## Eighth SSC CGL level Solution Set, topic Algebra

This is the eighth solution set of 10 practice problem exercise on topic Algebra for SSC CGL exam. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set.

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

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 of the topics
- is adequately fast in mental math calculation
- should first try to solve each problem using the most basic 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 happens in last step above. How to do that?

You need to use your **your problem solving abilities** only. There is no other recourse.

If you have not taken the corresponding test yet, you should refer to * SSC CGL level Question Set 8 on Algebra* and then go through these solutions.

Watch the two-part video solutions below.

**Part I**

**Part II**

### Eighth solution set on Algebra - 10 problems for SSC CGL exam - time 12 mins

**Q1.** If $a = \sqrt{7 + 2\sqrt{12}}$ and $b = \sqrt{7 - 2\sqrt{12}}$, then $a^3 + b^3$ is,

- 52
- 40
- 44
- 48

**Solution:**

Expanding the given expression $E = a^3 + b^3$ we get,

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

$a^2 = 7 + 2\sqrt{12}$ and $b^2 = 7 - 2\sqrt{12}$, and so, $a^2 + b^2 = 14$.

Again, $ab = \sqrt{7^2 - 4\times{12}} = 1$, and so, $(a^2 - ab + b^2) = 13$.

Now we have to transform $a + b$ and find its value.

$a^2 + b^2 = 14$ and $ab=1$,

So,

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

$\hspace{33mm} = 14 + 2 $

$\hspace{33mm} = 16$

And so, $(a + b) = 4$ and $E = 4\times{13} = 52$.

**Answer:** Option a: 52.

**Key concepts used:**

Awareness of the factors of $a^3 + b^3$ -- squaring $a$ and $b$ and summing will eliminate the square roots as well as multiplying $a$ and $b$ together to $ab$ would also eliminate the square roots -- with $a^2$, $b^2$ and $ab$ known $a + b$ could be found out (a + b could not have been negative).

#### Alternate solution using double square root surd simplification

We call a surd with a square root under a second square root as a **double square root surd**.

$7+4\sqrt{3}$ is such a double square root surd.

We have found that a double square root surd can always be simplified by expressing the surd expression as a square of a second surd expression.

That eliminates the outer second square root and converts the original double square root surd as a normal surd expression.

In this problem, it is simple to express $7+4\sqrt{3}$ as a square expression,

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

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

Or, $\sqrt{7+4\sqrt{3}}=2+\sqrt{3}$.

Similarly,

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

This method (there are two methods) of double square rooot surd simplification is easy and if you have used the method a few times, you would automatically form the simplified values of $2+\sqrt{3}$ and $2-\sqrt{3}$ in a few seconds.

This results in,

$a+b=4$, and

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

$=4\times{(4^2-3)}=52$, where $ab=1$.

We have used the **two-factor modified sum of cubes expansion** that *gives the solution very quickly*,

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

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

The **advantage** with this form of two-factor sum of cubes expansion is—**you don't have to evaluate $a^2+b^2$**.

$ab=1$ can be evaluated automatically **because that is inherent** in $2+\sqrt{3}$ and $2-\sqrt{3}$. By inherent we mean, the **difference of squares of the two terms $2$ and $\sqrt{3}$ is 1** in this case. Let us show you,

$ab=(2+\sqrt{3})(2-\sqrt{3})=[2^2-(\sqrt{3})^2]=1$.

Many of the two-term surd expressions that we encounter have this property invariably and **this property aids surd rationalization and simplification** greatly.

**Knowing these concepts and with experience of use, the problem can be solved in this alternate method much faster.**

**Q2.** If $x + \displaystyle\frac{2}{x} = 1$, then $\displaystyle\frac{x^2 + x + 2}{x^2(1 - x)}$ is,

- 2
- -2
- 1
- -1

**Solution:**

The target expression having little similarity with the given expression at first glance, a safe approach is to transform the given expression in an expression of $x$ with $0$ as the value of expression.

The strategy in this approach is, wherever in the target expression we can form this $0$ equivalent expression we would be able to reduce the complexity of the target expression immediately by replacing the input expression by a $0$.

Transforming the given expression we have,

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

Or, $x^2 -x + 2 = 0$.

