Mathematics (Core), 2019, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2019

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

61 - 70 of 80 Questions

#	Question	Ans
61.	If the universal set μ = {x : 1 ≤ x ≤ 20} and A = {y : multiple of 3} B = \|z : odd numbers} Find A ∩ B A. {1, 3, 6} B. {3, 5, 9, 12} C. {3, 9, 15} D. {2, 3, 9} Show Content Detailed Solution μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} A ∩ B = {3, 9, 15} There is an explanation video available below.	C
62.	In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females? A. 144 ways B. 15 ways C. 185 ways D. 12 ways Show Content Detailed Solution For the committee to include 2 females, we must have 3 males, so that there should be 5 members. That is, \(^4C_3 \times ^3C_2\) = \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\) = 4 × 3 = 12 ways There is an explanation video available below.	D
63.	Find the value of k in the equation: \(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) A. \(\sqrt{28}\) B. 7 C. 28 D. \(\sqrt{7}\) Show Content Detailed Solution \(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) \(\sqrt{4 \times 7} + \sqrt{16 \times 7} - \sqrt{k} = \sqrt{25 \times 7}\) \(2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7}\) \(6\sqrt{7} - 5\sqrt{7} = \sqrt{k}\) \(\sqrt{k} = \sqrt{7}\) \(\implies k = 7\) There is an explanation video available below.	B
64.	Evaluate \(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) A. 4 B. -32 C. -8 D. 16 Show Content Detailed Solution \(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) = \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\) = \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\) = -8. There is an explanation video available below.	C
65.	Find the value of x for \(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) A. x \(\geq\) -5 B. x \(\geq\) -1 C. x \(\leq\) -1 D. x \(\leq\) 3 Show Content Detailed Solution \(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) \(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\) \(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\) \(5(2x - 4) \geq 3(4x - 6)\) \(10x - 20 \geq 12x - 18\) \(10x - 12x \geq -18 + 20\) \(-2x \geq 2\) \(x \leq -1\) There is an explanation video available below.	C
66.	Determine the values for which \(x^2 - 7x + 10 \leq 0\) A. 2 \(\leq\) x \(\geq\) 5 B. -2 \(\leq\) x \(\leq\) 3 C. -2 \(\leq\) x \(\geq\) 3 D. 2 \(\leq\) x \(\leq\) 5 Show Content Detailed Solution \(x^2 - 7x + 10 \leq 0\) Solve for \(x^2 - 7x + 10 = 0\) We have, (x - 5)(x - 2) \(\leq\) 0. Conditions: Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0. \(\implies\) x \(\leq\) 5; x \(\geq\) 2. 2 \(\leq\) x \(\leq\) 5. Choosing x = 3, 3\(^2\) - 7(3) + 10 = 9 - 21 + 10 = -2 \(\leq\) 0. \(\therefore\) 2 \(\leq\) x \(\leq\) 5. There is an explanation video available below.	D
67.	Find the polynomial if given q(x) = x\(^2\) - x - 5, d(x) = 3x - 1 and r(x) = 7. A. 3x\(^3\) - 4x\(^2\) - 14x + 12 B. 3x\(^2\) + 3x - 7 C. 3x\(^3\) + 4x\(^2\) + 14x - 12 D. 3x\(^2\) - 3x + 4 Show Content Detailed Solution Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder. i.e In this case, the polynomial = (x\(^2\) - x - 5)(3x - 1) + 7. = (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7 = (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7 = 3x\(^3\) - 4x\(^2\) - 14x + 12 There is an explanation video available below.	A
68.	If 2x\(^2\) + x - 3 divides x - 2, find the remainder. A. 7 B. 3 C. 5 D. 6 Show Content Detailed Solution When you divide a polynomial p(x) by (x - a), the remainder = p(a) i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2). = 2(2)\(^2\) + 2 - 3 = 8 + 2 - 3 = 7. There is an explanation video available below.	A
69.	If \(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\), find the value of x. A. 5 B. 8 C. 6 D. 3 Show Content Detailed Solution \(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\) \(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\) \(2(2 - 12) + 5(2x) + 3(3x) = 132\) \(-20 + 10x + 9x = 132\) \(19x = 152\) \(x = 8\) There is an explanation video available below.	B
70.	Given the matrix \(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\). Find the inverse of matrix A. A. \(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\) B. \(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\) C. \(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\) D. \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\) Show Content Detailed Solution \(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\) \|A\| = (3 x 6) - (-2 x 1) = 18 + 2 = 20. A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\) = \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) = \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) There is an explanation video available below.	D

