SSC CGL Tier II level Solution Set 9, Algebra 4 | SureSolv

SSC CGL Tier II level Solution Set 9, Algebra 4

9th SSC CGL Tier II level Solution Set, 4th on Algebra

SSC CGL Tier 2 solution set 9 algebra 4

This is the 9th solution set of 10 practice problem exercise for SSC CGL Tier II exam and the 4th 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 9 on Algebra 4 before going through the solution.


9th solution set - 10 problems for SSC CGL exam: 4th on topic Algebra - answering time 12 mins

Q1. If $x+\displaystyle\frac{2}{x}=1$, then the value of $\displaystyle\frac{x^2+x+2}{x^2(1-x)}$ is,

  1. $2$
  2. $-1$
  3. $1$
  4. $-2$

Solution 1 - Problem analysis and execution

The given equation needs to be simplified first to an expression that is similar to either the denominator or the numerator of the target expression,

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

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

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

Armed with these useful results we turn our attention to the target expression,

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

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

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

$=1$.

Answer: Option c: 1.

Key concepts used: Finding similarities with the numerator and denominator expressions in the target expression after transforming the given expression -- End state analysis -- input transformation.

Q2. If $4x+5y=83$, and $3x : 2y=21 : 22$, then $(y-x)$ equals to,

  1. 4
  2. 7
  3. 3
  4. 11

Solution 2 - Conventional solution

Transformation of the given ratio expression will give us a second linear relation between $x$ and $y$ in addition to the first,

$3x : 2y=21 : 22$,

Or, $x:y=7:11$,

Or, $11x-7y=0$,

Or, $\displaystyle\frac{55}{7}x-5y=0$, multiplying by $\displaystyle\frac{5}{7}$ to transform the $y$ term to $5y$ for cancelling out at the next step.

Summing this result with the first given expression, $4x+5y=83$,

$4x+\displaystyle\frac{55}{7}x=83$,

Or, $\displaystyle\frac{83}{7}x=83$

Or, $x=7$,

So,

$y=11$, and

$y-x=4$.

Answer: Option a : 4.

Key concepts used: Input transformation -- Solving linear equations.

Solution 2 - Elegant solution

Applying basic ratio concepts of reintroducing the cancelled out HCF, say $p$, as factors to ratio terms,

$3x : 2y=21 : 22$,

Or, $x:y=7:11=7p : 11p$,

where actual vales of $x$ and $y$ are, $x=7p$ and $y=11p$.

Substituting these two values in the first given equation,

$4x+5y=83$,

Or, $28p+55p=83$,

Or, $p=1$.

So,

$(y-x)=11p-7p=4p=4$.

Easy and quick, but conceptual. Here we avoided solving the two linear equations altogether as well as evluation of actual values of $x$ and $y$.

Answer: Option a: 4.

Key concepts used: Basic ratio concepts -- Efficient simplification.

Q3. If $\displaystyle\frac{3-5x}{2x} + \displaystyle\frac{3-5y}{2y} + \displaystyle\frac{3-5z}{2z}=0$, then the value of $\displaystyle\frac{2}{x}+\displaystyle\frac{2}{y}+\displaystyle\frac{2}{z}$ is,

  1. 10
  2. 5
  3. 20
  4. 15

Solution 3 - Problem analysis and execution

Separating out the two terms in the numerator of each of the three therms would result in,

$\displaystyle\frac{3-5x}{2x} + \displaystyle\frac{3-5y}{2y} + \displaystyle\frac{3-5z}{2z}=0$,

Or, $\displaystyle\frac{3}{2x} + \displaystyle\frac{3}{2y} + \displaystyle\frac{3}{2z}=\displaystyle\frac{15}{2}$.

We would now convert the LHS to the form of target expression,

$\displaystyle\frac{3}{2x} + \displaystyle\frac{3}{2y} + \displaystyle\frac{3}{2z}=\displaystyle\frac{15}{2}$,

Or, $\displaystyle\frac{2}{x}+\displaystyle\frac{2}{y}+\displaystyle\frac{2}{z}=\displaystyle\frac{15}{2}\times{\displaystyle\frac{4}{3}}=10$.

It is purely extracting the target expression from the given expression by key pattern identification and input expression transformation. It is done all in mind, no need for any written deduction.

