## Basic ratio concepts revisited, ratio terms are multiples of ratio values

The definition of a ratio is,

A ratio expresses how many times one quantity is in comparison to another quantity of same unit.

It is a comparison function. For example, we may express the fact that the age of the son is $\frac{2}{3}$rd of the age of the father as,

$Age_{son} : Age_{father}=2:3$,

where,

$Age_{son}=\text{Age of the son}$, and

$Age_{father}=\text{Age of the father}$.

We call $Age_{son}$ and $Age_{father}$ as the * ratio terms* or

*, and the corresponding values of the ratio, 2 and 3 as*

**ratio quantities***.*

**ratio values**The ratio terms represent the quantities being compared in the ratio and the ratio values represent how many times one ratio quantity is with respect to the other. The ratio terms must be of same unit.

For mathematical manipulation, such a ratio is needed to be expressed as fractions. In our example then the ratio is expressed in an acceptable form as,

$\displaystyle\frac{Age_{son}}{Age_{father}}=\frac{2}{3}$.

**Convention** is,

In a proper ratio, the ratio values are always expressed as integers, and as a proper minimized fraction with no common factor between them. This convention ensures uniqueness and clarity of understanding.

In a problem, * even if the ratio values are not integers, they need to be converted to integers* by multiplying the two values with suitable factors. This is needed to aid manipulation of the ratio values as numerator and denominator of a fraction.

For example, in our problem, the ratio could have been given as,

$Age_{son} : Age_{father}=0.4:0.6$.

To transform this ratio form we will multiply both the numerator and the denominator by the same factor 5 transforming the ratio to,

$Age_{son} : Age_{father}=2:3$.

### Ratio quantities are multiples of corresponding ratio values

This is the * first most basic concept* in the topic of ratio and proportion. It means, in a given ratio,

Actual values of the ratio quantities are not given. The ratio values do not represent the actual values of the ratio quantities, but

each of the ratio quantity must be a multiple of its corresponding ratio value.

For example in our problem, the ages of the son and the father could have been 36 years and 54 years or for that matter 50 years and 75 years. For both set of actual values the ratio expression given would have been valid and $2:3$.

In each case though * the age of the son would certainly be a multiple of the ratio value of 2* and

**that of the father a multiple of 3.**### Ratio quantities are multiples of corresponding ratio values with a common factor

The second most basic concept is,

The two ratio quantities will always have a common factor 1 or more than 1 which when canceled out, the ratio values are expressed as a minimized proper fraction with no common factor between them. This canceled out common factor will be the HCF between the two quantities being compared so that after canceling out the HCF, the ratio values won't have any common factor left between them.

Alternatively we can state,

Each of the two ratio quantities will always be a multiple of corresponding ratio value, the multiplying factor being 1 or more than 1 and equal for both the ratio quantities, being the HCF between them.

For example, if $Age_{son}=50$ and $Age_{facther}=75$, the HCF of the two will be 25 and after canceling out this HCF, the ratio will finally be expressed as,

$Age_{son}:Age_{facther}=50:75=2:3$.

When the ratio values are given as a proper minimized fraction we won't know what the canceled out HCF was, and so from a ratio we won't know the actual values of the quantities being compared.

Based on this concept we use the * technique of reintroducing the HCF* in solving majority of the ratio problems. After reintroducing the HCF as a factor of both the ratio values,

*before solving it. Though generally not complex, this takes a bit of time.*

**in the conventional method a linear equation is formed from a equation of fractions**Instead, using the technique of reintroducing the HCF * along with the concept that each ratio quantity must be a multiple of the corresponding ratio value*, we can

*; by applying*

**avoid the algebraic process of forming and solving an algebraic equation***to reach the solution much faster.*

**multiple enumeration**We will explore * conventional procedural method*,

*,*

**mental cross-multiplication method***and the*

**componendo technique use method***in solving two chosen problems and analytically evaluate the efficiency of these approaches.*

**multiple of ratio value technique**The * multiple of ratio value technique* is a

*and should give you the quickest and the*

**rich ratio concept***for this type of ratio problems on number or ages.*

**elegant solution**### Chosen Problem 1.

If 12 is added to two numbers, their ratio changes from $3:4$ to $13:16$. The larger number is,

- 48
- 144
- 52
- 36

**Problem solving by conventional method, writing down the steps as we do in school**

The original ratio of the two numbers $A$ and $B$ is, $A:B=3:4$. Applying the HCF reintroduction technique, we get the actual values of $A$ and $B$ as, $A=3x$ and $B=4x$. Here we have assumed canceled out HCF as $x$, the single unknown variable and the numbers as $A$ and $B$.

