Mathematics (Core), 2010, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2010

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 49 Questions

#	Question	Ans
31.	Solve for x and y if x - y = 2 and x2 - y2 = 8 A. (-1, 3) B. (3, 1) C. (-3, 1) D. (1, 3) Show Content Detailed Solution x - y = 2 ...........(1) x2 - y2 = 8 ........... (2) x - 2 = y ............ (3) Put y = x -2 in (2) x2 - (x - 2)2 = 8 x2 - (x2 - 4x + 4) = 8 x2 - x2 + 4x - 4 = 8 4x = 8 + 4 = 12 x = \(\frac{12}{4}\) = 3 from (3), y = 3 - 2 = 1 therefore, x = 3, y = 1 There is an explanation video available below.	B
32.	If x is inversely proportional to y and x = 2\(\frac{1}{2}\) when y = 2, find x if y = 4 A. 4 B. 5 C. 1\(\frac{1}{4}\) D. 2\(\frac{1}{4}\) Show Content Detailed Solution x \(\alpha\) \(\frac{1}{y}\) .........(1) x = k x \(\frac{1}{y}\) .........(2) When x = 2\(\frac{1}{2}\) = \(\frac{5}{2}\), y = 2 (2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\) giving k = 5 from (2), x = \(\frac{5}{y}\) so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\) There is an explanation video available below.	C
33.	For what range of values of x is \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)? A. x < \(\frac{3}{2}\) B. x > \(\frac{3}{2}\) C. x < -\(\frac{3}{2}\) D. x > -\(\frac{3}{2}\) Show Content Detailed Solution \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\) Multiply through by through by the LCM of 2, 3 and 4 12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\) 6x + 3 > 4x + 6 6x - 4x > 6 - 3 2x > 3 \(\frac{2x}{2}\) > \(\frac{3}{2}\) x > \(\frac{3}{2}\) There is an explanation video available below.	B
34.	Solve the inequalities -6 \(\leq\) 4 - 2x < 5 - x A. -1 < x < 5 B. -1 < x \(\leq\) 5 C. -1 \(\leq\) x \(\leq\) 6 D. -1 \(\leq\) x < 6 Show Content Detailed Solution -6 \(\leq\) 4 - 2x < 5 - x split inequalities into two and solve each part as follows: -6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x -10 \(\leq\) -2x \(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\) giving 5 \(\geq\) x or x \(\leq\) 5 4 - 2x < 5 - x -2x + x < 5 - 4 -x < 1 \(\frac{-x}{-1}\) > \(\frac{1}{-1}\) giving x > -1 or -1 < x Combining the two results, gives -1 < x \(\leq\) 5 There is an explanation video available	B
35.	Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + ..... A. \(\frac{5}{8}\) B. \(\frac{5}{7}\) C. \(\frac{5}{11}\) D. \(\frac{5}{9}\) Show Content Detailed Solution Using S\(\infty\) = \(\frac{a}{1 - r}\) r = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\) S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\) = \(\frac{0.5}{({\frac{9}{10}})}\) = \(\frac{0.5 \times 10}{9}\) = \(\frac{5}{9}\) There is an explanation video available below.	D
36.	Evaluate \(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) A. 5y - 2x -18 = 0 B. 102 C. -102 D. -42 Show Content Detailed Solution \(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) = 2(6 - 27) - 0(4 - 24) + 5(36 - 48) = 2(-21) - 0 + 5(-12) = -42 + 5(-12) = -42 - 60 = -102 There is an explanation video available below.	C
37.	If P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) , what is P\(^-1\) A. \(\begin{pmatrix} -{\frac{1}{5}} & -{\frac{3}{5}} \\ -{\frac{1}{5}} & -{\frac{2}{5}} \end{pmatrix}\) B. \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ {\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) C. \(\begin{pmatrix} -{\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) D. \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) Show Content Detailed Solution P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) \|P\| = 2 - 1 x -3 = 5 P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}\) = \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) There is an explanation video available below.	D
38.	The interior angles of a quadrilateral are (x + 15)°, (2x - 45)°, ( x - 30)° and (x + 10)°. Find the value of the least interior angle. A. 112^o B. 102^o C. 82^o D. 52^o Show Content Detailed Solution (x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o when n = 4 x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o 5x - 50o = (8 - 4)90o 5x - 50o = 4 x 90o = 360o 5x = 360o + 50o 5x = 410o x = \(\frac{410^o}{5}\) = 82o Hence, the value of the least interior angle is (x - 30o) = (82 - 30)o = 52o There is an explanation video available below.	D
39.	A cylindrical pipe 50m long with radius 7m has one end open. What is the total surface area of the pipe? A. 749\(\pi\)m² B. 700\(\pi\)m² C. 350\(\pi\)m² D. 98\(\pi\)m² Show Content Detailed Solution Total surface area of the cylindrical pipe = area of circular base + curved surface area = \(\pi\)r\(^2\) + 2\(\pi\)rh = \(\pi\) x 7\(^2\) + 2\(\pi\) x 7 x 50 = 49\(\pi\) + 700\(\pi\) = 749\(\pi\)m\(^2\) There is an explanation video available below.	A
40.	Find the distance between the points (\(\frac{1}{2}\), \(\frac{1}{2}\)) and (-\(\frac{1}{2}\), -\(\frac{1}{2}\)). A. 1 B. o C. √3 D. √2 Show Content Detailed Solution Let D denote the distance between (\(\frac{1}{2}\), -\(\frac{1}{2}\)) then using D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) = \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\) = \(\sqrt{(-1)^2 + (-1)^2}\) = \(\sqrt{1 + 1}\) = √2 There is an explanation video available below.	D

