Ratio Proportion Concepts and Methods

Submitted by Atanu Chaudhuri on Mon, 30/05/2016 - 20:49

ratio and proportion

Ratio proportion concepts play a very important role in solving questions on mixture and alligation, percentage, ages, numbers systems and so on.

Here, we lay a concept foundation for the important math topic of ratio proportion and its application areas. Solution of carefully selected example problems strengthen the learning.

Comparison between two quantities gives rise to ratios

Ratio, proportion and unitary method all three are the results of comparison between two quantities or change of quantities. Together these form a very important part of mathematics.

Prerequisites:

To understand the concepts to be explained here or to solve the problems that arise in this topic area, the student should have gone through the tutorials,

Numbers, number systems and arithmetic operations

Factorization or finding out factors

HCF and LCM

Basic concepts on fractions and decimals part 1

Definition of Ratio

A ratio defines how many times is one quantity with respect to another of same type. It is expressed as a minimized fraction and its numerator is called antecedent and denominator consequent. In practice these terms are not very important. Simply speaking, a ratio is expressed as a proper minimized fraction such as,

\[
\text{Ratio of the age of father and son is} = \frac{50}{20} = \frac{5}{2}
\]

Properties of a ratio

A ratio expresses how many times is one quantity with respect to another of same type.
A ratio is expressed as a minimized proper fraction with no common factor between the numerator and the denominator.
A ratio is always between two quantities of same unit such as ratio of two lengths, ratio of two ages, ratio of two volumes, ratio of two weights and so on.

But a ratio can also be defined between two counts like ratio between the number of boys and girls in a class, ratio of population in two countries and so on. Notice that even the counts have to be of same types. You cannot define a ratio of boys in a class to the number of cows owned by Ramnaresh.

In other word, the counts must be same set members. For example, boys and girls belong to the same class, countries belong to the same planet and are at similar level.

Being a fraction, a ratio has no unit.

While calculating the ratio between two quantities, the HCF between the numerator and the denominator is canceled out.
The fraction of a ratio is rarely evaluated. Otherwise it loses its meaning. Thus a ratio generally remains a fraction. In special cases when we encounter a ratio in the form of an evaluated decimal, we need to transform it back to a proper fraction for dealing with it as a ratio.
A ratio can be between three quantities A: B : C. For example, ratio of three numbers is 3: 4 : 7 and their product is 672. Find the numbers.

We can assume the numbers to be $3x$, $4x$ and $7x$. Thus product is,

\[
3\times{4}\times{7}{x^3}=84x^3=672, \quad\text{so, } x^3=8, \text{ and } x=2
\]

Thus the numbers are 6, 8 and 14 with the ratio, 3 : 4 : 7.

Basic technique in solving ratio problems

A ratio between the age of the father and a son is,

\[
\frac{\text{Age of father}}{\text{Age of son}} = \frac{5}{2}
\]

What are the actual ages then?

At this point of time we are in no position to find out two unknown values from one known relationship. But at least we know that, for arriving at the ratio $\frac{5}{2}$, the HCF between the two ages has been canceled out. Putting back this canceled out HCF as $x$, we then have a very useful relationship,

\[
\frac{\text{Age of father}}{\text{Age of son}} = \frac{5}{2}=\frac{5x}{2x}
\]

From this we can confidently say that the actual age of the father and son is $5x$ and $2x$ respectively. This enables us to get the various other relationships between these two quantities, for example,

\[
\text{Age of father} + \text{Age of son} = \text{Total age} = (5x + 2x) = 7x,
\]

where $x$ is the HCF between the father's age and the son's age.

Selected problem examples in ratio

Problem 1:

The rabbit population in community A increases at 25% per year while that in B increases at 50% per year. If the present populations of A and B are equal, the ratio of number of rabbits in B to that in A after 2 years will be

144
1.72
1.90
1.25

Solution:

Let us assume, P as the common starting rabbit population.

Then Rabbit population of A in 2 yrs,

$= P\times{1.25{\%}}\times{1.25{\%}}$.

Similarly, Rabbit population of B in 2 yrs

$= P\times{1.5{\%}}\times{1.5{\%}}$.

Thus ratio of rabbit populations in B and A is,

$= \displaystyle\frac{1.5^2}{1.25^2} = \left(\frac{150}{125}\right)^2 = 1.2^2 = 1.44$.

Answer: choice a: 1.44.

Problem 2:

The distance between two oil rigs is 6 km. What will be the distance between these two rigs in maps of 1 : 50000 and 1 : 5000 scales respectively?

