## Ninth SSC CGL level Solution Set, topic Algebra

This is the ninth Solution set of 10 practice problem exercise on topic Algebra for SSC CGL exam. ** Students must complete the corresponding SSC CGL Question set 9 on Algebra in prescribed time first and then only refer to this solution set**.

* Otherwise,* without attempting the question set with all seriousness in the prescribed time, if the learner goes through the solutions

*he or she won't be able to appreciate and retain the special concepts involved in the solutions.** A golden rule of math problem solving *always will remain true,

Math can't be learned by heart.

This is true for achieving excellence in learning any subject * but is truer especially in Maths.* Here, in Maths you have to

**understand the concepts**and

*within the scope of the topic.*

**acquire the ability to use the concepts with special problem solving strategies to deal with any math problem**Furthermore, **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 elaborate steps.

In MCQ test instead, you need basically to * deduce the answer in shortest possible time and select the right choice*. Solving process will

*rather than on scratch paper.*

**mostly be in your head****Based on our analysis and experience** we have seen that, **for accurate and lightning quick answering**, the student

- must have complete understanding of the
**basic concepts, along with rich concepts**on the topics, - is
**adequately fast in mental math**calculation,*one need not be superfast human calculator*, **should first examine each problem for using the most basic concepts**in the specific topic area and**then use the rich concepts**if required,**does most of the deductive reasoning and calculation in his or her head**rather than on paper.

Actual problem solving happens in the last step above. * The problem solving ability lies at the heart of excellence in performance* in this cutting-edge test.

This problem set containing 10 problems **highlights the need of solving the problems using powerful strategies** *rather than brute force conventional deduction methods*. We have tried to* bring out the underlying strategies, techniques and reasoning that went into solving each problem in about a minute's time*.

If you follow intelligent and dedicated preparation methods using this type of resources, * you should also be able to reach the desired level of competence* for completing such a set of 10 questions comfortably within 12 minutes' share of time.

Lastly, **these are problems rich with possibilities**. We couldn't cover all aspects of each problem. Later we would deal with a few selected class of such beautiful problems in details. You *may refer to these special problem solving strategy and technique oriented detailed treatments* under the subsection **Difficult Algebra Problem Solving in a few steps.**

Watch the **two-part videos** below.

**Part I**

**Part II**

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

**Q1.** If $a -2b = 4$, then the value of the expression $a^3 - 8b^3 -24ab - 64$ is,

- 0
- 3
- 2
- -1

**Solution:**

As a first step, we always **examine** in a quick scan the **similarities between the desired end state** (expression to be evaluated) and **the initial state** (given expression). This is applying our * End state analysis approach* that has most extensive use in math problem solving.

As a result of this analysis, **we form our first conclusion:**

If we raise the $a - 2b$ to its cube, the first and last terms of the four term cube expression will match the first two terms of the end state $a^3 - 8b^3$. With this favorable pattern discovery we resort to cubing the given expression without any further delay (you could have actually deduced all the four terms of $(a - 2b)^3$ using a special form, to find that the similarities between the end expression and given expression are much more).

$(a - 2b) = 4$

Or, $(a - 2b)^3 = 64$,

Or, $a^3 - 8b^3 - 3a\times{2b}(a - 2b) = 64$, this is where we have used the special form of a cube expression * as a rich concept* to reduce complexity.

Or, $a^3 - 8b^3 - 24ab = 64$

How lucky! We have actually reached the desired expression precisely with the simplest result of 0.

