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SSC CGL level Solution Set 10, Algebra

Tenth SSC CGL level Solution Set, topic Algebra

SSC CGL level Solution set 10 algebra

This is the tenth Solution set of 10 practice problem exercise on topic Algebra for SSC CGL exam. Students must complete the corresponding SSC CGL level Question set 10 in prescribed time first and then only refer to this solution set.

Otherwise, without attempting the question set with all seriousness in the prescribed time, if the learner goes through the solutions he or she won't be able to appreciate and retain the special concepts involved in the solutions.

A golden rule of math problem solving always will remain true,

Math can't be learned by heart.

This is true for achieving excellence in learning any subject but is truer especially in Maths. Here, in Maths you have to understand the concepts and acquire the ability to use the concepts with special problem solving strategies to deal with any math problem within the scope of the topic.

Furthermore, it is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in elaborate steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. Solving process will mostly be in your head rather than on scratch paper.

Based on our analysis and experience we have seen that, for accurate and lightning quick answering, the student

  • must have complete understanding of the basic concepts, along with rich concepts on the topics,
  • is adequately fast in mental math calculation, one need not be superfast human calculator,
  • should first examine each problem for using the most basic concepts in the specific topic area and then use the rich concepts if required,
  • does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving happens in last step above. This problem solving ability lies at the heart of excellence in performance in this cutting-edge test.

This problem 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 on an average in about a minute's time.

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.

These questions are a bit advanced in nature and require a few new types of algebraic patttern recognition and use abilities.

Lastly, these are problems rich with possibilities. We couldn't cover all aspects of each problem. Later we would deal with a few selected class of such beautiful problems in details. You may refer to these special problem solving strategy and technique oriented detailed treatments under the subsection Difficult Algebra Problem Solving in a few steps.

Watch the solutions of the ten problems in the two-part video below.

Part I

Part II

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

Q1. If $2a + \displaystyle\frac{1}{3a} = 6$, then the value of the expression $3a + \displaystyle\frac{1}{2a}$ is,

  1. 12
  2. 9
  3. 4
  4. 8


As a first step, we always examine in a quick scan the similarities between the desired end state (expression to be evaluated) and the initial state (given expression). This is applying our End state analysis approach that has most extensive use in math problem solving.

As a result of this analysis, we form our first conclusion:

The product of the direct and inverse terms of the given expression and the target expression is same, that is, $2a\times{3a}=3a\times{2a}=6a^2$. The coefficient or the power of $a$ of the product in both cases remain same.

It then seems a case of transforming coefficients. This is a technique used often, especially in case of terms involving inverses. Thus we transform,

$2a + \displaystyle\frac{1}{3a} = 6$

Or, $a + \displaystyle\frac{1}{6a} = 3$

Or, $3a + \displaystyle\frac{1}{2a} = 9$. 

How lucky! We have directly reached the answer.

Answer: Option b: 9.

Key concepts used:

End state analysis to detect similarities between expression to be evaluated and the given expression -- using the coefficient transformation technique on expressions involving inverses.

Q2. If $x^2 + y^2 - 2x + 6y + 10 = 0$, then $(x^2 + y^2)$ is,

  1. 6
  2. 4
  3. 10
  4. 8


At first glance we get no clue to the solution. Then when we look a little deeper we get a hint of the possibility of forming two square expressions in $x$ and $y$. 

This is a direct use of the Technique of collection. By this technique,

We collect friendly terms (or entities in general) together to form useful expressions, which otherwise were not easily recognizable.

This technique is especially suitable for applications in Algebra. Thus rearranging the terms we get.

$x^2 + y^2 - 2x + 6y + 10 = 0$,

Or, $(x^2 - 2x + 1) + (y^2 + 6y + 9) = 0$,

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

We frequently use this Algebraic property of sum of squares. We name this property as the Principle of sum of squares and define it as,

If sum of squares of two terms (involving real variables) is zero, each square term must also be zero.

Reason is obvious, square of no real term can be negative. If sum of two non-negative terms is zero, each of the non-negative term must also be zero. So,

$x - 1  = 0$, and $y + 3$ = 0,

Or, $x = 1$, and $y = -3$,

Or, $x^2 + y^2 = 10$

Answer: Option c: 10.

Key concepts used:

Identifying the possibility of collecting friendly terms together to form sum of two squares -- technique of collection -- property of sum of two squares or principle of sum of squares.

