## 58th SSC CGL level Solution Set, 14th on Algebra

This is the 58th solution set of 10 practice problem exercise for SSC CGL exam and 14th on topic Algebra.

In solving the problems in this session, analytical conceptual strategies and powerful techniques led to the solution in a few simple steps and with least amount of number crunching. To absorb the concepts though the student must first understand and then apply the concepts and techniques for actual problem solving during managed practice sessions.

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 * 58th SSC CGL question set and 14th on Algebra* before going through the solution.

### 58th solution set - 10 problems for SSC CGL exam: 14th on topic Algebra - answering time 12 mins

**Problem 1.**

If $x=5-\sqrt{21}$, then the value of $\displaystyle\frac{\sqrt{x}}{\sqrt{32-2x}-\sqrt{21}}$ is,

- $\displaystyle\frac{1}{\sqrt{2}}\left(\sqrt{3}-\sqrt{7}\right)$
- $\displaystyle\frac{1}{\sqrt{2}}\left(\sqrt{7}+\sqrt{3}\right)$
- $\displaystyle\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)$
- $\displaystyle\frac{1}{\sqrt{2}}\left(7-\sqrt{3}\right)$

** Solution 1 - Problem analysis**

First task must be to express the surd expression under the square root in the numerator of the target expression to a square of sum so that that we get the value of $\sqrt{x}$ as a surd expression without any square root binding. The driving concept here is,

Double square roots cannot be dealt with.

Let us do this task first. The given expression,

$x=5-\sqrt{21}$

$=\frac{1}{2}\left(10-2\sqrt{21}\right)$.

We have applied here the technique of multiplying the middle surd term in the target square of sum by 2. The transformed equation is then,

$x=\frac{1}{2}\left(10-2\sqrt{21}\right)$

$=\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)^2$

So,

$\sqrt{x}=\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)$.

Now only we will have a look at the denominator of the target expression and from the form of it decide to directly substitute the value of $x$.

**Solution 1 - Problem solving execution**

Target expression,

$E=\displaystyle\frac{\sqrt{x}}{\sqrt{32-2x}-\sqrt{21}}$

$=\displaystyle\frac{\sqrt{x}}{\sqrt{32-2(5-\sqrt{21})}-\sqrt{21}}$

$=\displaystyle\frac{1}{\sqrt{2}}\left(\displaystyle\frac{\sqrt{7}-\sqrt{3}}{\sqrt{22+2\sqrt{21}}-\sqrt{21}}\right)$

$=\displaystyle\frac{1}{\sqrt{2}}\left(\displaystyle\frac{\sqrt{7}-\sqrt{3}}{1+\sqrt{21}-\sqrt{21}}\right)$, as $22+2\sqrt{21}=(1+\sqrt{21})^2$

$=\displaystyle\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)$.

**Answer:** Option c: $\displaystyle\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)$.

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

*--*

**Surd square of sum conversion***.*

**surd simplification****Problem 2.**

If $a+b+c+d=4$, then find the value of $\displaystyle\frac{1}{(1-a)(1-b)(1-c)}+\displaystyle\frac{1}{(1-b)(1-c)(1-d)}+$

$\hspace{30mm}\displaystyle\frac{1}{(1-c)(1-d)(1-a)}+\displaystyle\frac{1}{(1-d)(1-a)(1-b)}$ is,

- 1
- 4
- 5
- 0

**Solution 2 - Problem analysis**

The key patterns here are the presence of a 4 on the RHS of the given expression, four single variable terms $a$, $b$, $c$, and $d$ on the LHS and the target expression in terms of $(1-a)$, $(1-b)$, $(1-c)$, $(1-d)$.

The situation clearly calls for **two problem solving techniques** to be applied one after the other,

- first
1 each between the four terms on the LHS to form the**share the secondary term 4 on the RHS,***sum of the terms as appearing in the target expression*, and - second, applying
technique, substitute, $p=(1-a)$, $q=(1-b)$, $r=(1-c)$ and $s=(1-d)$, with the new single variables being intermediate use**component expression substitution**.**dummy variables**

The problem is simplified by the two steps to,

"If $p+q+r+s=0$, the value of $\displaystyle\frac{1}{pqr}+\displaystyle\frac{1}{qrs}+\displaystyle\frac{1}{rsp}+\displaystyle\frac{1}{spq}$ is,".

The two techniques together result in a simpler problem. This is an instance of the * well known concept of Solve a simpler problem*.

