## 2nd SSC CGL Tier II level Solution Set, 2nd on Algebra

This is the 2nd solution set of 10 practice problem exercise for SSC CGL Tier II exam and also the 2nd 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 * SSC CGL Tier II level question set 2 on Algebra 2* before going through the solution.

### 2nd solution set - 10 problems for SSC CGL exam: 2nd on topic Algebra - answering time 12 mins

**Q1. **The number of possible values of $x$ in the equation, $\sqrt{x^2-x+1} +\displaystyle\frac{1}{\sqrt{x^2-x+1}}=2-x^2$ is,

- 1
- 2
- 0
- 4

** Solution 1 - Problem analysis and execution**

The LHS is a sum of inverses but on brief analysis we couldn't find any easy way to simplify the equation to an equation in $x$ from which we could find the values of $x$ or roots of the equation.

Having exhausted this possibility we remembered that according to minima of sum of inverses principle, the minimum value of a sum of inverses is 2. On the RHS as any real value of $x$ will reduce the value of the RHS from 2, the only possible value that $x$ can have is determined to be 0. So the answer is 1.

We repeat the mathematical mechanism of the minima of sum of inverses to drive home this important concept so that it can be used for solving such problems with confidence in future.

#### Solution 1 - Mechanism of minima of sum of inverses

In our case,

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

$\hspace{11mm}=\left(p+\displaystyle\frac{1}{p}\right)$, where $p=\sqrt{x^2-x+1}$

$\hspace{11mm}=\left(\displaystyle\frac{p^2+1}{p}\right)$.

As any real $p^2$ in the numerator will increase much faster than $p$ in the denominator for values of $p \geq 1$, the expression $E$ will reach its minimum value only when the $p=1$ and $E=2$.

To deal with the values of $p \leq 1$, we substitute $p$ by $\displaystyle\frac{1}{q}$ transforming the expression to,

$E=\left(p+\displaystyle\frac{1}{p}\right)$

$\hspace{5mm}=\left(\displaystyle\frac{1}{q} +q\right)$,

$\hspace{5mm}=\left(q+\displaystyle\frac{1}{q}\right)$.

Now when $p \leq 1$, $q \geq 1$ and the value of expression $E$ reaches its minima 2 also when $q=p=1$.

So for all real values of $p$, the sum of its inverses will have minimum value 2.

**Answer:** Option a: 1.

**Note:** We have first examined the equation for conventional solution evaluating and assessing the suitability of alternative paths and then only after failing to find any suitable path to solution, examined the equation more closely to identify the key pattern of sum of inverses equated to $2-x^2$, ideally suited for applying * minima of sum of inverses concept. *Minimum value of 2 on the LHS will necessitate a value of $x=0$ when RHS is $2-x^2$ is a result of mathematical reasoning using very basic concepts of

**variable value trend analysis.****Key concepts used:** *Alternative ** evaluation -- key pattern identification -- minima of sum of inverses -- variable value trend analysis -- mathematical reasoning*.

**Q2.** If $p+\displaystyle\frac{1}{p}=5$, then the value of $\displaystyle\frac{p^4+\displaystyle\frac{1}{p^2}}{p^2-3p+1}$ is,

- 50
- 55
- 70
- 110

**Solution 2 - Problem analysis**

We note that if we take $p$ out of the numerator as a factor it transforms to,

$p^4 +\displaystyle\frac{1}{p^2}=p\left(p^3 + \displaystyle\frac{1}{p^3}\right)$.

As we know we can easily get the value of sum of inverses in cubes from the given expression of sum of inverses applying well known concepts elaborated in * Principle of interaction of inverses*, we decide that we have virtually (we have to execute it yet) taken care of the numerator.

Focusing our attention to the denominator we don't find any easy way to convert it to a sum inverses form. It is then time to **use the given expression in the second way in expanded form.**

#### Solution 2 - Problem solving final stage

Let us take care of the numerator expression first.

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

Or, $p^2 -1 + \displaystyle\frac{1}{p^2}=25-3=22$.

So,

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

$=5\times{22}$

$=110$.

Thus numerator $=110p$.

Now we will use the given expression in a second way in expanded form. This is direct application of **Multiple input use technique.**

Expanding the given expression and rearranging we get,

$p^2 - 5p + 1=0$.

So denominator is,

$p^2 -3p + 1=2p$.

Finally then the desired value of the target expression is,

$E=\displaystyle\frac{110p}{2p}=55$.

