## 92nd SSC CGL Solved question set, 9th on Surds and indices

This is the 92nd set of 10 solved questions for SSC CGL and 9th on Surds and indices. All problems are solved by specially quick methods based on discovered patterns. The solved question set contains,

**Question set on Surds and indices for SSC CGL**to be answered in 15 minutes (10 chosen questions)**Answers**to the questions, and,**Detailed conceptual solutions**to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be read.

The answers and detailed solutions follow the questions. First try to solve the questions yourself and then go through the solutions. All problems are solved by specially quick methods based on discovered patterns.

**IMPORTANT:** To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Surds and indices quickly, 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.

### 92nd Question set - 10 problems for SSC CGL exam: 9th on topic Surds and indices - answering time 15 mins

**Q1.** If $x+\left(\displaystyle\frac{1}{x}\right)=\sqrt{13}$, then what is the value of $x^5-\left(\displaystyle\frac{1}{x^5}\right)$?

- $169$
- $507$
- $169\sqrt{3}$
- $393$

**Q2.** If $x=\displaystyle\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$, then what is the value of $\displaystyle\frac{x+2\sqrt{a}}{x-2\sqrt{a}}+\displaystyle\frac{x+2\sqrt{b}}{x-2\sqrt{b}}$?

- $0$
- $4$
- $2$
- $\displaystyle\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}$

**Q3.** What is the value of $x$ in the equation, $\sqrt{\displaystyle\frac{1+x}{x}}-\sqrt{\displaystyle\frac{x}{1+x}}=\displaystyle\frac{1}{\sqrt{6}}$?

- $3$
- $2$
- $-2$
- None of these

**Q4.** If $a=\displaystyle\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $b=\displaystyle\frac{\sqrt{5}-1}{\sqrt{5}+1}$, the value of $\displaystyle\frac{a^2+ab+b^2}{a^2-ab+b^2}$ is,

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

**Q5.** The value of $\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$ is,

- $\pm 2$
- $2$
- $-2$
- $4$

**Q6.** If $x=1+\sqrt{2}+\sqrt{3}$, then find the value of $x^2-2x+4$.

- $2(7+\sqrt{6})$
- $4+\sqrt{6}$
- $2(4+\sqrt{6})$
- $2(3+\sqrt{7})$

**Q7.** If $4x=\sqrt{5}+2$, then the value of $\left(x-\displaystyle\frac{1}{16x}\right)$ is,

- $-1$
- $1$
- $2\sqrt{5}$
- $4$

**Q8.** If $x=\sqrt{2}+1$, then the value of $x^4-\displaystyle\frac{1}{x^4}$ is,

- $8\sqrt{2}$
- $18\sqrt{2}$
- $6\sqrt{2}$
- $24\sqrt{2}$

**Q9.** If $a=\sqrt{2}+1$ and $b=\sqrt{2}-1$, then the value of $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}$ is,

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

**Q10.** If $a=\displaystyle\frac{\sqrt{3}}{2}$, then the value of $\sqrt{1+a}+\sqrt{1-a}$ is,

- $\displaystyle\frac{\sqrt{3}}{2}$
- $\sqrt{3}$
- $2-\sqrt{3}$
- $2+\sqrt{3}$

### Answers to the questions

**Q1. Answer:** Option d: $393$.

**Q2. Answer:** Option c: $2$.

**Q3. Answer:** Option b: $2$.

**Q4. Answer:** Option d: $\displaystyle\frac{4}{3}$.

**Q5. Answer:** Option a: $\pm 2$.

**Q6. Answer:** Option c: $2(4+\sqrt{6})$.

**Q7. Answer:** Option b: $1$.

**Q8. Answer:** Option d: $24\sqrt{2}$.

**Q9. Answer:** Option a: $1$.

**Q10. Answer:** Option b: $\sqrt{3}$.

### 92nd solution set - 10 problems for SSC CGL exam: 9th on topic Surds and indices - answering time 15 mins

**Q1.** If $x+\left(\displaystyle\frac{1}{x}\right)=\sqrt{13}$, then what is the value of $x^5-\left(\displaystyle\frac{1}{x^5}\right)$?