12 cm and 1.2 cm
2 cm and 12 cm
120 cm and 12 cm
12 cm and 120 cm

Solution:

The ratios show map distance is to actual distance. Thus for first map unknown map distance is

\[
\frac{1}{50000}=\frac{x}{6\text{km}}=\frac{x}{6\times(10)^5\text{cm}}
\]

So,

\[
x = \frac{60}{5}\text{ cm}= 12\text{ cm}
\]

Similarly for the second map with ratio 1 : 5000, the map distance would be,

\[
y=\frac{60}{0.5}=120\text{ cm}
\]

Answer: choice d: 12cm and 120cm.

Problem 3:

Two numbers are in the ratio 17 : 20. If the difference between the numbers is 90, find the two numbers.

Solution:

let the numbers be $17x$ and $20x$. So their difference $3x=90$, and $x=30$. Thus the numbers are $510$ and $600$.

Problem 4:

In a hostel, a certain quantity of rice is spent daily for 30 students. Some students were absent one day and that day the quantity of rice spent decreased by a ratio of 1 : 6 in comparison to a normal day. How many students were absent on that day?

Solution:

As the rice consumption decrease was 1 : 6, the number of absent students would be 1:6 of 30, that is, 5.

Percentage concepts and its use

Percentage represented by the symbol ${\%}$ is a special kind of ratio in which the denominator is converted to 100.

For example, if the ratio of son's age to father's age is 2 : 5, we have,

\[
\frac{\text{Son's age}}{\text{Father's age}}=\frac{2}{5}=\frac{2\times{20}}{5\times{20}}=\frac{40}{100}=40\text{%}
\]

We can then say, son's age is 40% of father's age. This is again equivalent to saying -- son's age is 0.4 times father's age. Thus percent symbol can be replaced by the number divided by 100.

If we reverse the ratio, we get,

\[
\frac{\text{Father's age}}{\text{Son's age}}=\frac{5}{2}=\frac{5\times{50}}{2\times{50}}=\frac{250}{100}=250\text{%}
\]

Now we would have to say -- father's age is 250% of son's age, or 2.5 times son's age. The percent value more than 100 means the antecedent value is larger than the consequent.

Use of percentage in daily life

In academic life we find frequent use of percentage concept when we say, "My niece got 85% in Maths in the finals." In this statement nowhere are the information about the total or actual marks available. If the total marks were 200 then the student's score would have been,

$200\times{85\text{%}}=170$.

How did we arrive at the actual marks?

To get the actual marks the Total marks is just multiplied with the percentage score.

Here two important aspects of percentage use comes out,

A percentage value is always expressed on a reference anchor value. In this case the reference anchor value is the Total score.
When a percentage value is multiplied with the reference anchor value, it is first converted to its equivalent decimal value and then multiplied.

The equivalent decimal value of $85\text{%}$ is,

$\displaystyle\frac{85}{100}=0.85$.

It is so because by definition of percentage, when the actual score is $85\text{%}$ of the total marks, $100\text{%}$ is the total marks. In other words, the actual marks is $0.85$ portion of the total marks. That is why to get the actual value expressed as percentage of a total value, the percentage is converted to its decimal equivalent form by dividing it by 100 first.

On many occasions in real life, the percentage is used to express a value without mentioning the reference anchor value. This is both an opportunity to hide information, a disadvantage of the concept, as well as a powerful means to express a value in its simplest form when the reference value is well known or universally accepted.

A very common example of the second intent of expressing value as a percentage is, "Profit this year has gone up by 20%."

Here it is generally known that the this year's profit percentage is expressed with reference to the Profit last year, whatever be that profit. This is a very apt use of percentage as a comparison tool. The expression of 20% increase in profit simply says, this year profit is more than last year's profit by one-fifth of last year's profit.

This brings us to the second way to use percentage when in multiplication. While calculating actual value as a given percentage of a reference value, instead of converting the percentage to its equivalent decimal by dividing it by 100, often we convert it to a fraction by dividing it by 100 alright, but transforming the result to a fraction. We do this because usually fraction arithmetic is easier than decimal arithmetic.

If the last year's profit were Rs.20000, this year's profit will be,

$20000\times{\displaystyle\frac{120}{100}}=20000\times{\displaystyle\frac{6}{5}}=\text{Rs.}24000$. 120 in the numerator indicates increase of 20% from 100%.

We can call these concepts as rich percentage concepts.

Though the percentage concept is simple it is one of the most universally used concepts specially in business environment.

Selected problem examples in percentage

Problem 1:

A student has to score 40% to get through. He scores 40 but he fails to meet the target by 40 marks. What was the total marks in the exam?