Now let us deal wwith the target expression $E$ (that's what we will call the complex target expression).

$ E = \displaystyle\frac{x^2 - x + 2 + 2x}{x^2(1 - x)} $

$\hspace{5mm} = \displaystyle\frac{2x}{x^2(1 - x)} $

$\hspace{5mm} = \displaystyle\frac{2}{x(1 - x)} $

$ \hspace{5mm} = \displaystyle\frac{2}{x - x^2} $

$ \hspace{5mm} = \displaystyle\frac{2}{x - x^2} - 1 + 1 $

$ \hspace{5mm} = \displaystyle\frac{x^2 - x + 2}{x - x^2} + 1 $

$ \hspace{5mm} = 1 $

**Answer:** Option c: 1.

**Key concepts used: **

Transformation of given expression to a zero valued expression in $x$ -- judiciously finding this expression in the complex target expression and replacing each such occurrence with $0$.

This is a standard problem solving technique in Algebra that results in quick solution in case of dealing with complex expressions.

#### Alternate faster solution by the principle of target expression simplification first

You should always attempt to simplify the target expression first.

This approach invariably speeds up the solution.

Taking up this approach, by grouping first and third terms of the target numerator and dividing both numerator and denominator by $x$, the target expression is simplified greatly at one shot to,

$\displaystyle\frac{x^2+\displaystyle\frac{2}{x}+1}{x(1-x)}$

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

While simplifying the numerator, we matched it with the LHS of the given expression to achieve significant simplification at one shot.

Now only we will change the given expression to,

$x-x^2=2$.

Substituting you get the answer as 1.

A simple and fast solution, but yes, you have to discover the possibility of matching the LHS of the given expression in the numerator of the target expression. Even that's not also difficult if you follow the "**simplify target expression first**" rule.

**Q3.** If $x^3 + \displaystyle\frac{3}{x} = 4(a^3 + b^3)$ and $3x + \displaystyle\frac{1}{x^3} = 4(a^3 - b^3)$, then $a^2 - b^2$ is,

- 1
- 0
- 4
- 2

**Solution:**

This is a bit tricky problem and without careful inspection and recognition of hidden pattern solving this problem may take quite a bit of time.

We recognize the LHSs of the two given expressions as two parts of a whole cube equation. That's the key to the solution.

First we add the two equations giving,

$ 8a^3 = x^3 + \displaystyle\frac{3}{x} + 3x + \displaystyle\frac{1}{x^3} $

$\hspace{6mm} = x^3 + 3\left(x^2\times{\displaystyle\frac{1}{x}}\right) + 3\left(x\times{\displaystyle\frac{1}{x^2}}\right) + \displaystyle\frac{1}{x^3} $

$ \hspace{6mm} = \left(x + \displaystyle\frac{1}{x}\right)^3$

Or, $\left(x + \displaystyle\frac{1}{x}\right) = 2a$

In the same way, we would get,

$\left(x - \displaystyle\frac{1}{x}\right) = 2b$

Squaring the two and subtracting we get,

$4(a^2 - b^2) = 4$,

Or, $a^2 - b^2$ = 1.

**Answer: **Option a: 1.

**Key concepts used: **

Identifying that the LHSs of the two equations are two halves of a cube expression of $x + \displaystyle\frac{1}{x}$ if added and of $x - \displaystyle\frac{1}{x}$ if subtracted -- in the second stage squaring and subtracting eliminates $x$ and $\displaystyle\frac{1}{x}$ altogether leaving only $a^2 - b^2$, our desired expression.

This deduction is possible only if you foresee possibilities of one step ahead.

Pattern recognition is the critical ability needed to solve this problem.

**Q4.** If $x^2 - 4x + 1 = 0$, then $x^3 + \displaystyle\frac{1}{x^3}$ is,

- 44
- 64
- 48
- 52

**Solution: **

By inspecting the target expression we find the need to use the value of $x + \displaystyle\frac{1}{x}$. We now attempt to get this value from the given expression by transforming it,

$x^2 - 4x + 1 = 0$

Or, $x^2 + 1 = 4x $

Or, $x + \displaystyle\frac{1}{x} = 4$

Now the target expression,

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

Or, $E = 4\left((x + \frac{1}{x})^2 - 3\right)$

Or, $E = 4\times{13} = 52$

**Answer:** Option d: 52.

**Key concepts used:**

Expanded form of $x^3 + \displaystyle\frac{1}{x^3}$ -- transforming given expression in terms of a value for $x + \displaystyle\frac{1}{x}$ -- using this value in the factors of the target expression.