**Answer:** Option b : 55.

**Key concepts used:** * Problem breakdown* in taking care of numerator and denominator separately after due analysis of target expression --

*Key pattern discovery*to use given expression in a second way to evaluate denominator

**-- principle of interaction of inverses -- output expression transformation -- target transformation -- input transformation to evaluate sum of inverse of cubes -- multiple input use technique**

**-- simplification.****Q3.** If $\sqrt{4x-9} + \sqrt{4x+9}=5 + \sqrt{7}$, find the value of $x$.

- 3
- 4
- 5
- 7

**Solution 3 - Problem analysis**

We can see from the form of the LHS and RHS that if we raise the equation to a square, the square root remains only with the middle term. Equating the square root terms or the non-square root terms between the two sides of the equation we will then reach the solution.

#### Solution 3 - Problem solving execution

Raising the given equation to the power of 2,

$\left(\sqrt{4x-9} + \sqrt{4x+9}\right)^2=\left(5 + \sqrt{7}\right)^2$,

Or, $8x +2\sqrt{16x^2-81}=32 +10\sqrt{7}$.

Equating the non-square-root terms of LHS and RHS,

$8x=32$,

Or, $x=4$.

This can also verified by equating the square root parts of LHS and RHS or substituting value of $x$.

**Answer:** Option b: 4.

**Key concepts used: **Deductive reasoning -- mathematical reasoning -- square of sum concept -- **coefficient comparison.**

**Q4. **If $\sqrt{2x} - \sqrt{3y}=0$ and $\sqrt{7x} + \sqrt{2y}=0$ then the value of $x+y$ is,

- 1
- 2
- 3
- 0

**Solution 4 - Problem analysis**

Though the two variables $x$ and $y$ are under square roots, essentially the two equations can be considered to two linear equations in two variables. The variables here turn out to be $\sqrt{x}$ and $\sqrt{y}$ instead of plain $x$ and $y$. This is use of abstraction. Instead of $\sqrt{x}$ and $\sqrt{y}$ the two variables could have been $x^2$ and $y^2$ and in this form we could have then solved for $x^2$ and $y^2$. The form of the equation is important not the form of the variables. The form of the equations here is linear.

To solve two linear equations in two variables we take the simple approach of equalizing the coefficients of one variable in two equations by changing the coefficient of the chosen variable in one equation suitably. This we do by multiplying the whole equation by a suitable factor.

In this case $\sqrt{y}$ calls for elimination because of its opposite signs in two equations. We select the first equation and multiply it by $\sqrt{\displaystyle\frac{2}{3}}$ to transform the coefficient of $\sqrt{y}$ from $\sqrt{3}$ to $\sqrt{2}$ thus making it equal to the corresponding coefficient in the second equation.

$\sqrt{2x} - \sqrt{3y}=0$

Or, $\sqrt{\displaystyle\frac{4x}{3}} - \sqrt{2y}=0$.

Adding this equation with the second equation $\sqrt{7x} + \sqrt{2y}=0$ we get,

$\sqrt{\displaystyle\frac{4x}{3}} + \sqrt{7x}=0$,

Or, $\sqrt{x}\left(\sqrt{\displaystyle\frac{4}{3}}+\sqrt{7}\right)=0$.

So,

$\sqrt{x}=x=0$ and substituting it in any of the two equations we get $y=0$ also.

Finally then, $x+y=0$.

**Answer:** Option d: 0.

**Key concepts used: **Abstraction -- solving two linear equations in two variables.

**Q5. **Find the remainder when $x^5-9x^2+12x-14$ is divided by $(x-3)$.

- 56
- 184
- 0
- 1

**Solution 5 - Problem analysis**

Finding no obvious pattern we decide to use the never failing method of * continued factor extraction*.

By this method, the factor of $(x-3)$ will be extracted from the larger given expression step by step, and at each step the factored part will be dropped as we need to find only the remainder here. This will reduce the degree of the expression by 1 at each step making it simpler and simpler finally yielding the remainder as an integer at the last step.

#### Solution 5 - Problem solving execution

So we apply the continued factor extraction on the pair of expressions,

$x^5-9x^2+12x-14$

$=x^4(x-3) +3x^4-9x^2+12x-14$

$\Rightarrow 3x^3(x-3) +9x^3 -9x^2+12x-14$

$\Rightarrow 9x^2(x-3) +27x^2-9x^2+12x-14$

$\Rightarrow 18x^2+12x-14$

$=18x(x-3)+54x+12x-14$

$\Rightarrow 66x-14$

$=66(x-3) +198 -14$

$\Rightarrow 184$.

**Answer:** Option b: 184.

**Key concepts used:** * Continued factor extraction technique* --

**division and remainder concepts.****Note:** Though this method may seem to be a bit time consuming when applying it first, with experience, understanding of the basic concept of mechanism and practice it can be done quickly.