#### Solution 2 - Problem solving execution

The simplified target expression is then,

$\displaystyle\frac{1}{pqr}+\displaystyle\frac{1}{qrs}+\displaystyle\frac{1}{rsp}+\displaystyle\frac{1}{spq}$

$=\displaystyle\frac{1}{pqrs}\left(p+q+r+s\right)$, the whole expression is summed up where LCM of the denominators is $pqrs$.

$=0$.

**Answer:** Option d : 0.

**Key concepts used:** * Sharing secondary resources* --

**component expression substitution -- dummy variable -- solving a simpler problem.****Problem 3.**

If $a(2+\sqrt{3})=b(2-\sqrt{3})=1$, then the positive value of $\displaystyle\frac{1}{a^2+1}+\displaystyle\frac{1}{b^2+1}$ is,

- $1$
- $4$
- $9$
- $-5$

**Solution 3 - Problem analysis**

The two surd expressions are suitable for inversion and rationalization as,

$2+\sqrt{3}=\displaystyle\frac{1}{2-\sqrt{3}}$, and vice versa.

The * more efficient way to proceed is then to evaluate sums of inverses rather than squaring and summing $a$ and $b$ when evaluating the target expression*.

**Solution 3 - Problem solving execution**

The first given expression is,

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

Or, $a=\displaystyle\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$,

and, $\displaystyle\frac{1}{a}=2+\sqrt{3}$.

Similarly,

$b(2-\sqrt{3})=1$,

Or, $b=\displaystyle\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$,

and, $\displaystyle\frac{1}{b}=2-\sqrt{3}$.

The target expression is,

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

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

$=(2+\sqrt{3})\displaystyle\frac{1}{4}+(2-\sqrt{3})\displaystyle\frac{1}{4}$

$=1$

**Answer:** Option a: 1.

**Key concepts used: ****Problem solving strategy formulation** -- **principle of interaction of inverses** -- **input transformation** -- **Target transformation** -- **surd rationalization** -- **efficient simplification.**

#### Important Recommendation

(if it is possible to convert it in that manner)Almost always it is a better option to deal with the target expression in terms of sum of inverse expressionswhich invariably will involverather than evaluating individual terms of target expressionas well asmore calculations that will waste your valuable secondsincrease chances of error.

Be an intelligent problem solver using concepts and strategies rather than be a human calculator.

**Problem 4.**

If $\displaystyle\frac{x}{xa+yb+zc}=\displaystyle\frac{y}{ya+zb+xc}=\displaystyle\frac{z}{za+xb+yc}$, and $x+y+z \neq 0$ then each ratio can be expressed as,

- $\displaystyle\frac{1}{a+b-c}$
- $\displaystyle\frac{1}{a+b+c}$
- $\displaystyle\frac{1}{a-b-c}$
- $\displaystyle\frac{1}{a-b+c}$

**Solution 4 - Problem analysis and solving**

The given expression forms a chain of three expressions and so calls for application of * Chained equation treatment technique.* By this technique, an additional dummy constant is appended to the end of the chain,

$\displaystyle\frac{x}{xa+yb+zc}=\displaystyle\frac{y}{ya+zb+xc}=\displaystyle\frac{z}{za+xb+yc}=p$.

The advantage is, the chained expressions are freed and are able to form three independent equations with the dummy constant as below,

$\displaystyle\frac{x}{xa+yb+zc}=p$,

Or, $x=p(xa+yb+zc)$.

$\displaystyle\frac{y}{ya+zb+xc}=p$,

Or, $y=p(ya+zb+xc)$, and

$\displaystyle\frac{z}{za+xb+yc}=p$,

Or, $z=p(za+xb+yc)$.

The solution is now visible.

The three equations are added up,

$x+y+z=p(a+b+c)(x+y+z)$,

Or, $p=\displaystyle\frac{1}{a+b+c}$.

**Answer:** Option b: $\displaystyle\frac{1}{a+b+c}$.

**Key concepts used: Problem analysis -- Chained equation treatment technique -- Collection of friendly terms.**

**Problem 5.**

If $3(a^2+b^2+c^2)=(a+b+c)^2$, then the relation between $a$, $b$ and $c$ is,

- $a=b=c$
- $a \neq b=c$
- $a=b \neq c$
- $a \neq b \neq c$

**Solution 5 - Problem analysis **

Expanding the RHS we will get one $(a^2+b^2+c^2)$ reducing three such in the LHS to 2. This transformation makes the problem state promising,

$3(a^2+b^2+c^2)=(a+b+c)^2$,

Or, $3(a^2+b^2+c^2)=a^2+b^2+c^2 +2(ab+bc+ca)$,

Or, $2(a^2+b^2+c^2)=2(ab+bc+ca)$.

From this point onwards it is a straightforward rearrangement of terms and application of another important algebraic concept.

