11 - 20 of 50 Questions
# | Question | Ans |
---|---|---|
11. |
If m : n = 2 : 1, evaluate \(\frac{3m^2 - 2n^2}{m^2 + mn}\) A. \(\frac{4}{3}\) B. \(\frac{5}{3}\) C. \(\frac{3}{4}\) D. \(\frac{3}{5}\) Detailed Solutionm = 2, n = 1\(\frac{3m^2 - 2n^2}{m^2 _ mn}\) = \(\frac{3(2)^2 - 2(1)^2}{2^2 + 2(1)}\) = \(\frac{12 - 2}{4 + 2} = \frac{10}{6}\) = \(\frac{5}{3}\) |
|
12. |
H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y. A. H = \(\frac{p}{4y^2}\) B. H = \(\frac{2p}{y^2}\) C. H = \(\frac{p}{2y^2}\) D. H = \(\frac{p}{y^2}\) Detailed SolutionH \(\propto\) \(\frac{p}{y^2}\)H = \(\frac{pk}{y^2}\) 1 = \(\frac{8k}{2^2}\) k = \(\frac{4}{8}\) = \(\frac{1}{2}\) H = \(\frac{p}{2y^2}\) |
|
13. |
Solve 4x^{2}\) - 16x + 15 = 0. A. x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\) B. x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) C. x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\) D. x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\) Detailed Solution4x\(^2\) - 16x + 15 = 0(2x - 3)(2x - 5) = 0 x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) |
|
14. |
Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\) 3 + \(\frac{2}{3} \log_{10} 27\) A. 3 \(\log_{10}^2\) B. \(\log_{10}^2\) C. \(\log_{10}^3\) D. 2 \(\log_{10}^3\) Detailed Solutionlog\(_{10}\) 6 - log\(_{10}\)3\(^3\) + log\(_{10}\) (\(\sqrt[3]{27}\))\(^2\)= log \(_{10}\) 6 - log \(_{10}\) 27 + log\(_{10}\) 9 = log\(_{10}\) \(\frac{6 \times 9}{27}\) = log\(_{10}\)2 |
|
15. |
Bala sold an article for #6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for N6,600.00. A. 5% B. 10% C. 12% D. 13% Detailed Solution15 = (\(\frac{6,900 - C.P \times 100}{C.P}\))15 C.P = 690000 - C.P 100 C.P = \(\frac{690000}{115}\) C.P = N6,000 %profit = \(\frac{6,600 - 6,000}{6,000}\) x 100 = \(\frac{600}{6,000}\) x 100 = 10% |
|
16. |
If 3p = 4q and 9p = 8q - 12, find the value of pq. A. 12 B. 7 C. -7 D. -12 Detailed Solution9p = 8q - 129p = 2(4q) - 12 9p = 2(3q) - 12 9p = 6p - 12 3p = -12 p = -4 \(\frac{3 \times -4}{4} = \frac{4q}{4}\) q = 13 pq = -3 x -4 = 12 |
|
17. |
If (0.25)\(^y\) = 32, find the value of y. A. y = - \(\frac{5}{2}\) B. y = -\(\frac{3}{2}\) C. y = \(\frac{3}{2}\) D. y = \(\frac{5}{2}\) Detailed Solution(0.25)\(^y\) = 322\(^{-2y}\) = 3\(^{2}\) 2 - 2y = 5 y = - \(\frac{5}{2}\) |
|
18. |
There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy? A. \(\frac{1}{6}\) B. \(\frac{1}{4}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) Detailed SolutionTotal number of people in the lift = 8 boys + 4 girls= 12 people probability that a boy comes out = \(\frac{8}{12}\) = \(\frac{2}{3}\) |
|
19. |
Simplify: \(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\) A. \(\frac{x - 7}{x + 7}\) B. \(\frac{x + 7}{x - 7}\) C. \(\frac{x - 2}{x + 4}\) D. \(\frac{x + 2}{x - 2}\) Detailed Solution\(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\)\(\frac{(x - 7)(x + 2)}{(x - 7)(x - 2)}\) = \(\frac{x + 2}{x - 2}\) |
|
20. |
Which of these values would make \(\frac{3p^-}{p^{2-}}\) undefined? A. 1 B. \(\frac{1}{3}\) C. -\(\frac{1}{3}\) D. -1 Detailed SolutionP\(^2\) - p = 0p(p - 1) = 0 p = 0 or p = 1 |
