3 - (2 x \(\frac{5}{4}\)) + \(\frac{1}{2}\) = 3 - \(\frac{10}{4}\) + \(\frac{1}{2}\) = 3 - \(\frac{5}{2}\) + \(\frac{1}{2}\) = \(\frac{6 - 5 + 1}{2}\) = \(\frac{2}{2}\) = 1	D
3.	If N560.70 is shared in the ratio 7 : 2 : 1, what is the smallest share? A. N392.49 B. N56.70 C. N113.40 D. N112.14 E. N56.07 Show Content Detailed Solution 7 + 2 + 1 = 10 \(\frac{1}{10}\) x 560.70 = N56.07	E
4.	Seven years ago, the age of a father was three times that of his son, but in six years time the age of the son will be half that of his father, representing the present ages of the father and son by x and y, respectively, the two equations relating x and y are A. 3y - x = 0; 2y - x = 0 B. 3y - x = 14; x - 2y = 6 C. 3y - x =7; x - 2y = 6 D. 3y - x = 14; y - 2x = 6 E. x + 3y = 7; x = 2y = 12 Show Content Detailed Solution 7 years ago, Father(x - 7) years old, Son (y - 7) years x - 7 = 3(y - 7) x - 7 = 3y - 21 3y - x = -7 + 21 = 14 3y - x = 14 ... (1) In six years time, x + 6 = 2(y + 6) x + 6 = 2y + 12 2y + 12 = x + 6 12 - 6 = x - 2y 6 = x - 2y ... (2)	B
5.	The factors of 6x - 5 - x2 are A. -(x + 3)(x + 2) B. (x + 5)(x + 1) C. (x - 5)(1 - x) D. (x + 1)(x + 5) Show Content Detailed Solution 6x - 5 - x2 = (-1)(-x2 - 5 + 6x) = x2 - 6x + 5 = (x - 5)(x - 1) -(x - 1) = 1 - x = (x - 5)(1 - x)	C
6.	The solution of the quadratic equation bx2 + cx + a = 0 is given by A. x = b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) B. x = c \(\pm\) \(\frac{\sqrt{b^2 - 4ab}}{2b}\) C. x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\) D. x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2b}\) Show Content Detailed Solution bx2 + cx + a = 0 a = b; b = c; c = a x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\)	C
7.	The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves A. y = x³ and y = 3x² + x - 28 B. y = x³ + 3x² + 4x + 4 and the line y = \(\frac{28}{x}\) C. y = x³ + 3x² + 4x and y D. y = x² + 3x + 4 and y = \(\frac{28}{x}\) E. y = x² + 3x + 4 and line y = 28x Show Content Detailed Solution The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves y = x2 + 3x + 4 and y = \(\frac{28}{x}\)`.	D
8.	Write the equation 2 log2x - x log2(1 + y) = 3 in a form not involving logarithms A. 2^x(1 + y) = 3 B. 2^x - x(1 + y) = 8 C. x² = 8(1 + y)^x D. x² - x(1 + y) = 8 E. x² - (1 + y)² = 8 Show Content Detailed Solution 2log2 x - x log2 (1 + y) = 3 log2 \(\frac{x^2}{(1 + y)^x}\) = 3 = \(\frac{x^2}{(1 + y)^x}\) = 23 = 8 = x2 = 8(1 + y)x	C
9.	Find \(\alpha\) and \(\beta\) such that x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) A. \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\) B. \(\alpha\)= 1, \(\beta\) = -\(\frac{5}{7}\) C. \(\alpha\)= \(\frac{3}{5}\), \(\beta\) = -6 D. \(\alpha\)= 1, \(\beta\) = -\(\frac{3}{5}\) Show Content Detailed Solution x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x y\(\frac{1}{7}\) = x\(\alpha\) = x\(\frac{3}{8}\) + \(\frac{5}{8}\) + y\(\frac{6}{7}\) + \(\frac{1}{7}\) = x\(\alpha\)y\(\beta\) x1y\(\frac{-5}{7}\) = x\(\alpha\)y\(\beta\) \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\)	A
10.	Which of the following lines is not parallel to the line 3y + 2x + 7 = 0? A. 3y + 2x - 7 = 0 B. 9y + 6x + 17 = 0 C. 24y + 16x + 19 = 0 D. 3y - 2x + 7 = 0 E. 15y + 10x - 13 = 0 Show Content Detailed Solution Two lines are said to be parallel if the slope of the two lines are equal. The equation : \(3y + 2x + 7 = 0\) \(3y = -2x - 7\) \(y = \frac{-2}{3} x - \frac{7}{3}\) \(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\) All the options have the same slope except \(3y - 2x + 7 = 0\).	D

1.	Rationalize the expression \(\frac{1}{\sqrt{2} + \sqrt{5}}\) A. \(\frac{1}{3}\)(\(\sqrt{5} - \sqrt{2}\) B. \(\frac{\sqrt{2}}{3}\) + \(\frac{\sqrt{5}}{5}\) C. \(\sqrt{2} - \sqrt{5}\) D. 5(\(\sqrt{2} - \sqrt{5}\) E. \(\frac{1}{3}\)(\(\sqrt{2} - \sqrt{5}\) Show Content Detailed Solution \(\frac{1}{\sqrt{2} + \sqrt{5}}\) \(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\) = \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\) = \(\frac{\sqrt{2} - \sqrt{5}}{-3}\) = \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\)	A