**Q6.** If $a + \displaystyle\frac{1}{b}=1$ and $b + \displaystyle\frac{1}{c}=1$, then value of $c + \displaystyle\frac{1}{a}$ is,

- $1$
- $0$
- $2$
- $\displaystyle\frac{1}{2}$

**Solution 6 - Problem analysis**

To us the easiest way to solve this problem seems to find the value of $b$ in terms of $a$ from the first equation and substitute it in the second equation to get rid of variable $b$ altogether leaving only the variables $c$ and $a$.

#### Solution 6 - Problem solving execution

Finding $b$ in terms of $a$ from the first equation,

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

Or, $\displaystyle\frac{1}{b}=1-a$,

Or, $b=\displaystyle\frac{1}{1-a}$.

Substituting this value in the second equation,

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

Or, $\displaystyle\frac{1}{1-a} + \displaystyle\frac{1}{c}=1$,

Or, $\displaystyle\frac{1}{c}=1-\displaystyle\frac{1}{1-a}=-\displaystyle\frac{a}{1-a}$,

Or, $c=-\displaystyle\frac{1-a}{a}$,

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

**Answer:** Option a : 1.

**Key concepts used: ** Variable elimination in two linear equations by substitution --

*--*

**substitution technique**

**efficient simplification.**** Q7.** If $a:b=\displaystyle\frac{2}{9}:\displaystyle\frac{1}{3}$, $b:c=\displaystyle\frac{2}{7}:\displaystyle\frac{5}{14}$ and $d:c=\displaystyle\frac{7}{10}:\displaystyle\frac{3}{5}$ find the value of $a:b:c:d$.

- $8:12:15:7$
- $4:6:7:9$
- $30:35:24:16$
- $16:24:30:35$

**Solution 7 - Problem analysis and strategy formulation**

In the first step, the three ratio values are to be transformed to ratio of integer values with no common factors. This is **normalization of the ratio values. **

In the second step the three two quantity ratios are to be joined together to form a four quantity ratio.

Joining will be done between the first pair of ratios resulting in a three quantity ratio which again will be joined with the third ratio to finally form the four quantity ratio.

To join two ratios, the common quantity will be the denominator of the first ratio and the numerator of the second ratio as well as the ratio values corresponding to this common quantity in the two ratios will be same. If not same, the values will have to equalized to the base value of the LCM of the two original values. This is **base equalization.**

This will be crux of the whole problem solving process following the basic and rich ratio concepts.

#### Solution 7 - Problem solving execution first stage - ratio normalization

Normalizing the first ratio values we have,

$a:b=\displaystyle\frac{2}{9}:\displaystyle\frac{1}{3}$,

Or, multiplying the ratio values by 9,

$a:b=2:3$.

Normalizing the second ratio values,

$b:c=\displaystyle\frac{2}{7}:\displaystyle\frac{5}{14}$

Or multiplying the values by 14,

$b:c=4:5$.

Similarly normalizing the third ratio values by multiplying the values by 10,

$d:c=\displaystyle\frac{7}{10}:\displaystyle\frac{3}{5}=7:6$.

#### Solution 7 - Problem solving execution second stage - ratio joining

To join $a:b=2:3$ and $b:c=4:5$ we identify $b$ as the common quantity. The values corresponding to $b$ in the two ratios being $3$ and $4$, the target equalization value of this quantity will be the LCM of the two values, that is, $3\times{4}=12$.

We now transform the two ratios to equalize the values corresponding to quantity $b$ as 12,

$a:b=2:3=8:12$, and

$b:c=4:5=12:15$.

Joining the two ratios we get,

$a:b:c=8:12:15$.

Now we have to join this ratio with the third ratio, $d:c=7:6$.

To join the first step we have to take is to bring the common quantity $c$ in this case to the numerator of the third ratio,

$d:c=7:6$,

Or, $c:d=6:7$.

The target value at which the two values of $c$ are to be equalized now will be the LCM of $6$ and $15$ which is $30$.

For base equalization, transforming the two ratios to be joined we get,

$a:b:c=8:12:15=16:24:30$, and

$c:d=6:7=30:35$.

Finally then the four quantity ratio is,

$a:b:c:d=16:24:30:35$.

**Answer:** Option d: $16:24:30:35$.

** Key concepts used:** * Ratio value normalization* expressing the values as integers with no common factors --

*step by step --*

**joining ratios***to the LCM of the values corresponding to the commmon quantity -- preparing and tranforming the ratios in the proper form suiitable for joining --*

**base equalization***-- LCM.*

**basic and rich ratio concepts**** Q8.** If $a$ and $b$ are the roots of the equation $3x^2+2x+1=0$, which of the equations will have the roots, $\displaystyle\frac{1-a}{1+a}$ and $\displaystyle\frac{1-b}{1+b}$?

