## 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.

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 these 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.

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

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### 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**

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