- $169$
- $507$
- $169\sqrt{3}$
- $393$

#### Solution 1: Quick solution by target expression generation multiplying $x^2+\left(\displaystyle\frac{1}{x^2}\right)$ and $x^3-\left(\displaystyle\frac{1}{x^3}\right)$ and interaction of inverses property

Identify that the product of following two factors will generate the target expression,

$\left(x^2+\displaystyle\frac{1}{x^2}\right)\left(x^3-\displaystyle\frac{1}{x^3}\right)=\left(x^5-\displaystyle\frac{1}{x^5}\right)+\left(x-\displaystyle\frac{1}{x}\right)$.

Let us first evaluate $\left(x-\displaystyle\frac{1}{x}\right)$.

Raise the given equation to power 2 and subtract 4 from both sides,

$x^2-2+\displaystyle\frac{1}{x^2}=\left(x-\displaystyle\frac{1}{x}\right)^2=9$,

Or, $\left(x-\displaystyle\frac{1}{x}\right)=3$, by the positive choice values, $\left(x-\displaystyle\frac{1}{x}\right)$ must be positive.

From the square of the given expression,

$\left(x^2+\displaystyle\frac{1}{x^2}\right)=13-2=11$.

So,

$\left(x^3-\displaystyle\frac{1}{x^3}\right)=\left(x-\displaystyle\frac{1}{x}\right)\left(x^2+1+\displaystyle\frac{1}{x^2}\right)=36$.

Substituting,

$11\times{36}=\left(x^5-\displaystyle\frac{1}{x^5}\right)+3$,

Or, $\left(x^5-\displaystyle\frac{1}{x^5}\right)=396-3=393$.

**Answer:** Option d: $393$.

**Key concepts used:** **Target expression generation -- Interaction of inverses.**

**Note:** If $\left(x-\displaystyle\frac{1}{x}\right)=-3$,

$\left(x^3-\displaystyle\frac{1}{x^3}\right)=-36$ and target expression value would have been negative.

As a safe rule, take the usual positive value of a square root.

**Q2.** If $x=\displaystyle\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$, then what is the value of $\displaystyle\frac{x+2\sqrt{a}}{x-2\sqrt{a}}+\displaystyle\frac{x+2\sqrt{b}}{x-2\sqrt{b}}$?

- $0$
- $4$
- $2$
- $\displaystyle\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}$

#### Solution 2: Quick solution by variable reduction technique and denominator equalization

Identifying the similarity in the form of the two target expression terms, it is decided to divide the numerator and denominator of the first term by $2\sqrt{a}$ and of the second term by $2\sqrt{b}$,

$E=\displaystyle\frac{x+2\sqrt{a}}{x-2\sqrt{a}}+\displaystyle\frac{x+2\sqrt{b}}{x-2\sqrt{b}}$

$=\displaystyle\frac{\displaystyle\frac{x}{2\sqrt{a}}+1}{\displaystyle\frac{x}{2\sqrt{a}}-1}+\displaystyle\frac{\displaystyle\frac{x}{2\sqrt{b}}+1}{\displaystyle\frac{x}{2\sqrt{b}}-1}$.

From given equation, form of the values of the compound variables that are very similar with **number of variables in both numerators reduced by 1**,

$\displaystyle\frac{x}{2\sqrt{a}}=\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}$, and,

$\displaystyle\frac{x}{2\sqrt{b}}=\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}$.

The denominator $(\sqrt{a}+\sqrt{b})$ cancels out on substitution,

$E=\displaystyle\frac{3\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}+\displaystyle\frac{\sqrt{b}+3\sqrt{a}}{\sqrt{a}-\sqrt{b}}$

$=\displaystyle\frac{2(\sqrt{b}-\sqrt{a})}{\sqrt{b}-\sqrt{a}}=2$

**Answer:** Option c: 2.

**Key concepts used:** **Denominator equalization by variable reduction in numerator of substituted values to $\sqrt{a}$ and $\sqrt{b}$ instead of $\sqrt{ab}$ -- Denominator equalization -- Key pattern identification -- Solving in mind.**

**Note:** By transforming $x$ to $\displaystyle\frac{x}{2\sqrt{a}}$ and $\displaystyle\frac{x}{2\sqrt{b}}$ in the target, the numerators of substituted values reduced in number of variables by 1 making it suitable for simplification and finally denominator equalization.

This key pattern is identified by analyzing the given value of $x$ along with the form of the target expression.

