## First SSC CGL Solution Set on Algebra

This is the first solution set of 10 practice problem exercise for SSC CGL exam on Algebra. Students must complete the corresponding timed Question set first and then only refer to this solution set.

In this solution set, not only the answer and steps to the solution are given for each question, the reasoning behind the steps and concepts used are also clarified.

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 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 head rather than on paper.

Actual problem solving happens in item 3 and 4 above. How to do that?

In these 10 problem solution set, **a number of powerful problem solving strategies** have been used effectively. Student should make efforts to understand these general approaches of problem solving and apply these to future problems.

Watch the solutions of these 10 problems in the **two part video below.**

**First part:** Solutions for first 5 questions from 1 to 5

**Second part:** Solutions for last 5 questions from 6 to 10

### First solution set - 10 problems for SSC CGL exam

**Q1.** The value of, $\frac{1}{a^2 +ax + x^2}- \frac{1}{a^2 - ax + x^2} +\frac{2ax}{a^4 + a^2x^2 + x^4}$ is,

- 2
- 1
- -1
- 0

**Solution**

When you meet such balanced nearly whole square equations in the denominator with one middle term plus and the other minus, you can straightway pair the first and third terms $a^2$ and $x^2$ together so that when you combine the first two terms of the expression, the denominator turns to $(a^2 + x^2)^2 - a^2x^2$.

Immediately your attention shifts to the third denominator and quickly you can transform it to the same expression mentally.

$\begin{align}

(x^4+a^2x^2+x^4) & = (x^4+2a^2x^2+x^4) - a^2x^2 \\

& = (a^2+x^2)^2 - a^2x^2

\end{align}$

Now only the task of evaluating the numerator remains.

First two terms in an instant produces $-2ax$ which cancels out with the third term numerator resulting in a zero in the numerator.

**Answer: **Option 4: 0.

**Key concepts used:** Use of $(a + x)^2$ formula and Basic formula $(a + x)\times(a - x) = a^2 - x^2$ -- identifying regularities in the denominator first and then the numerator.

**In this type of problems, key lies more often than not in usable regular common patterns in the denominators.** This is direct application of **pattern identification and use technique**, but in a specific manner.

Once denominator complexity is resolved, numerator complexity automatically gets resolved.

**Q2.** If $x^3 + y^3 = 9$ and $x + y = 3$ then the value of $x^4 + y^4$ is,

- 81
- 32
- 27
- 17

**Solution**

Whenever you meet $x^3 + y^3 = 9$ and $x + y = 3$ together straightway go for the expression $x^3 + y^3 = (x + y)\times(x^2 -xy + y^2)$ which results in,

$\begin{align}

9 & = 3\times{((x + y)^2 - 3xy)} \\

& = 3\times{(9 - 3xy)}\\

& = 27 - 9xy

\end{align}$

or, $xy = 2$.

Now easiest way to get $x^4 + y^4$ is to multiply the first two given expressions together, giving,

$ x^4 + y^4 + xy(x^2 + y^2) = 27$. So,

$ \begin{align}x^4 + y^4 & = 27 - 2\times{((x + y)^2 - 2xy)} \\

& = 27 - 2\times{(9 - 4)} \\

& = 17

\end{align}$

**Answer:** Option 4 : 17.

**Key concepts used:** Use of $x^3 + y^3 = 9$ and $x + y = 3$ together to get value of $xy$ -- get target value by direct multiplication of given two expressions and using value of $xy$.

**Q3.** For any real number $x$ the maximum value of $4 - 6x - x^2$ is,

- 4
- 7
- 9
- 13

**Solution 3.1: Standard method**

To maximize or minimize a quadratic expression, convert the expression to a form that contains a perfect square of sum (subtractive or additive) in $x$. Both the terms in $x$ in the original expression must be fully used in the perfect square of sum in $x$.

Logic of maximization and minimization is applied on this form of the quadratic expression and solution obtained immediately.

Let us show you how to apply the method in our problem.

The given quadratic expression,

$4-6x-x^2=4-(x+3)^2+9=13-(x+3)^2$.

$6x$ and $x^2$ are absorbed in the perfect square. $6x$ being the middle term, the numeric term of 3 is immediately identified. $3^2=9$ being the extra 3rd term in the expansion of the square of sum, it is compensated outside the brackets by $+9$.

The logic is now applied on, $13-(x+3)^2$.

**Logic of maximization:** As $x$ is real, for all values of $x$ except $x=-3$, the square of sum would have a positive value and so will reduce the value of the quadratic expression from 13. Only when $x=-3$, or, $x+3=0$, the expression value will attain its maximum value of 13.

**Answer:** Option 4: 13.

**Key concepts used:** Properties of expanded three term square of sum -- Method of compensation -- Properties of a square of sum in real $x$ -- Solving in mind.