31.	Solve for x and y if x - y = 2 and x2 - y2 = 8 A. (-1, 3) B. (3, 1) C. (-3, 1) D. (1, 3) Show Content Detailed Solution x - y = 2 ...........(1) x2 - y2 = 8 ........... (2) x - 2 = y ............ (3) Put y = x -2 in (2) x2 - (x - 2)2 = 8 x2 - (x2 - 4x + 4) = 8 x2 - x2 + 4x - 4 = 8 4x = 8 + 4 = 12 x = \(\frac{12}{4}\) = 3 from (3), y = 3 - 2 = 1 therefore, x = 3, y = 1 There is an explanation video available below.	B
32.	If x is inversely proportional to y and x = 2\(\frac{1}{2}\) when y = 2, find x if y = 4 A. 4 B. 5 C. 1\(\frac{1}{4}\) D. 2\(\frac{1}{4}\) Show Content Detailed Solution x \(\alpha\) \(\frac{1}{y}\) .........(1) x = k x \(\frac{1}{y}\) .........(2) When x = 2\(\frac{1}{2}\) = \(\frac{5}{2}\), y = 2 (2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\) giving k = 5 from (2), x = \(\frac{5}{y}\) so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\) There is an explanation video available below.	C
33.	For what range of values of x is \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)? A. x < \(\frac{3}{2}\) B. x > \(\frac{3}{2}\) C. x < -\(\frac{3}{2}\) D. x > -\(\frac{3}{2}\) Show Content Detailed Solution \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\) Multiply through by through by the LCM of 2, 3 and 4 12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\) 6x + 3 > 4x + 6 6x - 4x > 6 - 3 2x > 3 \(\frac{2x}{2}\) > \(\frac{3}{2}\) x > \(\frac{3}{2}\) There is an explanation video available below.	B
34.	Solve the inequalities -6 \(\leq\) 4 - 2x < 5 - x A. -1 < x < 5 B. -1 < x \(\leq\) 5 C. -1 \(\leq\) x \(\leq\) 6 D. -1 \(\leq\) x < 6 Show Content Detailed Solution -6 \(\leq\) 4 - 2x < 5 - x split inequalities into two and solve each part as follows: -6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x -10 \(\leq\) -2x \(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\) giving 5 \(\geq\) x or x \(\leq\) 5 4 - 2x < 5 - x -2x + x < 5 - 4 -x < 1 \(\frac{-x}{-1}\) > \(\frac{1}{-1}\) giving x > -1 or -1 < x Combining the two results, gives -1 < x \(\leq\) 5 There is an explanation video available	B
35.	Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + ..... A. \(\frac{5}{8}\) B. \(\frac{5}{7}\) C. \(\frac{5}{11}\) D. \(\frac{5}{9}\) Show Content Detailed Solution Using S\(\infty\) = \(\frac{a}{1 - r}\) r = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\) S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\) = \(\frac{0.5}{({\frac{9}{10}})}\) = \(\frac{0.5 \times 10}{9}\) = \(\frac{5}{9}\) There is an explanation video available below.	D

36.	Evaluate \(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) A. 5y - 2x -18 = 0 B. 102 C. -102 D. -42 Show Content Detailed Solution \(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) = 2(6 - 27) - 0(4 - 24) + 5(36 - 48) = 2(-21) - 0 + 5(-12) = -42 + 5(-12) = -42 - 60 = -102 There is an explanation video available below.	C
37.	If P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) , what is P\(^-1\) A. \(\begin{pmatrix} -{\frac{1}{5}} & -{\frac{3}{5}} \\ -{\frac{1}{5}} & -{\frac{2}{5}} \end{pmatrix}\) B. \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ {\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) C. \(\begin{pmatrix} -{\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) D. \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) Show Content Detailed Solution P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) \|P\| = 2 - 1 x -3 = 5 P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}\) = \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\) There is an explanation video available below.	D
38.	The interior angles of a quadrilateral are (x + 15)°, (2x - 45)°, ( x - 30)° and (x + 10)°. Find the value of the least interior angle. A. 112^o B. 102^o C. 82^o D. 52^o Show Content Detailed Solution (x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o when n = 4 x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o 5x - 50o = (8 - 4)90o 5x - 50o = 4 x 90o = 360o 5x = 360o + 50o 5x = 410o x = \(\frac{410^o}{5}\) = 82o Hence, the value of the least interior angle is (x - 30o) = (82 - 30)o = 52o There is an explanation video available below.	D
39.	A cylindrical pipe 50m long with radius 7m has one end open. What is the total surface area of the pipe? A. 749\(\pi\)m² B. 700\(\pi\)m² C. 350\(\pi\)m² D. 98\(\pi\)m² Show Content Detailed Solution Total surface area of the cylindrical pipe = area of circular base + curved surface area = \(\pi\)r\(^2\) + 2\(\pi\)rh = \(\pi\) x 7\(^2\) + 2\(\pi\) x 7 x 50 = 49\(\pi\) + 700\(\pi\) = 749\(\pi\)m\(^2\) There is an explanation video available below.	A
40.	Find the distance between the points (\(\frac{1}{2}\), \(\frac{1}{2}\)) and (-\(\frac{1}{2}\), -\(\frac{1}{2}\)). A. 1 B. o C. √3 D. √2 Show Content Detailed Solution Let D denote the distance between (\(\frac{1}{2}\), -\(\frac{1}{2}\)) then using D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) = \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\) = \(\sqrt{(-1)^2 + (-1)^2}\) = \(\sqrt{1 + 1}\) = √2 There is an explanation video available below.	D