Answer: Option a: 10.

Key concepts used: Key pattern identification -- input transformation.

Q4. If $\displaystyle\frac{2p}{p^2-2p+1}=\frac{1}{4}$ then the value of $\left(p+\displaystyle\frac{1}{p}\right)$ is,

  1. $1$
  2. $10$
  3. $\displaystyle\frac{2}{5}$
  4. $7$

Solution 4 - Problem analysis and execution

Transposing and simplifying the given expression,

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

Or, $8p=p^2-2p+1$,

Or, $p^2+1=10p$,

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

Answer: Option b: 10.

Key concepts used: Key pattern identification -- Input transformation.

Q5. If $2x=\sqrt{a}+\displaystyle\frac{1}{\sqrt{a}}$, and $a \gt 0$, then the value of $\displaystyle\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}$ is,

  1. $a+1$
  2. $a-1$
  3. $\frac{1}{2}(a-1)$
  4. $\frac{1}{2}(a+1)$

Solution 5 - Problem analysis

We need to first evaluate and simplify the problematic expression $\sqrt{x^2-1}$ in the target expression by transforming the input expression,

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

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

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

Or, $\sqrt{x^2-1}=\displaystyle\frac{a-1}{2\sqrt{a}}$.

Solution 5 - Final execution

We would substitute this value in the target expression,

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

$=\displaystyle\frac{a-1}{2\sqrt{a}}\times{\displaystyle\frac{1}{\displaystyle\frac{2}{2\sqrt{a}}}}$

$=\displaystyle\frac{a-1}{2\sqrt{a}}\times{\sqrt{a}}$

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

Answer: Option c: $\displaystyle\frac{1}{2}(a-1)$.

Key concepts used: Key pattern identification -- Deductive reasoning -- Efficient simplification.

Q6. If $x^2+y^2+z^2=xy+yz+zx$, then the value of $\displaystyle\frac{3x^4+7y^4+5z^4}{5x^2y^2+7y^2z^2+3z^2x^2}$ is,

  1. $2$
  2. $1$
  3. $0$
  4. $-1$

Solution 6 - Problem analysis

The given expression is balanced and symmetric in $x$, $y$, and $z$ (interchangeable) whereas the target is not. So we would expect to evaluate individual numeric values of the three variables.

Solution 6 - Problem solving execution

The given equation is,

$x^2+y^2+z^2=xy+yz+zx$,

Or, $2(x^2+y^2+z^2)-2(xy+yz+zx)=0$,

Or, $(x-y)^2+(y-z)^2+(z-x)^2=0$

As this is a zero sum of square terms (with real variables), each of the squares must be zero,

$x=y=z$.

So the target exspression,

$E=\displaystyle\frac{3x^4+7y^4+5z^4}{5x^2y^2+7y^2z^2+3z^2x^2}$

$=\displaystyle\frac{(3+7+5)x^4}{(5+7+3)x^4}$

$=1$.

Answer: Option b : $1$.

Key concepts used:  Symmetric expression -- End state analysis, by comparing the symmetric given expression and the asymmetric target expression we formed an expectation about the most feasible path to the solution -- deductive reasoning -- input transformation -- zero sum of square terms.

Q7. If $\displaystyle\frac{a^2-bc}{a^2+bc}+\displaystyle\frac{b^2-ca}{b^2+ca}+\displaystyle\frac{c^2-ab}{c^2+ab}=1$, then the value of $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}$ is,

  1. $2$
  2. $1$
  3. $0$
  4. $-1$

Solution 7 - Problem analysis

If we separate out the two terms in the numerator of each term of the given expression, the first term of each component expression summed up results in the target expression. In other words, the givem expression includes the target expression. So if we can evaluate the sum of the second separated term summed expression, we would get the value of the target expression.

On analyzing the given expression with this approach the simple path to the solution is revealed in no time.

Solution 7 - Problem analysis and execution

The given expression is,

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

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

Or, $3-\displaystyle\frac{2bc}{a^2+bc}-\displaystyle\frac{2ca}{b^2+ca}-\displaystyle\frac{2ab}{c^2+ab}=1$,

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

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

Substituting this value in the given expression,

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

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

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

Answer: Option a: $2$.