Applying the standard method, the problem could easily be solved in mind.

**Solution 3.2: By mathematical reasoning**

By analyzing the expression with the knowledge that $x$ can be either positive or negative, we find in both cases of nature of $x$, the third term being square of $x$, will remain to be negative.

So to maximize the value of the expression, the second term must be positive, that is, $x$ must be negative.

Next we see that the third term being a square it will increase faster than the second term. So combining this knowledge with the requirement of minimizing the third term (being negative), we conclude that absolute value of $x$ must be as small as possible.

Starting with $x=-1$, we get target expression $E = 4 + 6 - 1 = 9$.

With $x=-2$, expression $E = 4 + 12 - 4 =12$ - increasing.

Decreasing $x$ still further for $x=-3$, we get, expression $E = 4 + 18 - 9 = 13$.

We stop here **as we notice 13 to be the maximum value among the choices.**

**Answer:** Option 4: 13.

**Key concepts used:** Deductive reasoning using basic mathematical knowledge -- **technique of iteration** or experimenting with prospective variable values. Termination **uses the set of choice values as problem solving resource.**

**Q4.** If $5^{\sqrt{x}} + 12^{\sqrt{x}} = 13^{\sqrt{x}}$ then value of $x$ is,

- $\frac{25}{4}$
- 4
- 6
- 9

#### Solution

From basic mathematical sense, we can take the first decision as $\sqrt{x}$ must be an integer. Derivation of the value of $x$ and then the term values if $\sqrt{x}$ is a fraction would be too complicated in the expected time given for answer and also beyond the scope of syllabus.

Secondly, 5 and 12 having a large separation, if the power is large, increasing power for 13 cannot be compensated by same increasing power of 12. With this reasoning, we would start iteration with power as 1, and then at $\sqrt{x} = 2$ we get equality,

$5^2 + 12^2 = 13^2$.

So, $x=4$.

**Answer:** Option 2: 4.

**Key concepts used:** Analytical reasoning based on all resources -- the nature of the expression, the choice values, the complexity and knowledge level required -- iteration technique.

Answering many of these questions require intelligent analysis based reasoning for fastest solution.

Alternatively, if you remember the relationship, answer comes in about 5 seconds. But the strength of the analytical reasoning lies in its wide applicability. Even if the values are changed, you can get to the answer surely.

**Q5.** If $a + b + c = 0$ then the value of $\frac{a^2 + b^2 + c^2}{a^2 - bc}$ is,

- 0
- 1
- 2
- 3

#### Solution

Sensing that the denominator must be a factor of the numerator by noticing the nature of the choice values, we resort straightaway to algebraic manipulation of isolating the factor $a^2 - bc$ in the numerator of the target expression.

Thus at the first step we get, numerator,

$\begin{align}N & = a^2 - bc + (b + c)^2 - bc \\

& = a^2 - bc + (-a)^2 - bc \\

& = a^2 - bc + a^2 - bc=2(a^2-bc)\\

\end{align}$

With denominator $(a^2-bc)$ cancelling out against the numerator,

Target expression, $E = 2$.

**Answer:** Option 3: 2.

**Key concepts used:** Selecting the strategy of factoring out the denominator in the numerator expression arrived at by analysing the nature of the expression **along with the nature of choice values** -- use of the other given expression value $a + b + c = 0$, or $b + c = -a$.

Always try to use a given equation with RHS as 0, by trasposing or other techniques.

**Q6.** If $x^2 + 2 = 2x$ then the value of $x^4 - x^3 + x^2 + 2$ will be,

- 0
- 1
- -1
- $\sqrt{2}$

#### Solution

The very first step that we will take is to **transpose the equation 1 to have a numerical value on the right of the equality sign.** We intend to factor the target expression with this transposed expression, $x^2 - 2x = -2$.

Factoring this way we get expression,

$E = x^2(x^2 - 2x) + 2x^3 - x^3 + 2x = x^3 - 2x^2 + 2x$.

We have straightaway simplified the tail by using $x^2 + 2 = 2x$ for substitution.

So, $E = x(x^2 - 2x) + 2x = -2x + 2x = 0$

**Answer:** Option 1 : 0.

**Key concepts used:** Usable transformation of the first given expression -- factoring and substituion.

**Q7.** If $x = (\sqrt{2} + 1)^{-\frac{1}{3}}$, then the value of $\left(x^3 - \frac{1}{x^3}\right)$,

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

#### Solution

We see $x^3$ in target expression and power $\frac{1}{3}$ in the given expression of $x$. So we straightway, cube the given expression and use the value in the target expression, giving target expression,

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

So, $E = \frac{1 - (2 + 2\sqrt{2} + 1)}{\sqrt{2} + 1} = \frac{-2 - 2\sqrt{2}}{\sqrt{2} + 1}=-2$

**Answer: **Option 2: $-2$.

**Key concepts used:** Removing the **complexity of fractional power in the given expression which satisfies the target expression requirements** also -- **End state analysis approach** -- surd arithmetic.

