## 8th SSC CHSL Solved Question Set, 2nd on Surds and indices

This is the 8th solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd on topic Surds and Indices. It contains,

**Question set on Surds and Indices**for SSC CHSL 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.

**IMPORTANT:** To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving any surd problem 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.

### 8th Question set - 10 problems for SSC CHSL exam: 2nd on topic Surds and Indices - answering time 15 mins

**Q1. **Simplified value of $\sqrt{20}+\sqrt{12}+\sqrt[3]{729}-\displaystyle\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{81}$ is,

- $0$
- $\sqrt{2}$
- $\sqrt{3}$
- $2\sqrt{2}$

**Q2.** Let $a=\displaystyle\frac{1}{2-\sqrt{3}}+\displaystyle\frac{1}{3-\sqrt{8}}+\displaystyle\frac{1}{4-\sqrt{15}}$. Then which one of the following is correct?

- $a \gt 18$
- $a=18$
- $a \lt 18$ but $a \neq 9$
- $a=9$

**Q3. **The value of $\sqrt{40+\sqrt{9\sqrt{81}}}$ is,

- $7$
- $\sqrt{111}$
- $9$
- $11$

**Q4. **If $x=\sqrt{3}+\displaystyle\frac{1}{\sqrt{3}}$, then the value of $\left(x-\displaystyle\frac{\sqrt{126}}{\sqrt{42}}\right)\left(x-\displaystyle\frac{1}{x-\displaystyle\frac{2\sqrt{3}}{3}}\right)$ is,

- $5\displaystyle\frac{\sqrt{3}}{6}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{5}{6}$
- $2\displaystyle\frac{\sqrt{3}}{3}$

**Q5.** If $9\sqrt{x}=\sqrt{12}+\sqrt{147}$, then value of $x$ is,

- $5$
- $3$
- $4$
- $2$

**Q6.** The number which when multiplied with $(\sqrt{3}+\sqrt{2})$ gives $(\sqrt{12}+\sqrt{18})$ is,

- $2\sqrt{3} -3\sqrt{2}$
- $3\sqrt{2} -2\sqrt{3}$
- $\sqrt{6}$
- $3\sqrt{2} +2\sqrt{3}$

**Q7.** If $a=7-4\sqrt{3}$, then the value of $a^{\frac{1}{2}}+a^{-\frac{1}{2}}$ is,

- $4$
- $3\sqrt{3}$
- $7$
- $2\sqrt{3}$

**Q8.** If $0.42\times{100^k}=42$, then the value of $k$ is,

- $4$
- $1$
- $2$
- $3$

**Q9.** If $2^x=3^y=6^{-z}$, then $\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}\right)$ is equal to,

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

**Q10.** The smallest among the numbers, $2^{250}$, $3^{150}$, $5^{100}$, $4^{200}$ is,

- $2^{250}$
- $3^{150}$
- $5^{100}$
- $4^{200}$

### Answers to the questions

**Q1. Answer:** Option a: $0$.

**Q2. Answer:** Option c: $a \lt 18$ but $a \neq 9$.

**Q3. Answer:** Option a: $7$.

**Q4. Answer:** Option c: $\displaystyle\frac{5}{6}$.

**Q5. Answer:** Option b: $3$.

**Q6. Answer:** Option c : $\sqrt{6}$.

**Q7. Answer:** Option a: $4$.

**Q8. Answer:** Option b: $1$.

**Q9. Answer:** Option b: $0$.

**Q10. Answer:** Option c: $5^{100}$.

### 8th solution set - 10 problems for SSC CHSL exam: 2nd on topic Surds and Indices - answering time 15 mins

**Q1. **Simplified value of $\sqrt{20}+\sqrt{12}+\sqrt[3]{729}-\displaystyle\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{81}$ is,

- $0$
- $\sqrt{2}$
- $\sqrt{3}$
- $2\sqrt{2}$

** Solution 1: Problem analysis and solution by Surd term factoring, Surd rationalization and cube root concept**

Taking factor 4 out of first and the second term, the terms are simplified to,

$\sqrt{20}=2\sqrt{5}$, and,

$\sqrt{12}=2\sqrt{3}$.

Taking the cube root, the third term simplifies to,

$\sqrt[3]{729}=9$, and taking square root, the 5th term simplifies to,

$-\sqrt{81}=-9$.

To simplify the 4th term, eliminate the surd denominator expression by surd rationalization.

Multiply denominator and numerator both by $(\sqrt{5}+\sqrt{3})$. Result is,

$-\displaystyle\frac{4}{\sqrt{5}-\sqrt{3}}\times{\displaystyle\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}}$

$=-\displaystyle\frac{4(\sqrt{5}+\sqrt{3})}{2}$

$=-2\sqrt{5}-2\sqrt{3}$.

