Algebra questions for competitive exams SSC CGL Algebra set 6
Solve 10 algebra questions for competitive exams SSC CGL Set 6 in 12 minutes and verify from answers. Learn how to solve quick from paired solution set 6.
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10 Algebra questions for competitive exams SSC CGL Algebra Set 6 - time to solve 12 mins
Q1. The value of $\sqrt{(x - 2)^2} + \sqrt{(x - 4)^2}$, where $2\lt{x}\lt{3}$, is,
- $2x - 6$
- 3
- 2
- 4
Q2. If $x = 10^{0.48}$, $y = 10^{0.70}$ and $x^z = y^2$, then approximate value of $z$ is
- 1.88
- 2.9
- 3.7
- 1.45
Q3. If $\displaystyle\frac{a}{1 - a} + \displaystyle\frac{b}{1 - b} + \displaystyle\frac{c}{1 - c} = 1$, thenĀ the value of $\displaystyle\frac{1}{1 - a} + \displaystyle\frac{1}{1 - b} + \displaystyle\frac{1}{1 - c}$,
- 4
- 2
- 1
- 3
Q4. If $x$ is real then the minimum value of $4x^2 - x - 1$ is,
- $\displaystyle\frac{5}{8}$
- $-\displaystyle\frac{15}{16}$
- $\displaystyle\frac{15}{16}$
- $-\displaystyle\frac{17}{16}$
Q5. If $9\sqrt{x} = \sqrt{12} + \sqrt{147}$ then the value of $x$ is,
- 3
- 4
- 5
- 2
Q6. If $a$ and $b$ are positive integers such that $a^2 - b^2 = 19$ then $a$ is,
- 8
- 0
- 9
- 10
Q7. If $ax^2 + bx + c = a(x - p)^2$, then the relation between $a$, $b$ and $c$ can be expressed as,
- $b^2 = ac$
- $abc = 1$
- $2b = a + c$
- $b^2 = 4ac$
Q8. If $x = 5 - \sqrt{21}$, then value of $\displaystyle\frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$ is,
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{3} - \sqrt{7})$
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} - \sqrt{3})$
- $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} + \sqrt{3})$
- $\displaystyle\frac{1}{\sqrt{2}}(3 + \sqrt{7})$
Q9. If $a^{\frac{1}{3}} = 11$ then $a^2 - 331a$ is
- 1331331
- 1334331
- 1331000
- 1330030
Q10. If $a = \displaystyle\frac{xy}{x + y}$, $b = \displaystyle\frac{xz}{x + z}$ and $c = \displaystyle\frac{yz}{y + z}$, where $a$, $b$ and $c$ are all non-zero numbers, then the value of $x$ is,
- $\displaystyle\frac{2abc}{ab + ac - bc}$
- $\displaystyle\frac{2abc}{ab + bc - ac}$
- $\displaystyle\frac{2abc}{ab + bc + ac}$
- $\displaystyle\frac{2abc}{ac + bc - ab}$
Answers to 10 Algebra questions for competitive exams SSC CGL Algebra Set 6
Q1. Answer: Option c: 2.
Q2. Answer: Option b : 2.9.
Q3. Answer: Option b: 2.
Q4. Answer: Option d: $-\displaystyle\frac{17}{16}$.
Q5. Answer: Option a: 3.
Q6. Answer: Option d : 10.
Q7. Answer: Option d: $b^2=4ac$.
Q8. Answer: Option b: $\displaystyle\frac{1}{\sqrt{2}}(\sqrt{7} - \sqrt{3})$.
Q9. Answer: Option c: 1331000.
Q10. Answer: Option d: $\displaystyle\frac{2abc}{ac + bc - ab}$.
Solutions to the questions
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SSC CGL Algebra solutions for competitive exams Set 6.
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