**Q3.** What is the value of $x$ in the equation, $\sqrt{\displaystyle\frac{1+x}{x}}-\sqrt{\displaystyle\frac{x}{1+x}}=\displaystyle\frac{1}{\sqrt{6}}$?

- $3$
- $2$
- $-2$
- None of these

#### Solution 3: Quick solution by direct combining of two terms in LHS of given equation and solving a quadratic equation

Directly combine the two terms in LHS of given equation,

$\sqrt{\displaystyle\frac{1+x}{x}}-\sqrt{\displaystyle\frac{x}{1+x}}=\displaystyle\frac{1}{\sqrt{6}}$,

Or, $\displaystyle\frac{(1+x)-x}{\sqrt{x(1+x)}}=\frac{1}{\sqrt{x(1+x)}}=\frac{1}{\sqrt{6}}$.

Raise to square and simplify to get the quadratic equation,

$x^2+x-6=0$,

Or, $(x+3)(x-2)=0$.

Though the value $x=-3$ satisfies the quadratic equation, it does not satisfy the given equation that has been squared up to get the quadratic equation.

The other root $x=2$ satisfies both the quadratic equation and the given equation.

**Answer:** Option b: $2$.

**Key concepts used:** **Numerator simplification -- Solving a quadratic equation -- Solving in mind.**

**Q4.** If $a=\displaystyle\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $b=\displaystyle\frac{\sqrt{5}-1}{\sqrt{5}+1}$, the value of $\displaystyle\frac{a^2+ab+b^2}{a^2-ab+b^2}$ is,

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

#### Solution 4: Quick solution by Key pattern identification that $ab=1$ and transforming numerator and denominator of target to $(a+b)^2$ and $(a-b)^2$

Identify the key pattern that, by the given values of $a$ and $b$, their product $ab=1$. If you add or subtract $ab$ in an expression, effectively, you are adding or subtracting $1$.

So, add and subtract $1=ab$ in numerator and denominator of target expression to convert both in terms of very convenient square of sum expressions,

$E=\displaystyle\frac{a^2+ab+b^2}{a^2-ab+b^2}$

$=\displaystyle\frac{(a+b)^2-1}{(a-b)^2+1}$.

From given values of $a$ and $b$,

$a+b=\displaystyle\frac{\sqrt{5}+1}{\sqrt{5}-1}+\displaystyle\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$=\displaystyle\frac{(\sqrt{5}+1)^2+(\sqrt{5}-1)^2}{5-1}$

$=\displaystyle\frac{12}{4}=3$.

$a-b=\displaystyle\frac{\sqrt{5}+1}{\sqrt{5}-1}-\displaystyle\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$=\displaystyle\frac{(\sqrt{5}+1)^2-(\sqrt{5}-1)^2}{5-1}$

$=\displaystyle\frac{4\sqrt{5}}{4}=\sqrt{5}$.

Substituting,

$E=\displaystyle\frac{9-1}{5+1}=\frac{4}{3}$.

**Answer:** Option d: $\displaystyle\frac{4}{3}$.

**Key concepts used:** **Key pattern identification -- Numerator and denominator simplification -- Solving in mind.**

The simplification steps are easy enough to carry out in mind after the identification of the key pattern of $ab=1$ and using the pattern to transform both numerator and denominator to the forms that were simple to evaluate.

**Q5.** The value of $\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$ is,

- $\pm 2$
- $2$
- $-2$
- $4$

#### Solution 5: Quick solution by double square root surd simplification of the surd expression $7+4\sqrt{3}$

To solve the problem, it is absolutely necessary to express the last expression under square root, $(7+4\sqrt{3})$ as a whole square to get rid of the double square root.

Splitting $7$ into two parts,

$7+4\sqrt{3}=(\sqrt{3})^2+2.2.\sqrt{3}+(2)^2=(\sqrt{3}+2)^2$.

This is the key to solving the problem.

The target expression is simplified now as,

$E=\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$

$=\sqrt{-\sqrt{3}+\sqrt{3+8(\sqrt{3}+2)}}$

$=\sqrt{-\sqrt{3}+\sqrt{19+8\sqrt{3}}}$.

Split again numeric term $19$ of double square root surd $(19+8\sqrt{3})$ into two parts for expressing it as a whole square,

$19+8\sqrt{3}=(4)^2+2.4.\sqrt{3}+(\sqrt{3})^2=(4+\sqrt{3})^2$.

