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SSC CGL Tier II level Solution Set 2, Algebra 2

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

SSC CGL Tier 2 solution set 2 algebra2

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.

Watch quick solutions in two-part video.

Part I: Q1 to Q5

Part II: Q6 to Q10


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. 1
  2. 2
  3. 0
  4. 4

Solution 1: Quick solution by pattern discovery, minima of sum of inverses and maxima minima equalization

The two terms on the LHS being inverses to each other, if you raise the equation to its square, you would get rid of the square root, but then the resulting equation would be too complex to deal with quickly and elegantly. Which way to go then?

At this point you remember the property of minimum value of sum of inverses of positive variable as 2. For convenience you replace,

$p= \sqrt{x^2−x+1}$ and convert the LHS to,

$p+\displaystyle\frac{1}{p}$.

A quick check ensures that $p$ is indeed positive for any value of $x$.

For all negative values and all values of $x$ greater than or equal to 1, it is obvious that $\sqrt{x^2−x+1}$ will be positive.

And when x is positive and less than 1, $1-x$ will make $\sqrt{x^2−x+1}$ positive.

LHS $p+\displaystyle\frac{1}{p}$ with positive $p$ will then have its minimum value of 2 when $p=1$, or $x=0$.

Your attention shifts now to the RHS and you find that its maximum value is 2, and that also occurs when $x=0$. For all non-zero values of $x$, $2-x^2$ is less than 2.

Simple conclusion is, only when $x=0$, RHS can become equal to the LHS and the equation is satisfied, that is, number of possible values of $x$ is just 1.

Answer: option a: 1.

Mechanism of minima of sum of inverses of positive variable

Let us explain briefly how $p+\displaystyle\frac{1}{p}$ gets its minimum value of 2 with positive real $p$.

You have sum of inverses as,

$p+\displaystyle\frac{1}{p}=\displaystyle\frac{p^2+1}{p}$.

When $p \geq 1$, $p^2$ in the numerator will increase much faster than $p$ in the denominator starting from the value of $p=1$ when minimum value of expression is 2.

For dealing with values of $p \leq 1$, let us assume, $p=\displaystyle\frac{1}{q}$ and the original sum of inverses is changed to,

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

The form of the sum of inverses expression remains unchanged.

In this second case consider all values of $p$ where $p \leq 1$ and $q \geq 1$.

By the same logic as before, the expression now will increase indefinitely from initial value 2 when $p$ goes on decreasing starting from 1 and $q$ goes on increasing starting from also 1.

In this case also then the minimum value of sum of inverses remains as 2.

Thus, for all possible positive real values of $p$, the sum of inverses $p+\displaystyle\frac{1}{p}$ will have its minimum value as 2.

Let's summarize now the important concepts and techniques that we have used for solving this awkward problem.

Key concepts used: Pattern discovery that LHS is a sum of inverses -- Dummy variable substitution -- Pattern identification of RHS maxima -- Minima of sum of inverses -- Maxima minima equalization -- 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,

  1. 50
  2. 55
  3. 70
  4. 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 -- principle of interaction of inverses -- output expression transformation -- target transformation --  input transformation to evaluate sum of inverse of cubes -- multiple input use technique to use given expression in a second way to evaluate denominator -- simplification.

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

  1. 3
  2. 4
  3. 5
  4. 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. 1
  2. 2
  3. 3
  4. 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 $(x-3)$.

  1. 56
  2. 184
  3. 0
  4. 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. $1$
  2. $0$
  3. $2$
  4. $\displaystyle\frac{1}{2}$

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 $a:b:c:d$.

  1. $8:12:15:7$
  2. $4:6:7:9$
  3. $30:35:24:16$
  4. $16:24:30:35$

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 -- joining ratios step by step -- base equalization to the LCM of the values corresponding to the commmon quantity -- preparing and tranforming the ratios in the proper form suiitable for joining -- basic and rich ratio concepts -- LCM.

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 $\displaystyle\frac{1-b}{1+b}$?

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

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,

  1. $3:2$
  2. $5:6$
  3. $2:3$
  4. $3:1$ 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.

  1. $\pm{1}$
  2. $0$
  3. $\pm{2}$
  4. $\pm{7}$

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.


Additional help on Suresolv Algebra

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

5 step Suresolv Algebra Reading and Practice Guide for SSC CGL Tier II.

You will get step by step recommended actions for success in SSC CGL Tier II and other relevant competitive exams along with the full list of articles on Algebra with links.

The list includes: ALL Concept articles on Algebra, ALL articles on How to solve difficult Algebra problems in a few steps, ALL Question and Solution Sets for SSC CGL and ALL Question and Solution sets for SSC CGL Tier II.

Important points to remember

  • SSC CGL problem solving builds the foundation for excellence on SSC CGL Tier II exam.
  • Concept articles are particularly valuable for learning the concepts and techniques, whereas How to solve in a few steps articles show how the techniques are applied to actually solve specially selected problems from real exam.
  • Finally, each of the large number of question sets should certainly be used as a timed mini-mock test for analysis of your difficulties in answering followed by clarification of doubts from the corresponding solution set.

Wish you all the sure success.


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