## How to solve difficult algebra questions quickly by Componendo dividendo

Componendo dividendo is a **proportion based algebraic method** to simplify complex relations quickly.

We'll show *how to apply Componendo and Dividendo for solving difficult Algebra* and other quantitative math problems in a few steps lightning quick.

Quick solution is achieved by applying the method like a formula (Componendo dividendo formula) obtaining a simplified result from a more complex proportion relation wholly in mind.

An extensive range of competitive test questions on Algebra are solved by Componendo dividendo to show how the method can be applied directly or in modified form for quick solution.

This session on Componendo dividendo algebra is divided into the sections:

- Componendo dividendo
**concepts**in brief, **Conditions for applying Componendo Dividendo:**How to decide that a problem can be simplified quickly by componendo dividendo,- Large number of
**solved competitive test questions on algebra**that show how Componendo dividendo to be applied normally or suitably modified to solve the problem quickly.

If you wish you may **skip** the following brief on concepts and go straight to the solutions by clicking **here.**

### What is Componendo dividendo and how it works

ComponendoandDividendoare two separate but similar and closely related proportion based techniques (or operations) for simplifying NUMERATORS of a proportion.These single-step techniques are combined by taking the ratio of their results to form the three-step method of Componendo dividendo that sometimes is known as Componendo and dividendo.

For simplicity we will call this combined three-step method as Componendo dividendo.

#### Componendo

Componendo is an operation of simply ADDING unity 1 to a proportion equation to simplify the NUMERATOR of the

target fraction ratio on LHS. Thenumber of terms of the simplified numerator is reducedby the operation.

#### How Componendo is used

How to apply Componendo and dividendo will be explained by solving the simple problem,

**Problem so solve:** If $\displaystyle\frac{x+y}{x-y}=\frac{7}{5}$ what is the value of $\displaystyle\frac{x}{y}$?

By Componendo unity 1 is ADDED to both sides of the proportion equation reducing the numerator to simply $2x$. Result of the operation is,

$\displaystyle\frac{2x}{x-y}=\frac{7+5}{5}=\frac{12}{5}$........................(1)

#### Proportion

By definition, a proportion is a ratio equated to the product of *constant of proportionality* and a second ratio. Usually, the *proportionality constant is unity* so that expressed in fraction form, a proportion is one ratio fraction equal to a second ratio fraction. Example of a proportion,

$a:b::c:d$,

Or, $a:b=c:d$, with unity proportionality constant,

Or, $\displaystyle\frac{a}{b}=\frac{c}{d}$, where the ratio of two quantities $a$ and $b$ is proportional to the second ratio of two other quantities $c$ and $d$.

By the property of a ratio, the unit of $a$ and $b$ is same. Unit of $c$ is also same as that of $d$ but this unit is different from that of ratio $a:b$.

#### Dividendo

Dividendo is also a simple of operation of SUBTRACTING unity 1 from the proportion equation of two ratios. The simple operation reduces the number of terms in the numerator of the result.

#### How to apply Dividendo

To apply dividendo, unity 1 is SUBTRACTED from both sides of the proportion equation. Result of dividendo on our problem equation is,

$\displaystyle\frac{2y}{x-y}=\frac{7-5}{5}=\frac{2}{5}$............................(2)

Same denominator in the results of both Componendo and dividendo is ensured by the two single-action operations.

#### Componendo dividendo

Componendo dividendo combines the results of Componendo and Dividendo by taking the ratio of the two. As both results have same denominator, it is eliminated on taking the ratio of the two results. In the final result, the ratio of SIMPLIFIED NUMERATORS of the two results is formed as a new ratio on the LHS.

#### How to apply Componendo Dividendo

Taking the ratio of results of Componendo and Dividendo in equation (1) and equation (2) respectively the final result obtained is,

$\displaystyle\frac{x}{y}=\frac{12}{2}=6$.

Advantage of Componendo Dividendo: As the three operations are very simple and well-defined, the final simplified result can directly be deduced (in mind) using Componendo dividendo just like a formula.

The same result can also be reached by the conventional method of cross-multiplication, collecting like terms together and finally forming the ratio.

**Think:** Which of the two ways of solving the example problem is more convenient and quicker?

### Conditions for applying Componendo dividendo

Componendo dividendo can be applied to simplify quickly any problem in any math topic area that *has a proportion (or even a ratio)* with,

The terms in the numerator and denominator same but one term differing in sign between the two.

Examine the similarities and differences between the numerator and denominator of the ratio $\displaystyle\frac{x+y}{x-y}$ that has been the **target of simplification.**

Both have the two terms $x$ and $y$ with only $y$ differing in sign between the numerator and denominator.