**Solution 5 - Problem solving execution**

We have,

$2(a^2+b^2+c^2)=2(ab+bc+ca)$,

Or, $(a-b)^2+(b-c)^2+(c-a)^2=0$.

We have used the concept of * Principle of collection of friendly terms* and rearranged the terms to the fruitful result.

By the well known algebraic concept of * zero sum of square expressions*,

If a sum of terms each of which is square of an expression involving real variable values is zero, the individual terms must be zero.

By applying this Zero sum of square terms concept we have,

$(a-b)=(b-c)=(c-a)=0$,

Or, $a=b=c$.

**Answer:** Option a: $a=b=c$.

**Key concepts used:** * Problem analysis* --

**Key pattern identification -- Collection of friendly terms -- Zero sum of square terms.****Problem 6.**

If $x=\sqrt{5} + \sqrt{3}$ and $y=\sqrt{5} - \sqrt{3}$, then the value of $(x^4-y^4)$ is,

- $16$
- $544$
- $64\sqrt{15}$
- $32\sqrt{15}$

**Solution 6 - Problem analysis and solving**

We find $x^2+y^2$ will eliminate the term $2\sqrt{15}$, resulting in,

$x^2+y^2=2\times{8}=16$.

Similarly the other factor of $x^4-y^4$, namely, $x^2-y^2$ results in the square of surd terms canceled out,

$x^2-y^2=4\sqrt{15}$.

So,

$x^4-y^4=(x^2+y^2)(x^2-y^2)$

$=16\times{4\sqrt{15}}$

$=64\sqrt{15}$.

This is the shortest calculaion path.

**Answer:** Option c : $64\sqrt{15}$.

**Key concepts used: **

*--*

**Basic algebraic concepts***.*

**Surd simplification -- efficient simplification****Problem 7.**

Let $p=\displaystyle\frac{5}{18}$, then $27p^3-\displaystyle\frac{1}{216} - \displaystyle\frac{9}{2}p^2 + \displaystyle\frac{1}{4}p$ is equal to,

- $\displaystyle\frac{4}{27}$
- $\displaystyle\frac{10}{27}$
- $\displaystyle\frac{5}{27}$
- $\displaystyle\frac{8}{27}$

**Solution 7 - Problem analysis**

The challenge in this problem is to find the easiest path to the solution that involves least amount of calculation. The only way to do this is to transform the given expression to a sum of cubes, if possible. Let us try.

**Solution 7 - Problem solving execution**

The given expression,

$27p^3-\displaystyle\frac{1}{216} - \displaystyle\frac{9}{2}p^2 + \displaystyle\frac{1}{4}p$

$=(3p)^3-3(3p)^2\times{\displaystyle\frac{1}{6}}+3(3p)\times{\left(\displaystyle\frac{1}{6}\right)^2} - \left(\displaystyle\frac{1}{6}\right)^3$

$=\left(3p-\displaystyle\frac{1}{6}\right)^3$

$=\left(\displaystyle\frac{5}{6}-\displaystyle\frac{1}{6}\right)^3$

$=\displaystyle\frac{8}{27}$.

**Answer:** Option d: $\displaystyle\frac{8}{27}$.

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

*--*

**Deductive reasoning**

*Hypothesis testing --*

*Term rearrangement -- Cube of sum.*#### Deductive Reasoning chain:

- By our experience, most of the SSC CGL problems are
*solvable by elegant concept based methods*within a minute.*The problems never demand time consuming calculations*. With this knowledge we were fairly sure that the expression can be transformed to a compact form. - With cubes of $3p$ and $\displaystyle\frac{1}{6}$ in the expression, the hypothesis formed immediately for compact form of $\left(3p-\displaystyle\frac{1}{6}\right)^3$.
- A little bit of term rearrangement and transformation of the terms in expanded cube of sum form produced the final confirmation. All could be done in mind, without any calculation, which is the hallmark of practically any SSC CGL problem, however hard it looks.

**Problem 8.**

For real $x+y+z=6$, then the value of $(x-1)^3+(y-2)^3+(z-3)^3$ is,

- $3xyz$
- $3(x-1)(y-2)(z-3)$
- $2(x-1)(y-2)(z-3)$
- $(x-1)(y-2)(z-3)$

** Solution 8 - Problem analysis**

Comparing the cube terms in the target expression with the given expression we decided immediately to transform the given expression as,

$x+y+z=6$,

Or, $(x-1)+(y-2)+(z-3)=0$.