Q3. If $x^2 = 2$, then $x + 1$ is,

  1. $x - 1$
  2. $\displaystyle\frac{2}{x - 1}$
  3. $\displaystyle\frac{x + 1}{3 - 2x}$
  4. $\displaystyle\frac{x - 1}{3 - 2x}$


The given and the target expression being very simple we decide to use the free resource of the choice values and test each one by one.

Without actual deduction we can see the first choice expression to be mathematically invalid. In case of second choice, $x^2 - 1 = 2$, or $x^2 = 3$, an invalid result.

For the third choice, $2x = 3$, again an invalid choice.

For the fourth expression we get,

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

Or, $(x + 1)(3 - 2x) = x - 1$

Or, $3x - 2x^2 + 3 - 2x = x - 1$

Or, $2x^2 = 4$,

$x^2 = 2$, a confirmation of given expression.

Answer: Option d: $\displaystyle\frac{x - 1}{3 - 2x}$.

Key concepts used:

Deciding to use the choice expressions as a resource and testing each choice expression to see whether it satisfies the given value of   $x$.

Q4. If $a^2 + b^2 + \displaystyle\frac{1}{a^2} + \displaystyle\frac{1}{b^2} = 4$  then $a^2 + b^2$ is,

  1. $1$
  2. $2\displaystyle\frac{1}{2}$
  3. $1\displaystyle\frac{1}{2}$
  4. $2$


This again requires a collection of friendly terms towards forming sum of two squares, but this time we use the resource of the value on the RHS also. 

$a^2 + b^2 + \displaystyle\frac{1}{a^2} + \displaystyle\frac{1}{b^2} = 4$,

Or, $\left(a^2 - 2 + \displaystyle\frac{1}{a^2}\right) $

$\hspace{8mm} + \left(b^2 - 2 + \displaystyle\frac{1}{b^2}\right) = 0$,

Or, $\left(a - \displaystyle\frac{1}{a}\right)^2 + \left(b - \displaystyle\frac{1}{b}\right)^2$.

And so, $a = \displaystyle\frac{1}{a}$,  or, $a^2 = 1$, and,

$b = \displaystyle\frac{1}{b}$,  or, $b^2 = 1$,

Or, $a^2 + b^2 = 2$

Answer: Option d: $2$.

Key concepts used:

Again collection of friendly terms to form a sum of two squares. It is a powerful technique if you can apply it.

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


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

$a^3 + b^3 + c^3 = (a + b + c)\times \\{(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc}$.

It is to some extent similar to,

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

So we go into our problem case,

$a^3 + b^3 + c^3 = (a + b + c)\times\\ {(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc}$,

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

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

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

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

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

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

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

Or, $3abc = 18$,

Or, $abc = 6$.

Answer: Option b: 6.

Key concepts used:

Rich concepts to establish relationship between the given expressions.

Q6. If $x + \displaystyle\frac{1}{16x} = 1$, then the value of $64x^3 + \displaystyle\frac{1}{64x^3}$ is,

  1. 64
  2. 76
  3. 52
  4. 4


We find the terms of the target expression are cubes of $4x$ and $\displaystyle\frac{1}{4x}$. Though the given expression looks different, again we observe, the coefficient and power of $x$ in the product of $x\times{16x}=16x^2$ is same as $4x\times{4x}$.

So we start transforming the coefficients of the input expression terms,

$x + \displaystyle\frac{1}{16x} = 1$,

Or, $4x + \displaystyle\frac{1}{4x} = 4$,

Or, $\left(4x + \displaystyle\frac{1}{4x}\right)^2 = 16$,

Or, $\left(16x^2 + \displaystyle\frac{1}{16x^2}\right) = 14$.


$64x^3 + \displaystyle\frac{1}{64x^3} $

$\hspace{5mm} = \left(4x + \displaystyle\frac{1}{4x}\right)\left(16x^2 + \displaystyle\frac{1}{16x^2} - 1\right) $

$\hspace{5mm} = 4\times{13} $

$\hspace{5mm} = 52$

Answer: Option c: 52.

Key concepts used:

End state analysis to decide that we need to transform the given expression in the form of the expression $\left(4x + \displaystyle\frac{1}{4x}\right)$ -- coefficient transformation of input terms.