The target expression is simplified once more to,

$E=\sqrt{-\sqrt{3}+4+\sqrt{3}}$

$=\sqrt{4}$

$=\pm 2$.

**Answer:** Option a: $\pm 2$.

**Key concepts used:** **Double square root surd simplification in two stages -- Key pattern identification.**

**Q6.** If $x=1+\sqrt{2}+\sqrt{3}$, then find the value of $x^2-2x+4$.

- $2(7+\sqrt{6})$
- $4+\sqrt{6}$
- $2(4+\sqrt{6})$
- $2(3+\sqrt{7})$

#### Solution 6: Quick solution by matching target with given expression, End state analysis

By taking 1 from RHS to LHS of the given equation and raising $(x-1)$ to its square, it easy to see that both the $x$ terms are obtained as in the target expression.

By comparing the given equation with the target expression, the given equation is so modified as to generate the target expression,

$x=1+\sqrt{2}+\sqrt{3}$,

Or, $(x-1)=\sqrt{2}+\sqrt{3}$,

Or, $(x-1)^2=5+2\sqrt{6}$,

Or, $x^2-2x+4=8+2\sqrt{6}=2(4+\sqrt{6})$.

By **target matching** the problem is solved easily in mind.

**Answer:** Option c: $2(4+\sqrt{6})$.

**Key concepts used:** **End state analysis -- Matching target expression with given equation to change the given equation suitably -- Solving in mind.**

**Q7.** If $4x=\sqrt{5}+2$, then the value of $\left(x-\displaystyle\frac{1}{16x}\right)$ is,

- $-1$
- $1$
- $2\sqrt{5}$
- $4$

#### Solution 7: Solve in only a few steps by matching target expression with given equation and changing the target expression suitably, Key pattern identification and End state analysis

By comparing the target expression with the given equation **with the objective of making the two as similar as possible**, the **key pattern identified** is,

Take out $\displaystyle\frac{1}{4}$ as a factor of the two terms of the target expression and convert both the $x$ terms to $4x$ instead of $x$ and $16x$, so that given value can be directly substituted in simplest form.

Target expression,

$E=\left(x-\displaystyle\frac{1}{16x}\right)$

$=\displaystyle\frac{1}{4}\left(4x-\displaystyle\frac{1}{4x}\right)$.

Now substitute $4x=\sqrt{5}+2$,

$E=\displaystyle\frac{1}{4}\left[(\sqrt{5}+2)-\displaystyle\frac{1}{(\sqrt{5}+2)}\right]$.

Rationalize the denominator of the second term by multiplying and dividing by $(\sqrt{5}-2)$,

$E=\displaystyle\frac{1}{4}\left[(\sqrt{5}+2)-(\sqrt{5}-2)\right]$

$=1$.

**Answer:** Option b: $1$.

**Key concepts used:** **End state analysis -- Comparing the given equation with the target expression, the target expression is modified to match the given equation -- Substitution -- Target matching -- Surd rationalization -- Solving in mind.**

**Q8.** If $x=\sqrt{2}+1$, then the value of $x^4-\displaystyle\frac{1}{x^4}$ is,

- $8\sqrt{2}$
- $18\sqrt{2}$
- $6\sqrt{2}$
- $24\sqrt{2}$

#### Solution 8: Quick solution by key pattern identification of simple values of $\left(x+\displaystyle\frac{1}{x}\right)$ and $\left(x-\displaystyle\frac{1}{x}\right)$

From the target expression it is clear that we need to get the values of $\left(x+\displaystyle\frac{1}{x}\right)$ and $\left(x-\displaystyle\frac{1}{x}\right)$.

From the given expression, $x=\sqrt{2}+1$,

$\left(x+\displaystyle\frac{1}{x}\right)$

$=\sqrt{2}+1+\displaystyle\frac{1}{\sqrt{2}+1}$

$=\sqrt{2}+1+\sqrt{2}-1$, rationalizing the second term

$=2\sqrt{2}$.

And,

$\left(x-\displaystyle\frac{1}{x}\right)$

$=\sqrt{2}+1-\displaystyle\frac{1}{\sqrt{2}+1}$

$=\sqrt{2}+1-\sqrt{2}+1$, rationalizing the second term

$=2$.

Multiplying the two,

$x^2-\displaystyle\frac{1}{x^2}=4\sqrt{2}$.