Solution:

He got 40 which is less than the target by 40. So, Target marks = 40 + 40 = 80. But as the target marks percentage is 40%, we get, $40{\%}=80$, or, $100{\%}=80\div{40}\times{100}=200$.

Problem 2:

In an exam, 10% students failed Physics, 20% failed in Maths and 10% failed in both the subjects. How much percentage of students passed in both the subjects?

Solution:

Together $10{\%} + 20{\%}=30{\%}$ students failed in either or both in Physics and mathematics. In this sum, we have counted the students who failed in both subjects, that is 10%, twice (set union - Venn diagram). Thus the percentage of students who passed in both subjects is, $100{\%}-(30{\%} - 10{\%})=80{\%}$.

Problem 3:

Salary of an employee is first raised by 10% and then decreased by 10%. What was the effective increase or decrease percentage?

Solution:

It is a decrease because the second action of reduction of same 10% amount was on the amount 110%, that is decrease is more than the increase.

Final decrease is $10{\%}\times{10{\%}}=0.1\times{0.1}=0.01=1{\%}$.

Reason:

$\text{Final value}=110\text{%}- 110\text{%}\times{10\text{%}}$

$=110\text{%} - (100 + 10)\text{%}\times{10\text{%}}$

$=100\text{%}-10\text{%}\times{10\text{%}}$

Thus, decrease is $10{\%}\times{10{\%}}$.

Problem 4:

If x% of A = y% of B, then

z% of A = $\left(\displaystyle\frac{yz}{x}\right)$% of B.

This is so because "of" indicates multiplication and "%" indicates $\displaystyle\frac{1}{100}$.

Unitary method

In ratio, for the first time we compared two quantities of same kind. Unitary method goes one step ahead. In straightforward and simple way it helps to find the changed value of one quantity that is related to the change of a second quantity of different kind. This method works when the change in both the quantities are uniform and so proportionate.

Example:

If 10 pencils cost Rs. 45 what will be the cost of 8 pencils?

Solution:

Step 1: 10 pencils cost Rs. 45.

Step 2: So 1 pencil costs Rs. $\displaystyle\frac{45}{10}$=Rs.4.5

Step 3: and 8 pencils will cost Rs.$4.5\times{8}$=Rs.36.

This is the Unitary method. As you have noticed, the basis of the method lies in the relationship between a pencil and its price. It is fixed and proportionate. So if you increase or decrease the number of pencils the total cost increases or decreaes proportionately.

In all cases where such relationship exists, you can apply unitary method. Let us know a bit more about this proportionality.

Proportion

When one ratio of one type of quantities and a second ratio of a different type of quantities are equal to each other we say that the quantities are proportionate to each other. We call the relationship as a proportion. For example, if ratios $a\colon{b}=c\colon{d}$ we have a proportion.

A proportion is mathematically expressed as, $a\colon{b}\colon\colon{c}\colon{d}$.

The double colon signifies a proportion, but is equivalent to equality of two ratios.

When a and c represents the initial and final values of quantity 1 and b and d of quantity 2, we say quantity 1 is directly proportional to quantity 2. We can write this as,

\[
\frac{q_{1i}}{q_{1f}}=\frac{q_{2i}}{q_{2f}}
\]

where ${q_{1i}}$ and ${q_{1f}}$ represents initial and final values of quantity type 1 and ${q_{2i}}$ and ${q_{2f}}$ represents initial and final values of quantity type 2. in this relationship as the value of quantity 1 increases the value of quantity 2 also increases and same with the decrease.

If any three values of this relationship is given, the fourth value can be calculated,

\[
q_{2f}=q_{2i}\times{\frac{q_{1f}}{q_{1i}}}
\]

For example in the problem of pencils above, initial and final number of pencils are 10 and 8, whereas intial price was Rs.45. So the Final price is given by,

Final price of 8 pencils in Rs.: $q_{2f}=q_{2i}\times{\displaystyle\frac{q_{1f}}{q_{1i}}}=45\times\displaystyle\frac{8}{10}=36$.

Thus we can see Unitary method embodies the concept of proportion and provides us a simple and intuitive method for finding the unknown quantity involved in the proportional change of two quantities. Here we have instead used an algebraic method that is equivalent to Unitary method.

Proportions can be of two types: direct proportion and indirect or inverse proportion.

Direct proportion

When one quantity changes with the change of another quantity uniformly and in the same direction, then we say that both quantities are proportionate to each other. It means that when quantity 1 increases, quantity 2 also increases; when quantity 1 decreases, quantity 2 also decreases and all the time the rate of change remains constant. Mathematically we express this as,

\[
Quantity 1 \varpropto Quantity 2
\]

Examples:

Problem 1:

If 8 dozen bananas can be bought at Rs. 180 then how many bananas can be bought in Rs. 30?