Adding up the terms now the result becomes,

$2\sqrt{5}+2\sqrt{3}+9-2\sqrt{5}-2\sqrt{3}-9$

$=0$.

**Answer:** Option a: $0$.

**Key concepts used: Surd term factoring -- Square root of common numbers -- Cube root of common numbers -- Surd rationalization **--

*.*

**Solving in mind****Q2.** Let $a=\displaystyle\frac{1}{2-\sqrt{3}}+\displaystyle\frac{1}{3-\sqrt{8}}+\displaystyle\frac{1}{4-\sqrt{15}}$. Then which one of the following is correct?

- $a \gt 18$
- $a=18$
- $a \lt 18$ but $a \neq 9$
- $a=9$

**Solution 2: Problem solving by Surd rationalization, Key pattern identification, Inequality analysis and Mathematical reasoning**

To evaluate $a$, you have to rationalize the three component surd expressions by multiplying and dividing the first, second and third terms by $(2+\sqrt{3})$, $(3+\sqrt{8})$ and $(4+\sqrt{15})$ respectively.

Each of the denominators results in 1 and the sum is simplified to,

$a=(2+\sqrt{3})+(3+\sqrt{8})+(4+\sqrt{15})$

In each of these three surd expressions, **square of first term exceeds the square of second term by 1.**

So * each of the first term must be greater than the corresponding second term*. This is the

**key pattern identification.**As a result, * the sum of the three first terms is larger than the sum of the three second terms*,

$(2+3+4=9) \gt (\sqrt{3}+\sqrt{8}+\sqrt{15})$.

Adding $(2+3+4=9)$ to both sides of the inequality, the LHS becomes 18 and the RHS becomes $a=9+(\sqrt{3}+\sqrt{8}+\sqrt{15})$ without disturbing the inequality.

So, sum of the three surd expressions,

$a \lt 18$.

But as $a=9+(\sqrt{3}+\sqrt{8}+\sqrt{15})$,

$a \gt 9$, that is, $a \neq 9$.

Combining the two inequalities you get the final result as,

$a \lt 18$ but $a \neq 9$.

**Answer:** Option c: $a \lt 18$ but $a \neq 9$.

**Key concepts used:** **Surd rationalization -- Key pattern identification -- Inequality analysis -- Inequality algebra -- Mathematical reasoning -- Solving in mind****.**

Using the key pattern, the problem is simple to solve quickly in mind.

**Q3. **The value of $\sqrt{40+\sqrt{9\sqrt{81}}}$ is,

- $7$
- $\sqrt{111}$
- $9$
- $11$

**Solution 3: Problem solution by square root in series**

Last of the square root $\sqrt{81}$ gives 9. Product of 9 and 9 is again 81. Square root 9 is added to 40 this time to result in 49. Final result is 7.

Main concept is, **you have to start with the rightmost or innermost square root first.**

Let us show you,

$\sqrt{40+\sqrt{9\sqrt{81}}}$

$=\sqrt{40+\sqrt{9\times{9}}}$

$=\sqrt{40+\sqrt{81}}$

$=\sqrt{40+9}=\sqrt{49}=7$.

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

**Key concepts used: Rule for evaluating Square roots in series, the last or innermost to be evaluated first -- Solving in mind.**

**Q4. **If $x=\sqrt{3}+\displaystyle\frac{1}{\sqrt{3}}$, then the value of $\left(x-\displaystyle\frac{\sqrt{126}}{\sqrt{42}}\right)\left(x-\displaystyle\frac{1}{x-\displaystyle\frac{2\sqrt{3}}{3}}\right)$ is,

- $5\displaystyle\frac{\sqrt{3}}{6}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{5}{6}$
- $2\displaystyle\frac{\sqrt{3}}{3}$

**Solution 4: Problem solving using Problem breakdown, Surd term factoring and Surd arithmetic**

Let's solve this problem mentally showing you the steps for clarity.

**Problem breakdown technique** is applied,

Breakdown the larger problem into smaller problems and take up simplying the components that can be simplified directly to reduce the overall complexity of the expression.

**First factor of target expression** simplifies to,

$\left(x-\displaystyle\frac{\sqrt{126}}{\sqrt{42}}\right)$

$=\left(x-\sqrt{3}\right)$

$=\displaystyle\frac{1}{\sqrt{3}}$, from given expression.

The **denominator of the second term of the second factor** simplifies to,

$\left(x-\displaystyle\frac{2\sqrt{3}}{3}\right)$

$=\left(x-\displaystyle\frac{2}{\sqrt{3}}\right)$

$=\sqrt{3}-\displaystyle\frac{1}{\sqrt{3}}$, from given expression.

And the **second factor** becomes,

$x-\displaystyle\frac{1}{\sqrt{3}-\displaystyle\frac{1}{\sqrt{3}}}$.