Q7. If $a^4 + a^2b^2 + b^4 = 8$ and $a^2 + ab + b^2 = 4$, then the value of $ab$ is,

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


This seems to be a bit laborious problem to solve.

Solution using basic concepts:

As a first step we square the second expression to reach a position similar to the first expression.

$a^4 + a^2b^2 + b^4 $

$\hspace{15mm} + 2ab(a^2 + ab + b^2) = 16$,

Or, $8 + 2ab\times{4} = 16$,

Or, $8ab = 8$,

Or, $ab = 1$.

Answer: Option d: 1.

You can solve this using a rich concept in the form of the expression,

$a^4 + a^2b^2 + b^4 $

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

But we find the basic approach comfortably quick needing also no memory load of another obscure formula.

Key concepts used:

From the end state reaching the form of the intial state judiciously collecting the terms in the expression of the square so that the given expression values can be used as much as possible.

Q8. If $x^2 + y^2 + z^2 = xy + yz + zx$, then the value of, $\displaystyle\frac{4x +2y -3z}{2x}$ is,

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


The target expression being totally assymmetric but the choice values being simple, we suspect a possible high degree of simplification in the given expression.

It is not easily apparent but also not so difficult to discover.

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

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

Again the use of Principle of sum of squares.

So, $(x - y) = (y - z) = (z - x) = 0$

Or, $x = y = z$

So the target expression is simplied as,

$\displaystyle\frac{4x +2y -3z}{2x} = \frac{3x}{2x} = \frac{3}{2}$

Answer: Option d: $\displaystyle\frac{3}{2}$.

Key concepts used:

The idea of symmetry of the given equation hinted at judicious collection of terms.

Q9. If $x\left(3 - \displaystyle\frac{2}{x}\right) = \displaystyle\frac{3}{x}$, and $x\neq{0}$ then $x^2 + \displaystyle\frac{1}{x^2}$ is,

  1. $2\displaystyle\frac{5}{9}$
  2. $2\displaystyle\frac{4}{9}$
  3. $2\displaystyle\frac{1}{3}$
  4. $2\displaystyle\frac{2}{3}$


Target expression being in the familiar form of sum of square of inverses, we feel we should be able to transform the given expression in the form of sum of inverses. With this reasonable objective we proceed in a focused manner.

$x\left(3 - \displaystyle\frac{2}{x}\right) = \displaystyle\frac{3}{x}$,

Or, $3 - \displaystyle\frac{2}{x} = \displaystyle\frac{3}{x^2}$

Or, $3 - \displaystyle\frac{2}{x} - \displaystyle\frac{3}{x^2} = 0$

Or, $3x - \displaystyle\frac{3}{x} - 2 = 0$

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

Now we  are in familiar grounds.

Squaring the last equation,

$x^2 + \displaystyle\frac{1}{x^2} - 2 = \displaystyle\frac{4}{9}$

Answer: Option c: $2\displaystyle\frac{4}{9}$.

Key concepts used:

Transforming the given expression in the form of sum of inverses (the minus sign is also welcome and considered similar in this type of problems).

Q10. If $(x -a)(x - b) = 1$ and $(a - b) + 5 = 0$, then $(x - a)^3 - \displaystyle\frac{1}{(x - a)^3}$ is

  1. 140
  2. 125
  3. -125
  4. 1


To eliminate $x$, in the very beginning we subtract $x - b$ from $x - a$ to get $b - a$ which equals $5$.

In other words, this relationship can be expressed as,

$(x - a) - (x - b) = 5$

We are a bit uncomfortable with the absence of $x - b$ in the target expression as this is an assymmetry. To set right the symmetry, we use the first given expression,

$(x -a)(x - b) = 1$,

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

We use this to transform the simple subtracted relationship to,

$(x - a) - (x - b) = 5$,

Or, $(x - a) - \displaystyle\frac{1}{(x - a)} = 5$,

If we look at this equation and compare it with the target expression, we see no reason as to why we should not substitute any variable $p$ for $(x - a)$. We thus get,

$p - \displaystyle\frac{1}{p} = 5$,


$p^2 + \displaystyle\frac{1}{p^2} = 25 + 2 = 27$

The target expression thus is simplified as,

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

$\hspace{5mm} = 5\times{\left(27 + 1\right)} = 140$

Answer: Option a: 140.

Key concepts used:

Converting terms at every stage to transform the given expression in the form of sum of inverses.

Guided help on Algebra in Suresolv

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