21 - 30 of 49 Questions
# | Question | Ans |
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21. |
Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\) A. \(\sqrt{3 - 2}\) B. 2 - \(\sqrt{3}\) C. \(\sqrt{3}\) D. -2 Detailed SolutionTan 60 = 3; Tan 30 = 1\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\) = \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{3}\) |
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22. |
The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group A. 10.1 years B. 9.3 years C. 8.7 years D. 3 years |
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23. |
In what number base was the addition 1 + nn = 100, where n > 0, done? A. n - 1 B. n + 2 C. n + 1 D. n |
C |
24. |
Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\) A. 4 B. \(\sqrt{3} + 4\) C. 4 \(\sqrt{2}\) D. 4\(\sqrt{3} + 4\) Detailed Solution\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\) 2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4 = 4 |
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25. |
Three exterior angles of a polygon are 30\(^o\), 40\(^o\) and 60\(^o\). If the remaining exterior angles are 46\(^o\) each, name the polygon. A. decagon B. nonagon C. octagon D. hexagon Detailed SolutionSum of all exterior angles is 360\(^o\)360\(^o\) (30\(^o\) - 40\(^o\)) 360 - (130\(^o\)) 230\(^o\) remaining is 46\(^o\) = \(\frac{230}{46}\) = 5 5 + 3 = 8 sides; Octagon |
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26. |
Simplify the expression \(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\) A. \(a^2 - b^2\) B. \(b^2 - a^2\) C. \(a^2b - ab^2\) D. \(ab^2 - a^2b\) |
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27. |
Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),... A. -\(\frac{1}{3}\) B. -\(\frac{1}{5}\) C. -\(\frac{1}{15}\) D. \(\frac{1}{9}\) Detailed Solutiona = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)= 7 - 10 = \(\frac{-3}{15}\) d = - \(\frac{-1}{5}\) T6 = a + 5d = \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\) = \(\frac{2}{3}\) - 1 = \(\frac{2 - 3}{3}\) = \(\frac{-1}{3}\) |
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28. |
The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation. A. \(6x^2 - x + 2 = 0\) B. \(6x^2 - x - 2 = 0\) C. \(6x^2 + x - 2 = 0\) D. \(6x^2 + x + 2 = 0\) Detailed Solution(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\) \(6x^2 - 4n + 3n - 2 = 0\) \(6x^2 - x - 2 = 0\) |
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29. |
Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\) A. x = \(\frac{6 + 12}{d^2 + y}\) B. x = \(\frac{12}{d^2 - y}\) C. x = \(\frac{12}{y} - 2d^2\) D. x = \(\frac{12}{2d^2 + y}\) |
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30. |
Consider the statements: p it is hot, q: it is rainingWhich of the following symbols correctly represents the statement "It is raining if and only if it it is cold"? A. p \(\iff\) \(\sim\)q B. p \(\iff\) q C. \(\sim\)p \(\iff\) \(\sim\)q D. q \(\iff\) \(\sim\)p |
D |
21. |
Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\) A. \(\sqrt{3 - 2}\) B. 2 - \(\sqrt{3}\) C. \(\sqrt{3}\) D. -2 Detailed SolutionTan 60 = 3; Tan 30 = 1\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\) = \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{3}\) |
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22. |
The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group A. 10.1 years B. 9.3 years C. 8.7 years D. 3 years |
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23. |
In what number base was the addition 1 + nn = 100, where n > 0, done? A. n - 1 B. n + 2 C. n + 1 D. n |
C |
24. |
Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\) A. 4 B. \(\sqrt{3} + 4\) C. 4 \(\sqrt{2}\) D. 4\(\sqrt{3} + 4\) Detailed Solution\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\) 2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4 = 4 |
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25. |
Three exterior angles of a polygon are 30\(^o\), 40\(^o\) and 60\(^o\). If the remaining exterior angles are 46\(^o\) each, name the polygon. A. decagon B. nonagon C. octagon D. hexagon Detailed SolutionSum of all exterior angles is 360\(^o\)360\(^o\) (30\(^o\) - 40\(^o\)) 360 - (130\(^o\)) 230\(^o\) remaining is 46\(^o\) = \(\frac{230}{46}\) = 5 5 + 3 = 8 sides; Octagon |
26. |
Simplify the expression \(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\) A. \(a^2 - b^2\) B. \(b^2 - a^2\) C. \(a^2b - ab^2\) D. \(ab^2 - a^2b\) |
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27. |
Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),... A. -\(\frac{1}{3}\) B. -\(\frac{1}{5}\) C. -\(\frac{1}{15}\) D. \(\frac{1}{9}\) Detailed Solutiona = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)= 7 - 10 = \(\frac{-3}{15}\) d = - \(\frac{-1}{5}\) T6 = a + 5d = \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\) = \(\frac{2}{3}\) - 1 = \(\frac{2 - 3}{3}\) = \(\frac{-1}{3}\) |
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28. |
The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation. A. \(6x^2 - x + 2 = 0\) B. \(6x^2 - x - 2 = 0\) C. \(6x^2 + x - 2 = 0\) D. \(6x^2 + x + 2 = 0\) Detailed Solution(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\) \(6x^2 - 4n + 3n - 2 = 0\) \(6x^2 - x - 2 = 0\) |
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29. |
Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\) A. x = \(\frac{6 + 12}{d^2 + y}\) B. x = \(\frac{12}{d^2 - y}\) C. x = \(\frac{12}{y} - 2d^2\) D. x = \(\frac{12}{2d^2 + y}\) |
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30. |
Consider the statements: p it is hot, q: it is rainingWhich of the following symbols correctly represents the statement "It is raining if and only if it it is cold"? A. p \(\iff\) \(\sim\)q B. p \(\iff\) q C. \(\sim\)p \(\iff\) \(\sim\)q D. q \(\iff\) \(\sim\)p |
D |