This is use of * End state analysis approach* along with

*term on the RHS between the terms on the LHS. Usually the sharing is equal for each term on LHS. But here the sharing is determined by the end state of target expression.*

**Sharing of secondary resource**The next step is to use * Component expression substitution* by taking,

$p=(x-1)$,

$q=(y-2)$, and

$r=(z-3)$.

This simplifies the problem into a known form, and is an instance of * solve a simpler problem concept*.

**Solution 8 - Problem solving execution**

By the substitution of the intermediate * dummy variables*, the problem is transformed into,

"If $p+q+r=0$, then the value of $p^3+q^3+r^3$ is,".

We have then,

$p+q+r=0$,

Or, $(p+q)=-r$

Or, $(p+q)^3=-r^3$,

Or, $p^3+q^3+r^3+3pq(p+q)=0$,

Or, $p^3+q^3+r^3=3pqr$.

The LHS is the target expression. We finally get its value by * reverse substitution* of the values of the dummy variables,

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

**Answer:** Option b: $3(x-1)(y-2)(z-3)$.

**Key concepts used:** **End state analysis approach -- Component expression substitution -- Sharing of secondary resources -- Dummy variables -- Solve a simpler problem -- Problem transformation -- Basic algebra concepts -- Reverse substitution****.**

**Problem 9.**

If $x+\displaystyle\frac{1}{x}=\sqrt{3}$, then the value of $x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$ is,

- $1$
- $\sqrt{3}$
- $-\sqrt{3}$
- $0$

**Solution 9 - Problem analysis**

Looking at the given expression and the high power terms of the target expression, we decide that the key lies in transforming the given expression to sum of higher powers of inverse terms with value 0.

If we can find such an expression, we can start factorizing the target expression so that the zero valued tranformed given expression is factored out at each step.

This is what we call as * Zero valued factor search technique*.

* In simplifying any long target expression to a simple resultant output value*, the surest way to solution is to

*by transforming the given expression and use it for*

**search for a suitable expression with zero value***from the long target expression.*

**continued factor extraction**We felt this as the surest way to simplify the target expression fast.

#### Solution 9 - Problem solving execution

We have the given expression,

$x+\displaystyle\frac{1}{x}=\sqrt{3}$.

Squaring and rearranging,

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

Or, $x^2+\displaystyle\frac{1}{x^2}-1=0$.

With this result we know we have found the key to the solution.

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

Or, $x^3+\displaystyle\frac{1}{x^3}=0$.

This is the expression form we desired. We will now take out this sum of inverse cubes from the target expression, and at each such step, the number of terms of the target expression will reduce because of the zero value of the factor taken out.

Target expression,

$E=x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$

$=x^{27}\left(x^3+\displaystyle\frac{1}{x^3}\right)+x^{18}+x^{12}+x^6+1$

$=x^{15}\left(x^3+\displaystyle\frac{1}{x^3}\right)+x^6+1$

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

$=0$.

**Answer:** Option d: 0.

**Key concepts used:** **End state analysis -- Zero valued factor search -- Principle of sum of inverses -- Continued factor extraction -- Efficient simplification.**

**Problem 10.**

If $\displaystyle\frac{p}{a}+\displaystyle\frac{q}{b}+\displaystyle\frac{r}{c}=1$, and $\displaystyle\frac{a}{p}+\displaystyle\frac{b}{q}+\displaystyle\frac{c}{r}=0$ where $a$, $b$, $c$ and $p$, $q$, $r$ are non-zero, the value of $\displaystyle\frac{p^2}{a^2}+\displaystyle\frac{q^2}{b^2}+\displaystyle\frac{r^2}{c^2}$ is,

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

**Solution 10 - Problem analysis and solving**

Though there are six variables, we find **division of*** three pairs of variables repeat unchanged* in their occurrence in both the given and target expressions. So we decide, this is a suitable situation for

*by taking,*

**Component expression substitution**$\displaystyle\frac{p}{a}=x$,

$\displaystyle\frac{q}{b}=y$, and

$\displaystyle\frac{r}{c}=z$.

Here $x$, $y$ and $z$ are the **interim dummy variables.**

By these three substitutions the given problem is simplified to,

"If $x+y+z=1$, and $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$, find the value of, $x^2+y^2+z^2$".

This is thus a case of * solving a simpler problem*.

From the second equation, summing up the three terms we get,

$\displaystyle\frac{xy+yz+zx}{xyz}=0$,

Or, $xy+yz+zx=0$.

Thus,

$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=1$.

**Answer: **Option a: 1.

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

*--*

**Component expression substitution***--*

**Principle of representative***--*

**Dummy variable***--*

**Problem transformation**

**Solving a simpler problem --**

**Basic algebra concepts***.*

**-- efficient simplification**### 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**

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

**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 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 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**