**Answer:** Option a: 0.

#### Key concepts used:

* End state analysis* to detect

*between*

**similarities***expression to be evaluated*and

*the given expression when cubed*-- using the

*of a cube expression of $(x - y)^3 = x^3 - y^3 -3xy(x - y)$.*

**rich concept****Q2.** If $x + \displaystyle\frac{1}{x} = 4$, then $x^4 + \displaystyle\frac{1}{x^4}$ is,

- 124
- 194
- 64
- 81

**Solution:**

This is a * problem involving inverses,* $x$ and $\displaystyle\frac{1}{x}$. If we square a sum of such inverse terms, the variable $x$ is always eliminated in the middle term. This is our

*.*

**rich concept to be used in this type of problems**So we square the given expression,

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

Or, $\left(x + \displaystyle\frac{1}{x}\right)^2 = 16$,

Or, $x^2 + \displaystyle\frac{1}{x^2} + 2 = 16$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 14$.

We look at the expression to be evaluated and decide that we are progressing on the right track. We would need one more stage of squaring.

$x^2 + \displaystyle\frac{1}{x^2} = 14$,

Or, $\left(x^2 + \displaystyle\frac{1}{x^2}\right)^2 = 196$,

Or, $x^4 + \displaystyle\frac{1}{x^4} = 194$

**Answer:** Option b: 194.

#### Key concepts used:

* Concept of transformation of the middle term of a square of sum of inverses to purely numeric form* -- continuous end state analysis.

**Q3.** If $x^2 = y + z$, $y^2= z + x$ and $z^2 = x + y$, then the value of $\displaystyle\frac{1}{x + 1} + \displaystyle\frac{1}{y + 1} + \displaystyle\frac{1}{z + 1} $ is,

- 2
- -1
- 4
- 1

#### Solution:

Before starting *we express our appreciation of the beauty of this cute problem*.

As always, *especially when in a tight spot, as we are with this awkward looking problem, we must take recourse to end state analysis*.

But in this case, *our attempt is not only to find similarities between the expression to be evaluated and the given expression*, * we try actively examining how such a similarity can be created by transforming the given expression*. This is where

*. By any means*

**we use the active mode of End State analysis approach***between the two expressions.*

**we must create maximum similarity***If we can, our problem is solved in a flash*, if not God bless you!.

Looking at the left hand side of the first given expression more closely, we think, what happens if we add a $x$ to it! Won't we get a $x + 1$? This in fact is the key discovery.

$x^2 = y + z$,

Or, $x^2 + x = x + y + z$,

Or, $x(x + 1) = x + y + z$,

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

With this result, we easily see through the whole problem. Though we reach the complete solution mentally, we have to show you the process as we don't have recourse to inter-mind communication!

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

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

$ = 1$

**Answer:** Option d: 1.

#### Key concepts used:

* Active mode of End state analysis approach* in establishing desired similarity by transformation of given expression -- identifying the hidden key information for the transformation.

**Q4.** If $x = 7 - 4\sqrt{3}$, then $x^{\frac{1}{2}} + x^{\frac{-1}{2}}$ is,

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

**Solution:**

With a quick look at the end state (we are doing end state analysis as usual), that is, our expression to be evaluated, we find it to contain $\sqrt{x}$, whereas the given expression is in $x$. We decide that our task is cut out, we must find the square root of $x$.

To do this, our basic concept on surds extends its help. We remember that,

Surd expressions in only two terms many times hide a three term expression representing a square of sum of two terms.

Our deductive reasoning also supports this action. We reason, unless the surd expression on the RHS of the given expression is transformed to a square expression, how can we get $\sqrt{x}$, and if we do not get it easily, we won't have a lucid solution as in the choice values! So we transform the surd expression with confidence,

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

$\hspace{5mm} = 2^2 - 2\times{2}\times{\sqrt{3}} + (\sqrt{3})^2$

$ \hspace{5mm} = (2 - \sqrt{3})^2$.

So, $\sqrt{x} = 2 - \sqrt{3}$.

