## 60th SSC CGL level Solution Set, 4th on topic fractions indices and surds

This is the 60th solution set of 10 practice problem exercise for SSC CGL exam and 4th on topic fractions indices and surds. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

- must have complete understanding of the basic concepts in the topic area
- is adequately fast in mental math calculation
- should try to solve each problem using the basic and rich concepts in the specific topic area and
- does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use **your problem solving abilities** to gain an edge in competition.

If you have not taken the companion test yet, first take the test on * SSC CGL level Question set 60 on fractions indices surds 4* and then go through the solutions for gaining maximum benefit.

### 60th solution set- 10 problems for SSC CGL exam: 4th on topic Fractions indices and surds - time 12 mins

**Problem 1.**

The sum of cubes of the numbers $22$, $-15$ and $-7$ is equal to,

- 3
- 9630
- 0
- 6930

**Solution 1: Problem analysis and solving**

The first cube is of 22 which is a sum of other two cube base values, 15 and 7, $22=15+7$. Moreover, the second two cubes will evaluate to negative values. So we decide to use cube of sum expression,

$(22)^3=(15)^3+7^3+3\times{15}\times{7}\times{22}$

$=(15)^3+7^3+105\times{66}$

$=(15)^3+7^3+6930$.

So,

$(22)^3+(-15)^3+(-7)^3=6930$.

**Answer:** Option d: 6930.

**Key concepts used:** * Key pattern identification* --

*--*

**basic algebraic concepts***.*

**Efficient simplification****Problem 2.**

Which of the following is the correct relation?

- $\sqrt{5}+\sqrt{3} \lt \sqrt{6}+\sqrt{2}$
- $\sqrt{5}+\sqrt{3} = \sqrt{6}+\sqrt{2}$
- $\sqrt{5}+\sqrt{3} \gt \sqrt{6}+\sqrt{2}$
- $\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)=1$

**Solution 2: Problem analysis**

In *surd expression comparison*, one of the most important concepts that we use frequently is what we call, the **Equal difference surd comparison concept**.

Let us briefly state and explain the mechanism of this important concept.

#### Equal difference surd comparison concept

Let the two surd expressions we need to compare be,

$\sqrt{a}-\sqrt{b}$, and

$\sqrt{c}-\sqrt{d}$

where $a-b=c-d$.

If $a \gt c$, by * equal difference surd comparison concept*,

$\sqrt{a}-\sqrt{b} \lt \sqrt{c}-\sqrt{d}$.

#### Proof of equal difference surd comparison concept

$a-b=c-d$,

Or, $a-c=b-d$.

As $a \gt c$, $b \gt d$.

Taking up the surd expressions,

$\sqrt{a}-\sqrt{b}=\displaystyle\frac{a-b}{\sqrt{a}+\sqrt{b}}$, by rationalization.

Similarly,

$\sqrt{c}-\sqrt{d}=\displaystyle\frac{c-d}{\sqrt{c}+\sqrt{d}}$.

Taking the ratio,

$\displaystyle\frac{\sqrt{a}-\sqrt{b}}{\sqrt{c}-\sqrt{d}}=\frac{\sqrt{c}+\sqrt{d}}{\sqrt{a}+\sqrt{b}}$.

As $a \gt c$ and $b \gt d$,

$\sqrt{a}+\sqrt{b} \gt \sqrt{c}+\sqrt{d}$.

So,

$\sqrt{a} - \sqrt{b} \lt \sqrt{c}-\sqrt{d}$.

#### Solution 2: Problem solving execution

In all four options the same two surd expressions appear. Let us evaluate the comparative relation between these two expressions.

By equal difference surd comparison concept,

$\sqrt{6}-\sqrt{5} \lt \sqrt{3}-\sqrt{2}$,

Or, $\sqrt{5}+\sqrt{3} \gt \sqrt{6}+\sqrt{2}$.

So only Option: $c$ is true.

**Answer:** Option c: $\sqrt{5}+\sqrt{3} \gt \sqrt{6}+\sqrt{2}$.

**Key concepts used:** * Surd rationalization* --

*--*

**inequality concepts***.*

**equal difference surd comparison concept****Problem 3.**

$\sqrt[3]{(333)^3+(333)^3+(334)^3-3\times{333}\times{333}\times{334}}$ is closest to,

- 11
- 10
- 12
- 15

**Solution 3: Problem analysis and solving**

As direct calculation will be inordinately lengthy and as the numeric cubed terms have a patttern, we will use sum of cubes expression, $x^3-y^3=(x-y)(x^2+xy+y^2)$.

$E=\sqrt[3]{(333)^3+(333)^3+(334)^3-3\times{333}\times{333}\times{334}}$

$=\sqrt[3]{(334)^3-(333)^3-3\times{(333)^2}(334-333)}$

$=\sqrt[3]{(334)^2+333\times{334}+(333)^2-3(333)^2}$

$=\sqrt[3]{(334)^2-(333)^2+333}$, the 334 in product is split as $334=(333+1)$ cancelling two 333 squares

$=\sqrt[3]{667+333}$, using $a^2-b^2=(a-b)(a+b)$ where $334+333=667$

$=\sqrt[3]{1000}$

$=10$.

Answer: Option b: 10.

**Key concepts used:** * Basic algebra concepts* --

*--*

**efficient simplification***--*

**abstraction, we have used the numeric constant terms of $333$ and $334$ as variables in algebraic relations***.*

**delayed evaluation, we have carried out numeric calculation only at the last step****Problem 4.**

The value of $\sqrt{0.00060516}$ is equal to,

- 0.246
- 0.0246
- 0.00246
- 0.000246

**Solution 4: Problem analysis and solving execution**

It can be assumed from the values that,

$(24.6)^2=605.16$.