This pattern (that we call Componendo Dividendo signature) only is exploited by Componendo and then Dividendo to reduce the numerator first as one of the two terms and next as the second term. The denominators remaining unchanged, it is eliminated in the ratio of results of Componendo and Dividendo. In the final result just the RATIO OF THE TWO TERMS in either the numerator or the denominator is obtained as LHS.

Though essentially simple, the operations of Componendo Dividendo can be modified in a number ways to quickly solve large number of problems that share the single criterion of having a Ratio with Componendo Dividendo pattern or signature (or even a variation of this pattern).

### Techniques of applying Componendo dividendo in solving difficult Algebra problems quickly

#### Example problem 1: Application of Componendo only

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

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

#### Solution Example problem 1

All three terms on the LHS satisfy the basic condition of **applying Componendo** as the variable terms are $a$ for both and differ by sign. The variable in the numerator of each will immediately be eliminated by adding 1 to each term. Best of all, the result of applying Componendo is exactly what is desired.

Compensating the addition of 3 in the LHS, 3 is added also to the RHS getting LHS as the target expression and answer 4 on the RHS.

**Note:** Though an LHS term in this problem doesn't have two terms in the numerator, applying combined three-step Componendo dividendo would give the simple enough result of $\displaystyle\frac{1}{a}$ on the LHS, say for the first term.

If the

result of dividendo is divided by the result of componendo, final result on the LHS would be $a$ instead. The choice of divisor and dividend in the third step would depend on the form of the final result that is desired.

**Answer:** Option a: $4$.

#### Example problem 2: A second application of only Componendo

If $\displaystyle\frac{a^2-bc}{a^2+bc}+\displaystyle\frac{b^2-ca}{b^2+ca}+\displaystyle\frac{c^2-ab}{c^2+ab}=1$, find the value of $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}$.

- $2$
- $1$
- $0$
- $-1$

#### Solution to Example problem 2

All three ratios have the right form for applying Componendo Dividendo.

But from the target expression, it is clear that if only componendo is applied on each term of the equation, result obtained would essentially be the terms of the target expression. A compensating 3 is to be added on the RHS and the resulting common factor of 2 on the LHS would be cancelled out.

The final result is,

$\displaystyle\frac{2a^2}{a^2+bc}+\displaystyle\frac{2b^2}{b^2+ca}+\displaystyle\frac{2c^2}{c^2+ab}=4$

Or, $\displaystyle\frac{a^2}{a^2+bc}+\displaystyle\frac{b^2}{b^2+ca}+\displaystyle\frac{c^2}{c^2+ab}=2$.

**Answer:** Option a: $2$.

Answer obtained practically in a single step all in mind.

#### Example problem 3: Applying Modified Componendo dividendo to form each term of the target expression and then their product

If $x=\displaystyle\frac{a-b}{a+b}$, $y=\displaystyle\frac{b-c}{b+c}$, and $z=\displaystyle\frac{c-a}{c+a}$, then the value of $\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is,

- $0$
- $1$
- $\displaystyle\frac{1}{2}$
- $2$

#### Solution to Example problem 3

**First conclusion:** Three separate equations for $x$, $y$ and $z$ means that the target expression also needs to be considered as a product of three separate ratios in $x$, $y$ and $z$.

**Second conclusion:** So the problem of evaluating $\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is broken down into evaluation of three simpler fraction ratios of $\displaystyle\frac{(1-x)}{(1+x)}$, $\displaystyle\frac{(1-y)}{(1+y)}$ and $\displaystyle\frac{(1-z)}{(1+z)}$. The product of these three fraction ratios would form the target expression automatically.

And the product of the values of the three fraction ratios would be the value of the target expression. Simple.

Only the three fragmented fraction ratios are to be evaluated. As these three are very similar, method of evaluation of one would also be applicable for evaluation of the other two.

**Third conclusion:** Each of the fragmented target expressions satisfies the condition for applying three-step componendo dividendo, but we cannot apply the quick method on the expression, can we? We'll see an unusual answer to this question later.

**Question 1:** Can we apply componendo dividendo on the first given equation $x=\displaystyle\frac{a-b}{a+b}$ to produce the first target fragment ratio, $\displaystyle\frac{1-x}{1+x}$ as the LHS?

**Answering conclusion 3:** Yes, we can. Carefully comparing the target fraction ratio with the first given LHS ratio $\displaystyle\frac{x}{1}$, it is clear that the three usual steps of componendo dividendo are to be modified.

