Mathematics (Core), 2008, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2008

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 49 of 49 Questions

#	Question	Ans
41.	Find the mean deviation of 2, 4, 5, and 9 A. 1 B. 2 C. 5 D. 7 Show Content Detailed Solution X = 20/4 = 5 mean deviation = 8/4 = 2 There is an explanation video available below.	B
42.	In how many ways can the letters of the word ACCEPTANCE be arranged? A. 10! / (2!2!3!) B. 10! / ( 2!3!) C. 10! / (2!2!) D. 10! Show Content Detailed Solution ACCEPTANCE = 10 Letters A = 2 letters C = 3 letters E = 2 letters Can be arranged in 10! / (2!3!2!) ways There is an explanation video available below.	A
43.	Find the number of ways of selecting 6 out of 10 subjects for an examination A. 128 B. 216 C. 215 D. 210 Show Content Detailed Solution \(^{10}C_6 = \frac{10!}{(10-6)!6!}\\ =\frac{10!}{4!6!}\\ =\frac{10\times 9\times 8 \times 7 \times 6!}{4\times 3\times 2\times 1 \times 6!}\\ =210\) There is an explanation video available below.	D
44.	The probability of picking a letter T from the word OBSTRUCTION is? A. ¹/₁₁ B. ²/₁₁ C. ³/₁₁ D. ⁴/₁₁ Show Content Detailed Solution OBSTRUCTION Total possible outcome = 11 Number of chance of getting T = 2 P(picking T) = 2/11 There is an explanation video available below.	B
45.	The result of rolling a fair die 150 times is as summarized in the table above. What is the probability of obtaining a 5? A. ³/₁₀ B. ¹/₅ C. ¹/₆ D. ¹/₁₀ Show Content Detailed Solution Total possible outcome 12+18+x+30+2x+45 = 105+3x ∴105+3x = 150 3x = 150-105 3x = 45 x = 15 P(obtaining 5) = \(\frac{2x}{(105+3x)}But x= 15\\ =\frac{2(15)}{(105+3(15))}\\ =\frac{30}{(105+45)}\\ =\frac{30}{150}\\ =\frac{1}{5}\) There is an explanation video available below.	B
46.	Find the values of x and y respectively if \(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) A. -3, -2 B. -5, -3 C. -2, -5 D. -3, -5 Show Content Detailed Solution \(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) therefore, (x, y) = (-3, -5) respectively There is an explanation video available below.	D
47.	\(\begin{pmatrix} -2 & 1 \\ 2 & 3 \end{pmatrix}\) \(\begin{pmatrix}p & q \\ r & s\end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\). What is the value of r? A. -\(\frac{1}{8}\) B. \(\frac{3}{8}\) C. \(\frac{5}{8}\) D. \(\frac{1}{4}\) Show Content Detailed Solution -2p + r = 1.......(i) 2p + 3r = 0.......(ii) r - 1 + 2p ........(iii) 2p + 3(1 + 2p) = 0 2p + 3(1 + 2p) = 0 2p + 3 + 6p = 0 3 - 8p = 0 \(\to\) 8p = 3 p = \(\frac{3}{8}\) 6 = 1 - 2 \(\frac{3}{8}\) = 1 - \(\frac{6}{8}\) \(\frac{8 - 6}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\) There is an explanation video available below.	D
48.	Calculate the distance between points L(-1, -6) and M(-3, -5) A. √5 B. 2√3 C. √20 D. √50 Show Content Detailed Solution L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\) D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\) D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\) D = \(\sqrt{(-2)^2 + 1^2}\) D = \(\sqrt{4 + 1}\) D = \(\sqrt{5}\) There is an explanation video available below.	A
49.	A student sitting on a tower 68 metres high observes his principal's car at the angle of depression of 20o. How far is the car from the bottom of the tower to the nearest metre? A. 184m B. 185m C. 186m D. 187m Show Content Detailed Solution Tan 20o = \(\frac{68m}{x}\) x tan 20o = 68 x = \(\frac{68}{tan 20}\) = \(\frac{68}{0.364}\) x = 186.8 = 187m There is an explanation video available below.	D

