## 45th SSC CGL level Solution Set, 11th on Algebra

This is the 45th solution set of 10 practice problem exercise for SSC CGL exam and 11th on topic Algebra.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through these solutions, one must solve many problems in a systematic manner using this conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

If you have not yet taken this test you may take it by referring to the * 45th SSC CGL question set and 11th on Algebra* before going through the solution.

Watch the **video solutions in the two-part video** below.

**Part I: Q1 to Q5**

**Part II: Q6 to Q10**

### 45th solution set - 10 problems for SSC CGL exam: 11th on topic Algebra - answering time 15 mins

**Problem 1.**

If $\displaystyle\frac{5x}{2x^2+5x+1}=\frac{1}{3}$, then the value of $\left(x+\displaystyle\frac{1}{2x}\right)$ is,

- 5
- 10
- 15
- 20

** Solution 1: Quick solution by mathematical reasoning and matching target expression with the given expression**

The target expression is a modified sum of inverses with unequal coefficients. Simplify the given expression and you are certain to get some form of sum of inverses.

Divide numerator and denominator of the given expression both by x and cross-multiply to get,

$2x+\displaystyle\frac{1}{x}+5=15$,

Or, $2x+\displaystyle\frac{1}{x}=10$.

So you got your sum of inverses. At this second stage, you just have to fine tune the coefficients of the two terms to match the target and it is easy.

Multiply through by $\displaystyle\frac{1}{2}$ to get,

$x+\displaystyle\frac{1}{2x}=5$.

**Answer:** option a: 5.

Solution is in just two steps, and all in mind.

**Answer:** Option a: 5.

**Key concepts used:** * Mathematical reasoning that given expression can be changed to a sum of inverses* --

**Given expression simplification to obtain the sum of inverses -- equalization of coefficients of sum of inverses.****Problem 2.**

If $x^2+1=2x$, then the value of $\displaystyle\frac{x^4+\displaystyle\frac{1}{x^2}}{x^2-3x+1}$ is,

- $1$
- $-2$
- $0$
- $2$

**Solution 2 - Problem analysis**

Again we take the * End state analysis approach* but this time in conjunction with

*first to compare transformed input expression, $x^2-2x+1=0$ and then find that we can very well use it to simplify the denominator of target expression greatly,*

**input transformation**$E=\displaystyle\frac{x^4+\displaystyle\frac{1}{x^2}}{x^2-2x+1-x}$,

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

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

This is a very encouraging result as we can now apply the * principle of interaction of inverses* to evaluate the

*easily.*

**sum of inverse of cubes**#### Solution 2 - Problem solving execution

To evaluate the sum of inverses of cubes in the target expression we now feel the need to use the given expression in a second way using * multiple input use technique*,

$x^2+1=2x$,

Or, $x+\displaystyle\frac{1}{x}=2$, **a pure sum of inverses.**

Squaring the equation and simplifying we get,

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

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

Immediately we recognize this expression as the second factor in the two factor expanded expression for sum of cubes,

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

$=2$.

So the target expression value,

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

Again using the rich algebraic techniques and the rich general problem solving technique we have reached the solution in a few steps wholly menatally without writing down anything using pen and paper. We could use the * rich concept of mental maths* on this problem also.

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

**Key concepts used:** * End state analysis approach* --

*--*

**input transformation technique***, we always look for ways to simplify the denominator for immediate positive results --*

**denominator simplification***, we have further simplified the already simplified target expression immediately without any further delaye and obtained the sum of cubes expression --*

**simplification technique***in transforming the input expression into the*

**multiple input use technique***-- taking recourse to the well known steps embodied in the*

**second form of a pure sum of inverses***, reaching the*

**principle of interaction of inverses**

**solution wholly mentally -- mental maths.****Problem 3.**

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 equal to,

- $1$
- $0$
- $4$
- $2$

**Solution 3 - Problem analysis and solving**

As we examined the two given expressions, we could recognize the well known * expanded four term expression of a cube of sum split into two parts* in the two LHS expressions. Sometimes we meet with this ingenious form of problem formation. This is

**split expression problem structure.**It immediately urges us to add up the two equations giving,

$\left(x+\displaystyle\frac{1}{x}\right)^3=8a^3$,

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

Squaring and rearranging the terms we get,

$\left(x^2+\displaystyle\frac{1}{x^2}\right)=4a^2-2$.

