## 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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