41.	Find the mean deviation of 2, 4, 5, and 9 A. 1 B. 2 C. 5 D. 7 Show Content Detailed Solution X = 20/4 = 5 mean deviation = 8/4 = 2 There is an explanation video available below.	B
42.	In how many ways can the letters of the word ACCEPTANCE be arranged? A. 10! / (2!2!3!) B. 10! / ( 2!3!) C. 10! / (2!2!) D. 10! Show Content Detailed Solution ACCEPTANCE = 10 Letters A = 2 letters C = 3 letters E = 2 letters Can be arranged in 10! / (2!3!2!) ways There is an explanation video available below.	A
43.	Find the number of ways of selecting 6 out of 10 subjects for an examination A. 128 B. 216 C. 215 D. 210 Show Content Detailed Solution \(^{10}C_6 = \frac{10!}{(10-6)!6!}\\ =\frac{10!}{4!6!}\\ =\frac{10\times 9\times 8 \times 7 \times 6!}{4\times 3\times 2\times 1 \times 6!}\\ =210\) There is an explanation video available below.	D
44.	The probability of picking a letter T from the word OBSTRUCTION is? A. ¹/₁₁ B. ²/₁₁ C. ³/₁₁ D. ⁴/₁₁ Show Content Detailed Solution OBSTRUCTION Total possible outcome = 11 Number of chance of getting T = 2 P(picking T) = 2/11 There is an explanation video available below.	B
45.	The result of rolling a fair die 150 times is as summarized in the table above. What is the probability of obtaining a 5? A. ³/₁₀ B. ¹/₅ C. ¹/₆ D. ¹/₁₀ Show Content Detailed Solution Total possible outcome 12+18+x+30+2x+45 = 105+3x ∴105+3x = 150 3x = 150-105 3x = 45 x = 15 P(obtaining 5) = \(\frac{2x}{(105+3x)}But x= 15\\ =\frac{2(15)}{(105+3(15))}\\ =\frac{30}{(105+45)}\\ =\frac{30}{150}\\ =\frac{1}{5}\) There is an explanation video available below.	B

46.	Find the values of x and y respectively if \(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) A. -3, -2 B. -5, -3 C. -2, -5 D. -3, -5 Show Content Detailed Solution \(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) therefore, (x, y) = (-3, -5) respectively There is an explanation video available below.	D
47.	\(\begin{pmatrix} -2 & 1 \\ 2 & 3 \end{pmatrix}\) \(\begin{pmatrix}p & q \\ r & s\end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\). What is the value of r? A. -\(\frac{1}{8}\) B. \(\frac{3}{8}\) C. \(\frac{5}{8}\) D. \(\frac{1}{4}\) Show Content Detailed Solution -2p + r = 1.......(i) 2p + 3r = 0.......(ii) r - 1 + 2p ........(iii) 2p + 3(1 + 2p) = 0 2p + 3(1 + 2p) = 0 2p + 3 + 6p = 0 3 - 8p = 0 \(\to\) 8p = 3 p = \(\frac{3}{8}\) 6 = 1 - 2 \(\frac{3}{8}\) = 1 - \(\frac{6}{8}\) \(\frac{8 - 6}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\) There is an explanation video available below.	D
48.	Calculate the distance between points L(-1, -6) and M(-3, -5) A. √5 B. 2√3 C. √20 D. √50 Show Content Detailed Solution L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\) D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\) D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\) D = \(\sqrt{(-2)^2 + 1^2}\) D = \(\sqrt{4 + 1}\) D = \(\sqrt{5}\) There is an explanation video available below.	A
49.	A student sitting on a tower 68 metres high observes his principal's car at the angle of depression of 20o. How far is the car from the bottom of the tower to the nearest metre? A. 184m B. 185m C. 186m D. 187m Show Content Detailed Solution Tan 20o = \(\frac{68m}{x}\) x tan 20o = 68 x = \(\frac{68}{tan 20}\) = \(\frac{68}{0.364}\) x = 186.8 = 187m There is an explanation video available below.	D