Mathematics (Core), 2011, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2011

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

21 - 30 of 46 Questions

#	Question	Ans
21.	If \(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\), find the value of x. A. -6 B. 6 C. -12 D. 12 Show Content Detailed Solution \(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\) (2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1) 6x - 15 = 8x - 3 6x - 8x = 15 - 3 -2x = 12 x = \(\frac{12}{-2}\) = -6 There is an explanation video available below.	A
22.	Evaluate \(\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\) A. 25 B. 45 C. 15 D. 55 Show Content Detailed Solution There is an explanation video available below.	A
23.	The inverse of matrix N = \(\begin{vmatrix} 2 & 3 \\ 1 & 4\end{vmatrix}\) is A. \(\frac{1}{5}\) \(\begin{vmatrix} 2 & 1 \\ 3 & 4\end{vmatrix}\) B. \(\frac{1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2\end{vmatrix}\) C. \(\frac{1}{5}\) \(\begin{vmatrix} 2 & -1 \\ -3 & 4\end{vmatrix}\) D. \(\frac{1}{5}\) \(\begin{vmatrix} 4 & 3 \\ 1 & 2\end{vmatrix}\) Show Content Detailed Solution There is an explanation video available below.	B
24.	What is the size of each interior angle of a 12-sided regular polygon? A. 120^o B. 150^o C. 30^o D. 180^o Show Content Detailed Solution There is an explanation video available below.	B
25.	A chord of circle of radius 7cm is 5cm from the centre of the circle.What is the length of the chord? A. 4√6 cm B. 3√6 cm C. 6√6 cm D. 2√6 cm Show Content Detailed Solution From Pythagoras theorem \|OA\|2 = \|AN\|2 + \|ON\|2 72 = \|AN\|2 + (5)2 49 = \|AN\|2 + 25 \|AN\|2 = 49 - 25 = 24 \|AN\| = \(\sqrt {24}\) = \(\sqrt {4 \times 6}\) = 2√6 cm \|AN\| = \|NB\| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts) \|AN\| + \|NB\| = \|AB\| 2√6 + 2√6 = \|AB\| \|AB\| = 4√6 cm There is an explanation video available below.	A
26.	A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold A. 48 cm³ B. 33 cm³ C. 60 cm³ D. 27 cm³ Show Content Detailed Solution There is an explanation video available below.	B
27.	The perpendicular bisector of a line XY is the locus of a point A. whose distance from X is always twice its distance from Y B. whose distance from Y is always twice its distance from X. C. which moves on the line XY D. which is equidistant from the points X and Y Show Content Detailed Solution There is an explanation video available below.	D
28.	The midpoint of P(x, y) and Q(8, 6) is (5, 8). Find x and y. A. (2, 10) B. (2, 8) C. (2, 12) D. (2, 6) Show Content Detailed Solution P(x, y) Q(8, 6) midpoint = (5, 8) x + 8 = 5 \(\frac{y + 6}{2}\) = 8 x + 8 = 10 x = 10 - 8 = 2 y + 6 = 16 y + 16 - 6 = 10 therefore, P(2, 10) There is an explanation video available below.	A
29.	Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2). A. 5y - 2x -18 = 0 B. 5y + 2x - 18 = 0 C. 5y - 2x + 18 = 0 D. 5y + 2x - 2 = 0 Show Content Detailed Solution 2y = 5x + 4 (4, 2) y = \(\frac{5x}{2}\) + 4 comparing with y = mx + e m = \(\frac{5}{2}\) Since they are perpendicular m1m2 = -1 m2 = \(\frac{-1}{m_1}\) = -1 \(\frac{5}{2}\) = -1 x \(\frac{2}{5}\) The equator of the line is thus y = mn + c (4, 2) 2 = -\(\frac{2}{5}\)(4) + c \(\frac{2}{1}\) + \(\frac{8}{5}\) = c c = \(\frac{18}{5}\) y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\) 5y = -2x + 18 &l	B
30.	In a right angled triangle, if tan \(\theta\) = \(\frac{3}{4}\). What is cos\(\theta\) - sin\(\theta\)? A. \(\frac{2}{3}\) B. \(\frac{3}{5}\) C. \(\frac{1}{5}\) D. \(\frac{4}{5}\) Show Content Detailed Solution tan\(\theta\) = \(\frac{3}{4}\) from Pythagoras tippet, the hypotenus is T i.e. 3, 4, 5. then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\) cos\(\theta\) - sin\(\theta\) \(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\) There is an explanation video available below.	C

