## 35th SSC CGL level Question Set, 10th on Algebra

This is the 35th question set of 10 practice problem exercise for SSC CGL exam and 10th on topic Algebra.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through the solutions, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

Before taking the test you may like to refer to our **concept tutorials** on Algebra and other related topics,

**Basic and rich Algebraic concepts for elegant solutions of SSC CGL problems,**

**More Basic and rich Algebraic concepts for elegant solutions of SSC CGL problems**

* SSC CGL level difficult Algebra problem solving by Componendo dividendo*.

### 35th question set - 10 problems for SSC CGL exam: 10th on topic Algebra - answering time 15 mins

**Q1. **Two numbers $a$ and $b$ (where $a \gt b$) are such that their sum is three times their difference. The value of $\displaystyle\frac{3ab}{4(a^2-b^2)}$ is then,

- $\frac{1}{3}$
- $\frac{1}{2}$
- $1\frac{1}{4}$
- $\frac{5}{6}$

**Q2.** If $\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$, then $x$ is,

- $\displaystyle\frac{12}{5}$
- $\displaystyle\frac{5}{12}$
- $\displaystyle\frac{5}{7}$
- $\displaystyle\frac{7}{5}$

**Q3.** If $a+b=1$, $c+d=1$ and $a-b=\displaystyle\frac{d}{c}$ then the value of $c^2-d^2$ is,

- $1$
- $\displaystyle\frac{a}{b}$
- $\displaystyle\frac{b}{a}$
- $-1$

**Q4. **If $x + \displaystyle\frac{1}{x} = \sqrt{3}$, then the value of $x^{18} + x^{12} + x^6 + 1$ is,

- 1
- 3
- 0
- 2

**Q5. **If $a$ and $b$ are two real numbers and the expression $ax^3 + 3x^2 -8x + b$ is exactly divisible by the expressions $(x + 2)$ and $(x-2)$ then the values of $a$ and $b$ are,

- $a=-2$; $b=12$
- $a=2$; $b=12$
- $a=12$; $b=2$
- $a=2$; $b=-12$

**Q6.** If $(a^2+b^2)^3 = (a^3+b^3)^2$ then $\displaystyle\frac{b}{a} + \displaystyle\frac{a}{b}$ will be equal to,

- $\displaystyle\frac{2}{3}$
- $-\displaystyle\frac{1}{3}$
- $-\displaystyle\frac{2}{3}$
- $\displaystyle\frac{1}{3}$

** Q7.** $\left(x + \displaystyle\frac{1}{x}\right)\left(x - \displaystyle\frac{1}{x}\right)\left(x^2 + \displaystyle\frac{1}{x^2} -1\right)\left(x^2 + \displaystyle\frac{1}{x^2} +1\right)$ is equal to,

- $\left(x^6 + \displaystyle\frac{1}{x^6}\right)$
- $\left(x^6 - \displaystyle\frac{1}{x^6}\right)$
- $\left(x^8 + \displaystyle\frac{1}{x^8}\right)$
- $\left(x^8 - \displaystyle\frac{1}{x^8}\right)$

** Q8.** If $x+y=a$ and $xy=b^2$ then the value of $x^3 - x^2y - xy^2 + y^3$ in terms of $a$ and $b$ is,

- $a^3 - 4b^2a$
- $a^3 + 3b^2$
- $(a^2 + 4b^2)a$
- $a^3 - 3b^2$

**Q9.** If $x=99999$, the value of $\displaystyle\frac{4x^3-x}{(2x+1)(6x-3)}$ is,

- 11111
- 66666
- 33333
- 22222

** Q10.** The length of the intercept of the graph of the equation $9x -12y = 108$ between x and y axes is,

- 12 units
- 18 units
- 9 units
- 15 units

### Solutions to the problems

For detailed conceptual solutions with answers you should refer to the companion * SSC CGL level Solution Set 35 on Algebra* where you will get detailed explanations on easiest path to the solutions.

Watch the **quick solutions in the two-part video**.

**Part I: Q1 to Q5**

**Part II: Q6 to Q10**

### Additional help on SSC CGL Algebra

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