61.	If the universal set μ = {x : 1 ≤ x ≤ 20} and A = {y : multiple of 3} B = \|z : odd numbers} Find A ∩ B A. {1, 3, 6} B. {3, 5, 9, 12} C. {3, 9, 15} D. {2, 3, 9} Show Content Detailed Solution μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} A ∩ B = {3, 9, 15} There is an explanation video available below.	C
62.	In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females? A. 144 ways B. 15 ways C. 185 ways D. 12 ways Show Content Detailed Solution For the committee to include 2 females, we must have 3 males, so that there should be 5 members. That is, \(^4C_3 \times ^3C_2\) = \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\) = 4 × 3 = 12 ways There is an explanation video available below.	D
63.	Find the value of k in the equation: \(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) A. \(\sqrt{28}\) B. 7 C. 28 D. \(\sqrt{7}\) Show Content Detailed Solution \(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) \(\sqrt{4 \times 7} + \sqrt{16 \times 7} - \sqrt{k} = \sqrt{25 \times 7}\) \(2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7}\) \(6\sqrt{7} - 5\sqrt{7} = \sqrt{k}\) \(\sqrt{k} = \sqrt{7}\) \(\implies k = 7\) There is an explanation video available below.	B
64.	Evaluate \(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) A. 4 B. -32 C. -8 D. 16 Show Content Detailed Solution \(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) = \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\) = \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\) = -8. There is an explanation video available below.	C
65.	Find the value of x for \(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) A. x \(\geq\) -5 B. x \(\geq\) -1 C. x \(\leq\) -1 D. x \(\leq\) 3 Show Content Detailed Solution \(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) \(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\) \(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\) \(5(2x - 4) \geq 3(4x - 6)\) \(10x - 20 \geq 12x - 18\) \(10x - 12x \geq -18 + 20\) \(-2x \geq 2\) \(x \leq -1\) There is an explanation video available below.	C

66.	Determine the values for which \(x^2 - 7x + 10 \leq 0\) A. 2 \(\leq\) x \(\geq\) 5 B. -2 \(\leq\) x \(\leq\) 3 C. -2 \(\leq\) x \(\geq\) 3 D. 2 \(\leq\) x \(\leq\) 5 Show Content Detailed Solution \(x^2 - 7x + 10 \leq 0\) Solve for \(x^2 - 7x + 10 = 0\) We have, (x - 5)(x - 2) \(\leq\) 0. Conditions: Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0. \(\implies\) x \(\leq\) 5; x \(\geq\) 2. 2 \(\leq\) x \(\leq\) 5. Choosing x = 3, 3\(^2\) - 7(3) + 10 = 9 - 21 + 10 = -2 \(\leq\) 0. \(\therefore\) 2 \(\leq\) x \(\leq\) 5. There is an explanation video available below.	D
67.	Find the polynomial if given q(x) = x\(^2\) - x - 5, d(x) = 3x - 1 and r(x) = 7. A. 3x\(^3\) - 4x\(^2\) - 14x + 12 B. 3x\(^2\) + 3x - 7 C. 3x\(^3\) + 4x\(^2\) + 14x - 12 D. 3x\(^2\) - 3x + 4 Show Content Detailed Solution Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder. i.e In this case, the polynomial = (x\(^2\) - x - 5)(3x - 1) + 7. = (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7 = (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7 = 3x\(^3\) - 4x\(^2\) - 14x + 12 There is an explanation video available below.	A
68.	If 2x\(^2\) + x - 3 divides x - 2, find the remainder. A. 7 B. 3 C. 5 D. 6 Show Content Detailed Solution When you divide a polynomial p(x) by (x - a), the remainder = p(a) i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2). = 2(2)\(^2\) + 2 - 3 = 8 + 2 - 3 = 7. There is an explanation video available below.	A
69.	If \(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\), find the value of x. A. 5 B. 8 C. 6 D. 3 Show Content Detailed Solution \(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\) \(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\) \(2(2 - 12) + 5(2x) + 3(3x) = 132\) \(-20 + 10x + 9x = 132\) \(19x = 152\) \(x = 8\) There is an explanation video available below.	B
70.	Given the matrix \(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\). Find the inverse of matrix A. A. \(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\) B. \(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\) C. \(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\) D. \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\) Show Content Detailed Solution \(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\) \|A\| = (3 x 6) - (-2 x 1) = 18 + 2 = 20. A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\) = \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) = \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) There is an explanation video available below.	D

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