How to solve Ratio and Proportion Mixture Problems 2

Submitted by Atanu Chaudhuri on Tue, 05/04/2016 - 14:31

How to solve Ratio and Proportion Mixture Problems at lightning speed

Use conceptual reasoning to solve Ratio and proportion mixture problems in a flash

Learn how to solve ratio and proportion mixture problems conceptually at lightning speed using basic concepts on mixture and alligation.

Just like a few other types of arithmetic problems, mixture and alligation problems also are heavily dependent on ratio and proportion concepts.

By solving three liquid ratio and proportion mixture problems in a few simple steps using conceptual reasoning we will showcase the power of conceptual reasoning in solving ratio and proportion mixture problems quickly and elegantly in a few steps.

Introduction to Ratio and proportion mixture of liquids or solids

Nature of the operation of mixing

Ratio and proportion Mixture problems invariably involves ratios of amount of two or more liquids that can be mixed homogeneously. One of the basic properties of mixing is, once you mix two liquids you won't be able to separate out the two from the mixture. That's inherent in mixing homogeneously.

For example, a milkman might sell diluted milk, but when one of his important customers objects, the most he can do is to put in more milk in the diluted milk to make it more concentrated milk, but he won't be able to take out the water from the initial diluted milk that he had sold his customer.

Repeated mixing of one or the other liquid in a container increases complexity, because in a mixing problem you might know the ratio of amounts of two liquids mixed, but you may not know the actual amounts mixed.

Substitution of some amount of mixture with a component liquid

A frequent operation in mixing problems is to take out, say 5 liters, of diluted milk and then replace it with the same volume of water. By this operation, the starting volume remains unchanged but the ratio of amounts of milk and water in the mixture changes.

The core concept in dealing with such problems is, when you take out the 5 liters of diluted milk, you take out milk and water both in certain proportions. While replacing the 5 liters though you put in only water. The milk portion taken out in 5 liters of mixture is wholly lost, but the 5 liters of water that you put in won't increase the amount of water by the whole of 5 liters as, while taking out 5 liters of diluted milk you took out a portion of water also that was lost.

Repeated substitution creates complexity and more difficulties.

Ratio and proportion mixture and alligation problems on Mixing of two mixtures

Instead of carrying out the mixing operations on one mixture, two mixtures with two different proportions of same liquids in them may also be mixed, that too in different proportions for increasing the complexity level.

We recommend,

Be clear about the basic and rich concepts and use the concepts in solving all problems rather than use specific procedures for specific types of problems.

Basic concepts on ratios

You need to have clear understanding about ratios and what does a ratio really represent, because ratios form the base concept layer where you may encounter difficulties in many mixing and other types of problems.

Definition of a ratio expression

A ratio represents comparison of one common characteristic in same units of two or more entities and is expressed as a series of numbers or variables separated by the symbol of colon ':'.

This is an abstract but exhaustive definition of ratios. This abstraction coupled with exhaustiveness will help you to deal with different types of situations in a specific area, say of ratios.

A simple example of use of ratios follows as a problem example.

Problem example: If ratio of ages of son and father is 1 : 3 and 12 years back 1 : 7, what are their ages?

Basic Ratio concepts

Fractional form: A ratio is conventionally shown in minimized fraction form. In fact an important characteristic of ratios is,

A ratio can always be expressed as a fraction and conventionally is expressed in minimized fraction form (with all common factors cancelled out).

The HCF of the two actual quantities in ratio is canceled out between the two quantities in ratio: All the common factors cancelled out between the two terms of a ratio is the HCF of the actual values of the two original quantities involved in the ratio.
Re-introduction of HCF technique: For convenience, you may always re-introduce this canceled out HCF by multiplying each of the two ratio terms with the HCF. In our example, the present ratio of ages of 1 : 3 can very well be expressed as $x : 3x$, with $x$ as the HCF of the two ages of son and father. With this re-introduction of a single variable $x$ in the relationship, we may now represent the problem state 12 years back as, $(x - 12) : (3x - 12) = 1 : 7$, Or, $7x - 84 = 3x - 12$, or, $x = 18$, and get the present ages of son and the father respectively, as 18 years, and 54 years.
You won't introduce two variables of two unknowns in ratio sums, just use one variable to get the two ratio terms in the single variable expressions. This is the key technique in ratio sums.
Expressing amount of each component in terms of portion of total amount in the mixture: This re-introduction of the canceled out HCF as a single variable $x$ enables us to express the total age as, $x + 3x = 4x$, and more importantly, each ratio term as a portion of the total - son's age is a single portion of total age and father's age automatically the rest three portions of total age. This is a very important way of looking at changing ratios especially in mixing sums where total volume is fixed. If we express each liquid component as a portion of the fixed total volume, we don't have to deal with two varying quantities, our thinking load clearly diminishes significantly.

Basic mixture and alligation concepts

Expressing amount of liquids as portions of total mixed amount

Two mixable liquids (oil and water can't easily be mixed) milk and water are mixed thus forming a diluted milk with milk to water ratio, say 4 : 1.

