How to solve a tricky algebra problem by adapted componendo dividendo in a few steps | SureSolv

How to solve a tricky algebra problem by adapted componendo dividendo in a few steps

Sometimes you need to change and adapt a powerful method to fit the problem

How to solve a tricky algebra problem by adapted componendo dividendo

Our main focus in problem solving, be it a real life problem or SSC CGL math, is to find a way to solve the problem as quickly as possible with minimum labor. This translates for math problem solving to,

  1. Solve the MCQ math without writing any deduction steps at all if we can. It doesn't mean we will load our memory and mind with lots of calculations and deductions, that we would have had to write on paper otherwise. It actually means, we would avoid deductive steps as far as possible.
  2. Find a way to the solution that is as short and as quick as possible.

How is this possible, you may ask. You may even think we have a bag of tricks that we would use. No, we avoid the word trick completely. Instead we rely on,

  1. Clear basic concepts on the topic involved. With honest disciplined effort anyone can acquire it.
  2. Pattern based rich concepts that are derived from basic concepts and methods that use the rich concepts to solve problems quickly. These rich concepts and pattern based methods actually hold the power to solve problems quickly, cleanly and easily without giving any load to your mind. Some of these rich concepts you may find elsewhere, but to a large extent these are derived, explained and used here while solving problems.
  3. But, first of all, when you are in the process of solving a problem, you should be able to identify the useful pattern on which you will apply the suitable rich concept. It is kind of a tightly bound pair, a specific pattern in the problem and suitable rich concept based method for solving it. The only way to acquire this skill and ability is to solve many problems, all the while trying to approach it for solving quickly and in mind, and absorb the knowledge that are available on this special basic and rich concept based way of problem solving.

Essentially this is a different way of thinking that does not require great intelligence, but it is also true you can acquire this ability only with honest and disciplined effort over some time.

The ideal situation would be when you would be able to derive your own pattern based rich concepts as you meet a new pattern in a problem that you haven't seen earlier.

We would try to give you some idea about these elusive and vague sounding words through solving an apparently tricky SSC CGL level algebra problem.

First we will give you the problem and you should try to solve it before going ahead further.

Let us remind you a very important aspect of successful solving of a series of many math MCQ problems in a hard competitive test within a relatively tight time schedule,

There is no place for guessing the answer or trying to remember tricks that you can apply. When you solve a problem you would solve it surely and quickly, otherwise you will skip it. No way you can afford to wrestle or struggle with a problem in such a competitive test. As soon as you go through the problem and analyze it evaluating the possible paths to the solution, 90% of cases, you should be able to see up to the end of the problem in a few tens of secs. With a few others, when no clear path to the solution is visible you may explore with educated guesses, but those should be rare.

SSC CGL level Algebra problem to be solved in mind

If $\displaystyle\frac{a}{b}=\frac{2}{3}$ and $\displaystyle\frac{b}{c}=\frac{4}{5}$ then $\displaystyle\frac{a+b}{b+c}$ is,

  1. $\displaystyle\frac{6}{8}$
  2. $\displaystyle\frac{8}{6}$
  3. $\displaystyle\frac{27}{20}$
  4. $\displaystyle\frac{20}{27}$

Solution—the analytical thinking part

Yes, analytical thinking is the basis of all efficient problem solving except intuitive problem solving. This is a higher stage, where analysis and method selection have become part of your mind. Solution comes immediately and intuitively. We are not discussing that stage.

Our habit is to look at the required expression and the given expressions and analyze and compare the two sets. We have given a name to it, End state analysis approach. In its entirety it is by far one of the most powerful problem solving approaches we know of. If you are interested you may go through the article on End state analysis approach. It is not related to math at all, but it should be useful to anyone interested in new quick way to solve problems.

On brief analysis, finding no clear and short deduction path (yes, by habit we search for clear and short deduction paths also), we started on a different path of useful patterns and immediately detected the possibility of getting the numerator of the target, $a+b$ just by adding 1 to the first given equation.

$\displaystyle\frac{a}{b}=\frac{2}{3}$,

Or, $\displaystyle\frac{a+b}{b}=\frac{5}{3}$.

We still have to deal with the denominator $b$, but now we remembered, isn't it the case for componendo dividendo? After the first step on the first equation, numerator takes the desired form, but denominator remains untouched. Only after the second step is taken on the second equation (usually subtractive), we eliminate the untouched denominator by dividing the two resulting equations.


If you want to know more about the highly effective and popular method of Componendo dividendo, refer to our articles,

Componendo dividendo explained, and

Componendo dividendo applied to number system and ratio and proportion problems.


Can we transform the numerator of the second equation to the same form as the first? We turned our attention to the second equation. Yes we can, just by inverting it and adding 1 to it,

$\displaystyle\frac{b}{c}=\frac{4}{5}$,

Or, $\displaystyle\frac{c}{b}=\frac{5}{4}$.

What a surprise! It has turned out to be exactly as we wanted. We just have to add 1 to this result and divide the two results.

Adding 1 to the second inverted equation,

$\displaystyle\frac{c}{b}=\frac{5}{4}$,

Or, $\displaystyle\frac{b+c}{b}=\frac{9}{4}$.

Just divide the first result by this second result to eliminate the common denominator $b$,

$\displaystyle\frac{a+b}{b+c}=\frac{20}{27}$

Though we have described the thinking process, it was not exactly so. As soon as we rejected deductive approach, and seen that the numerator of the target can be formed by adding 1 to the first equation, automatically we went to the second equation, inverted it, added 1 and divided the two results.

All in mind.

But we must admit—it was all because of our confidence on the ease and and power of componendo dividendo, the three step process, and experience of using it. One of the steps of proper componendo dividendo is subtraction. Instead in both cases we have added 1. That's where we have adapted the popular and powerful method to suit our purpose perfectly.

Time to time you need to adapt known power-packed methods, or even take only a part of it to create a new method.

Before signing off we would say, you may try your own method to see if it is faster.


Other resources on Componendo dividendo

Componendo dividendo explained

Componendo dividendo in Algebra

Componendo dividendo applied on number system and ratio proportion problems

Componendo dividendo adapted to solve difficult algebra problems quickly 4

Componendo dividendo uncovered to solve difficult algebra problems quickly 5

Hidden Componendo dividendo uncovered to solve difficult algebra problems quickly 6

Partially hidden Componendo dividendo revealed to solve a not so difficult algebra problem in minimum time 7.

Additionally, each article under category Efficient Math Problem solving under Exams contains use of such concept based rich power methods for unusually quick math problem solving. If you are interested you can explore by referring to Efficient math problem solving.


Resources on Algebra problem solving

The list of Difficult algebra problem solving in a few steps quickly is available at, Quick algebra.

To go through the extended resource of powerful concepts and methods to solve difficult algebra problems, you may click on,

Algebra questions, quick solutions and especially powerful methods to solve difficult algebra problems.