Arithmetic problems on mixing liquids and based on ages

Application of Ratio concepts - mixing liquids and problems on ages

Arithmetic problems on mixing liquids and ages

Mixing liquids and problems based on ages form two important application areas of the ratio concept. We will treat mixing liquids or alligation first and then the problems based on ages.

Mixing liquids or Aliigation

When two materials are mixed together in a certain ratio homogeneously, a mixture is formed. Various operations on such a mixture give rise to the topic of Mixing liquids or alligation.

Usually liquids that can be mixed are used in mixtures. But there are different types of mixing. In this discipline one characteristic will remain valid whatever be the type of mixing,

Before and after mixing, the state of the components mixed need necessarily have to be homogeneous.

This homogeneity means, you can't separate out the mixed components from the mixture easily. But more importantly, the condition that must be valid in any mixture is,

The components mixed in a mixture in a particular ratio will be present in the same ratio throughout the volume of the mixture. If we take out any volume of the mixture, both the components will be taken out in their original ratio.

For example, from a milk-water mixture with milk to water ratio as 6 : 1, if we take out 7 litres of mixture, 6 litres in it will be milk and 1 litre water.

Varieties of mixing by mixed components

Examples are,

  • Water and milk, Water and fruit juice.
  • Alloys give rise to mixing metals through specific processes.
  • Grains of same cereal of different price can be mixed giving rise to a grain mixture of a different price.

Types of mixing operations - Replacing a certain quantity of one material repeatedly

In a container of $a$ litres filled with milk, as first operation, $b$ litres of milk is replaced by water. Amount of milk, water and their ratio becomes,

Milk $= a-b = a\left(1 - \displaystyle\frac{b}{a}\right)$

Water $ = b$

Ratio of milk to water $ = \displaystyle\frac{a-b}{b}$.

The first stage was easy with no confusion, as the initial liquid state was pure milk with no water. But after the first substitution of milk by water, the mixture is no longer pure, from now onward it always will contain some amount of milk and some amount of water, homogeneously mixed throughout its volume, in a particular ratio.

At the second stage again a volume of $b$ litres of the mixture is replaced by water. Now the amount of milk becomes,

Milk$ = a\left(1 - \displaystyle\frac{b}{a}\right) - b\left(1 - \displaystyle\frac{b}{a}\right)$

$\hspace{8mm}= (a - b)\left(1 - \displaystyle\frac{b}{a}\right)$

$\hspace{8mm}=a\left(1 - \displaystyle\frac{b}{a}\right)^2$

If this process of replacing mixture by $b$ litres of water is continued $n$ times, the amount of milk in the mixture becomes,

Milk$ = a\left(1 - \displaystyle\frac{b}{a}\right)^n $.

Problem 1:

In a 30 litre can filled with milk, 3 litres of milk is replaced thrice. Find the remaining amount of milk in the mixture.

Solution 1:

First replacement of 3 litres of milk ny 3 litres of water - volume remains unchanged

After first operation the amount of milk and water in the diluted mixture are,

$\hspace{6mm}$ milk $=30-3=27$ litres, and water 3 litres.

Thus, amount of milk in each litre of mixture, $=27\div30=\displaystyle\frac{9}{10}$ litres.

Second replacement of 3 litres  of mixture by 3 litres of water - volume remains unchanged

Milk reduction in second replacement of 3 litres is not 3 litres it is less than three litres.

To be exact, the milk amount reduction after the second replacement operation

$\hspace{6mm}=$ milk contained in 3 litres of diluted milk taken out


$\hspace{6mm}=2.7$ litres.

Thus remaining total amount of milk in the diluted mixture after the second operation,

$\hspace{6mm}= 27 - 2.7$

$\hspace{6mm}=24.3$ litres.

Third replacement of 3 litres of diluted milk by 3 litres of water - volume remains unchanged

Milk in 1 litre of diluted milk after second repplacement $= \displaystyle\frac{24.3}{30} = 0.81$ litres.

Thus milk in 3 litres of mixture $= 2.43$ litres.

So, after the third replacement remaining milk = $24.3 - 2.43=21.87$ litres.

By applying formula, remaining milk after third replacement,

Remaining Milk$ = 30\left(1 - \displaystyle\frac{3}{30} \right)^3 = 30(0.9)^3=30\times{0.729}=21.87$ litres.


Always understand the mechanism of arriving at the formula so that with variations in problem descriptions, you are able to adjust to the new situation and use the basic concepts to solve the problem. Formula comes last in peference, as in all tests judging problem solving ability, usually you won't get an opportunity to apply a formula routinely.

Other mixing operations

Problem 1:

Milk and water is mixed in a ratio of 2:1 in a mixture. How much milk is to be added to the mixture to make the ratio of milk to water 5 : 2?