- $y^2-2y+3=0$
- $y^2+2y-3=0$
- $y^2-2y-3=0$
- $y^2+2y+3=0$

** Solution 8 - Problem analysis:**

In a quadratic equation, $x^2-px+q=0$ with roots $a$ and $b$ we know that the sum of the roots, $a+b=p$ and product of the roots, $ab=q$. This follows from expansion of the product of two factors and equating the coefficients,

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

$=x^2-px+q$

$=0$.

This is the main concept to be used in two stages to solve this problem here.

#### Solution 8 - Problem solving execution first stage

First we normalize the quadratic equation to remove the coefficient of $x^2$ and make it 1,

$3x^2+2x+1=0$,

Or, $x^2+\displaystyle\frac{2}{3}x+\displaystyle\frac{1}{3}=0$.

So sum and product of the roots $a$ and $b$,

$a+b=-\displaystyle\frac{2}{3}$ and,

$ab=\displaystyle\frac{1}{3}$.

Mark here, we don't have the values of the roots $a$ and $b$, but only have the values of their sum and product.

#### Solution 8 - Problem solving execution second stage

Let us assume,

$p=\displaystyle\frac{1-a}{1+a}$ and,

$q=\displaystyle\frac{1-b}{1+b}$.

So,

$p+q=\displaystyle\frac{(1-a)(1+b) +(1-b)(1+a)}{(1+a)(1+b)}$

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

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

$=\displaystyle\frac{2-\frac{2}{3}}{1-\frac{2}{3}+\frac{1}{3}}$

$=\displaystyle\frac{\frac{4}{3}}{\frac{2}{3}}$

$=2$

Similarly,

$pq=\displaystyle\frac{(1-a)(1-b)}{(1+a)(1+b)}$

$=\displaystyle\frac{1-(a+b)+ab}{\frac{2}{3}}$

$=\displaystyle\frac{1+\frac{2}{3}+\frac{1}{3}}{\frac{2}{3}}$

$=\displaystyle\frac{2}{\frac{2}{3}}$

$=3$

So the desired equation will be,

$y^2-2y+3=0$.

**Answer:** Option a: $y^2-2y+3=0$.

**Key concepts used:** Relationships between the roots of a quadratic equation -- * product of roots and sum of roots concept* -- simplification.

**Q9.** If $2x^2-7xy+3y^2=0$, then the value of $x:y$ is,

- $3:2$
- $5:6$
- $2:3$
- $3:1$ and $1:2$

**Solution 9 - Problem analysis and execution**

By mid-term splitting and observation we factorize the given quadratic equation in two variables as,

$2x^2-7xy+3y^2=(2x-y)(x-3y)=0$.

This gives us two relations between $x$ and $y$ satisfying the quadratic equation,

$2x=y$ and

$x=3y$.

In the first case the desired ratio is,

$x:y=1:2$ and in the second case,

$x:y=3:1$.

Finally then the desired ratio have two values for two root value pairs,

$1:2$ and $3:1$.

**Ans:** Option d: $3:1$ and $1:2$.

**Key concepts used:** Factorization of quadratic equation -- basic ratio concepts.

** Q10.** Find the value of $\alpha$ when the expression $\displaystyle\frac{x^2}{y^2} + {\alpha}x+\displaystyle\frac{y^2}{4}$ is a perfect square.

- $\pm{1}$
- $0$
- $\pm{2}$
- $\pm{7}$

**Solution 10 - Problem analysis and execution**

Examining the quadratic expression in two variables we detect that both the square terms are in proper square forms,

$a^2=\left(\displaystyle\frac{x}{y}\right)^2$ and

$b^2=\left(\displaystyle\frac{y}{2}\right)^2$.

For the quadratic equation to be a perfect square then the mid-term must be,

$2ab=2\times{\displaystyle\frac{x}{y}}\times{\displaystyle\frac{y}{2}}=x$.

Thus for the given equation to be a perfect square,

$\alpha x=x$,

Or, $\alpha=1$.

But we have considered only the form $(a+b)^2$. For the form $(a-b)^2$, the mid-term will be $-2ab=-x$ and the desired value of $\alpha$ will be,

$\alpha=-1$.

Finally then for all possibilities the desired value of $\alpha$ will be,

$\alpha=\pm {1}$.

**Answer: **Option a: $\pm {1}$.

**Key concepts used:** **Square of sum concepts -- principle of exhaustivity -- we have exhaustively examined all possibilities.**

### Other resources that you may find valuable

#### Concept tutorials on Algebra

**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**

#### General guidelines for successful performance in SSC CGL test

**7 Steps for sure success in SSC CGL tier 1 and tier 2 competitive tests**

#### Elegant math problem solving group of articles

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

#### Other question and solution sets on Algebra

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**SSC CGL level Solution Set 1, 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**