Many times this statement may be expressed in more usable forms of,

A total 5 portions of diluted milk, water is one fourth of milk amount.
Milk in diluted mixture is four-fifths of total volume and water one-fifth of total volume of mixture.
In 5 portions of diluted milk, 4 portions of total mixture is milk and 1 portion of total mixture is water.

This way of looking at the same piece of concept in three different ways forms what we call a Rich concept. Depending on the problem situation, we may think amounts of mixed liquids in any of these forms that gives us the most elegant solution.

This is direct application of powerful problem solving 360 degree approach on a single piece of concept to convert it to a rich concept. In learning and problem solving this ability is invaluable.

Operation of replacement of certain amount of mixture with only one liquid

Most frequently a fixed amount, say 3 liters, of a mixture, say our diluted milk of milk to water ratio 4 : 1, is taken out of the mixture and replaced by the same amount of a single liquid (it could be any of the two liquids or may even be another mixture), say by water.

We can perceive that by this operation, the diluted milk get further diluted, but what exactly does happen in the replacement operation?

Result of first operation of taking out 3 liters of mixture

The 3 liters of diluted milk that we take out contained both milk and water in the ratio of 4 : 1. So by our third statement expressing amount of each liquid as a portion of total mixture volume, we may be clear that by taking out this 3 liters of diluted milk,

$3\times{\displaystyle\frac{4}{5}}=\displaystyle\frac{12}{5}$ liters of milk and $3\times{\displaystyle\frac{1}{5}}=\displaystyle\frac{3}{5}$ liters of water are taken out.

This is dealing with exact amounts of milk and water taken out in the first part of replacement operation.

Result of second operation of putting in 3 liters of water

The first conclusion is obvious: by adding 3 liters of water, total volume of mixture returns to its earlier value. It effectively remains unchanged. But what happens with portions of milk and water and the final result of the two part operation? Just note that volume of mixture is not known.

It is clear that, milk amount is reduced by exactly $\displaystyle\frac{12}{5}$ liters but water amount has not increased by 3 liters, in fact it is increased by, $3 - \displaystyle\frac{3}{5} = 3\left(1 - \displaystyle\frac{1}{5}\right)$ litres.

There are more to mixture and alligation concepts. We will elaborate on these concepts as required.

Let us now solve a few mixing liquid problems in as few steps as possible using the basic ratio and mixing liquids concepts and suitable problem solving strategies and techniques.

Problem 1.

In 40 liters of a mixture of milk and water, the ratio of milk to water is 7 : 1. To make the ratio of milk to water 3 : 1, the amount of water to be added in liters is,

$6$
$6\frac{2}{3}$
$6\frac{3}{4}$
$6\frac{1}{2}$

Solution using conceptual reasoning:

Milk in the beginning is seven eighth of 40 liters, that is, 35 liters and rest 5 liters was water.

Milk amount is not changed and by the desired ratio it has to be 3 times the water in the new mixture. This volume of water in the new mixture is then,

$\displaystyle\frac{1}{3}\times{35} = 11\displaystyle\frac{2}{3}$ liters.

As the original water amount was 5 liters, $6\displaystyle\frac{2}{3}$ liters of additional amount of water needs to be added to the original mixture to make the milk to water ratio in the new mixture $3 : 1$.

Answer: Option b: $6\displaystyle\frac{2}{3}$.

Key concepts used:

Finding the exact amount of milk and water to start with.
From the quantity relationship of milk and water of the desired ratio getting the final water amount.
Finding additional amount of water required as a difference of the final amount and initial amount.
Use of most basic concepts of ratio and mixing liquids.

Conventional solution using variable $x$

Let the water to be added be $x$ liters.

In the beginning the 40 liters of diluted milk contained,

$\displaystyle\frac{7}{8}\times{40} = 35$ liters of milk and rest $5$ liters of water.

As after addition of $x$ liters of water, ratio of milk to water becomes $3 : 1$, we have,

$\displaystyle\frac{35}{x + 5} = \displaystyle\frac{3}{1}$,

Or, $35 = 3x + 15$,

Or, $3x = 20$,

Or, water to be added, $x = \frac{20}{3} = 6\frac{2}{3}$ liters.

Remarks: This is a deductive process suitable for formal deductive and descriptive answering environment. As such nothing is wrong in this solution except that it mechanically follows procedures to the solution and thereby takes more time.

In case of the conceptual solution, the approach is to use the volume relationships of the milk and water in the ratio of the final mixture directly thus cutting short all deductions. Naturally this way you reach the solution much quicker and this approach is ideally suitable for MCQ based tests.

A note for school students:

In school tests students may have to produce the routine deductive solution by the existing rules of solving such problems. Use of mathematical or conceptual reasoning may not be allowed. But undoubtedly, as use of conceptual reasoning forms the heart of learning and problem solving, students would invariably increase their problem solving abilities if they think in the ways of efficient problem solving using basic and rich subject concepts and powerful general problem solving strategies and techniques.