Solution 1:

In any ratio, the most basic concept is, the common factors between the two quantities in ratio are canceled out to form a proper fraction. The ratios we encounter in problems are in this form.

Using this concept, for ease of working out ratio problems, the most frequently used technique is to reintroduce the canceled out $HCF$ as $x$ in the numerator and denominator of the proper fraction of the ratio.

So we assume. original volume of mixture was $3x$, with $2x$ milk and $x$ water, where $x$ is the common factor reintroduced.

$x$ amount of water remains unchanged in the operation, so to make the ratio of milk to water 5 : 2, the milk is to become, $\displaystyle\frac{5x}{2}$. The percentage increase is then,

$ \displaystyle\frac{\displaystyle\frac{5x}{2} - 2x}{2x}\times{100}=\frac{1}{4}\times{100}= 25{\%} $.

Problem 2:

In what proportion must wheat at Rs.9.40 per kg be mixed with wheat at Rs.8.2 per kg, so that the mixture is worth Rs.8.6 per kg?

Solution 2:

Let the amount of first lot of wheat to be mixed be $x$ kg and that of second lot $y$ kg.

So total cost of the wheat before mixture,

$\hspace{6mm} =(9.4)x + (8.2)y$

$\hspace{6mm}= (8.6)(x + y)$, after the mixing

$\hspace{6mm}=(8.6)x +(8.6)y$.

Thus we get,

 $(9.4 - 8.6)x = (8.6 - 8.2)y$

Or, $ (0.8)x = (0.4)y$.


$\displaystyle\frac{x}{y}=  \displaystyle\frac{0.4}{0.8} = \frac{1}{2}$.

Problems based on ages

Basic Principle

In this problem area, there will be two or more than two persons of certain ages. In years past and future their age ratios or relationships will be of interest. Most important point to remember is:

As years pass, age of all concerned change by the same amount of years.

Problem examples

Problem 1:

The ratio of ages of two brothers 4 years ago was 11 : 14. After 4 years the ratio will be 13 : 16. What are their present ages?

Solution 1:

Let the ages 4 years ago be $11x$ and $14x$. So 8 years later we have the ratio of their ages as,

 $\displaystyle\frac{11x + 8}{14x + 8} = \displaystyle\frac{13}{16}$

Or, $16(11x + 8) = 13(14x + 8)$

Or, $176x + 128= 182x + 104$

Or, $ 6x  =  24$

Or, $x  =  4$.

So 4 years ago ages of the two brother were 44 years 56 years and present ages are, 48 years and 60 years.

Problem 2:

Ratio involving three numbers: The ratio of ages of two brothers Bittu and Tuku is 3 : 2 and brothers Bittu and Kamal 6 : 7. What is the ratio of ages of Tuku and Kamal?

Solution 2:

Though this problem involves ages, it actually is a special problem involving ratio of three numbers $a : b : c$.

Given a ratio of values of two quantities $A$ and $B$ as $a : b$ and of $B$ and $C$ as $b: c$, it follows that the values of $A$, $B$ and $C$ can be related in a ratio of $a: b : c$.

Rule of joining two ratios:

The two ratios have four values. Firstly all the values must of same type, and secondly, two of the values must represent values of a single common entity.

In the problem, the common entity is the brother Bittu. So, the values, 3 as denominator of the first ratio (now numerator) and 6 as numerator of the second ratio must be made equal. Thus, we transform the first ratio as ratio of ages between Tuku and Bittu $2 : 3 = 4 : 6$ and can say now, the ratio of ages of Tuku and Kamal is $4 : 7$.

Problem 3:

A brother is twice as old as his younger brother. After 6 years ratio of their ages will be 3 : 4. What was the ratio of their ages two years ago?

Solution 3:

Let their ages now $x$ and $2x$ years. So after 6 years,

$\displaystyle\frac{x + 6}{2x + 6}=\frac{3}{4}$

Or, $ 4x + 24 = 6x + 18$,

Or, $ 2x = 6$,

Or, $ x = 3 $.

Thus the present age of the brothers are 3 and 6 years, and 2 years ago the ratio of their ages was 1 : 4.

You may refer to the following related resources of question and solution sets which will give you some idea of type of problems that appear in standard MCQ tests on these topics.

Efficient Math Problem Solving in a few simple steps

How to solve Arithmetic mixture problems in a few simple steps 1

How to solve Arithmetic mixture problems in a few simple steps 2

Question and Solution sets on SSC CGL competitive exam

SSC CGL Solution Set 5 Arithmetic Ratios

SSC CGL Question Set 5 Arithmetic Ratios

SSC CGL Solution Set 4 Arithmetic Ratios

SSC CGL Question Set 4 Arithmetic Ratios