Raising the first sum of inverses to square and simplifying,

$x^2+\displaystyle\frac{1}{x^2}=8-2=6$.

Again take the product,

$x^4-\displaystyle\frac{1}{x^4}=24\sqrt{2}$.

**Answer:** Option d: $24\sqrt{2}$.

**Key concepts used:** **Key pattern identification -- Surd rationalization -- Solving in mind.**

**Q9.** If $a=\sqrt{2}+1$ and $b=\sqrt{2}-1$, then the value of $\displaystyle\frac{1}{a+1}+\displaystyle\frac{1}{b+1}$ is,

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

#### Solution 9: Quick solution by direct substitution and surd term factoring

From given values,

$a+1=\sqrt{2}+2=\sqrt{2}(\sqrt{2+1})$, this is use of **surd term factoring**, and,

$b+1=\sqrt{2}$.

The target expression,

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

$=\displaystyle\frac{1}{\sqrt{2}(\sqrt{2+1})}+\displaystyle\frac{1}{\sqrt{2}}$

$=\displaystyle\frac{1+\sqrt{2}+1}{2+\sqrt{2}}$

$=\displaystyle\frac{2+\sqrt{2}}{2+\sqrt{2}}$

$=1$.

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

**Key concepts used:** **Direct substitution -- Surd term factoring to make fraction addition easier -- Solving in mind.**

**Q10.** If $a=\displaystyle\frac{\sqrt{3}}{2}$, then the value of $\sqrt{1+a}+\sqrt{1-a}$ is,

- $\displaystyle\frac{\sqrt{3}}{2}$
- $\sqrt{3}$
- $2-\sqrt{3}$
- $2+\sqrt{3}$

#### Solution 10: Quick solution by substitution and second method of double square root surd simplification

Substitute the value of $a$ in the target expression,

$E=\sqrt{1+a}+\sqrt{1-a}$

$=\sqrt{1+\displaystyle\frac{\sqrt{3}}{2}}+\sqrt{1-\displaystyle\frac{\sqrt{3}}{2}}$.

Both the terms involve double square root surd simplification that is not so straightforward to express in the form of a square.

Let us first simplify the the expressions one step more,

$E=\sqrt{\displaystyle\frac{2+\sqrt{3}}{2}}+\sqrt{\displaystyle\frac{2-\sqrt{3}}{2}}$.

Essentially we need to express $(2+\sqrt{3})$ and $(2-\sqrt{3})$ as whole square expressions.

The problem lies in both the cases in the middle surd term having no factor of 2.

So introduce the factor 2 with the surd term by multiplying and dividing the expressions under square roots by 2,

$E=\sqrt{\displaystyle\frac{4+2\sqrt{3}}{4}}+\sqrt{\displaystyle\frac{4-2\sqrt{3}}{4}}$.

Now we can express both $(4+2\sqrt{3})$ and $(4-2\sqrt{3})$ as whole square expressions by splitting numeric term 4,

$4+2\sqrt{3}=(\sqrt{3})^2+2.\sqrt{3}.1+1=(\sqrt{3}+1)^2$, and,

$4-2\sqrt{3}=(\sqrt{3})^2-2.\sqrt{3}.1+1=(\sqrt{3}-1)^2$.

The target expression,

$E=\displaystyle\frac{\sqrt{3}+1+\sqrt{3}-1}{2}=\sqrt{3}$.

**Answer:** Option b: $\sqrt{3}$.

**Key concepts used:** **Double square root surd simplification by forced introduction of middle term coefficient of 2 -- Solving in mind.**

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

**Concept based pattern and method formation**, and,**Identification of the key pattern**and**use of the method**associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the *concept based pattern identification and use of quick problem solving method.*

**Note:****If you are an SSC CGL aspirant**, you may refer to our useful article * 7 steps for sure success in SSC CGL tier 1 and tier 2 tests* in which we have included all our student resources on SSC CGL topics as links.

**Watch out for more student resources on SSC CGL in this article.**

You may also like to go through the **Fraction and surd related article** directly by referring to the following valuable articles.

### Concept tutorials on Fractions, Surds, decimals and related topics

**Fractions and Surds concepts part 1**

**Base equalization technique is indispensable for solving fraction problems**

**How to solve surds part 1, Rationalization**

**How to solve surds part 2, Double square root surds and surd term factoring**

**How to solve surds part 3, Surd expression comparison and ranking**

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