#### Modified Componendo dividendo

**First modified step: Modified dividendo:** *Subtract the proportion equation from unity 1*. Result,

$\displaystyle\frac{1-x}{1}=\frac{2b}{a+b}$.....................................(3}

**Second modified step: Normal componendo:** Add 1 to the proportion equation. Result is,

$\displaystyle\frac{1+x}{1}=\frac{2a}{a+b}$.....................................(4)

**Third step: Normal ratio of first step result and second step result:** Dividing equation (3) by equation (4) get the target fragment ratio as LHS,

$\displaystyle\frac{1-x}{1+x}=\frac{b}{a}$.......................................(5)

Applying exactly same three operations of modified componendo dividendo on the second and third proprtion equations, results obtained are respectively,

$\displaystyle\frac{1-y}{1+y}=\frac{c}{b}$........................................(6), and,

$\displaystyle\frac{1-z}{1+z}=\frac{a}{c}$........................................(7).

Multiply equations (5), (6) and (7) together to get target expression as LHS and 1 on the RHS. $a$, $b$ and $c$ cancel each other out in the product of three RHS ratios.

$\displaystyle\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)91+z)}=1$.

**Answer:** Option b: $1$.

**Note:** There was never any concern about the results that would be obtained on the RHS because it is enough to get the target expression on the LHS of an equation. The RHS would automatically be the value of the target expression.

#### Steps in Modified Componendo dividendo

*Dividendo in modified form is used first.* Instead of subtracting 1 from the proportion, *the proportion equaion is subtracted from 1.* This is **modified dividendo**.

Also, *instead of componendo first, dividendo is done first* followed by componendo and ending in combining step of componendo dividendo.

Frequently the sequence or form of three simple steps need to be modified suitably to arrive at the

result that the target expression demands.

This is **modified componendo dividendo. **

**Ability to modify the simple componendo dividendo process suitably** is *the most important skill for solving hard problems by the process easily.*

Just remember, in solving tough problems, *you need to think analytically rather than apply a method mechanically.*

Though it was difficult to imagine at the beginning, the problem could be solved wholly in mind without writing on paper, which is the ideal situation for answering competitive test questions.

#### Example problem 4: Determine the ratio on which Componendo dividendo is to be applied to form the target expression

If $1.5a=0.04b$ then the value of $\displaystyle\frac{b-a}{b+a}$ will be equal to,

- $\displaystyle\frac{73}{77}$
- $\displaystyle\frac{2}{75}$
- $\displaystyle\frac{75}{2}$
- $\displaystyle\frac{77}{33}$

#### Solution to Example problem 4: Rule for determining the Source Ratio, componendo dividendo on which will produce the target expression

The target expression fully satisfies the conditions for applying componendo dividendo.

Question is,

**Question 1:** What will be the **source proportion** on which application of componendo dividendo will **produce the target expression** on the LHS of the result?

Following is the rule for getting the answer to this important question. **Frequently this rule needs to be used in solving problems.**

**The RULE for DETERMINING SOURCE RATIO that produces COMPONENDO DIVIDENDO READY TARGET RATIO by applying componendo dividendo is,**

To produce the result of componendo dividendo ready target expression $\displaystyle\frac{b-a}{b+a}$, apply componendo dividendo on the LHS of the proportion,

$\displaystyle\frac{\text{positive term of target numerator}}{\text{second term of target numerator}}=\displaystyle\frac{b}{a}=\frac{1.5}{0.04}$.

Result of applying componendo dividendo would be,

$\displaystyle\frac{b-a}{b+a}=\frac{1.46}{1.54}=\frac{73}{77}$.

**Answer: ** Option a: $\displaystyle\frac{73}{77}$.

Notice that standard form componendo dividendo steps are modified to produce the desired result in this problem.

This again is an example using MODIFIED COMPONENDO DIVIDENDO.

The modified steps are,

**First modified step:** dividendo: subtraction of unity 1 from the proportion equation.

**Second modified step:** Normal componendo: addition of unity 1 to proportion equation.

**Third step of taking ratio:** Just take ratio of result of first step and the result of the second step.

**Modifying the sequence of componendo and dividendo** ensured that the divisor and dividend in the third step would remain unchanged from usual sequence of first result to second result ratio.

#### More on How to identifiy the resultant ratio when componendo dividendo is applied on a proportion

When componendo dividendo is applied on $\displaystyle\frac{a+b}{a-b}$ resultant ratio will be $\displaystyle\frac{a}{b}$.

When componendo dividendo is applied on $\displaystyle\frac{b+a}{b-a}$ resultant ratio will be $\displaystyle\frac{b}{a}$.

**Rule:** By simple componendo dividendo, numerator of resultant ratio will be the variable that is positive in both numerator and denominator.