Key concepts used:  Key pattern identificatoion -- End state analysis -- deductive reasoning -- input transformation.

Solution 7: Elegant solution

We expose you to the elegant solution after the more involved solution so that you can appreciate the saving in time and steps in execution as well as the concepts and strategies involved in achieving the elegant solution.

At the outset we identify that each of the terms of the given expression is in the form of $\displaystyle\frac{p-q}{p+q}$ whereas each of the terms in the target expression is in the form of $\displaystyle\frac{p}{p+q}$.

This is an ideal case for applying Componendo technique of adding 1 to each such given expression term and forming the target expression terms in 1 simple step,

$\displaystyle\frac{a^2-bc}{a^2+bc}+1+\displaystyle\frac{b^2-ca}{b^2+ca}+1+\displaystyle\frac{c^2-ab}{c^2+ab}+1=1+3$,

Or, $\displaystyle\frac{2a^2}{a^2+bc}+\displaystyle\frac{2b^2}{b^2+ca}+\displaystyle\frac{2c^2}{c^2+ab}=4$,

Or, $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}=2$, eliminating the extra factor of 2.

Answer: Option a: $2$.

Key concepts used: Key pattern identification, this time we have identified a different more powerful pattern -- Componendo technique -- Efficient simplification -- Many ways technique -- Problem solving skillset enhancement.

Note: However complex the expression is, a result bearing pattern can simplify it in a single step, that need not even be deduced (actually deduced in mind in a few seconds).

Q8. If $2\left(x^2+\displaystyle\frac{1}{x^2}\right) - \left(x-\displaystyle\frac{1}{x}\right) -7 =0$ then the two valid values of $x$ are,

  1. $0$, $1$
  2. $\frac{1}{2}$, $1$
  3. $1$, $2$
  4. $2$, $-\frac{1}{2}$

Solution 8 - Problem analysis and execution

The given expression is in power 4 of $x$ and so has four valid values of $x$ out of which two we have to choose that match with one of the choices given.

With this knowledge we simplify the given equation in the first stage.

We identify the key pattern in the given expression that $\left(x^2+\displaystyle\frac{1}{x^2}\right)=\left(x-\displaystyle\frac{1}{x}\right)^2+2$ which immediately transforms the given expression to,

$2\left(x^2+\displaystyle\frac{1}{x^2}\right) - \left(x-\displaystyle\frac{1}{x}\right) -7 =0$,

Or, $2\left(p^2+2\right)-p-7=0$, where we have used component expression substitution of $\left(x-\displaystyle\frac{1}{x}\right)$ by $p$

Or, $2p^2-p-3=0$,

Or, $p^2-\frac{1}{2}p-\frac{3}{2}p=0$

Or, $(p+1)(p-\frac{3}{2})=0$

So the first root of $p$ is,

$p=x-\displaystyle\frac{1}{x}=-1$,

Transposing and simplifying,

$x^2+x-1=0$.

This produces two real surd roots of $x$ involving $\pm\sqrt{b^2-4ac}=\pm\sqrt{1^2+4}=\pm\sqrt{5}$ from the Sreedhar Acharya's formula. As these cannot be in the choice value set we would consider the second root of $p$ for the second valid pair of values of $x$,

$p=x-\displaystyle\frac{1}{x}=\displaystyle\frac{3}{2}$,

Or, $2x^2-3x-2=0$,

Or, $(2x+1)(x-2)=0$,

Or, $x=-\frac{1}{2}$, and $x=2$.

Answer: Option d: $2$, $-\frac{1}{2}$.

Key concepts used: Problem analysis -- Deductive reasoning -- Principle of interaction of inverses -- Key pattern identification -- Component expression substitution -- Solving a simpler problem -- Free resource use of the choice value set -- Sreedhar Acharya's relation -- Solving quadratic equation -- Roots of a quadratic equation.