Now we will subtract the second given expression from the first getting,

$\left(x-\displaystyle\frac{1}{x}\right)^3=8b^3$,

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

Or, $\left(x^2+\displaystyle\frac{1}{x^2}\right)=4b^2+2$.

Equating the two derived results,

$4a^2-2=4b^2+2$,

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

**Answer:** Option a: $1$.

In this case also the simple steps could be carried out wholly in mind.

**Key concepts used: Principle of collection of friendly terms** in adding and subtracting the LHSs of the two given expressions to get the cube of additive sum of inverses and subtractive sum of inverses in $x$ respectively --

*to derive the sum of inverse of squares from two derived expressions -- equating the same expressions obtained from two sources we get the solution quickly --*

**principle of interaction of inverses**

**mental maths.****Problem 4.**

If $ab + bc+ca=0$, then the value of $\displaystyle\frac{1}{a^2-bc} +\displaystyle\frac{1}{b^2-ac}+\displaystyle\frac{1}{c^2-ab}$ is,

- $0$
- $1$
- $-1$
- $2$

**Solution 4 - Problem analysis**

Though the denominators of the three fraction terms of the target expression are symmetric around $a$, $b$ and $c$, at the first look, there seemed to be no easy way to resolve the problem of equalizing the denominators to the same form so that the sum of three fractions can be reduced to a simple number.

At the second look, we discovered the slightly deviant partially hidden key pattern using the given equation resource. We just need to replace $-bc$ in $a^2-bc$ by $ab+ac$ so that the denominator of the first term is simplified to, $a(a+b+c)$. Likewise the second and the third.

Let us show you how.

**Solution 4 - Problem solving execution**

Given,

$ab+bc+ca=0$,

Or, $-bc=ab+ca$,

Or, $-ac=ab+bc$, and

$-ab=bc+ca$.

Substituting these three products in the three denominator expressions we transform the target expression as,

$\displaystyle\frac{1}{a^2-bc} +\displaystyle\frac{1}{b^2-ac}+\displaystyle\frac{1}{c^2-ab}$

$=\displaystyle\frac{1}{a^2+ab+ca} +\displaystyle\frac{1}{b^2+ab+bc}+\displaystyle\frac{1}{c^2+bc+ca}$

$=\displaystyle\frac{1}{a+b+c}\left(\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}\right)$

$=\displaystyle\frac{1}{a+b+c}\left(\displaystyle\frac{ab+bc+ca}{abc}\right)$

$=0$.

This is the second use of the given input resource.

There has been no need to to derive the steps using pen and paper; once the key pattern was discovered in about 15 seconds rest of the steps took in total well below a minute, wholly by mental manipulation of expressions. * Mental maths* at work to solve the elegantly in a very short time.

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

**Key concepts used: Denominator simplification** --

*with a target to*

**input transformation***to same value as far as possible --*

**equalize the denominators***-- base equalization --*

**denominator equalization***on algebraic fractions --*

**fraction arithmetic***-- the*

**multiple input use***broke the back of the problem handing us the elegant solution --*

**key pattern discovery**

**mental maths.****Problem 5.**

If $x$ is a rational number and $\displaystyle\frac{(x+1)^3-(x-1)^3}{(x+1)^2-(x-1)^2}=2$ then the sum of numerator and denominator of $x$ is,

- $7$
- $4$
- $5$
- $3$

**Solution 5 - Problem analysis and execution**

We need to cancel out the common factor between the numerator and denominator and then simplify. As the principal terms are $(x+1)$ and $(x-1)$ it is much better to temporarily use representative substituted single variables for these two expressions. * Abstraction technique*,

*and*

**principle of representative***at work here. Goal is to transform a more complex expression to a simple form so that we can easily and quickly deal with it and simplify it further.*