Multiply $\displaystyle\frac{1}{\sqrt{3}}$, the simplified result of first factor, through the second factor. Result is,

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

Now evaluate $\displaystyle\frac{x}{\sqrt{3}}$ from given expression.

Multiply given expression by $\displaystyle\frac{1}{\sqrt{3}}$, result is,

$\displaystyle\frac{x}{\sqrt{3}}=1+\displaystyle\frac{1}{3}=\displaystyle\frac{4}{3}$.

Target expression is simplified finally to,

$\displaystyle\frac{4}{3}-\displaystyle\frac{1}{2}=\displaystyle\frac{5}{6}$.

*We have broken down the simplification problem into smaller problems and simplified each problem fragment using given expression, surd term factoring and surd arithmetic.*

**Answer:** Option c: $\displaystyle\frac{5}{6}$.

**Key concepts used: Problem breakdown -- Surd arithmetic -- Surd term factoring -- Proper use of given information in suitable form multiple times -- Solving in mind.**

**Q5.** If $9\sqrt{x}=\sqrt{12}+\sqrt{147}$, then value of $x$ is,

- $5$
- $3$
- $4$
- $2$

**Solution 5: Problem analysis and solving using surd term factoring and key pattern identification**

Factor 4 and 9 out the two terms in RHS to simplify the expression,

$9\sqrt{x}=\sqrt{12}+\sqrt{147}=2\sqrt{3}+7\sqrt{3}=9\sqrt{3}$.

So, $\sqrt{x}=\sqrt{3}$, and

$x=3$.

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

*Key concepts used:* **Surd term factoring -- Key pattern identification -- Surd arithmetic in adding coefficients of $\sqrt{3}$ -- Solving in mind****.**

**Q6.** The number which when multiplied with $(\sqrt{3}+\sqrt{2})$ gives $(\sqrt{12}+\sqrt{18})$ is,

- $2\sqrt{3} -3\sqrt{2}$
- $3\sqrt{2} -2\sqrt{3}$
- $\sqrt{6}$
- $3\sqrt{2} +2\sqrt{3}$

**Solution 6: Problem analysis and solution by mathematical reasoning, goal formation, key pattern identification and surd term factoring**

Comparing the first surd expression with the second surd expression, **try to form the first expression by taking a surd factor out of both the terms** $\sqrt{12}$ and $\sqrt{18}$ of second surd expression.

In other words, can we form $(\sqrt{3}+\sqrt{2})$ from $(\sqrt{12}+\sqrt{18})$ by taking a common factor out of both $\sqrt{12}$ and $\sqrt{18}$? If we can, this common factor taken out would be our answer.

With properly analyzed goal, solution comes easily. You should be able to identify quickly that taking out $\sqrt{6}$ from the first and second surd terms of the second given expression produces the first expression. This is a good example of **surd term factoring technique.**

Let's show you,

$\sqrt{12}+\sqrt{18}=\sqrt{6}\sqrt{2}+\sqrt{6}\sqrt{3}=\sqrt{6}(\sqrt{3}+\sqrt{2})$.

You get the answer in a single step that too mentally.

**Answer:** Option c : $\sqrt{6}$.

**Key concepts used:** *Mathematical reasoning -- Goal formation -- Key pattern identification -- Surd term factoring -- Solving in mind.*

**Q7.** If $a=7-4\sqrt{3}$, then the value of $a^{\frac{1}{2}}+a^{-\frac{1}{2}}$ is,

- $4$
- $3\sqrt{3}$
- $7$
- $2\sqrt{3}$

**Solution 7: Problem analysis and Solving by goal formation, double square root surd simplification and surd rationalization**

From the target expression form the clear goal that you have to express $a$ as a square of a two term surd expression so that $a^{frac{1}{2}}=\sqrt{a}$ becomes a normal 1st level surd expression with only one square root in the terms that you can deal with.

This is a problem of **double square root surd simplification** as *taking square root of $a$ that already includes a square root would result in one square root under a second square root.*

Let's show you how to convert expression of $a$ to a square expression.

$a=7-4\sqrt{3}$

$=2^2-2.2.\sqrt{3}+(\sqrt{3})^2$, in the form of $a^2-2.a.b +b^2=(a-b)^2$

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

So, $a^{\frac{1}{2}}=2-\sqrt{3}$.

And the target expression,

$a^{\frac{1}{2}}+a^{-\frac{1}{2}}$

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

$=(2-\sqrt{3})+(2+\sqrt{3})$, rationalizing the second term denominator by multiplying and dividing the term by $(2+\sqrt{3})$

$=4$.

**Answer:** Option a: 4.

** Key concepts used:** *Problem analysis -- Goal formation -- Double square root surd simplification -- Surd rationalization** -- *Solving in mind.