Q9. If $x=\sqrt[3]{a+\sqrt{a^2+b^3}}+\sqrt[3]{a-\sqrt{a^2+b^3}}$, then $x^3+3bx$ is equal to,

  1. $0$
  2. $1$
  3. $a$
  4. $2a$

Solution 9 - Problem analysis

With an objective to free the two RHS terms from the cube roots, we raise $x$ to the power of 3 at the very start,

$x=\sqrt[3]{a+\sqrt{a^2+b^3}}+\sqrt[3]{a-\sqrt{a^2+b^3}}$,

Or, $x^3=\left(a +\sqrt{a^2+b^3}\right)+\left(a -\sqrt{a^2+b^3}\right)$

$\hspace{20mm}+3\left(\sqrt[3]{a+\sqrt{a^2+b^3}}\right)\left(\sqrt[3]{a-\sqrt{a^2+b^3}}\right)$

$\hspace{20mm}\times{\left(\sqrt[3]{a+\sqrt{a^2+b^3}}+\sqrt[3]{a-\sqrt{a^2+b^3}}\right)}$,

$=2a+3x\sqrt[3]{\left(a+\sqrt{a^2+b^3}\right)\left(a-\sqrt{a^2+b^3}\right)}$

$=2a+3x\sqrt[3]{a^2-a^2-b^3}$

Or, $x^3=2a-3bx$.

So,

$x^3+3bx=2a$.

Answer: Option d: $2a$.

Key concepts used: Key pattern identification -- Deductive reasoning -- Efficient simplification.

Q10. If $bc+ca+ab=abc$, then the value of $\displaystyle\frac{b+c}{bc(a-1)} + \displaystyle\frac{c+a}{ca(b-1)}+\displaystyle\frac{a+b}{ab(c-1)}$ is,

  1. $0$
  2. $1$
  3. $-\frac{1}{2}$
  4. $-\frac{3}{2}$

Solution 10 - Problem analysis and execution

By End state analysis we compare the similarities between the target expression and the given expression and identify the key pattern that will lead us to the solution. Usually we need to transform the given expression to move towards the target expression.

Occasionally though we need to transform the target expression so that we can use the given expression in simplifying the target expression.

In this problem, we multiply and divide the first, second and third terms of the target expression by $a$, $b$ and $c$ respectively,

$E=\displaystyle\frac{b+c}{bc(a-1)} + \displaystyle\frac{c+a}{ca(b-1)}+\displaystyle\frac{a+b}{ab(c-1)}$,

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

$=\displaystyle\frac{abc-bc}{abc(a-1)} + \displaystyle\frac{abc-ca}{abc(b-1)}+\displaystyle\frac{abc-ab}{abc(c-1)}$, we have used the given expression

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

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

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

$=\displaystyle\frac{ab+bc+ca}{abc}$

$=\displaystyle\frac{abc}{abc}$

$=1$.

Answer: Option b: $1$.

We have used the given expression in two stages to simplify the target to the largest extent. At each stage of simplification, we not only have used the given expression also focused on the primary objective of denominator simplification.

The simplification possibility could be sensed in the beginning itself because of the perfectly symmetric formation of the target expression in terms of $x$, $y$ and $z$ which is confirmed by the choice values.

Key concepts used: Symmetric expression -- Free resource use -- Key pattern identification -- End state analysis -- Target expression transformation -- numerator transformation -- Substitution of transformed given expression -- Denominator simplification -- multiple input use technique, input expression has been used in two different ways on two occasions -- efficient simplification.

Solution 10: More efficient solution

In the above solution we have modified the numerator of each term by a factor multiplication. Instead of this extra step we detect that the useful relation already exists in the denominator and use it straightaway,

$E=\displaystyle\frac{b+c}{bc(a-1)} + \displaystyle\frac{c+a}{ca(b-1)}+\displaystyle\frac{a+b}{ab(c-1)}$

$=\displaystyle\frac{b+c}{abc-bc} + \displaystyle\frac{c+a}{abc-ca}+\displaystyle\frac{a+b}{abc-ab}$

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

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

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

$=\displaystyle\frac{bc+ca+ab}{abc}$

$=\displaystyle\frac{abc}{abc}$

$=1$.

This is a more efficient solution as one transformative step is saved in this solution compared to the previous one.

Answer: Option b: $1$.

Key concept used instead of Numerator transformation: Denominator transformation, using existing useful pattern. In the earlier solution we have created the useful pattern by multiplication, an additional step. Other concepts used are same -- Using existing key pattern -- Many ways technique -- Problem solving skill enhancement.

Note: The more you are able to use existing useful pattern instead of creating the pattern by tranformation, faster will be your solution.


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