**substitution technique**Thus, at the outset we use, $p=x+1$ and $q=x-1$, to simplify the given expression,

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

Or, $\displaystyle\frac{p^3-q^3}{p^2-q^2}=2$,

Or, $\displaystyle\frac{p^2+pq+q^2}{p+q}=2$, eliminating factor $(p-q)$ between numerator and denominator,

Or, $\displaystyle\frac{(p+q)^2-pq}{p+q}=2$, expression transformation in terms of $(p+q)$ and $pq$ as both gives simple results, $(p+q)=2x$ and $pq=x^2-1$; this is * efficient simplification*,

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

Or, $3x^2-4x+1=0$

Or, $(3x-1)(x-1)=0$

So one root of $x$ is $\displaystyle\frac{1}{3}$ giving an answer of 4.

The other root of $x=1$ produces an answer value 2 which is not in among the given choices. Using the free resource of the choice values we have eliminated this second value of $x$ as a lead to a possible solution.

Thus 4 is the answer for the sum of numerator and denominator of $x$.

**Answer:** Option b: $4$.

**Key concepts used:** Basic algebraic concepts -- abstraction -- substitution -- * principle of representative* -- factorization of subtractive sum of cubes and subtractive sum of squares --

*--*

**factorization of quadratic equation***-- deductive reasoning --*

**mathematical reasoning***.*

**free resource use****Problem 6.**

If $a+b+c=0$ then the value of $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2$ will be equal to,

- $4(a^2+b^2+c^2)$
- $4(ab+bc+ca)$
- $0$
- $8abc$

**Solution 6 - Problem analysis and solving**

We replace,

$a+b=-c$,

$b+c=-a$, and

$c+a=-b$ to get the target expression as,

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

$=4(a^2+b^2+c^2)$.

Solution is in just in two steps wholly mentally. The key was using the given expression for simplifying substitutions. Mental maths at work here as the solution could be reached wholly mentlly. The key was identifying the simplifying substitutions.

**Answer:** Option a : $4(a^2+b^2+c^2)$.

**Key concepts used: **

*--*

**Key pattern identification***.*

**substitution technique****Problem 7.**

If $x^2+y^2=5xy$, ,then the value of $\left(\displaystyle\frac{x^2}{y^2}+\displaystyle\frac{y^2}{x^2}\right)$ is equal to,

- $32$
- $16$
- $-23$
- $23$

**Solution 7 - Problem analysis and solving **

Examining the target expression we find it to be essentially in terms of a single compound variable, namely, $\displaystyle\frac{x}{y}$,

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

$=\left(p^2+\displaystyle\frac{1}{p^2}\right)$, where we have represented $\displaystyle\frac{x}{y}$ as compound variable $p$.

So the problem is transformed to evaluation of this sum of inverse of squares. From the objective we are fairly sure that we will be able to transform the imput expression into a sum of inverses, from which getting the value of sum of inverses of squares is just a simple step more.

Now examining the given input expression we get a clear idea of how we can transform the given expression into a sum of inverses.

$x^2+y^2=5xy$,

Or, $\displaystyle\frac{x^2+y^2}{xy}=5$,

Or, $p+\displaystyle\frac{1}{p}=5$.

We got our desirable sum of inverses from which we can easily evaluate the sum of inverse of squares.

First the sum of inverses is squared and then simplified to get the final result,

$p+\displaystyle\frac{1}{p}=5$,

Or, $p^2+2+\displaystyle\frac{1}{p^2}=25$,

Or, $p^2+\displaystyle\frac{1}{p^2}=25-2=23$, our desired solution.

**Answer:** Option d: $23$.

** Key concepts used:** * Pattern identification* -- Identifying the two square terms in the target expression as primarily the

*-- to ease the manipulation and simplification of expressions we resorted to*

**same term appearing in direct and inverse form***,*

**abstraction technique***and*

**substitution technique***-- thus transformed the target expression became a sum of inverses of squares -- with this result we felt we should be able to transform the input expression into a sum of inverses in single power from which getting a sum of inverses in squares is just a simple step more --*

**principle of representative***--*

**deductive reasoning***--*

**input transformation**

**principle of interaction of inverses.****Problem 8.**

If $x+\displaystyle\frac{2}{x}=1$, then tne value of $\displaystyle\frac{x^2+x+2}{x^2(x-1)}$ is,