#### Example problem 5: Repeated Componendo dividendo to solve surd algebra problem

If $\displaystyle\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}=2$, then $x$ is,

- $\displaystyle\frac{7}{5}$
- $\displaystyle\frac{5}{12}$
- $\displaystyle\frac{5}{7}$
- $\displaystyle\frac{12}{5}$

#### Solution to Example surd algebra problem 5

It is easy to see that the given LHS ratio is to be simplified in two stages,

- first by converting the ratio to a ratio of two component surd expressions, $\sqrt{3+x}$ and $\sqrt{3-x}$, and,
- after removing the square root, again forming a ratio involving just $x$ and $3$.

Apply normal three-step componendo dividendo on the given proportion equation to simplify it to,

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

Remove the square roots by raising both sides of the equation to squares,

$\displaystyle\frac{3+x}{3-x} = 9$.

The LHS of the proportion is again ready for applying Componendo dividendo,

$\displaystyle\frac{3}{x} = \frac{9+1}{9-1}=\frac{5}{4}$

Or, $x = \displaystyle\frac{12}{5}$.

**Answer:** Option d: $\displaystyle\frac{12}{5}$.

In this case instead of the target expression, the given ratio itself satisfies conditions for applying componendo dividendo.

This is an example where * Componendo dividendo rule could be applied twice*, and as is usually true with componendo dividendo, solution obtained wholly mentally. This is one of the biggest advantages of Componendo dividendo—you can solve problems using Componendo dividendo in mind.

#### Example problem 6: Componendo dividendo conditions are satisfied by a part of the target expression

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

- $1$
- $2$
- $\displaystyle\frac{3}{7}$
- $\displaystyle\frac{1}{7}$

#### Solution to Example problem 6: Evaluating the target expression component that satisfies conditions of componendo dividendo

Not the given expression or the target expression, the componendo dividendo ready ratio is found only as a part of of the tarrget expression. That is the point made highlighted by this problem,

Wherever it occurs, a fraction ratio expression that satisfies conditions of componendo dividendo, should be explored first for simplification by componendo dividendo.

So it is decided that the second part of the target expression $\displaystyle\frac{2y-x}{2y+x}$ will be evaluated first by componendo dividendo.

In the next step, by source ratio identification rule, the source ratio identified for generation of this target expression component,

$\displaystyle\frac{2y}{x}=\frac{5}{2}$ from given expression.

Result of application of modified componendo dividendo (dividendo first, componendo second) is,

$\displaystyle\frac{2y-x}{2y+x}=\frac{5-2}{5+2}=\frac{3}{7}$.

Substituting this value for the second component of the target expression, the final answer becomes,

$\displaystyle\frac{3}{7}+\displaystyle\frac{4}{7}=1$.

**Answer:** Option a: $1$.

Though this is a simple problem, we would choose componendo dividendo instead of conventional method of substitution simply because componendo dividendo not only is quick, but also **it creates practically no stress load on the mind** that even a simple cross-multipplication and substitution creates.

#### Example problem 7: Possibility of componendo dividendo application is hidden

If $\displaystyle\frac{x^2-x+1}{x^2+x+1}=\frac{2}{3}$ find the value of $x+\displaystyle\frac{1}{x}$.

- $6$
- $5$
- $8$
- $4$

#### Solution to Example problem 7: The componendo dividendo hint in given expression forced exploring the possiblity in the target

Really, the LHS ratio in the given expression being a three term one, a moment more was needed to identify it as a componendo dividendo ready fraction ratio.

In the next step, the result of applying componendo dividendo on this LHS formed automatically as,

$\displaystyle\frac{x^2+1}{x}=\frac{{5}{3}}{\frac{1}{3}}=5$, the two terms are $(x^2+1)$ and $x$, identification of which is critical for quick solution of the problem.

Executing the two steps in mind takes little time.

But even this little time would be wasted if the target can't be deduced from the resultant fraction ratio of $\displaystyle\frac{x^2+1}{x}=5$.

So next the target expression is examined with specific intention of matching it with $\displaystyle\frac{x^2+1}{x}$.

As it always happens, when with specific intent something hidden in front is searched for, it is invariably is discovered easily.

Target expression,

$x+\displaystyle\frac{1}{x}=\displaystyle\frac{x^2+1}{x}=5$.

**Answer:** Option b: $5$.

Always look for the possibility of applying componendo dividendo even if the problem has just a hint of it.

