## 17th SSC CGL Tier II level Question Set, 6th on Algebra

This is the 17th question set of 10 practice problem exercise for SSC CGL Tier II exam and the 6th on topic Algebra. To solve such problems quickly, identification of inherent patterns and use of associated methods are necessary. The corresponding solution set encapsulates the approach.

It is recommended that you take the test first and then only refer to its solution set. Taking the test will enable you to appreciate the pattern based solution process better. The answers and link of the corresponding solution set are available at the end.

### 17th question set - 10 problems for SSC CGL Tier II exam: 6th on topic Algebra - answering time 15 mins

**Q1. **If $\displaystyle\frac{x+y}{z}=2$, then the value of $\displaystyle\frac{y}{y-z}+\displaystyle\frac{x}{x-z}$ is,

- 1
- 2
- 0
- -1

**Q2.** If $3\sqrt{\displaystyle\frac{1-a}{a}}+9=19-3\sqrt{\displaystyle\frac{a}{1-a}}$, the values of $a$ are,

- $\displaystyle\frac{1}{5}$, $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{2}{5}$, $\displaystyle\frac{3}{5}$
- $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$
- $\displaystyle\frac{3}{10}$, $\displaystyle\frac{7}{10}$

**Q3. **If $x+y+z=0$, then the value of $\displaystyle\frac{3y^2+x^2+z^2}{2y^2-xz}$ is,

- $2$
- $1$
- $\displaystyle\frac{5}{3}$
- $\displaystyle\frac{3}{2}$

**Q4. **If $3x+4y-2z+9=17$, $7x+2y+11z+8=23$ and $5x+9y+6z-4=18$ then the value of $x+y+z-34$ is,

- $-28$
- $-31$
- $-14$
- $-45$

**Q5. **If $x$, $y$, and $z$ are three factors of $a^3-7a-6$, then the value of $x+y+z$ is,

- 3a
- a
- 3
- 6

**Q6.** If $3x+5y+7z=49$ and $9x+8y+21z=126$, then what is the value of $y$?

- 3
- 5
- 2
- 4

** Q7.** $x$, $y$ and $z$ are real numbers. If $x^3+y^3+z^3=13$, $x+y+z=1$ and $xyz=1$, then the value of $xy+yz+zx$ is,

- $1$
- $-1$
- $3$
- $-3$

** Q8.** Three numbers are in arithmetic progression. The sum of the numbers is 30 and the product is 910. The greatest of the three numbers is,

- 17
- 10
- 13
- 15

**Q9.** The value of $x$ which satisfies the equation, $\displaystyle\frac{x+a^2+2c^2}{b+c}+\displaystyle\frac{x+b^2+2a^2}{c+a}+\displaystyle\frac{x+c^2+2b^2}{a+b}=0$ is,

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

** Q10.** If $ax+by+cz=20$, $a^2+b^2+c^2=16$ and $x^2+y^2+z^2=25$, then the value of $\displaystyle\frac{a+b+c}{x+y+z}$ is,

- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{5}$
- $\displaystyle\frac{5}{4}$
- $\displaystyle\frac{5}{3}$

To know how to solve the problems quickly in a few steps, refer to the corresponding solution set,

**SSC CGL Tier II Solution set 17, Algebra 6.**

Or watch **two-part video solutions** below.

**Part1: Q1 to Q5**

**Part 2: Q6 to Q10**

### Answers to questions

**Problem 1.** **Answer:** Option b: 2.

**Problem 2.** **Answer:** Option c : $\displaystyle\frac{1}{10}$, $\displaystyle\frac{9}{10}$.

**Problem 3.** **Answer:** Option a: $2$.

**Problem 4.** **Answer:** Option b: $-31$.

**Problem 5.** **Answer:** Option a: 3a.

**Problem 6.** **Answer:** Option a : 3.

**Problem 7.** **Answer:** Option d: $-3$.

**Problem 8.** **Answer:** Option c: 13.

**Problem 9.** **Answer:** Option d: $-(a^2+b^2+c^2)$.

**Problem 10.** **Answer:** Option a: $\displaystyle\frac{4}{5}$.

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