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 13

Deductive reasoning and use of symmetry in algebraic expressions enable quick solution in a few steps

Usually we solve difficult Algebra problems in a few steps using basic and rich Algebra concepts and techniques together with powerful general problem solving techniques. This off-the-grid problem required more of deductive reasoning, enumeration, mathematical reasoning, general problem solving techniques and core algebraic concepts such as symmetry in algebraic expressions and less of algebraic expression manipulation.

In this session we will show how problem analysis and deductive reasoning help to choose the appropriate strategy of problem solving and the more fundamental and abstract concept of symmetry in algebraic expressions provides the clinching breakthrough.

Let us go through the problem solving process to appreciate the use of the highly effective strategies and techniques.


Chosen Problem 1.

If $\displaystyle\frac{1}{a+1} +\displaystyle\frac{1}{b+1} +\displaystyle\frac{1}{c+1}=2$, then value of $a^2+b^2+c^2$ is,

  1. $\displaystyle\frac{3}{4}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{9}{16}$
  4. $\displaystyle\frac{2}{5}$

Problem analysis - first observation

On a brief look the problem seemed to be incomplete. It was felt that more given expressions should have been there.

Problem analysis - Second observation

On a quick but deeper second stage analysis, it became clear that reaching the end state of $a^2+b^2+c^2$ from the given expression through deduction of algebraic expressions would be impracticable.

Note: As problem solvers, we do not assume a path of solution as impossible. The potential solution paths are simply categorized as nearly impossible or impracticable that might take inordinately long time defeating the main objective of quick elegant solution, more favorable, most promising and the such. This is solution path evaluation or alternative evaluation.

The decision to reject this commonly followed approach of deductive steps was clear and it could be taken based on experience in solving many problems. This was an important decision and about half of the total solution time was spent in crossing this mental barrier.

Problem analysis - Final choice of strategy

Apart from concept and technique based deductive transformation, the only other most favorable strategy for these problems is solution through enumeration. We formally state the strategy or technique of enumeration as,

Substituting suitable values of variables in a planned manner and analyzing the effects substitution so that the series of trials will converge on to the actual solution quickly.

Choice of suitable values for substitution must be such that calculations take minimal time and the trials home in to the actual solution quickly. This is possible primarily through mathematical reasoning, even if you are not aware of the exact nature of the process.

Occasionally as a variation of the strategy, the choice values are used as trial values of variables for substitution. This turns out then to be the application of free resource use principle.

Problem analysis - nature of the key expression is symmetric

The symmetry in problem expression is formally stated as,

When an algebraic expression involved in a problem is symmetric, coefficient and power of the individual variables are equal and the structure of the corresponding terms are same.

In our problem the given expression satisfy these criteria and so is symmetric.

Problem analysis - the conclusions derived from the criterion of symmetric expression

The symmetry in problem expression produces two important results or conclusions,

  • The variables are fully interchangeable without changing any outcome. This means, in a symmetric expression we can interchange variables $a$, $b$, and $c$ between each other in any way without changing nature and value of expression. By observation we can believe that this indeed is the fact under the circumstances.
  • The second more specific conclusion follows from the first. Because of the symmetry in the problem expression, the assumption that the values of all the variables are equal will be a valid and the most balanced assumption. It means we can assume, $a=b=c$ in our problem. This result follows from mathematical reasoning.

Problem solving execution - Choice of trial values for the variables

At this stage instead of choosing trial values of the variables directly, the nature of the given expression is analyzed again, but this time we have the balanced assumption of,

$a=b=c$.

The expression symmetry and variable equality dictated that all the three terms of the given expression must have equal values. As there are 3 terms, the sum must have 3 as a factor. Again as the final result on the RHS is 2, the sum must also have a 2 as a factor and a 3 as a divisor giving individual terms as, $\displaystyle\frac{2}{3}$.

Inversing a term, we get the denominator, say, $(a+1)=\displaystyle\frac{3}{2}$.

This immediately results in,

$a=b=c=\displaystyle\frac{1}{2}$.

And,

$a^2+b^2+c^2=\displaystyle\frac{3}{4}$.

Answer: Option a : $\displaystyle\frac{3}{4}$.

Note: By mathematical reasoning and expression analysis we have avoided even the multiple trial steps. In reality, after crossing the mental barrier of solving through deductive steps, solution was nearly instantaneous totaling less than a minute's time.

Mathematical analysis

After solving the problem when we analyzed it again, it became clear that the problem without the choice value set has not one but infinite many solutions. For example, if we assume the values of $a$ and $b$ as any pair of real numbers, the value of $c$ can be deduced from the given equation and then the value of target equation can be calculated as a real number. If $a=b=2$, the value of $c$ becomes, $-\displaystyle\frac{1}{4}$. If $a=b=1$, $c=0$. So our initial conclusion of impracticability of deductive solution and rejection of the approach was right. In fact deductive solution would have been impossible. The given problem is then a malformed problem.

In enumeration of trial values approach, if out of infinite many possible solutions, one particular solution appears among the choice values, it becomes our target solution. If not, we might have to continue repeating our trials, taking inordinately long time. Under these conditions the quickest and most elegant solution would be reached only when the values of all the variables are equal.

Our take is, in SSC CGL or similar competitive tests, all math problems have under a minute elegant solutions. It is up to us to find it. That's why we assumed the equality of the variable values.