How to solve difficult SSC CGL Algebra problems in a few steps 8

How to solve difficult SSC CGL algebra problems in a few simple step -8

Numerator denominator addition and Target as a variable techniques help to reach elegant solutions in a few steps

The school Algebra also form an important part of most of the competitive job tests such as SSC CGL, Bank POs etc.

We'll not repeat the usual reasons behind the difficulties faced by students in solving Algebra problems and the set of basic and rich concepts that are invaluable in reaching elegant solutions for seemingly difficult Algebra problems.

As a refresher on these, you may refer to our posts on basic and rich algebra concepts and Algebra problem simplification in a few steps 1, in a few steps 2, in a few steps 3 and in a few steps 4.

In this session we will highlight the thinking process searching for the elegeant solutions through two carefully selected problems.

In the first problem,

By strong application of End state analysis and free resource use principle, comparing the choices with the given expression, it was evident that the three given terms need not be summed up but each needs to be transformed. We have achieved this by numerator denominator addition technique.

And in the second problem,

We have shown how using Many ways technique a problem can be solved in more than one way. In the first conceptual method we have applied target as variable technique to isolate a part of the target expression suitable for application of Componendo dividendo technique. This is application of Target as a variable technique.

For best results, you should attempt to solve each of these problems yourself and measure your time and number of steps to the solution before you go ahead with the explanation of the solution.

Problem 1.

If $\displaystyle\frac{b - c}{a} +\displaystyle\frac{a + c}{b}+ \displaystyle\frac{a - b}{c} = 1$ and $a - b + c \neq 0$, then which of the following is correct?

  1. $\displaystyle\frac{1}{a} = \displaystyle\frac{1}{b} + \displaystyle\frac{1}{c}$
  2. $\displaystyle\frac{1}{c} = \displaystyle\frac{1}{a} + \displaystyle\frac{1}{b}$
  3. $\displaystyle\frac{1}{b} = -\displaystyle\frac{1}{c} - \displaystyle\frac{1}{a}$
  4. $\displaystyle\frac{1}{b} = \displaystyle\frac{1}{c} + \displaystyle\frac{1}{a}$

Solution: First stage Problem analysis:

The very first thing that strikes us is the second condition of inequality. It indicates that on transformation of the given expression,

the expression, $(a - b + c)$ will be factored out and as it is not zero, its second factor will be equated to 0 leading us to the solution.

This is a reasonable way of arriving at a conclusion using deductive reasoning.

The second conclusion we draw is, the three terms of the given expression are not to be summed up, as the choice values also have the three variables $a$, $b$ and $c$ in the denominator of the three terms.

Combining these two analytical results, we decide our task is to somehow transform each of the numerators of the three terms in the given expression to $(a - b + c)$ without changing the denominators at all.

First stage action: transforming the numerators

A standard technique of transforming numerator without changing the denominator is to add or subtract 1 from the fraction term. Let us call this general technique as Numerator denominator addition technique. In Componendo and dividendo this technique is applied twice and then the ratio taken.

Examining the numerators of the three terms we find the third term to be ready for the transformation,

$\displaystyle\frac{a - b}{c} + 1 = \displaystyle\frac{a - b + c}{c}$, just as we wanted. But from where would this 1 come? We need to compensate this addition by a subtraction of 1, that is, addition of -1.

Again we find this $-1$ to work perfectly with the second term of the expression,

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

Now only we take care of the 1 on the right hand side of the equation not by bringing it to the left hand side, but instead moving the third term to the RHS,

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

Summing up the three terms now,

$(a - b + c)\times{\left(\displaystyle\frac{1}{b} + \displaystyle\frac{1}{c} - \displaystyle\frac{1}{a}\right)} = 0$

Canceling the common factor $(a - b + c)$,  which is not equal to zero, we get.

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

Answer: Option a: $\displaystyle\frac{1}{a} = \displaystyle\frac{1}{b} + \displaystyle\frac{1}{c} $.