11. |
If m : n = 2 : 1, evaluate \(\frac{3m^2 - 2n^2}{m^2 + mn}\) A. \(\frac{4}{3}\) B. \(\frac{5}{3}\) C. \(\frac{3}{4}\) D. \(\frac{3}{5}\) Detailed Solutionm = 2, n = 1\(\frac{3m^2 - 2n^2}{m^2 _ mn}\) = \(\frac{3(2)^2 - 2(1)^2}{2^2 + 2(1)}\) = \(\frac{12 - 2}{4 + 2} = \frac{10}{6}\) = \(\frac{5}{3}\) |
|
12. |
H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y. A. H = \(\frac{p}{4y^2}\) B. H = \(\frac{2p}{y^2}\) C. H = \(\frac{p}{2y^2}\) D. H = \(\frac{p}{y^2}\) Detailed SolutionH \(\propto\) \(\frac{p}{y^2}\)H = \(\frac{pk}{y^2}\) 1 = \(\frac{8k}{2^2}\) k = \(\frac{4}{8}\) = \(\frac{1}{2}\) H = \(\frac{p}{2y^2}\) |
|
13. |
Solve 4x^{2}\) - 16x + 15 = 0. A. x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\) B. x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) C. x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\) D. x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\) Detailed Solution4x\(^2\) - 16x + 15 = 0(2x - 3)(2x - 5) = 0 x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\) |
|
14. |
Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\) 3 + \(\frac{2}{3} \log_{10} 27\) A. 3 \(\log_{10}^2\) B. \(\log_{10}^2\) C. \(\log_{10}^3\) D. 2 \(\log_{10}^3\) Detailed Solutionlog\(_{10}\) 6 - log\(_{10}\)3\(^3\) + log\(_{10}\) (\(\sqrt[3]{27}\))\(^2\)= log \(_{10}\) 6 - log \(_{10}\) 27 + log\(_{10}\) 9 = log\(_{10}\) \(\frac{6 \times 9}{27}\) = log\(_{10}\)2 |
|
15. |
Bala sold an article for #6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for N6,600.00. A. 5% B. 10% C. 12% D. 13% Detailed Solution15 = (\(\frac{6,900 - C.P \times 100}{C.P}\))15 C.P = 690000 - C.P 100 C.P = \(\frac{690000}{115}\) C.P = N6,000 %profit = \(\frac{6,600 - 6,000}{6,000}\) x 100 = \(\frac{600}{6,000}\) x 100 = 10% |
16. |
If 3p = 4q and 9p = 8q - 12, find the value of pq. A. 12 B. 7 C. -7 D. -12 Detailed Solution9p = 8q - 129p = 2(4q) - 12 9p = 2(3q) - 12 9p = 6p - 12 3p = -12 p = -4 \(\frac{3 \times -4}{4} = \frac{4q}{4}\) q = 13 pq = -3 x -4 = 12 |
|
17. |
If (0.25)\(^y\) = 32, find the value of y. A. y = - \(\frac{5}{2}\) B. y = -\(\frac{3}{2}\) C. y = \(\frac{3}{2}\) D. y = \(\frac{5}{2}\) Detailed Solution(0.25)\(^y\) = 322\(^{-2y}\) = 3\(^{2}\) 2 - 2y = 5 y = - \(\frac{5}{2}\) |
|
18. |
There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy? A. \(\frac{1}{6}\) B. \(\frac{1}{4}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) Detailed SolutionTotal number of people in the lift = 8 boys + 4 girls= 12 people probability that a boy comes out = \(\frac{8}{12}\) = \(\frac{2}{3}\) |
|
19. |
Simplify: \(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\) A. \(\frac{x - 7}{x + 7}\) B. \(\frac{x + 7}{x - 7}\) C. \(\frac{x - 2}{x + 4}\) D. \(\frac{x + 2}{x - 2}\) Detailed Solution\(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\)\(\frac{(x - 7)(x + 2)}{(x - 7)(x - 2)}\) = \(\frac{x + 2}{x - 2}\) |
|
20. |
Which of these values would make \(\frac{3p^-}{p^{2-}}\) undefined? A. 1 B. \(\frac{1}{3}\) C. -\(\frac{1}{3}\) D. -1 Detailed SolutionP\(^2\) - p = 0p(p - 1) = 0 p = 0 or p = 1 |