SSC CGL level Solution Set 11, Algebra

Eleventh SSC CGL level Solution Set, topic Algebra

SSC-CGL-Algebra-Solution-set11

This is the eleventh Solution set of 10 practice problems on topic Algebra for SSC CGL exam as well as school students interested in elegant problem solving. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set.

This exercise set containing 10 problems highlights the need of solving the problems using powerful strategies rather than brute force conventional deduction methods. We have tried to bring out the underlying strategies, techniques and reasoning that went into solving a problem in about a minute's time on an average.

If you follow intelligent and dedicated preparation methods using this type of resources, you should also be able to reach the desired level of competence for completing such a set of 10 questions comfortably within 12 minutes' share of time.

Before going ahead you should refer to the companion SSC CGL level Question Set 11 on Algebra, take the test and then only return to these conceptual solutions for gaining maximum benefits.


Recommendation: You may also refer to the Algebra oncepts in details under the subsection of Efficient Math Problem Solving as well as the tutorial on Basic and rich Alegebra concepts


Eleventh solution set on Algebra - 10 problems for SSC CGL exam - time 12 mins

Q1. If $a$, $b$ and $c$ are non-zero and $a + \displaystyle\frac{1}{b} = 1$ and  $b + \displaystyle\frac{1}{c} = 1$, the value of $abc$ is,

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

Solution:

From first expression,

$ab + 1 = b$ and from second expression,

$bc + 1 = c$.

Multiplying the first transformed equation by $c$,

$abc + c = bc$,

Or, $abc = bc - c = -1$, from the second transformed equation.

Answer: Option b: -1.

Key concepts used:

Transformation of both the given equations to get rid of the inverses to see how the three variables are related to each other -- Just by comparing the two resulting equations, the key idea of multiplying the first transformed equation by $c$ becomes apparent.

The goal state $abc$ we can't forget, can we?

Q2. If $a + \displaystyle\frac{1}{a - 2} = 4$, then $(a - 2)^2 + \displaystyle\frac{1}{(a - 2)^2}$ is,

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

Solution:

Transforming the given equation to make it similar to the target expression we have,

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

In this transformed equation as well as the target expression we now have only $(a - 2)$ appearing as a variable. This is the situation when we can think of more complex form of variable $(a - 2)$ in a simpler form, say $p$. This won't change the solution in any way.

So using the substitution technique we transform the problem itself to,

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

Being aware of the principle of inverses, this should be easy for us. We just raise the transformed and substituted given expression to power 2,

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

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

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

Answer: Option d: 2.

Key concepts used:

Using end state analysis, comparing target expression with given expression and transforming given expression to a form similar to the target expression -- use of input transformation technique -- using substitution technique to simplify things -- use of principle of inverses to reach the solution easily with confidence.

Q3. If $a + b + c = 2s$, then $\displaystyle\frac{(s - a)^2 + (s - b)^2 + (s - c)^2}{a^2 + b^2 + c^2}$ is,

  1. $0$
  2. $a^2 + b^2 + c^2$
  3. $1$
  4. $2$

Solution:

Let us assume the target expression as,

$E = \displaystyle\frac{E_{1}}{E_{2}}$ for ease of dealing with the complex fractional form of the target expression. Our intention is to deal with the numerator first following the well known Principle of more complex first that says,

The more complex expression can be simplified more and should be taken up first for math problem solving.

We observe that if we square up the given expression we get $4s^2$ on the RHS, and the numerator of target expression also would have $4s^2$ when the three square terms are expanded. This similarity is the key to the solution of the problem. With this hint of a possible solution we are at first urged towards the action of squaring up the given expression for a start.

But again, following the golden principle of more complex first, we decide to expand the most complex of the three expression, the numerator of the target, first.

Our intention is to use the information contained in the simpler expressions in simplifying the numerator.

Thus, expanding the three squares of the numerator of target expression we get,

$E_1 = 4s^2 + a^2 + b^2 + c^2 - 2s(a + b + c)$

$\hspace{6mm} = 4s^2 + a^2 + b^2 + c^2 - 2s\times{2s}$, just substituting the value of $a + b + c = 2s$,

$\hspace{6mm} = 4s^2 + a^2 + b^2 + c^2 - 4s^2$

$\hspace{6mm} = a^2 + b^2 + c^2$

Denominator of target expression $E_2$ also being $a^2 + b^2 + c^2$, answer is 1.

Answer: Option c: $1$.

Key concepts used:

First detect the chance of eliminating $s$ from the numerator of the target expression by using the given expression with the help of end state analysis -- taking up expansion of the numerator of target expression and not squaring the given expression following the golden principle of more complex first -- collection of friendly terms together -- using the value of given expression now finally eliminates $s$ and results in the simple answer to us.

Q4. If $xy(x + y) = 1$, then $\displaystyle\frac{1}{x^3y^3} - x^3 - y^3$ is,

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

Solution:

In Algebra, identifying the troublesome part of the target expression and how to solve this troubling part in target expression usually holds the key to the problem.