Key concepts and techniques used:

  • The inequality condition gave us the first clue of a factor that will be canceled out as it is not zero. Our attempt should be to form this factor in the numerator as from the choice values we see the variables appearing individually separate and unchanged.
  • To transform the numerators to the same factor of $(a + b - c)$ we add and subtract 1 suitably to each of the three terms to form the factor in the three numerators. Once the factor $(a - b + c)$ is formed we could cancel it out as it is not zero as given. The second factor handed out our answer. For transforming the numerator we have used the numerator denominator addition technique. Generally by adding or subtracting a 1 to the term this method is implemented.
  • All through pattern analysis and deductive reasoning have been used along with End state analysis.

Problem 2.

If $\displaystyle\frac{x}{y} = \displaystyle\frac{4}{5}$, then the value of $\displaystyle\frac{4}{7} + \displaystyle\frac{2y - x}{2y + x}$ is,

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

Problem analysis and elegant solution

On analysis of the target expression we find the second term pefectly suitable for application of Componendo dividendo technique resulting in $\displaystyle\frac{x}{y}$, value of which is given. It would have been an ideal condition for a quick solution except for the extra term in the target expression.

To get over this difficulty we express the target itself as a dummy variable,

$E = \displaystyle\frac{4}{7} + \displaystyle\frac{2y - x}{2y + x}$,

Or, $E - \displaystyle\frac{4}{7} = \displaystyle\frac{2y - x}{2y + x}$.

Now we add  1 to both sides,

$E + \displaystyle\frac{3}{7} = \displaystyle\frac{4y}{2y + x}$,

and subtract both sides from 1,

$\displaystyle\frac{11}{7} - E = \displaystyle\frac{2x}{2y + x}$.

Taking a ratio,

$\displaystyle\frac{E + \displaystyle\frac{3}{7}}{\displaystyle\frac{11}{7} - E}= 2\displaystyle\frac{y}{x} = \displaystyle\frac{5}{2}$,

Or, $2E + \displaystyle\frac{6}{7} = \displaystyle\frac{55}{7} - 5E$,

Or, $7E = 7$,

Or $E = 1$.

This is a special technique of expressing target expression itself as a variable to isolate a convenient portion of it for manipulation. We name this rich concept and technique as Target expression as a variable technique.

Answer: Option a: 1.

Alternate method

Instead of isolating the convenient part of the target we would now substitute the given value directly in the target after transforming it a little bit,

$E = \displaystyle\frac{4}{7} + \displaystyle\frac{2y - x}{2y + x}$

$\hspace{5mm}=\displaystyle\frac{4}{7} + \displaystyle\frac{\displaystyle\frac{2y}{x} - 1}{\displaystyle\frac{2y}{x} + 1}$

$\hspace{5mm}=\displaystyle\frac{4}{7} + \displaystyle\frac{\displaystyle\frac{5}{2} - 1}{\displaystyle\frac{5}{2} + 1}$

$\hspace{5mm}=\displaystyle\frac{4}{7} + \displaystyle\frac{\displaystyle\frac{3}{2}}{\displaystyle\frac{7}{2}}$

$\hspace{5mm}=\displaystyle\frac{4}{7} + \displaystyle\frac{3}{7}$


This is a standard mechanical deduction while the first method is more of conceptual nature.

Key concepts and techniques used in the first method:

  • From analysis of the target deciding to apply componendo dividendo on the part target.
  • Declaring the target itself as variable separating out the part expression, applying the componendo and dividendo and final simplification.

Key concepts and techniques used in the second method:

  • Directly substituting the given value in the target after transformation and routine simplification.
  • This is the conventional method.


In this case again we find the problem could be done in more than one way. This is a case of Many ways technique, practice of which enhances the power of your problem solving skill in general.

Regarding choice of method, there is not much difference between the two in terms of time to solve. If you are conceptually strong you should go for the first method which can be done wholly mentally in a short time. Otherwise adopt the routine method.