So squaring only the value in the second choice, $0.0246$ will shift the decimal point in comparison to the square of $24.6$ six places left and thus will generate three leading zeros before $605.16$.

**Answer:** Option b: 0.0246.

**Key concepts used:** * Basic decimal concepts* --

*--*

**place value mechanism***--*

**free resource use***.*

**choice value test****Problem 5.**

The value of $(3+2\sqrt{2})^{-3}+(3-2\sqrt{2})^{-3}$ is,

- 108
- 189
- 180
- 198

#### Solution 5: Problem analysis and solving execution

The problem involves cube of surd terms, and so we need to use the surd rationalization technique after expressing the negative powers of terms in fraction form.

The target expression,

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

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

$=\left(3-2\sqrt{2}\right)^3+\left(3+2\sqrt{2}\right)^3$

This is in the form of $a^3+b^3$, where $a=3-2\sqrt{2}$ and $b=3+2\sqrt{2}$.

So the target expression is,

$E=(a+b)(a^2-ab+b^2)$

$=6[(a+b)^2-3ab]$

$=6\times{[(6)^2-3(9-8)]}$

$=6\times{33}$

$=198$.

**Answer:** Option d: 198.

**Key concepts used:** * Basic algebraic concepts* --

*--*

**sum of cubes***--*

**abstraction***--*

**delayed evaluation***--*

**surd rationalization***--*

**basic indices concepts***.*

**efficient simplification**#### Problem 6.

$4^{61}+4^{62}+4^{63}+4^{64}$ is always divisible by,

- 3
- 11
- 13
- 17

**Solution 6: Problem analysis and solving execution**

We need to take the largest common factor $4^{61}$ out of the four terms to simplify the expression.

The target expression is thus,

$E=4^{61}+4^{62}+4^{63}+4^{64}$

$=4^{61}(1+4+16+64)$

$=4^{61}(85)$.

Out of the given choice values 17 as a factor satisfies the target expression.

**Answer:** Option d: 17.

**Key concepts used:** * Simplification by taking common factor out of each term* --

*.*

**basic factors and multiples concepts****Problem 7.**

The value of $\sqrt{30+\sqrt{30+\sqrt{30+...}}}$ is,

- 6
- 5
- 7
- $3\sqrt{10}$

#### Solution 7: Problem analysis and solving execution

With such an additive non-terminating square root series, we need to assume the target as $p$ and square it to take out one 30 from the target leaving $p$,

$p=\sqrt{30+\sqrt{30+\sqrt{30+...}}}$,

Or, $p^2=30+p$,

Or, $p^2-p-30=0$,

Or, $(p-6)(p+5)=0$.

As $p$ cannot be negative, $p=6$.

**Answer:** Option a: 6.

**Key concepts used:** * Basic number system concepts* --

*--*

**Non-terminating expression conversion technique***--*

**basic algebraic concepts***.*

**factorization of quadratic equations****Problem 8.**

The value of $\sqrt{2\sqrt[3]{4\sqrt{2\sqrt[3]{4...}}}}$ is,

- $2$
- $2^3$
- $2^5$
- $2^2$

**Solution 8: Problem analysis and solving execution**

To access the repeating sequence within the square roots we need to raise the target expression first to square and then to cube. Assuming the target expression as $p$,

$p^2=2\sqrt[3]{4\sqrt{2\sqrt[3]{4...}}}$.

Again cubing,

$p^6=32\sqrt{2\sqrt[3]{4...}}$

Or, $p^6=32p$,

Or, $p^5=32$,

Or, $p=2$.

Here the expression has been multiplicative that aided final evaluation.

**Answer:** Option a: 2.

**Key concepts used:** * Basic indices concepts* --

*--*

**basic algebraic concepts***.*

**non-terminating expression conversion technique****Problem 9.**

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

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

**Problem analysis and solving execution**

We need to factorize the second expression so that we get one factor as the first expression. So let us factorize the second expression,

$(\sqrt{12}+\sqrt{18})$

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

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

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

So it is simply $\sqrt{6}$.

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

**Key concepts used:** **Surd factorization concepts.**

**Problem 10.**

When it is given that $\sqrt{3}=1.732$, the value of $\displaystyle\frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}$ is,

- 1.414
- 1.732
- 2.551
- 4.899

#### Solution 10: Problem analysis and solving

We need to simplify the target expression first.

The target expression,

$E=\displaystyle\frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}$

$=\displaystyle\frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}$

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

$=\sqrt{3}$

$=1.732$.

**Answer:** Option b: 1.732.

**Key concepts used:** * Surd simplification* --

*.*

**Surd factorization****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 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**

#### Question sets and Solution Sets in Fractions, Surds, Indices and related topics

**SSC CGL level Solved Question set 93 on Surds and Indices 10**

**SSC CGL level Solved Question set 92 on Surds and Indices 9**

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**SSC CGL level Question Set 70 on fractions surds 6**

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**SSC CGL level Question Set 61 on fractions indices surds 5**

**SSC CGL level Solution Set 61 on fractions indices surds 5**

**SSC CGL level Question Set 60 on fractions indices surds 4**

**SSC CGL level Solution Set 60 on fractions indices surds 4**

**SSC CGL level Question Set 59 on fractions square roots and surds 3**

**SSC CGL level Solution Set 59 on fractions square roots and surds 3**

**SSC CGL level Question Set 47 on fractions decimals and surds 2**

**SSC CGL level Solution Set 47 on fractions decimals and surds 2**

**SSC CGL level Question Set 17 on fractions decimals and surds 1**

**SSC CGL Level Solution set 17 on fractions decimals and surds 1**