## 3rd SSC CGL Tier II level Question Set, 3rd on Algebra

This is the 3rd question set of 10 practice problem exercise for SSC CGL Tier II exam and also the 3rd 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 this conceptual analytical approach.

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

### 3rd question set - 10 problems for SSC CGL exam: 3rd on topic Algebra - answering time 12 mins

**Q1. **If $4y-3x=13$ and $xy=14$, then $64y^3-27x^3$ is,

- 8479
- 8400
- 8740
- 8749

**Q2.** If $x=(0.09)^2$, $y=\displaystyle\frac{1}{(0.09)^2}$ and $z=(1- 0.09)^2 - 1$, then which of the following relations is true?,

- $y \lt x$ and $x=z$
- $y \lt z \lt x$
- $z \lt x \lt y$
- $x \lt y$ and $x=z$

**Q3.** Minimum value of $x^4 +\displaystyle\frac{1}{x^4+1} - 3$ is,

- $0$
- $-1$
- $-2$
- $-3$

**Q4. **If $\displaystyle\frac{p}{3}=\frac{q}{2}$ then the value of $\displaystyle\frac{2p+3q}{3p-2q}$ is,

- $1$
- $\displaystyle\frac{12}{5}$
- $\displaystyle\frac{5}{12}$
- $\displaystyle\frac{12}{7}$

**Q5. **If $(a-4)^2 + (b-9)^2+(c-3)^2=0$, then the value of $\sqrt{a+b+c}$ is,

- $\pm{4}$
- $4$
- $-4$
- $\pm{2}$

**Q6.** If $\displaystyle\frac{1}{x+y}=\displaystyle\frac{1}x+\displaystyle\frac{1}{y}$, where, $x \neq 0$, $y \neq 0$ and $x \neq y$, the value of $x^3-y^3$ is,

- $0$
- $2$
- $1$
- $-1$

** Q7.** If $a^x=(x+y+z)^y$, $a^y=(x+y+z)^z$ and $a^z=(x+y+z)^x$ find the value of $a+b+c$ where $a \neq 0$.

- $0$
- $1$
- $a^3$
- $a$

** Q8.** If $\displaystyle\frac{x-a^2}{b^2+c^2} + \displaystyle\frac{x-b^2}{c^2+a^2} +\displaystyle\frac{x-c^2}{a^2+b^2} =3$ then the value of $x$ is,

- $a^2+b^2$
- $a^2+b^2+c^2$
- $a^2+b^2-c^2$
- $a^2-b^2-c^2$

**Q9.** If $ax^2+bx+c=a(x-p)^2$, then the correct relation between $a$, $b$ and $c$ would be,

- $2b=a+c$
- $b^2=4ac$
- $b^2=ac$
- $abc=1$

** Q10.** If $\displaystyle\frac{y}{x} - \displaystyle\frac{x}{y}=3$, then find the value of $\displaystyle\frac{x^3}{y^3} + \displaystyle\frac{y^3}{x^3}$.

- $10\sqrt{3}$
- $10\sqrt{13}$
- $13\sqrt{3}$
- $13\sqrt{10}$

For detailed conceptual solution to the questions, refer to **SSC CGL Tier II level Solution set 3 Algebra 3.**

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

**Part I: Q1 to Q5**

**Part II: Q6 to Q10**

### Answers to the questions

**Problem 1:** Answer: Option d: 8749.

**Problem 2:** Answer: Option c : $z \lt x \lt y$.

**Problem 3:** Answer: Option c: $-2$.

**Problem 4:** Answer: Option b: $\displaystyle\frac{12}{5}$.

**Problem 5:** Answer: Option b: 4.

**Problem 6:** Answer: Option a : 0.

**Problem 7:** Answer: Option d: $a$.

**Problem 8:** Answer: Option b: $a^2+b^2+c^2$.

**Problem 9:** Answer: Option b: $b^2=4ac$.

**Problem 10:** Answer: Option b: $10\sqrt{13}$.

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