21.	If \(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\), find the value of x. A. -6 B. 6 C. -12 D. 12 Show Content Detailed Solution \(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\) (2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1) 6x - 15 = 8x - 3 6x - 8x = 15 - 3 -2x = 12 x = \(\frac{12}{-2}\) = -6 There is an explanation video available below.	A
22.	Evaluate \(\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\) A. 25 B. 45 C. 15 D. 55 Show Content Detailed Solution There is an explanation video available below.	A
23.	The inverse of matrix N = \(\begin{vmatrix} 2 & 3 \\ 1 & 4\end{vmatrix}\) is A. \(\frac{1}{5}\) \(\begin{vmatrix} 2 & 1 \\ 3 & 4\end{vmatrix}\) B. \(\frac{1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2\end{vmatrix}\) C. \(\frac{1}{5}\) \(\begin{vmatrix} 2 & -1 \\ -3 & 4\end{vmatrix}\) D. \(\frac{1}{5}\) \(\begin{vmatrix} 4 & 3 \\ 1 & 2\end{vmatrix}\) Show Content Detailed Solution There is an explanation video available below.	B
24.	What is the size of each interior angle of a 12-sided regular polygon? A. 120^o B. 150^o C. 30^o D. 180^o Show Content Detailed Solution There is an explanation video available below.	B
25.	A chord of circle of radius 7cm is 5cm from the centre of the circle.What is the length of the chord? A. 4√6 cm B. 3√6 cm C. 6√6 cm D. 2√6 cm Show Content Detailed Solution From Pythagoras theorem \|OA\|2 = \|AN\|2 + \|ON\|2 72 = \|AN\|2 + (5)2 49 = \|AN\|2 + 25 \|AN\|2 = 49 - 25 = 24 \|AN\| = \(\sqrt {24}\) = \(\sqrt {4 \times 6}\) = 2√6 cm \|AN\| = \|NB\| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts) \|AN\| + \|NB\| = \|AB\| 2√6 + 2√6 = \|AB\| \|AB\| = 4√6 cm There is an explanation video available below.	A

26.	A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold A. 48 cm³ B. 33 cm³ C. 60 cm³ D. 27 cm³ Show Content Detailed Solution There is an explanation video available below.	B
27.	The perpendicular bisector of a line XY is the locus of a point A. whose distance from X is always twice its distance from Y B. whose distance from Y is always twice its distance from X. C. which moves on the line XY D. which is equidistant from the points X and Y Show Content Detailed Solution There is an explanation video available below.	D
28.	The midpoint of P(x, y) and Q(8, 6) is (5, 8). Find x and y. A. (2, 10) B. (2, 8) C. (2, 12) D. (2, 6) Show Content Detailed Solution P(x, y) Q(8, 6) midpoint = (5, 8) x + 8 = 5 \(\frac{y + 6}{2}\) = 8 x + 8 = 10 x = 10 - 8 = 2 y + 6 = 16 y + 16 - 6 = 10 therefore, P(2, 10) There is an explanation video available below.	A
29.	Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2). A. 5y - 2x -18 = 0 B. 5y + 2x - 18 = 0 C. 5y - 2x + 18 = 0 D. 5y + 2x - 2 = 0 Show Content Detailed Solution 2y = 5x + 4 (4, 2) y = \(\frac{5x}{2}\) + 4 comparing with y = mx + e m = \(\frac{5}{2}\) Since they are perpendicular m1m2 = -1 m2 = \(\frac{-1}{m_1}\) = -1 \(\frac{5}{2}\) = -1 x \(\frac{2}{5}\) The equator of the line is thus y = mn + c (4, 2) 2 = -\(\frac{2}{5}\)(4) + c \(\frac{2}{1}\) + \(\frac{8}{5}\) = c c = \(\frac{18}{5}\) y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\) 5y = -2x + 18 &l	B
30.	In a right angled triangle, if tan \(\theta\) = \(\frac{3}{4}\). What is cos\(\theta\) - sin\(\theta\)? A. \(\frac{2}{3}\) B. \(\frac{3}{5}\) C. \(\frac{1}{5}\) D. \(\frac{4}{5}\) Show Content Detailed Solution tan\(\theta\) = \(\frac{3}{4}\) from Pythagoras tippet, the hypotenus is T i.e. 3, 4, 5. then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\) cos\(\theta\) - sin\(\theta\) \(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\) There is an explanation video available below.	C