After addition of 12 to $A$ and $B$ then the problem status is,

$\displaystyle\frac{3x+12}{4x+12}=\frac{13}{16}$,

Or, $16\times{(3x+12)}=13\times{(4x+12)}$,

Or, $48x+192=52x+156$,

Or, $4x=36$,

Or, $x=9$.

So the larger number $B=4x=36$.

**Answer:** Option d: 36.

**Method analysis and improvement by applying delayed evaluation technique**

In any case, we need to form the changed status equation of two fractions after addition of 12 to both the numbers. Just like above if you follow the exact steps, it will take you the longest time. **This is a fully procedural method that in most cases is the slowest and most inefficient.**

We can improve the method by applying * delayed evaluation technique* which forms a part of broader

*by avoiding the multiplication,*

**efficient simplification concepts**$\displaystyle\frac{3x+12}{4x+12}=\frac{13}{16}$,

Or, $16\times{(3x+12)}=13\times{(4x+12)}$,

Or, $16\times(3x) +16\times(12)=13\times(4x)+13\times{12}$,

Or, $13\times{4x}-16\times{3x}=12\times{(16-13)}$,

Or, $4x=12\times{3}$,

Or, $x=9$.

So,

$B=4x=36$.

Here we have avoided the multiplication of 12 by 16 and 13, saving valuable seconds.

**Note:** If you write the steps it will always waste valuable time. That's why we always recommend,

Solve the problem in mind as far as possible.

If you have visualization problem, just write down the changed situation status equation of fractions after addition of 12 to the two numbers and do the rest in your mind.

#### Problem solving by cross-multiplication and simplification mentally

In this method if you are comfortable with mental manipulation of simple expressions and also mental math at simple level, just write down the changed status equation first as,

$\displaystyle\frac{3x+12}{4x+12}=\frac{13}{16}$.

**Without forming the equation,**

$16\times{(3x+12)}=13\times{(4x+12)}$,

* form each term* in your mind by

*from the fraction,*

**cross-multiplication***(in this case subtraction) and*

**collection of similar terms**

**then form the simple equation in your mind at the last step.****An effective tactic in this process is to form each term separately.**

One can easily arrive at the $x$ term as $52x-48x=4x$, the LHS,

And then $3\times{12}=36$ as RHS.

**This will give you the solution much quicker if you are adept in this process.**

#### Further improvement of cross-multiplication method

* If you avoid writing down even the changed status information forming it in your mind* and carry out the rest of the steps, the solution will come still faster. But be aware, unless you are quite skillful in visualization, mental manipulation of expressions, and mental math, you may make mistakes. Our recommendation,

Write down the changed status equation and do the rest mentally.

Conceptually though we are following the conventional method only.

#### Problem solving by componendo technique

Componendo dividendo is a powerful algebraic technique that frequently can be applied to simplify a particular form of expression in a few seconds. The form of expression it is applied to is,

$\displaystyle\frac{a+b}{a-b}=\frac{5}{2}$, say.

By adding 1 to both sides, then subtracting 1 from both sides, and finally taking the ratio of the two, we get a quick simple result of,

$\displaystyle\frac{a}{b}=\frac{5+2}{5-2}=\frac{7}{3}$.

Many times we apply the first part of the technique, that is componendo technique, to simplify the numerator,

$\displaystyle\frac{a+b}{a-b}+1=\frac{5}{2}+1$,

Or, $\displaystyle\frac{2a}{a-b}=\frac{7}{2}$.

We have simplified the numerator by eliminating $b$.

Using this technique on our problem we get,

$\displaystyle\frac{3x+12}{4x+12}=\frac{13}{16}$,

Or, $\displaystyle\frac{x}{4x+12}=\frac{3}{16}$,

Inverting the LHS and simplifying,

$\displaystyle\frac{4x+12}{x}=\frac{16}{3}$,

Or, $4 + \displaystyle\frac{12}{x}=\frac{16}{3}$,

Or, $\displaystyle\frac{12}{x}=\frac{4}{3}$,

Or $x=9$.

These tasks of adding 1 to fractions, inverting the fraction and so on are simple processes and can be quickly done in mind without using pen and paper. If you are skilled in this type of mental manipulation, this approach should reach you to the solution pretty quick.

#### Problem solving using Multiple of ratio value technique and multiple enumeration technique, the Elegant Solution

The starting point is nevertheless the equation of fractions representing the status after addition of 12 to both the numbers,

$\displaystyle\frac{3x+12}{4x+12}=\frac{13}{16}$.

In this equation, $3x+12$ must be a multiple of 13 greater than 12. For multiple 26 of 13, we don't get a feasible value of $x$, but for 39, three times 13, we get $x=9$ in a few seconds. Testing it against the denominator takes a few more seconds. With $x=9$, the denominator $4x+12=48$ which is again three times 16 thus satisfying the ratio term relation of $\displaystyle\frac{13}{16}$.