In this problem, the term $\displaystyle\frac{1}{x^3y^3}$ is the awkward one. To get a clue about how to remove it, we look at the given expression and immediately identify the possibility by transforming the given expression and moving further ahead,

$x + y = \displaystyle\frac{1}{xy}$,

Or, $(x + y )^3 = \displaystyle\frac{1}{x^3y^3}$, 

Or, $x^3 + y^3 + 3xy(x + y) = \displaystyle\frac{1}{x^3y^3}$,

Or, $\displaystyle\frac{1}{x^3y^3} - x^3 - y^3 = 3xy(x + y) = 3$.

We have used the given expression in two ways - first to solve the inverse of $x^3y^3$ problem by just splitting up the LHS of given expression into two parts. Secondly, we have used the value of whole of the LHS to simplify things greatly.

Answer: Option c: $3$.

Key concepts used:

Identifying the troublesome part of the target expression and finding the easiest way to resolve this problem by using the transformed given equation -- use of the given expression again but in its simplest form in the expanded cube of sum of two terms.

Q5. If $a + b + c = 6$, $a^2 + b^2 + c^2 = 14$ and $a^3 + b^3 + c^3 = 36$, then the value of $abc$ is,

  1. 3
  2. 6
  3. 9
  4. 12

Solution:

Just looking at the three given expressions we decide to use the enhanced rich expression for the first time,

$a^3 + b^3 + c^3 - 3abc$

$\hspace{5mm} = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

It is to some extent similar to,

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

So in our problem case we get,

$a^3 + b^3 + c^3 - 3abc = 36 - 3abc$

$\hspace{5mm} = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$,

Or, $36=6(14 - (ab + bc + ca)) + 3abc$.

Again we are well aware of this three term expression $xy + yz + zx$ appearing in $(x + y + z)^2$. Let us evaluate this expression first.

$(a + b + c)^2 = 36$

$\hspace{5mm} = a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

Or, $ab + bc + ca = 11$.

So, $36=6(14 - (ab + bc + ca)) + 3abc$,

Or, $36 = 6(14 - 11) + 3abc$,

Or, $3abc = 18$,

Or, $abc = 6$.

Answer: Option b: 6.

Key concepts used:

Use of enhanced basic concept to establish relationship between the three given expressions.

Q6. The minimum value of $(a - 2)(a - 9)$ is,

  1. $\displaystyle\frac{-11}{4}$
  2. $0$
  3. $\displaystyle\frac{-49}{4}$
  4. $\displaystyle\frac{49}{4}$

Solution:

This problem falls under a special class of Algebraic problems, namely, finding maxima or minima of a quadratic equation.

The core concept in both the cases of finding maxima and minima lies in transforming the given quadratic expression in such a way that a square of a two term sum in $x$ is formed. As this square will always be positive, any nonzero value of this square term will increase the value of the expression in our case of finding minima.

With this logic, it can be concluded that only when the square of the two term sum in $x$ is zero, we get the minima.

So we transform the input expression with a goal to express it into two parts, one part being a square of a two term sum in $x$,

$(a - 2)(a - 9) = a^2 - 11a + 18$

$\hspace{5mm} = \left(a^2 - 2\times{\displaystyle\frac{11}{2}}a + \left(\displaystyle\frac{11}{2}\right)^2\right) - \left(\displaystyle\frac{11}{2}\right)^2 + 18$

$\hspace{5mm} = \left(a - \displaystyle\frac{11}{2}\right)^2 - \displaystyle\frac{49}{4}$.

This is our desired form of expression. From this expression we can conclude that when $a=\displaystyle\frac{11}{2}$ only then the square term will be zero and the given expression will have its minima of $-\displaystyle\frac{49}{4}$.

Answer: Option c: $-\displaystyle\frac{49}{4}$.

Key concepts used:

Use of the specific rich concept of finding minima based on the more fundamental concept that a square term will always be positive and increase the value of the expression unless it is zero where the minima occurs.

Q7. The terms $a$, $1$, and $b$ are in AP and the terms $1$, $a$ and $b$ are in GP. Find the values of $a$ and $b$, where $a\neq{b}$.

  1. 4, 1
  2. 2, 4
  3. -2, 1
  4. -2, 4

Solution:

AP or Arithmetic Progression of three terms specifies that the difference between any two adjacent terms will be a constant. Thus from the first expression we get,

$1 - a = b - 1$, or, $a + b = 2$.

Similarly, the concept of GP or Geometric progression specifies that the ratio of any two adjacent terms will be a constant. By this concept we can transform the second statement to the relation,

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

Or, $a^2 = b$.

Substituting this value of $b$ in the first expression we get a quadratic equation in $a$,

$a + a^2 = 2$,

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

This gives, $a = -2$, or, $a = 1$.

If $a = 1$, from the given equations, $b = 1 = a$ which is an invalid result.

So, $a = -2$, and,

$b = 4$.