**Q5.** If $x^4 + \displaystyle\frac{1}{x^4} = 119$ and $x \gt 1$, then positive value of $x^3 - \displaystyle\frac{1}{x^3}$ is,

- 27
- 36
- 25
- 49

**Solution:**

Comparing the target expression with the given expression we understand the possibility of the needed value of $\left(x - \displaystyle\frac{1}{x}\right)$ to be extracted from the given expression.

We have,

$x^4 + \displaystyle\frac{1}{x^4} = 119$

Or, $x^4 + 2 + \displaystyle\frac{1}{x^4} = 121$

Or, $\left(x^2 + \displaystyle\frac{1}{x^2}\right)^2 = 121$

Or, $x^2 + \displaystyle\frac{1}{x^2} = 11$, as both the terms on the LHS are in square it cannot be $(-11)$.

Again transforming,

$x^2 + \displaystyle\frac{1}{x^2} = 11$

Or, $x^2 - 2 + \displaystyle\frac{1}{x^2} = 9$

Or, $\left(x - \displaystyle\frac{1}{x}\right) = 3$, as $x \gt 1$, $\displaystyle\frac{1}{x} \lt x$ and $x - \displaystyle\frac{1}{x}$ is positive (it could have been $-3$).

Now from the target expression we have,

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

$ \hspace{5mm} = 3\times{(11 + 1)} = 36$

**Answer:** Option b: 36.

**Key concepts used:**

Factors of $x^3 - \displaystyle\frac{1}{x^3}$ -- step-down transformation of the given expression into squares on the LHS in two steps to get the required values of $x^2 + \displaystyle\frac{1}{x^2}$ and $x - \displaystyle\frac{1}{x}$.

**Q6.** If $x^3 + y^3 = 9$ and $x + y = 3$, then value of $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y}$ will be,

- $\displaystyle\frac{5}{2}$
- $\displaystyle\frac{3}{2}$
- $-1$
- $\displaystyle\frac{1}{2}$

**Solution: **

First we always examine the target expression against the given expression to find usable similarities. The given expressions directly gives the value of $x + y$ and also hints at giving the value of $xy$ on transformation. But even if you are not able to see this possibility at least you know that target expression is to contain $x + y$.

So first transforming target expression we have,

$\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} = \displaystyle\frac{x + y}{xy} = \frac{3}{xy}$. We need only to get the value of $xy$.

Now we turn our attention to the given expressions, especially the first one.

$x^3 + y^3 = 9 = (x + y)(x^2 - xy + y^2) $

$ \hspace{18mm} = 3(x^2 + 2xy + y^2 -3xy) $

$ \hspace{18mm} = 3((x +y)^2 -3xy)$

Or, $9 - 3xy = 3$,

Or, $xy = 2$.

So, $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} = \displaystyle\frac{3}{xy} = \frac{3}{2}$.

**Answer:** Option b: $\displaystyle\frac{3}{2}$.

**Key concepts used:**

Examining end state against initial information, transforming the target expression up to a point -- transforming the given expression of $x^3 + y^3 = 9$ to get the value of the desired left out expression $xy$.

**Q7.** If $ a : b = 2 : 3$ and $b : c = 4 : 5$, then the value of $a^2 : b^2 : bc$ is,

- 16 : 36 : 20
- 16 : 36 : 45
- 4 : 9 : 45
- 4 : 36 : 40

**Solution: **

On inspecting the given ratio expressions against the target ratio we decide that ratio joining by base equalization is required to be done. But this is to be done not in a straightforward manner.

we have $a : b = 2 : 3$ which gives, $a^2 : b^2 = 4 : 9$.

But the second ratio we don't square. Instead we multiply numerator and denominator by $b$ to get, $b^2 : bc = 4 : 5$.

Now we have the common middle term of $b^2$ same in both the transformed ratios.

To join these two ratios, the ratio values corresponding to $b^2$ have to be equalized to the LCM of their values in two ratios, which is $4\times{9}=36$.