- $2$
- $1$
- $-1$
- $-2$

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

Comparing the given expression with the two expressions in the target fraction, we find similarity with the numerator expression. But to know the extent of similarity precisely, we proceeded to * linearize* (getting rid of the fraction form) the given expression in the form of a equation with 0 in the RHS,

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

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

Now we use this equation to simplify the numerator of the target expression first,

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

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

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

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

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

Again we feel the need to use the given expression but in a second form,

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

Or, $x^2-x=-2$

Using this result on the transformed and reduced target expression finally we get,

$E=\displaystyle\frac{2}{x^2-x}=-1$.

**Answer:** Option c: $-1$.

Here also the solution could be reached quickly through mental processing only.

**Key concepts used:** * Pattern identification* -- input transformation --

*--*

**numerator simplification***--*

**multiple input use**

**mental math.****Problem 9.**

If $\displaystyle\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} +1}=a\sqrt[3]{4}+b\sqrt[3]{2}+c$ and $a$, $b$ and $c$ are rational numbers, then $a+b+c$ is equal to,

- 1
- 3
- 0
- 2

**Solution 9 - Problem analysis**

The cube roots are only two and appear in the same form in the LHS and also in the RHS.

Secondly, the cube root is applied basically on 2; the cube root on 4 can be expressed in terms of cube root of 2.

Thirdly it is apparent from the RHS of the given equation that the coefficients $a$, $b$ and $c$ are to be compared with corresponding coefficients on the LHS.

With this knowledge now we will go ahead with the solution.

#### Solution 9 - Problem solving execution

We can assume $p=\sqrt[3]{2}$ to simplify the given equation to,

$\displaystyle\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} +1}=a\sqrt[3]{4}+b\sqrt[3]{2}+c$,

Or, $\displaystyle\frac{1}{p^2 + p +1}=ap^2+bp+c$.

Now the challenge is transformed to simplifying the denominator in the LHS so that it is eliminated altogether and we can compare the coefficients of like terms in $p$, our new variable, on the RHS and LHS to get the values of $a$, $b$ and $c$.

On examining we find the denominator to be the second factor of the factors of sum of cubes,

$x^3-y^3=(x-y)(x^2+x+1)$.

We then multiply both the numerator and denominator with the other factor $(x-y)$ to simplify as desired,

$\displaystyle\frac{1}{p^2 + p +1}=ap^2+bp+c$,

Or, $\displaystyle\frac{p-1}{(p-1)(p^2 + p +1)}=ap^2+bp+c$

Or, $p-1=ap^2+bp+c$, as $p^3-1=\left(\sqrt[3]{2}\right)^3-1=2-1=1$.

Now we can compare the coefficients of like terms on both LHS and RHS.

Comparing coefficients of $p^2$, $a=0$,

Comparing coefficients of $p$, $b=1$.

Lastly comparing the constants, $c=-1$.

Finally, $a+b+c=0$.

**Answer:** Option c: 0.

**Key concepts used:** * Key pattern of similarity of terms identified* -- abstraction -- substitution --

*--*

**Principle of representative***--*

**factors of sum of cubes**

**comparison of coefficients of like terms of an equation.****Problem 10.**

If $x=7+4\sqrt{3}$ then the value of $\left(\sqrt{x}+\displaystyle\frac{1}{\sqrt{x}}\right)$ is,

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

**Solution 10 - Problem analysis and solving**

As the input expression involves $x$ and the target expression $\sqrt{x}$, the surd expression on the RHS of the given equation as value of $x$ must be transformed to a square of a surd sum. Unless this is achieved the problem can't be solved.

#### Solution 10 - Problem solving execution

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

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

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

Or, $\displaystyle\frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$, by surd rationalization.

So the target experession,

$\left(\sqrt{x}+\displaystyle\frac{1}{\sqrt{x}}\right)=4$

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

**Key concepts used:** * Deductive reasoning* --

*--*

**key pattern recognition***--*

**transformation to square of sum of surd term expression***.*

**surd rationalization technique**### 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 other related topics

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

**More rich algebraic concepts and techniques 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**

#### 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 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 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**

### 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**