## Multiple of ratio value technique based on basic factors multiples concept gives you lightning fast solution

Use of the rich ratio technique of multiple of ratio value helps to solve a broad spectrum of ratio problems and specifically many standard number or age ratio problems elegantly in a few steps much faster than conventional procedural methods. The chosen problem here is a tricky one and we will solve it using multiple of ratio value technique as well as conventionally to highlight the power of the rich ratio technique.

We will take up the problem solving straightaway skipping the explanation of ratio concepts. For concept details, you may refer to our earlier article,

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

### The Chosen Problem.

When the son was born, the ratio of ages of his father and mother was 11 : 10. When the son will be twice his present age, the ratio of ages of his father and mother will be 19 : 18. What is then the ratio of the ages of his father and mother at present?

- 16 : 15
- 17 : 16
- 15 : 14
- 14 : 13

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

Let us assume the ages of the father and the mother when their son was born to be, $F$ and $M$ so that by the first given statement,

$F:M=11:10$.

Expressing as fractions,

$\displaystyle\frac{F}{M}=\frac{11}{10}$.

Applying the HCF reintroduction technique, we get the actual values of $F$ and $M$ as, $F=11x$ and $M=10x$. Here we have assumed canceled out HCF as $x$, the single unknown variable and the actual age variable values as $F$ and $M$.

Assuming the present age of the son to be $a$, when he is twice his present age, that is, of age $2a$, the number of years elapsed from the time when he was born would be $2a$. The changed ratio of ages of his father and mother can then be expressed as,

$\displaystyle\frac{11x+2a}{10x+2a}=\frac{19}{18}$.

The difficulty in this problem arises from the presence of two unknown variables, $x$ and $a$. We don't have two linear algebraic relations required for finding these two variable values exactly.

**Think again. Do we really need the exact values of two variables?**

Let us proceed in the familiar way of cross-multiplying the two sides and resolve this basic doubt.

Cross-multiplying,

$198x+36a=190x+38a$,

Or, $8x=2a$,

Or, $4x=a$.

We need to evaluate the ratio of ages of father and mother at present, that is,

$\displaystyle\frac{F+a}{M+a}=\frac{11x+a}{10x+a}$.

At this point we are able to see why we don't need the exact values of $x$ and $a$. In this ratio situation, when $x$ with coefficient suitably modified, is replaced with $a$, the variable $a$ will be cancelled out between numerator and denominator and we will get the desired answer of the ratio.

Multiplying numerator and denominator by 4,

$\displaystyle\frac{44x+4a}{40x+4a}=\frac{11a+4a}{10a+4a}=\frac{15}{14}$.

Answer is option c: 15 : 14.

**Answer:** Option c: 15 : 14.

Now we will solve the problem using the quick method of multiple of ratio value.

#### Problem solving elegantly using multiple of ratio value based on basic factors multiples concept and enumeration

As before, using HCF reintroduction technique assuming cancelled out HCF as $x$ and present age of son as $a$, we get the ratio relation as,

$\displaystyle\frac{11x+2a}{10x+2a}=\frac{19}{18}$.

In multiple of ratio value technique based on basic factors multiples concept, we will try out multiples of one ratio value, say 19 in the numerator of RHS, and examine whether any feasible positive integer combination of numerator variables $x$ and $a$ in the LHS results in the multiple of 19 on trial.

We will review a simpler problem before going ahead further.

In a simpler problem, a ratio relation such as the expression below may need to be solved,

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

In this simpler case, we try out multiples of 13 and see if any feasible value of $x$ results in the same multiple value of 13 in the LHS numerator also. This is the first step, **the trial step**. In this simpler problem trial value 39 with factor of 3 with 13 results in a feasible value of $x=9$.

At the second step in which we verify whether this feasible value of $x=9$ obtained in the first step, indeed satisfies the relationship of the denominators in the LHS and RHS. This is the **verification step.**

**In the verification step**, conveniently we find the LHS to result in 48, with the same factor of 3 with existing denominator ratio value of 16. The ratio 13 : 16 thus remains unchanged with $x=9$. This is the crucial breakthrough point and is achieved with practice in only a few seconds wholly in mind.

Coming back to our problem, now we can appreciate the difficulty involved in trying values of two variables in the LHS numerator instead of one to match resulting LHS value with the trial multiple of RHS numerator ratio value.

Interestingly, existence of two variable values to try in combination restricts the possibilities, and the difficulty as apprehended is not really that much.

After failing with first trial value of 19, second trial value of $19\times{2}=38$ is given easily by $x=2$ and $a=8$ in the LHS numerator. This combination also results in $36=18\times{2}$ in the LHS denominator thus satisfying the ratio $19:18$.

With these two values, the desired ratio relation is evaluated to be,

$\displaystyle\frac{11x+a}{10x+a}=\frac{30}{28}=\frac{15}{14}$.

As the method is strongly conceptual, with a bit of confidence on numbers and factors, the problem can be solved wholly in mind. This is especially so because the numbers or factors involved are small.

We will show you a third method that is deductive but very quick and mental because of use of componendo dividendo method. Conceptual dependence is less in this method.

#### Third solution to problem 1: By componendo dividendo and its variation

Assume $x$ to be the present age of the son and that of the father and mother as $f$ and $m$. When he was born ratio of ages of his father and mother was,

$\displaystyle\frac{f-x}{m-x}=\frac{11}{10}$, and when the son will be twice his present age the ratio will be,

$\displaystyle\frac{f+x}{m+x}=\frac{19}{18}$.

At first stage, to simplify the equation handling we will subtract 1 from each of the equations and then take a ratio.

The first subtraction result will be,

$\displaystyle\frac{f-m}{m-x}=\frac{1}{10}$ and the second subtraction result will be,

$\displaystyle\frac{f-m}{m+x}=\frac{1}{18}$.

The ratio of the two will then be,

$\displaystyle\frac{m+x}{m-x}=\frac{18}{10}=\frac{9}{5}$

We will apply componendo dividendo on this to get,

$\displaystyle\frac{m}{x}=\frac{7}{2}$,

Or, $x=\displaystyle\frac{2m}{7}$.

Substituting in the first equation we get,

$\displaystyle\frac{f-\frac{2m}{7}}{\frac{5m}{7}}=\frac{7f-2m}{5m}=\frac{11}{10}$,

Or, $55m=70f-20m$,

Or, $75m=70f$,

Or, $f : m=15 : 14$.

This method is also not difficult to carry out in your mind at all.

**Answer: **Option c: 15 : 14.

**Key concepts used:** * Basic ratio concepts* --

**Age ratio problem -- Componendo dividendo method -- Multiple of ratio value technique -- Basic factors multiples concept -- Solving in mind.**Solving in three ways is the application of **many ways technique** practice of which enhances your ability to see new possibilities in a problem solving situation, mathematical or otherwise.

You may now compare the conventional method of solution, the second elegant method and the third componendo dividendo based method of solution and choose the one that suits you. Our recommendation will nevertheless be always to absorb the concepts and develop the skillsets to follow the problem solver's approach as in the second method.

**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**

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