Armed with this key result we attack the target expression with confidence,

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

$\hspace{5mm} = 2 - \sqrt{3} + \displaystyle\frac{1}{2 - \sqrt{3}}$

$\hspace{5mm} = 2 - \sqrt{3} + \displaystyle\frac{2 + \sqrt{3}}{(2 - \sqrt{3})\times{(2 + \sqrt{3})}}$

$ \hspace{5mm} = 2 - \sqrt{3} + 2 + \sqrt{3}$, the denominator turns to 1,

$ \hspace{5mm} = 4$

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

#### Key concepts used:

Deciding to form square root of the given surd expression by examining the target expression -- basic surd concept of expressing a two term surd expression as a whole square -- rationalization of surds concept.

**Q5.** $(y -z)^3 + (z -x)^3 + (x -y)^3$ is equal to,

- $(x - y)(y + z)(x - z)$
- $(y - z)(z + x)(y - x)$
- $(y - z)(z - x)(x - y)$
- $3(y - z)(z - x)(x - y)$

**Solution:**

Just look at the given expression, which also is the target expression in this special case, for a few seconds to realize that the sum of the original uncubed terms $(y - z)$, $(z - x)$ and $(x - y)$ is zero.

This is your key information discovery.

Now you just forget about $x$'s and $y$'s and straightaway use the much used substitution technique to define,

$(y - z) = a$

$(z - x) = b$

$(x - y) = c$,

So, we have, $a + b + c = 0$ and we have to find out effectively the simplified value of the equivalent form of, $a^3 + b^3 + c^3$.

This is our well known comfortable territory and in a few seconds we have our solution.

$a + b + c = 0$,

Or, $a + b = -c$,

Or, $a^3 + b^3 + 3ab(a + b) = -c^3$,

Or, $a^3 + b^3 - 3abc + c^3 = 0$,

Or, $a^3 + b^3 + c^3 = 3abc $

$\hspace{20mm} = 3(y - z)(z - x)(x - y)$.

**Answer:** Option d: $3(y - z)(z - x)(x - y)$.

#### Key concepts used:

Key pattern discovery in the summation of the three uncubed terms in the given expression. Here it helps to notice that nowhere these terms have been expressed in any expanded form. That is another push toward thinking each of these terms as a substituted single variable. Rest is standard procedures we are familiar with.

**Q6.** The sum of $\displaystyle\frac{a}{b}$ and its reciprocal is 1 and $a\neq{0}$, $b\neq{0}$. The value of $a^3 + b^3$ is,

- 2
- 0
- $-1$
- 1

#### Solution:

As the target expression has no inverses we straightway simplify the given expression.

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

Or, $\displaystyle\frac{a^2 + b^2}{ab} = 1$,

Or, $a^2 - ab + b^2 = 0$.

This we recognize the second term of the expansion,

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

**Answer:** Option b: 0.

#### Key concepts used:

* End state analysis* to transform the given expression -- concept of expanded form of $x^3 + y^3$.

**Q7.** If $ x = b + c - 2a$, $y = c + a - 2b$ and $z = a + b -2c$, then the value of $x^2 + y^2 - z^2 + 2xy$ is,

- $a + b + c$
- $0$
- $a - b + c$
- $a + b - c$

**Solution:**

This again is a special problem where we use our bird's eye view, or * Principle of Collection*, and think of

*.*

**summing up all of the three equations**Sometimes we need to employ this strategy, though in this case we were aware of the positive prospects in this approach by just looking at the problem.

Summing up the three equations,

$x + y + z = 0$,

Or, $x + y = -z$,

Or, $x^2 + y^2 + 2xy = z^2$,

Or, $x^2 + y^2 - z^2 + 2xy = 0$.

**Answer:** Option b: 0.

#### Key concepts used:

* Principle of collection, we collect all the equations together* -- by summing up the three equations we eliminate $a$, $b$ and $c$ altogether and get a nice easy expression in the bargain.

**Q8.** If $x = \displaystyle\frac{4ab}{a + b}$, where $a\neq{b}$, then the value of, $\displaystyle\frac{x + 2a}{x - 2a} + \displaystyle\frac{x + 2b}{x - 2b}$ is,

- a
- 2
- 2ab
- b

We will follow the **golden rule of simplifying the target expression first.**

To do this, the best approach would be first to add 1 to each of the fraction terms of the target expression and reduce both numerators to $2x$. We would have to subtract compensating 2. Let us show you.