2.	Simplify 3 - 2 \(\div\) \(\frac{4}{5}\) + \(\frac{1}{2}\) A. 1\(\frac{3}{4}\) B. -1 C. 1\(\frac{3}{10}\) D. 1 E. 1\(\frac{9}{10}\) Show Content Detailed Solution 3 - 2 \(\div\) (\(\frac{4}{5}\)) + \(\frac{1}{2}\) 3 - (2 x \(\frac{5}{4}\)) + \(\frac{1}{2}\) = 3 - \(\frac{10}{4}\) + \(\frac{1}{2}\) = 3 - \(\frac{5}{2}\) + \(\frac{1}{2}\) = \(\frac{6 - 5 + 1}{2}\) = \(\frac{2}{2}\) = 1	D
3.	If N560.70 is shared in the ratio 7 : 2 : 1, what is the smallest share? A. N392.49 B. N56.70 C. N113.40 D. N112.14 E. N56.07 Show Content Detailed Solution 7 + 2 + 1 = 10 \(\frac{1}{10}\) x 560.70 = N56.07	E
4.	Seven years ago, the age of a father was three times that of his son, but in six years time the age of the son will be half that of his father, representing the present ages of the father and son by x and y, respectively, the two equations relating x and y are A. 3y - x = 0; 2y - x = 0 B. 3y - x = 14; x - 2y = 6 C. 3y - x =7; x - 2y = 6 D. 3y - x = 14; y - 2x = 6 E. x + 3y = 7; x = 2y = 12 Show Content Detailed Solution 7 years ago, Father(x - 7) years old, Son (y - 7) years x - 7 = 3(y - 7) x - 7 = 3y - 21 3y - x = -7 + 21 = 14 3y - x = 14 ... (1) In six years time, x + 6 = 2(y + 6) x + 6 = 2y + 12 2y + 12 = x + 6 12 - 6 = x - 2y 6 = x - 2y ... (2)	B
5.	The factors of 6x - 5 - x2 are A. -(x + 3)(x + 2) B. (x + 5)(x + 1) C. (x - 5)(1 - x) D. (x + 1)(x + 5) Show Content Detailed Solution 6x - 5 - x2 = (-1)(-x2 - 5 + 6x) = x2 - 6x + 5 = (x - 5)(x - 1) -(x - 1) = 1 - x = (x - 5)(1 - x)	C

6.	The solution of the quadratic equation bx2 + cx + a = 0 is given by A. x = b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) B. x = c \(\pm\) \(\frac{\sqrt{b^2 - 4ab}}{2b}\) C. x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\) D. x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2b}\) Show Content Detailed Solution bx2 + cx + a = 0 a = b; b = c; c = a x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\)	C
7.	The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves A. y = x³ and y = 3x² + x - 28 B. y = x³ + 3x² + 4x + 4 and the line y = \(\frac{28}{x}\) C. y = x³ + 3x² + 4x and y D. y = x² + 3x + 4 and y = \(\frac{28}{x}\) E. y = x² + 3x + 4 and line y = 28x Show Content Detailed Solution The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves y = x2 + 3x + 4 and y = \(\frac{28}{x}\)`.	D
8.	Write the equation 2 log2x - x log2(1 + y) = 3 in a form not involving logarithms A. 2^x(1 + y) = 3 B. 2^x - x(1 + y) = 8 C. x² = 8(1 + y)^x D. x² - x(1 + y) = 8 E. x² - (1 + y)² = 8 Show Content Detailed Solution 2log2 x - x log2 (1 + y) = 3 log2 \(\frac{x^2}{(1 + y)^x}\) = 3 = \(\frac{x^2}{(1 + y)^x}\) = 23 = 8 = x2 = 8(1 + y)x	C
9.	Find \(\alpha\) and \(\beta\) such that x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) A. \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\) B. \(\alpha\)= 1, \(\beta\) = -\(\frac{5}{7}\) C. \(\alpha\)= \(\frac{3}{5}\), \(\beta\) = -6 D. \(\alpha\)= 1, \(\beta\) = -\(\frac{3}{5}\) Show Content Detailed Solution x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x y\(\frac{1}{7}\) = x\(\alpha\) = x\(\frac{3}{8}\) + \(\frac{5}{8}\) + y\(\frac{6}{7}\) + \(\frac{1}{7}\) = x\(\alpha\)y\(\beta\) x1y\(\frac{-5}{7}\) = x\(\alpha\)y\(\beta\) \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\)	A
10.	Which of the following lines is not parallel to the line 3y + 2x + 7 = 0? A. 3y + 2x - 7 = 0 B. 9y + 6x + 17 = 0 C. 24y + 16x + 19 = 0 D. 3y - 2x + 7 = 0 E. 15y + 10x - 13 = 0 Show Content Detailed Solution Two lines are said to be parallel if the slope of the two lines are equal. The equation : \(3y + 2x + 7 = 0\) \(3y = -2x - 7\) \(y = \frac{-2}{3} x - \frac{7}{3}\) \(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\) All the options have the same slope except \(3y - 2x + 7 = 0\).	D