## 51st SSC CGL level question Set, 12th on Algebra

This is the 51st question set of 10 practice problem exercise for SSC CGL exam and 12th on topic Algebra.

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

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

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

### 51st question set - 10 problems for SSC CGL exam: 12th on topic Algebra - answering time 15 mins

**Problem 1.**

If $x^2=y+z$, $y^2=z+x$ and $z^2=x+y$, the value of $\displaystyle\frac{1}{x+1}+\displaystyle\frac{1}{y+1}+\displaystyle\frac{1}{z+1}$ is,

- $1$
- $4$
- $-1$
- $-4$

**Problem 2.**

If $a^2+b^2+c^2+3=2(a+b+c)$ then the value of $(a+b+c)$ is,

- 2
- 3
- 5
- 4

**Problem 3.**

If $a^2-4a-1=0$, then the value of $a^2+\displaystyle\frac{1}{a^2}+3a-\displaystyle\frac{3}{a}$ is,

- 40
- 35
- 30
- 25

**Problem 4.**

If $x+\displaystyle\frac{1}{x}=99$, find the value of $\displaystyle\frac{100x}{2x^2+102x+2}$.

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

**Problem 5.**

If $\sqrt{1+\displaystyle\frac{x}{961}}=\displaystyle\frac{32}{31}$, then the value of $x$ is,

- 63
- 64
- 61
- 65

**Problem 6.**

If $1.5a=0.04b$ then the value of $\displaystyle\frac{b-a}{b+a}$ will be equal to,

- $\displaystyle\frac{73}{77}$
- $\displaystyle\frac{75}{2}$
- $\displaystyle\frac{2}{75}$
- $\displaystyle\frac{77}{33}$

**Problem 7.**

The value of the expression, $\displaystyle\frac{(a-b)^2}{(b-c)(c-a)}+\displaystyle\frac{(b-c)^2}{(a-b)(c-a)}+\displaystyle\frac{(c-a)^2}{(a-b)(b-c)}$ is,

- $2$
- $3$
- $0$
- $\displaystyle\frac{1}{3}$

**Problem 8.**

If $9\sqrt{x}=\sqrt{12}+\sqrt{147}$, then $x$ is,

- 5
- 2
- 3
- 4

**Problem 9.**

If $p:q=r:s=t:u=2:3$ then $(mp+nr+ot):(mq+ns+ou)$ is equal to,

- 2 : 3
- 3 : 2
- 2 : 1
- 1 : 2

**Problem 10.**

If $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}+\displaystyle\frac{1}{c+1}=2$ then $a^2+b^2+c^2$ is,

- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{27}{16}$
- $\displaystyle\frac{4}{3}$

### Answers to the questions

Problem 1. **Answer:** Option a: $1$.

Problem 2. **Answer:** Option b : 3.

Problem 3. **Answer:** Option c: 30.

Problem 4. **Answer:** Option b: $\displaystyle\frac{1}{3}$.

Problem 5. **Answer:** Option a: 63.

Problem 6. **Answer:** Option a : $\displaystyle\frac{73}{77}$.

Problem 7. **Answer:** Option b: $3$.

Problem 8. **Answer:** Option c: 3.

Problem 9. **Answer:** Option a: 2 : 3.

problem 10. **Answer: **Option a: $\displaystyle\frac{3}{4}$.

### Solutions to the problems

For detailed conceptual solutions with answers you should refer to the companion **S*** SC CGL level Solution Set 51 on Algebra 12* where you will get detailed explanations on easiest path to the solutions.

Watch **quick solutions on two-part video**.

**Part I: Q1 to Q5**

**Part II: Q6 to Q10**

### Guided help on Algebra in Suresolv

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