In the first step we have found out a feasible value of $x=9$ that results in a multiple of 13, specifically with factor 3. At the second step, it is checked whether this value of $x$ results in a multiple of 16 that has the same factor of 3. This requirement is a stringent condition for keeping the ratio $\displaystyle\frac{13}{16}$ unchanged.

This should be * the fastest* and

*if you are skilled in number manipulation and factor multiple concept.*

**the elegant method**The * important point to remember* is,

It is up to you to choose the method that suits you, but remember

, and to save seconds and achieve lightning speed, not only would you need to adopt the right method but also you must be highly skilled in applying it.saving seconds counts towards your overall success

The many approaches to solve a problem as shown above encourage you to search for alternatives and evaluate them. It is a direct application of the powerful problem solving skill improving **Many ways technique.**

Let us highlight the power of the Multiple of ratio value technique by solving the second chosen problem.

### Chosen Problem 2.

The ratio of age of Avi and Ron 9 years ago was 2:5, and today it is 5:8. What will be the ratio after 9 years?

- $\displaystyle\frac{11}{18}$
- $\displaystyle\frac{9}{14}$
- $\displaystyle\frac{8}{11}$
- $\displaystyle\frac{2}{3}$

** Elegant Problem solving by Multiple of ratio value technique**

The present age ratio is,

$\displaystyle\frac{Age_{Avi}}{Age_{Ron}}=\frac{5}{8}=\frac{5x}{8x}$, where, $x$ is the HCF of the two ages.

Nine years ago then the age relation was,

$\displaystyle\frac{5x-9}{8x-9}=\frac{2}{5}$.

So $5x-9$ must be a multiple of 2 so that $5x \gt 9$. The first such multiple of 2 in the RHS numerator is 6 with factor of 3 that results in a feasible value of $x$.

For this multiple value 6 in the RHS numerator, $5x-9=6$ and $x=3$.

With $x=3$, denominator $8x-9=15$. As this value of 15 also has the factor 3 with 5, the ratio relation $\displaystyle\frac{2}{5}$ remains unchanged. So $x=3$ satisfies all conditions of the ratio relation of both sides of the equation.

With $x=3$, age ratio after nine years will then be,

$\displaystyle\frac{15+9}{24+9}=\frac{8}{11}$.

**Answer:** Option C: $\displaystyle\frac{8}{11}$.

### Basic concepts application approach

* Multiple of ratio value technique uses one of the two basic concepts inherent in a ratio and that's why it usually works faster than other methods for a specific type of number or age problems if properly applied. It works on the most basic conceptual level and is a part of the broader and more powerful Basic concepts application approach *which says,

The more basic a concept you use for solving a problem satisfactorily, the more efficient your solution will be.

*The method uses mathematical reasoning and is not conventional procedure based.*

**Resources that should be useful for you**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Concept Tutorials on related topics**

**Basic concepts on fractions and decimals part 1**

**Basic concepts on Ratio and Proportion**

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

**How to solve problems on mixing liquids and based on ages**

**Basic and Rich Percentage Concepts for solving difficult SSC CGL problems**

#### Efficient solution techniques on related topics

**How to solve SSC CGL level arithmetic mixture problems in a few simple steps 1**

**How to solve SSC CGL level arithmetic mixture problems in a few simple steps 2**

**How to solve SSC CGL level number and age ratio problems lightning fast**

**SSC CGL level solved question sets on mixture or alligation**

**SSC CGL level solved questions sets 78 on mixture or alligation 1**

**Other SSC CGL question and solution sets on Ratio & Proportion and Percentage**

**SSC CGL level Solution Set 76, Percentage 4**

**SSC CGL level Question Set 76, Percentage 4**

**SSC CGL level Solution Set 69, Percentage 3**

**SSC CGL level Question Set 69, Percentage 3**

**SSC CGL level Solution Set 68, Ratio Proportion 6**

**SSC CGL level Question Set 68, Ratio Proportion 6**

**SSC CGL level Solution Set 31, Ratio and Proportion 5**

**SSC CGL level Question Set 31, Ratio and Proportion 5**

**SSC CGL Level Solution Set 25, Percentage, Ratio and Proportion 4**

**SSC CGL level Question Set 25, Percentage, Ratio and Proportion 4**

**SSC CGL level Solution Set 24, Arithmetic Ratio & Proportion 3**

**SSC CGL level Question Set 24, Arithmetic Ratio & Proportion 3**

**SSC CGL level Solution Set 5, ****Arithmetic Ratio & Proportion 2**

**SSC CGL level Question Set 5, ****Arithmetic Ratio & Proportion 2**

**SSC CGL level Solution Set 4, Arithmetic Ratio & Proportion 1**

**SSC CGL level Question Set 4, ****Arithmetic Ratio & Proportion 1**

**If you like, **you may * subscribe* to get latest content from this place.