**Alternatively** we will use rationalization of surd denominator on,

$x^3=\displaystyle\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$, by multiplying numerator and denominator by $\sqrt{2}-1$.

Then,

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

**Q8.** If $4b^2 + \frac{1}{b^2}=2$, then value of $8b^3 + \frac{1}{b^3}$ is,

- 0
- 2
- 1
- 5

#### Solution

If we forget the given expression for the moment and consider the target expression in the form of $x^3 + y^3$, we get,

$E = (2b + \frac{1}{b})(4b^2 - 2 + \frac{1}{b^2})$

Now if we use the given expression in the second factor, it turns out to be 0, resulting in the answer as 0.

**Answer: **Option 1: 0.

**Key concepts used:** The basic problem in this sum was the faint invitation to find the the value of $2b + \frac{1}{b}$. Avoiding that, it is necessary to use the most basic formula of expanding $x^3 + y^3$ sensing ahead that $-xy$ in the second factor $x^2 -xy + y^2$ will be $-2$.

**Q9.** If $x^\frac{1}{3} + y^\frac{1}{3} - z^\frac{1}{3} = 0$ then value of $(x + y - z)^3 + 27xyz$ is,

- $-1$
- 1
- 0
- 27

**Solution**

In this type of sums always use transposition to avoid dealing with cubing or squaring a three variable expression. Thus from the given expression we get,

$x^\frac{1}{3} + y^\frac{1}{3} = z^\frac{1}{3}$

Now cubing both sides we get,

$x + 3x^\frac{1}{3}y^\frac{1}{3}(x^\frac{1}{3} + y^\frac{1}{3}) + y = z$

or, $(x + y - z) = -3x^\frac{1}{3}y^\frac{1}{3}z^\frac{1}{3}$

Cubing again both sides, $(x + y - z)^3 = -27xyz$.

So answer is 0.

Answer: Option 3: 0.

**Key concepts used:** Transposition -- moving towards target powers by cubing and using simplifying again by using given expression -- cubing again.

**Q10.** If $x^{x\sqrt{x}} = (x\sqrt{x})^x$ then $x$ is equal to,

- $\frac{4}{9}$
- $\frac{2}{3}$
- $\frac{9}{4}$
- $\frac{3}{2}$

**Solution**

In indices sums, first explore * equalization of base technique*. The LHS is already in powers of $x$. Let's then bring the RHS in powers of $x$,

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

Now equating powers on both sides, we get,

$x\sqrt{x}=\frac{3x}{2}$, or,$\sqrt{x}=\frac{3}{2}$,

or $x=\frac{9}{4}$.

**Answer:** Option 3: $\frac{9}{4}$.

**Key concepts used:** Equalization of base to $x$ -- equating the powers.

### 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 Algebra that you may find valuable

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 other related tutorials

**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 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 Solved Question Set 30, Algebra 7**

**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**

#### Efficient solutions for difficult SSC CGL problems on Algebra in a few steps

**How to solve a difficult surd algebra question by repeated componendo dividendo in a few steps 17**

**How to solve difficult SSC CGL level problem mentally using patterns and methods 16**

**How to solve a difficult SSC CGL Algebra problem mentally in quick time 15**

**How to solve a tricky Algebra problem by adapted componendo dividendo in a few steps**

**How to solve difficult SSC CGL Algebra problems in a few steps 14**

**How to solve difficult SSC CGL Algebra problems in a few steps 13**

**How to solve difficult SSC CGL Algebra problems in a few steps 12**

**How to solve difficult SSC CGL Algebra problems in a few steps 11**

**How to solve difficult SSC CGL Algebra problems in a few assured steps 10**

**How to solve difficult SSC CGL Algebra problems in a few steps 9**

**How to solve difficult SSC CGL Algebra problems in a few steps 8**

**How to solve difficult SSC CGL Algebra problems in a few steps 7**

**How to solve difficult Algebra problems in a few simple steps 6**

**How to solve difficult Algebra problems in a few simple steps 5**

**How to solve difficult surd Algebra problems in a few simple steps 4**

**How to solve difficult Algebra problems in a few simple steps 3**

**How to solve difficult Algebra problems in a few simple steps 2**

**How to solve difficult Algebra problems in a few simple steps 1**

#### 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 CPO CAPF CISF level Solved Question sets on Algebra

**SSC CPO level Solved Question Set 1 Algebra 1**

**SSC CPO level Solved Question Set 2 Algebra 2**

#### 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**

**SSC CHSL level Solved Question set 14 on Algebra 4**

### 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**