#### Example problem 8: Adapting the problem for solving it quickly by componendo

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

- $\displaystyle\frac{6}{8}$
- $\displaystyle\frac{8}{6}$
- $\displaystyle\frac{27}{20}$
- $\displaystyle\frac{20}{27}$

#### Solution to Example problem 8: Modifying the problem to solve it quickly by componendo two times

The problem is a bit tricky and doesn't have the tell-tale pattern of componendo dividendo in it.

So most times it would be solved without using componendo dividendo.

But being used to the great comfort of using componendo dividendo to solve a problem quickly and cleanly, the problem is examined one more time **with the specific intent of finding a way to use componendo dividendo to solve it.**

Now on closer examination it seems, the two-variable sums in the numerator and denominator of the target may perhaps be evaluated by componendo from the look of the two given expressions.

The target fraction ratio does have the

pattern of componendoin it. Both the numerator and denominator are two variable sums and a two variable sum is formed in the result numerator by a normal componendo operation, isn't it!.

Explore the possibility yourself.

Yes, it is very much possible to get $(a+b)$ as numerator of the result of componendo on the first given equation and $(b+c)$ as the numerator of result of componendo on inverted second given equation. Denominators in both results would be $b$. That's it, the ratio of results two operations of componendo would

produce value of the target.

First **apply componendo** on the first proportion equation. The result is,

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

Now **invert the second given equation** to bring $b$ in the denominator so that after applying componendo $(b+c)$ in numerator and $b$ in denominator are obtained in the result. This would ensure cancelling out $b$ as well as forming the target ratio on the LHS of the division operation.

The inverted second equation is,

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

**Apply componendo on the proportion equation.** This is **the second time componendo is applied.** Result is,

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

The LHSs are in right form now for taking the ratio of the two.

Divide the first result of componendo by the second. Result is,

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

**Answer:** Option d: $\displaystyle\frac{20}{27}$.

Instead of Componendo Dividendo,

it is Componendo Componendo, anew way of using the three step operation.

With the right objective, a way out is usually found. Problem solving is basically discovering and using hidden patterns. Though a bit tricky, the problem is solved in a few tens of seconds, all in mind.

In this case, not only

the problem itself is modified to adapt it for applying componendo, the concept of componendo dividendo has been radically changed to apply componendo twice and take the ratio of the two results.

The **idea is valuable** because there would be many problems that may be modified for applying the clean method of componendo dividendo.

It creates a new approach of problem solving by THE CONCEPT OF COMPONENDO DIVIDENDO.

#### Example problem 9: When not to apply Componendo dividendo in a problem with pattern of Componendo dividendo

If $x=\displaystyle\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$, and $y=\displaystyle\frac{1}{x}$, then the value of $3x^2-5xy+3y^2$ is,

- 1771
- 1717
- 1177
- 1171

#### Solution to problem example 9: Quick solution NOT by Componendo dividendo but by using simple result of $(a+b)^2+(a-b)^2=2(a^2+b^2)$

The LHS in the given equation fully satisfies the conditions for applying componendo dividendo. But resist the temptation for a moment and examine the relation between $x$ and $y$.

As the two variables are mutually inverse to each other, their sum would have the result,

$x+y=\displaystyle\frac{a+b}{a-b}+\displaystyle\frac{a-b}{a+b}$, where $a=\sqrt{13}$ and $b=\sqrt{11}$ are used temporarily just for the simple look of it,

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

$=\displaystyle\frac{2(a^2+b^2)}{a^2-b^2}$, the numerator is a very useful algebraic identity that is the key to quick solution of this awkward problem

Or, $x+y=\displaystyle\frac{2\times{24}}{2}=24$.

This extremely simple result could be obtained mentally in a few tens of seconds.

Turning focus on the target it is easily expressed in terms of $(x+y)=24$ and $xy=1$ as,

$3x^2-5xy+3y^2=3[(x+y)^2-2xy]-5xy$

$=3(24^2-2)-5=1717$.

You may try the componendo dividendo way and compare the two approaches.

**Remember:** With the pattern $x=\displaystyle\frac{1}{y}$ and $x=\displaystyle\frac{a+b}{a-b}$ in a problem it is always easier to solve the problem using the algebraic identity of $(a+b)^2+(a-b)^2=2(a^2+b^2)$.

You may see the solution using componendo dividendo in our own early article, **How to solve a difficult surd algebra question by repeated componendo dividendo in a few steps 17.**

**This shows that we couldn't decide what is the simplest way to the solution to this awkward problem that time. Now we have learned and know for sure.**

### Other resources on Componendo dividendo

You may like to go through the related **tutorials**,

*Componendo dividendo uncovered to solve difficult algebra problems quickly 5*

**Componendo dividendo applied in number system and ratio proportion problems**

**Componendo dividendo in Algebra**

**Componendo dividendo explained**

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