If you are comfortable with double square root surd simplification and rationalization, you can solve this problem easily in mind.

**Q8.** If $0.42\times{100^k}=42$, then the value of $k$ is,

- $4$
- $1$
- $2$
- $3$

** Solution 8: Problem analysis and solving by base equalization of term bases that would result in equal powers of two sides of equation**

$0.42=42\times{10^{-2}}=42\times{100^{-1}}$ so that the give equation is simplified to,

$0.42\times{100^k}=42$

Or, $42\times{100^{-1}}\times{100^k}=42\times{100^{k-1}}=42$, as $a^x\times{a^y}=a^{x+y}$

Cancelling out 42 from both sides of the equation,

$100^{k-1}=1=100^0$, as $a^0=1$, where $a$ is any real number.

As bases 100 on both sides of the equation are equal, the powers of the two terms are also equal,

$k-1=0$, if $a^x=a^y$, then $x=y$,

$k=1$.

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

**Key concepts used:** **Conversion of decimals to product of integer and negative power of 10, and then of 100 -- Indices problem -- Base equalization technique -- Product of two terms with equal bases adds up the powers -- $1=100^0$ -- If bases of two terms in RHS and LHS of an equation are same powers must be equal -- Solving in mind.**

**Q9.** If $2^x=3^y=6^{-z}$, then $\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}\right)$ is equal to,

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

**Solution 9: Problem analysis and Solution by End state analysis, key pattern identification, freeing up the bases by equating Chained equation to a dummy variable**

Let's solve the problem mentally while showing the steps for clarity.

Identify the key pattern that $2\times{3}=6$.

So free up each of 2, 3 and 6 of their powers by converting the powers to inverted powers of a dummy variable equated to the chained equation,

$2^x=3^y=6^{-z}=p$

So, $2^x=p$,

Or, $2=p^{\displaystyle\frac{1}{x}}$,

$3^y=p$,

Or, $3=p^{\displaystyle\frac{1}{y}}$, and,

$6^{-z}=p$,

Or, $6=p^{-\displaystyle\frac{1}{z}}$.

Multiply the first two equations (LHS with LHS and RHS with RHS and maintain the equality of the products),

$2\times{3}=p^{\frac{1}{x}+\frac{1}{y}}$

Or, $6=p^{-\frac{1}{z}}=p^{\frac{1}{x}+\frac{1}{y}}$.

As bases on both sides of the equation, dummy variable $p$, is same, the powers must be equal,

$\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}=-\displaystyle\frac{1}{z}$

Or, $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$.

The inverted powers in the target expression gives a clear indication of the way to the solution. This is **End state analysis approach.**

**Answer:** Option b: $0$.

**Key concepts used:** * End state analysis -- Key pattern identification -- Freeing up the bases 2, 3 and 6 of their powers by equating the chained equation to a dummy variable -- Equal bases on two sides of an equation must have powers equal *--

**Solving in mind.****Q10.** The smallest among the numbers, $2^{250}$, $3^{150}$, $5^{100}$, $4^{200}$ is,

- $2^{250}$
- $3^{150}$
- $5^{100}$
- $4^{200}$

**Solution 10: Problem analysis and Solution by base equalizaion with power as base, HCF of powers as target equal power and comparison of numbers with equal powers**

To compare the four given numbers with large powers, the **powers must be made equal** by transforming the powers suitably. Out of these four transformed numbers with equal powers, the number with smallest base must be the smallest number.

Question is—**what should be the right value of power to which powers of all the terms are to be transformed?**

The easiest option is, * the HCF of the four powers, 250, 150, 100 and 200.* The HCF of these four numbers is 50 and the terms are changed to,

$2^{250}=(2^5)^{50}=32^{50}$, 5 times 50 is 250,

$3^{150}=(3^3)^{50}=27^{50}$, 3 times 50 is 150

$5^{100}=(5^2)^{50}=25^{50}$, 2 times 50 is 100,

$4^{200}=(4^4)^{50}=256^{50}$, 4 times 50 is 200.

So the third given number, $5^{100}$ is the smallest among the four as its base 25 is the smallest among 32, 27, 25 and 256 and all powers are same.

Here we have used **two indices rules**:

- $a^{xy}=(a^x)^y$, and
- If $a^x \lt b^x$, then $a \lt b$.

To apply the second rule, the powers must be equalized first. This is also * base equalization* but in this case the

**base in abstraction**is the power that is equalized.

**Answer: **Option c: $5^{100}$.

**Key concepts used:** **Base equalization technique applied to equalization of powers -- With equal powers, the number with smaller actual base would be the smaller one -- Target equal power is the HCF of the powers that will create the smallest term bases -- Indices principle that powers are multiplied if a base in power is again raised to a second power -- Solving in mind.**

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