Transforming thus, the two ratios are changed to,

$a^2 : b^2 = 16 : 36$, and $b^2 : bc = 36 : 45$.

Now we can join these two ratios to get the desired ratio,

$a^2 : b^2 : bc = 16 : 36 : 45$.

**Answer:** Option b: 16 : 36 : 45.

**Key concepts used:**

Ratio joining -- required individual ratio transformation.

**Q8.** If $ a : b = 3 : 2$, then the ratio of, $2a^2 + 3b^2 : 3a^2 - 2b^2$ is,

- 6 : 5
- 30 : 19
- 12 : 5
- 5 : 3

**Solution:**

$a : b = 3 : 2$

Or, $2a = 3b$,

Or, $4a^2 = 9b^2$

In the target ratio, the first expression,

$2a^2 + 3b^2 = \displaystyle\frac{1}{2}(4a^2 + 6b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{2}(9b^2 + 6b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{2}\times{15b^2}$.

The second expression of the ratio is transformed suitably as,

$3a^2 - 2b^2 = \displaystyle\frac{1}{4}(12a^2 - 8b^2) $

$ \hspace{24mm} = \displaystyle\frac{1}{4}(27b^2 - 8b^2) $

$\hspace{24mm} = \displaystyle\frac{1}{4}\times{19b^2}$.

Taking the ratio of the two,

$2a^2 + 3b^2 : 3a^2 - 2b^2 = 30 : 19$.

**Answer:** Option b: 30 : 19.

**Key concepts used:**

Ratio concepts -- substitution concepts.

#### Alternate simpler solution using ratio of two variables

Whenever a ratio of two variables is given and desired is also a ratio of two expressions in these two variables, usually solution can be reached much easily and quickly if you convert the target expression into expressions in ratio of two variables and substitute the ratio value from the given expression.

In this problem, by dividing both numerator and denominator by $b^2$ it is easy to convert it into expressions involving ratio $\displaystyle\frac{a}{b}$,

$\displaystyle\frac{2\left(\displaystyle\frac{a^2}{b^2}\right)+3}{3\left(\displaystyle\frac{a^2}{b^2}\right)-2}$

From given expression get, $\displaystyle\frac{a^2}{b^2}=\frac{9}{4}$ and substitute in the changed target expression,

$\displaystyle\frac{\displaystyle\frac{18}{4}+3}{\displaystyle\frac{27}{4}-2}$

$=\displaystyle\frac{30}{19}$.

Solution is simple and fast in this approach, especially when you contrast it with the earlier solution.

**Q9.** The expression $x^4 - 2x^2 + k$ will be a perfect square if value of $k$ is,

- 1
- 2
- -1
- -2

**Solution: **

Just by inspection, we can imagine the expression to be $(x^2 - 1)^2$, giving $k$ value 1 for the expression to be a perfect square.

More formally, by Sreedhar Acharya's formula for roots of quadratic equation in a single variable, both roots to be same, $b^2$ must be equal to $4ac$ where the equation is $ax^2 + bx + c$.

In our case, this comes down to, $4=4k$, that is, $k=1$.

Sreedhar Acharya's profound formula gives the two roots of a quadratic equation $ax^2 + bx + c = 0$ in a single variable $x$.

#### Sreedhar Acharys's formula

$\displaystyle\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

**Answer:** Option a: 1.

**Key concepts used:**

Sreedhar Acharya's formula for roots of quadratic equation in a single variable and the condition for equality of both roots.

**Q10.** If $a = 11$ and $b = 9$, then the value of, $\displaystyle\frac{a^2 + b^2 + ab}{a^3 - b^3}$ is,

- $20$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{20}$
- $2$

**Solution:**

Knowing that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ the given expression is transformed to,

$E = \displaystyle\frac{1}{a - b} = \displaystyle\frac{1}{2}$

**Answer:** Option b: $\displaystyle\frac{1}{2}$

**Key concepts used: **

Awareness of the factors of $a^3 - b^3$ for simplification of the target expression before using the given values of a and b.

**Recommendation: **

Always simplify the target expression before using given variables.