Problem 2.

A metal alloy has in it Copper, Nickel and Zinc in the ratio of 5 : 2 : 3. The amount of Nickel in kg that must be added to 100 kg of alloy to have a new ratio of these three metals as 5 : 3 : 3 is,

Solution using conceptual reasoning:

Total number of portions in 100 kg of alloy in the beginning was 10, and so each portion value was 10 kg. In the changed alloy only 1 portion of Nickel is added to the total, rest 10 portions remaining unchanged. So 10 kg Nickel was added.

If this conceptual reasoning doesn't seem to be good enough, we can otherwise observe that Copper and Zinc together form 80 kg combined amount in 100 kg of initial total amount of the alloy. After addition of only Nickel, this combined 80 kg of Copper and Zinc now forms $\frac{8}{11}$ of the new total volume,

Or, $80$ kg = $\frac{8}{11}$ of the new total volume.

So, total new volume of the alloy must then be, $=\frac{11}{8}\times{80} = 110$ kg.

As Nickel was 20 kg to start with, 10 kg of Nickel must have been added.

Answer: Option c: 10.

Conventional deductive solution using variable $x$

Initial amount of Nickel in 100 kg of alloy is,

$\displaystyle\frac{2}{10}\times{100} = 20$ kg

Let $x$ be the additional amount of Nickel mixed in kgs. So the proportion of Nickel in the total volume of the final alloy is,

$\displaystyle\frac{20 + x}{100 + x} = \displaystyle\frac{3}{11}$

Or, $220 + 11x = 300 + 3x$,

Or, $8x = 80$,

Or, $x = 10$ kg.

Not a very long process, but still the solution based on conceptual reasoning is not only faster, it also strengthens your concepts on the topic because of use of the concepts. In the deductive process you deduce mechanically, thinking takes a backseat.

Problem 3.

A can contains two liquids A and B mixed in ratio 4 : 1. When 10 liters of the mixture is replaced by equal volume of liquid B, the ratio of A and B changes to 2 : 3. The volume of liquid A in liters intially was,

**Solution using conceptual reasoning:**

Initially and finally the total number portions are equal and of value 5. The intial and final volumes also being same, the value of 1 portion in both the cases are same.

By the replacement, $\displaystyle\frac{4}{5}\times{10}= 8$ liters of liquid A was lost in the replacement. This is equivalent to loss of 2 portions in the ratio. So each portion value is 4 liters. Thus original amount of liquid A was 4 portions = 16 liters.

Answer: b : 16.

Key concepts used:

Drawing the conclusion by conceptual reasoning that initial and final value of each portion is same. This could happen because intial and final volumes as well as the totals of the ratio terms remained at same values. This is a very promising information that will be used for reaching solution quickly. This is the case of using Portion use technique.
From basic mixing liquids concepts, finding the exact amount of liquid A lost in replacement of 10 liters of mixture by 10 liters of liquid B.
As the value of one portion in both the cases is same, this loss in amount of liquid A can directly be equated to loss in number of portions of liquid A. By this we get exact value of one portion as 4 liters.
As liquid A was initially of 4 portions, its volume was $4\times{4} = 16$ liters.

Conventional deductive solution using variable $x$

Let initial volumes of liquid A and liquid B be $4x$ and $x$ respectively.

In 10 liters of this mixture taken out, volume of liquid A is then,

$\displaystyle\frac{4}{5}\times{10}=8$ liters and volume of liquid B as $2$ liters.

After addition of 10 liters of liquid B, the volume of liquid A changed to,

$4x - 8$ liters, and the volume of liquid B became,

$x - 2 + 10 = x + 8$ liters.

So, the new ratio of liquid A to liquid B being 2 : 3, we get,

$\displaystyle\frac{4x - 8}{x + 8} = \displaystyle\frac{2}{3}$,

Or, $12x - 24 = 2x + 16$,

Or, $10x = 40$,

Or, $x = 4$,

So original volume of liquid A was $4x=16$ liters.

Observations on conceptual reasoning based solutions

The conceptual reasoning is a type of deductive reasoning that is primarily based on basic and rich concepts on the topic of the subject.
The reasoning processes are carried out mostly in mind and reaching the solution in most cases should not take more than a minute.
This approach of solving problems is ideally suitable for MCQ based competitive tests.
But school students should also gain a lot of insight into the problem solving aspects of maths.
Continued use of this type of approach should considerably improve the agility and creativity of young minds.

Observations on conventional deductive solutions

This is the standard solution taught and known by most, but this solution,

is deductive and suitable for environments where descriptive procedure oriented solutions are preferred,
is error-prone because of the mechanical nature of the procedure (which is true for all routine deductions, and errors do happen),
takes more time because of more number of steps and use of pen and paper is necessary,
uses single aspect of the concept of mixing liquids topic,
is unsuitable and anti-productive for MCQ based test environment.