8 bananas
24 bananas
16 bananas
25 bananas
e. None of the above

Solution 1:

Rule: There are two quantities; in the three steps of unitary method, the unknown quantity that is to be found out is to mentioned second at every step.

At Rs. 180 bananas bought were 8 dozens

So, at Rs.1 bananas bought were $\displaystyle\frac{8}{180}$ dozens

So, at Rs. 30 bananas bought were $\displaystyle\frac{8}{180}\times{30}$ dozens=16

Problem 2:

A tree is 6 metres tall and casts a shadow of 4 metres long. At the same time, a flag pole casts a shadow 50 metres long. What is the height of the flag pole?

75 m
100 m
80 m
65 m
None of the above

Solution 2:

In this case as the day time is same, in parallel light,

\[
\text{length of shadow} \varpropto \text{height of the pole}
\]

Now, 4 metres shadow is cast by 6 metre high tree

so, 1 metre shadow is cast by $\displaystyle\frac{6}{4}$ metre high tree or pole

so, 50 metre shadow is cast by $\displaystyle\frac{6}{4}\times{50}=75$ metre high pole.

Indirect or inverse proportion

In this second type of proportionality of two quantities, change in one quantity takes place in the reverse direction with respect to the change in the second quantity. In other words, when one quantity increases the other decreases and vice versa. It is expressed mathematically as,

\[
Quantity 1 \varpropto \frac{1}{Quantity2}
\]

We say $Quantity 1$ is inversely proportional to $Quantity 2$.

Problem 1: 3 men can do a work in 24 days. How many days will be taken by 6 men to complete the same work?

This is a work and time problem which involves an inverse relationship.

Solution 1: Intuitively we can see that putting more people on the job reduces the time of completion. Following unitary method for inverse proportion, now we have,

Step 1: 3 men do the work in 24 days

Step 2: 1 man will do the work in $3\times{24}=72$ days, this we call as mandays

Step 3: 6 men will then do the work in $\displaystyle\frac{72}{6}=12$ days.

We can also arrive at the same solution by applying intuitive logic -- as number of men are doubled, the number of days must be halved.

Problem 2: A car covers a distance in 20 minutes traveling at a speed of 30km/hr. How long will it take to cover the same distance if it travels at a speed of 20km/hr?

Speed (S) and time (T) are inversely proportional when distance is constant

Solution 2:

Applying unitary method we get,

Step 1: At speed 30km/hr car takes 20 mins

Step 2: At speed 1km/hr car will take $30\times{20}=600$ mins

Step 3: At speed 20 km/hr car will take $\displaystyle\frac{600}{20}=30$ mins.

Naturally when the car moves very slow at 1km/hr it would take a very long time of 600 mins to cover the distance.

Problem 3: 6 spiders can catch 6 flies in 6 hours. How many flies can 90 spiders catch in 90 hours?

Spider (S) is directly proportional to Flies (F) but inversely proportional to time (T). Trick is, first hold one of the three constant and vary the other two desirably. At the next stage, then vary the third one.

\[
S \varpropto \frac{F}{T}
\]

Solution 3: Fly is kept at the last position

Step 1: 6 spiders in 6 hours catch 6 flies

Step 2: 1 spider in 6 hours catch 1 fly (holding Time constant)

Step 3: 1 spider in 1 hour catch $\displaystyle\frac{1}{6}$ fly (holding Spider constant)

Step 4: 1 spider in 90 hours catch $\displaystyle\frac{1}{6}\times{90}=15$ flies $(F \varpropto T)$

Step 5: 90 spiders in 90 hours catch $90\times{15}=1350$ flies $(S \varpropto F)$.

Resources that should be useful for you

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but 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

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Basic and Rich Percentage Concepts for solving difficult SSC CGL problems

You are here

Ratio and proportion

Comparison between two quantities gives rise to ratios

Prerequisites:

Definition of Ratio

Properties of a ratio

Basic technique in solving ratio problems

Selected problem examples in ratio

Percentage concepts and its use

Use of percentage in daily life

Selected problem examples in percentage

Unitary method

Proportion

Direct proportion

Indirect or inverse proportion

Resources that should be useful for you

Concept Tutorials on related topics

Efficient solution techniques on related topics

SSC CGL level solved question sets on mixture or alligation

SSC CHSL level solved question sets on Mixture or Alligation

SSC CGL Tier II level solved question sets on mixture or alligation

SSC CGL Tier II level question and solution sets on Ratio and Proportion

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