### Additional help on SSC CGL Algebra

Apart from a **large number of question and solution sets** and a valuable article on "* 7 Steps for sure success on Tier 1 and Tier 2 of SSC CGL*" rich with concepts and links, you may refer to our other articles specifically on Algebra listed on latest shown first basis,

#### First to read tutorials on Basic and rich Algebra concepts and related topics

**More rich algebraic concepts and techniques for elegant solutions of SSC CGL problems**

**Basic and rich algebraic concepts for elegant Solutions of SSC CGL problems**

**SSC CGL level difficult Algebra problem solving by Componendo dividendo**

**Proof of least value of sum of reciprocals for any number of positive variables**

**How to factorize 25 selected quadratic equations quickly by factor analysis**

#### SSC CGL Tier II level Questions and Solutions on Algebra

**SSC CGL Tier II level Solution Set 17, Algebra 6**

**SSC CGL Tier II level Question Set 17, Algebra 6**

**SSC CGL Tier II level Question Set 14, Algebra 5**

**SSC CGL Tier II level Solution Set 14, Algebra 5**

**SSC CGL Tier II level Question Set 9, Algebra 4**

**SSC CGL Tier II level Solution Set 9, Algebra 4**

**SSC CGL Tier II level Question Set 3, Algebra 3**

**SSC CGL Tier II level Solution Set 3, Algebra 3**

**SSC CGL Tier II level Question Set 2, Algebra 2**

**SSC CGL Tier II level Solution Set 2, Algebra 2**

**SSC CGL Tier II level Question Set 1, Algebra 1**

**SSC CGL Tier II level Solution Set 1, Algebra 1**

#### Difficult algebra problem solving in a few steps

**How to solve difficult algebra problems in a few steps**

#### SSC CGL level Question and Solution Sets on Algebra

**SSC CGL level Solved Question Set 90, Algebra 18**

**SSC CGL level Question Set 81, Algebra 17**

**SSC CGL level Solution Set 81, Algebra 17**

**SSC CGL level Question Set 74, Algebra 16**

**SSC CGL level Solution Set 74, Algebra 16**

**SSC CGL level Question Set 64, Algebra 15**

**SSC CGL level Solution Set 64, Algebra 15**

**SSC CGL level Question Set 58, Algebra 14**

**SSC CGL level Solution Set 58, Algebra 14**

**SSC CGL level Question Set 57, Algebra 13**

**SSC CGL level Solution Set 57, Algebra 13**

**SSC CGL level Question Set 51, Algebra 12**

**SSC CGL level Solution Set 51, Algebra 12**

**SSC CGL level Question Set 45 Algebra 11**

**SSC CGL level Solution Set 45, Algebra 11**

**SSC CGL level Solution Set 35 on Algebra 10**

**SSC CGL level Question Set 35 on Algebra 10**

**SSC CGL level Solution Set 33 on Algebra 9**

**SSC CGL level Question Set 33 on Algebra 9**

**SSC CGL level Solution Set 23 on Algebra 8**

**SSC CGL level Question Set 23 on Algebra 8**

**SSC CGL level Solution Set 22 on Algebra 7**

**SSC CGL level Question Set 22 on Algebra 7**

**SSC CGL level Solution Set 13 on Algebra 6**

**SSC CGL level Question Set 13 on Algebra 6**

**SSC CGL level Question Set 11 on Algebra 5**

**SSC CGL level Solution Set 11 on Algebra 5**

**SSC CGL level Question Set 10 on Algebra 4**

**SSC CGL level Solution Set 10 on Algebra 4**

**SSC CGL level Question Set 9 on Algebra 3**

**SSC CGL level Solution Set 9 on Algebra 3**

**SSC CGL level Question Set 8 on Algebra 2**

**SSC CGL level Solution Set 8 on Algebra 2**

**SSC CGL level Question Set 1 on Algebra 1**

**SSC CGL level Solution Set 1 on Algebra 1**

#### SSC CHSL level Solved Question Sets on Algebra

**SSC CHSL level Solved Question set 11 on Algebra 1**

**SSC CHSL level Solved Question set 12 on Algebra 2**

**SSC CHSL level Solved Question set 13 on Algebra 3**

### Getting content links in your mail

#### You may get link of any content published

- from this site by
or,**site subscription** - on competitive exams by
.**exams subscription**