## Surds and indices questions for SSC CHSL 2nd set, Answers and Quick Solutions

2nd set of surds and indices questions for SSC CHSL with answers and quick solutions by surds techniques. These are previous year SSC CHSL questions.

The set contains,

**Surds and Indices questions**for SSC CHSL to be answered in 15 minutes (10 chosen questions)**Answers**to the questions, and**Quick conceptual solutions**to the questions.

Take the timed test, verify answers and learn how to solve the questions quickly from solutions.

### 2nd set of 10 Surds and indices questions for SSC CHSL - 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 2nd set of 10 Surds and indices questions for SSC CHSL

**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: $5^{100}$.

### Solutions to the 2nd set of 10 Surds and indices questions for SSC CHSL - answering time was 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 the square roots of first and the second term, the terms are simplified to,

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

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

Third term simplified by taking cube root of value 729 under roots,

$\sqrt[3]{729}=9$,

And simplify the 5th term by taking square root of 81 under square root,

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

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

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 final result,

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

$=0$.

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

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

*.*

**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 Rationalization of surds, Key pattern identification, Inequality analysis and Mathematical reasoning**

To get a clear idea of the value of $a$, first thing to do is to **eliminate the denominators** in its RHS value.

And to eliminate the three denominators, apply** rationalization of surds** 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 turns to 1 and the given expression simplified to,

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

Or, $a=(2+3+4) +(\sqrt{3}+\sqrt{8}+\sqrt{15})=A+B$, where $A=2+3+4=9$, and $B=(\sqrt{3}+\sqrt{8}+\sqrt{15})$.

This is the time to examine the three terms of $A$ and those of $B$, the two components of $a$.

**Identify the key pattern** that **square of 2, 3 and 4 exceeds the square of 1st, 2nd and 3rd surd terms of $B$ respectively by 1,**

$2=\sqrt{4} \gt \sqrt{3}$, $3=\sqrt{9} \gt \sqrt{8}$, and $4=\sqrt{16} \gt \sqrt{15}$.

That means,

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

Or, $B \lt A$,

Or, $A+B \lt 2A$, add $A=9$ to both sides of the inequality,

Or, $a \lt 18$.

Also, with $a=A+B=9+B$, where $B$ is surds, $a \neq 9$.

Combine the two inequalities to get final result,

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

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

**Key concepts used:** **Rationalization of surds -- 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**

Rule to evaluate a series of square roots under square roots,

Evaluate first the innermost, that is, the rightmost square root.

Last of the square roots $\sqrt{81}$ gives 9. Product of 9 and 9 is again 81. Taking its 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.**

The detailed steps are,

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

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

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

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

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

**Key concepts used: Rule for evaluating Square roots in series, the rightmost 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,

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

*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**

**Goal:** To express $(\sqrt{12}+\sqrt{18})$ as a product of $(\sqrt{3}+\sqrt{2})$ and a second surd factor.

Expressed differently, job is to tranform $(\sqrt{12}+\sqrt{18})$ to get one factor as $(\sqrt{3}+\sqrt{2})$.

**Questions is:** How to get $(\sqrt{3}+\sqrt{2})$ extracted as a factor out of $(\sqrt{12}+\sqrt{18})$?

With properly analyzed goal, solution comes easily.

**Compare the two expression to easily identify** that taking out $\sqrt{6}$ from the first and second surd terms of $(\sqrt{12}+\sqrt{18})$ produces $(\sqrt{3}+\sqrt{2})$.

This is a good example of application of often underrated **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 double square root surd simplification and rationalization of surds**

As value of $a$ is a surd expression and target has $\sqrt{a}$, this is a problem of **double square root surd simplification.**

Have to express $a=7-4\sqrt{3}$ as a square of surds 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 -- Double square root surd simplification -- Rationalization of surds** -- *Solving in mind.

If you are comfortable with double square root surd simplification and rationalization of surds, 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),

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

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

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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