Target expression,

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

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

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

As both the terms have the specific pattern suitable for simplifying the numerator to same value by adding 1, this step can easily be carried out in mind.

Continuing simplifying the two terms of the target expression, divide both numerator and denominator of each term by $x$. This is also easy to visualize as,

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

Each term has only one $x$ and only now we would substitute the value of $x = \displaystyle\frac{4ab}{a + b}$,

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

$=\displaystyle\frac{4b}{b-a}+\displaystyle\frac{4a}{a-b}-2$

$=\displaystyle\frac{4(b-a)}{b-a}-2$

$=4-2=2$.

The factor $(b-a)$ could be canelled out as, $a \neq b$.

**Answer:** Option b: 2.

#### Key concepts used:

The never failing **strategic rule of simplifying the target expression first** by itself as much as possible before substituting the given value of $x$. The simplification steps are easy enough to carry out in mind. At the stage of substitution of value of $x$ only the step may be written for accuracy. The first simplification used **numerator equalization** by pattern discovery.

**Q9.** If $x = 3 + 2\sqrt{2}$ then the value of $\left(\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}\right)$ is,

- $2\sqrt{3}$
- $3\sqrt{3}$
- 2
- 1

**Solution:**

$x$ is given and $\sqrt{x}$ is required. So we transform the given surd expression to a square expression in one step.

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

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

So,

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

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

$ \hspace{5mm} = \sqrt{2} + 1 - \sqrt{2} + 1 = 2$

**Answer:** Option c: 2.

#### Key concepts used:

Concluding that given surd expression has to be transformed to a whole square expression.

**Q10.** If $x = 2 - 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$, then the value of $x^3 - 6x^2 + 18x + 18$ is,

- 40
- 33
- 22
- 45

#### Solution:

This we found to need some attention to details.

The first natural step is to transform the given expression.

$x = 2 - 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$,

Or, $x - 2 = 2^{\frac{2}{3}} - 2^{\frac{1}{3}}$.

Looking at the target expression already we have decided on this transformation and then cubing the equation. But at this point we need to do a second stage analysis.

Notice that we have to eliminate the one third power of 2 altogether and then only we would be able to evaluate the target expression.

With this alert in our mind, we look for opportunity of eliminating $2^\frac{1}{3}$ altogether from the RHS.

$(x - 2)^3 = (2^\frac{2}{3} - 2^{\frac{1}{3}})^3$,

Or, $x^3 - 6x^2 + 12x - 8$

$ \hspace{5mm} = 2^2 - 2 - 3\times{2^\frac{2}{3}}\times{2^{\frac{1}{3}}} (2^\frac{2}{3} - 2^{\frac{1}{3}})$,

Or, $x^3 - 6x^2 + 12x = 10 - 6(x - 2)$,

Or, $x^3 - 6x^2 + 18x = 10 + 12$,

Or, $x^3 - 6x^2 + 18x + 18 = 40$.

**Answer:** Option a: 40.

#### Key concepts used:

Constant alertness of the need for eliminating $2^\frac{1}{3}$ altogether -- transformation of given equation -- using rich concept $(x - y)^3 = x^3 - y^3 -3xy(x - y)$ on the numeric expression in terms of $2^\frac{1}{3}$.

### Guided help on Suresolv Algebra

All Suresolv Algebra articles are listed with links at the end, but it is unguided list.

To use this *extensive range of articles on quick algebra* problem solving **effectively**, *follow the guide,*

**5 step Suresolv Algebra Reading and Practice Guide for SSC CGL Tier II and Other competitive exams.**

*Basically, it is how to read and practice Suresolv Algebra guide.*

**This contains even list of puzzles or even high school math articles on Algebra. **

Wish you all the sure success.

### Resources on Suresolv 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,

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