Answer: Option d: -2, 4.

Key concepts used:

From the definition of AP and GP getting the two expressions in $a$ and $b$ -- from these two equations, eliminating $b$ to form a quadratic equation only in $a$ -- out of two roots of $a$ one is trivial and invalid by problem description -- the second value of $a$ gives us the value of $b$.

Q8. If $x\neq{0}$, $y\neq{0}$ and $z\neq{0}$, and $\displaystyle\frac{1}{x^2} + \displaystyle\frac{1}{y^2} + \displaystyle\frac{1}{z^2} = \displaystyle\frac{1}{xy} + \displaystyle\frac{1}{yz} + \displaystyle\frac{1}{zx}$, then the relation between $x$, $y$ and $z$ is,

  1. $x + y = z$
  2. $x + y + z = 0$
  3. $x=y=z$
  4. $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} + \displaystyle\frac{1}{z} = 0$

Solution:

In the very beginning we observe that only the inverses of the variables $x$, $y$ and $z$ appearing in all the terms. This is use of pattern recognition skill in action.

Thus we simplify the form of the given expression by substituting $\displaystyle\frac{1}{x} = a$, $\displaystyle\frac{1}{y} = b$, and $\displaystyle\frac{1}{z} = c$.

The tranformed given expression becomes then,

$\displaystyle\frac{1}{x^2} + \displaystyle\frac{1}{y^2} + \displaystyle\frac{1}{z^2} $

$\hspace{10mm} = \displaystyle\frac{1}{xy} + \displaystyle\frac{1}{yz} + \displaystyle\frac{1}{zx}$

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

This is a significant simplification of form and makes our task much easier.

Now we remember the very useful Principle of collection of friendly terms as well as the Principle of zero sum of square terms to transform the expression to,

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

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

Or, $a = b = c$,

Or, $x=y=z$.

Answer: Option c: $x=y=z$.

Key concepts used:

By transforming the complicated looking given expression to much simpler form using substitution technique, we observe a possiblity of using principle of collection of friendly terms with the final objective of result in a zero sum of square terms. By applying the principle of zero sum of square terms, answer is obtained easily.

Q9. If $a^2 - 4a - 1 = 0$, then $a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$ is,

  1. 25
  2. 35
  3. 40
  4. 30

Solution:

First two terms of the target expression form a very promising expression conforming to our well known small world governed by principle of inverses. We are confident of dealing with any problem in this small world of inverses if we can spot the inverse pattern, specifically in Algebraic problems.

Our task is now focused in finding an inverse pattern in the given expression,

$a^2 - 4a - 1 = 0$,

Or, $a - 4 - \displaystyle\frac{1}{a} = 0$,

Or, $a - \displaystyle\frac{1}{a} = 4$.

Carrying on further with the intention of finding the sum of squares of inverses, we square up the transformed equation,

Or, $a^2 - 2 +\displaystyle\frac{1}{a^2} = 16$,

Or, $a^2 +\displaystyle\frac{1}{a^2} = 18$.

With these two useful results we now deal with the target expression,

$a^2 + \displaystyle\frac{1}{a^2} + 3a - \displaystyle\frac{3}{a}$

$\hspace{5mm} = 18 + 3\left(a - \displaystyle\frac{1}{a}\right)$

$\hspace{5mm} = 18 + 3\times{4} = 18 + 12$

$\hspace{5mm} = 30$

Answer: Option d: 30.

Key concepts used:

Analysing the two term two parts of the target expression we identify sums of inverses in both the parts -- now sure of finding an inverse expression in the given expression we transform it to actually discover the useful form of expression -- the rest is routine procedures guided by the principle of inverses.

Q10. 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, find the value of $a + b + c$.

  1. 3
  2. 0
  3. 2
  4. 1

Solution:

To get rid of the cube roots for the time being let us substitute $\sqrt[3]{2} = p$ and get the transformed given equation as,

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

Now we detect a useful pattern in the expression of the denominator and straighaway utilize it by multiplying the fraction numerator and denominator by $(p - 1)$ getting,

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

Or, $\displaystyle\frac{p - 1}{p^3 - 1} = ap^2 + bp + c$,

Or, $p - 1 = ap^2 + bp + c$.

As $p$ is not a variable, but a constant, the coefficients of $p^2$, $p$ and the constant in both sides of the equation must be same.

Thus we get, $a=0$, $b=1$ and $c=-1$, so that $a + b + c = 0$.

Answer: Option b: 0.

A neat solution to a very awkward problem.

Key concepts used:

First simplification of form by substitution technique -- detecting the possibility of using a new factor technique to eliminate the denominator of the fraction altogether -- equating coefficients of like terms on two sides of the equation to get values of $a$, $b$ and $c$. We have used here three rich algebra concepts.


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

More rich algebraic concepts and techniques for elegant solutions of SSC CGL problems

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

SSC CGL Tier II level Questions and Solutions on Algebra

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 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 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