The three main fundamental pillars of Arithmetic are **Percentages, Averages, Variations, Ratio and Proportion**. So the aspirant should create an in-depth understanding about the concept of Ratio, Proportion and Variation'. Application of Ratio and Variation are broadly expanded in all the other Quantitative Topics. Hence it is a mandatory required topic for your serious preparation for the exams. In this article we provide important formulas, shortcuts and tricks to solve aptitude questions on ratio and proportion with help of examples.

- Definition and basic concept of ratio.
- Calculations on ratio values.
- Application of algebraic constant.
- Ratio applications in age problems.
- Ratio applications in Mixtures.
- Ratio application in Geometry.
- Income- Expenditure- Savings problems.
- Different ratio expressions and its effective transformations.
- Properties of ratio.
- Combining ratios
- Proportion
- Continued Proportion
- Re-arrangements of proportionals
- Variation
- Joint Variation

## Concept of ratio.

For example; if the salaries of Ajay and Bhavan are Rs.5000 and Rs. 6000 respectively. Then we can say that their respective salaries are in the ratio of 5:6. Here 5 and 6 are most relatively simplified representation of the original values 5000 and 6000, or 5 and 6 are the relatively prime numbers (doesn't have any common factors other than 1) which are the factors of the given values. Means, we can't simplify these values further as integers.

In a ratio expression, the order of values is very important. In the above example the required ratio is 5:6 instead 6:5 is wrong.

The ratio of any two same kinds quantities x and y can be expressed as either **x/y **or **x : y**. Here **x** is called **'antecedent'** and **y** is called **'consequent'**.

In a reverse approach, if it is given that the ratio of the salaries of Ajay and Bhavan is 5:6, then it doesn't mean that the salaries of Ajay and Bhavan are 5 rupee and 6 rupee respectively. Instead their respective salaries are a certain multiple of 5 and 6. So, from the given data, we can express the salaries of Ajay and Bhavan are in the following way;

Salary of Ajay = 5k

Salary of Bhavan = 6k, where k is a positive real number.

And 'k' is called the '**Multiplicative Constant'**.

**Calculation as per ratio expression:**

For finding the original quantities form a given ratio expression required at least one constant value related to the given data, which may be in any of the following manner;

- Sum of their individual salaries is Rs.11,000.
- Bhavan's salary is Rs.1,000 more than that of Ajay.

How to utilize these data for finding the individual salaries as per the pre mentioned example?

As per given ratio expression, Ajay's and Bhavan's salaries are in the ratio of 5:6.

And Bhavan's salary = 6k

**From data (i);**

11k = 11,000

K = 1,000

From the assumption,

Ajay's salary = 5k

= 5 * 1000

= Rs.5,000.

Bhavan's salary = 6k

= 6*1000

= Rs.6000.

From data (ii);

Bhavan's salary - Ajay's salary = 6k - 5k

= k =1,000

Therefore, Ajay's salary = 5k

= 5 *1000

= Rs.5,000.

Bhavan's salary = 6k

= 6*1000

= Rs.6000.

**Example: (practical understanding of Ratio)**

In Cahndu's birthday celebration, he cut a beautiful and delicious round chocolate cake and distributed it to his family members, consisting his father, mother and sister. The ratio of the quantity of cake received by Chandu, father, mother and sister is 3:1:2:4.

Means; if he divided the cake into ten equal pieces, Chandu got 3 out of the ten equal pieces.

His father got 1 out of the ten equal pieces.

His mother got 2 out of the ten equal pieces, and his sister got 3 out of the ten equal pieces.

We can express the same concept in another way.

Sum of the ratio values = 3 + 1 + 2 + 4 = 10

Consider the given full cake into ten equal pieces.

Then the share for Chandu = 3/10 of the cake.

Share for Chandu 's father = 1/10 of the cake.

Share for Chandu 's father = 2/10 of the cake.

Share for Chandu 's sister = 4/10 of the cake.

**Example:**

A total of 120 toffees are distributed among three friends A, B and C in a respective ratio of 3:4:5. Find the share of each member.

**Solution:**

Sum of ratio values = 3 + 4 + 5 = 12

Total 120 is divided in to 12 equal parts. Each part consists 10 toffees.

Share for A = 3/12 of 120 = 30 toffees.

Share for B = 4/12 of 120 = 40 toffees.

Share for C = 5/12 of 120 = 50 toffees.

## Effective mental calculation method for ratio related problems.

There are lots of various methods for finding the answer of a ratio related question. For better understanding, we will explain various concepts with using example.

**Example 1:**

If the number of students in three different classes A, B and C is in the respective ratio of 5:6:8. It is given that the strength in class C is 15 students lesser than the combined strength of the classes A and B together. Find the strength of class A.

For this question we can consider three different methods. Finally you can select one among them for finding the answer in the quickest way.

### Method 1: Algebraic method (Application of algebraic constant)

Let the strengths of classes A, B and C are 5k, 6k and 8k respectively, where 'k' is the algebraic constant.

Combined strength of A and B is 5k + 6k = 11k

Strength of C is 8k, this is 3k lesser than the combined strength of A and B, i.e. 11k - 8k = 3k

3k = 15

Therefore, k = 5

Strength of class A = 5k = 25 students.

### Method 2: Fractional approach

i.e. A + B + C = T

A = 5/19 of T

B = 6/19 of T

C = 8/19 of T

A + B - C = (5/19 + 6/19 - 8/19) of T

= 3/19 of T

Given 3/19 of T = 15

Therefore T = (15 * 9) / 3 = 95

Strength of A = 5/19 of 95 = 25 students.

### Method 3: Unitary method

This is most advisable type of approach towards almost all the different varieties of ratio problems.

**How this method is differing from the typical algebraic approach?**

Well, here we are not using any algebraic characters for the solution. Hence we can operate it by mind. No need of a little bit of rough work. However you have to improve in mental calculation and visualizing techniques.

Hence the answer is 25 students.

**Example 2:**

If p:q = 3 : 4, find;

- 2p + 3q
- (5p + 3q)/(5p - 3q)
- (3p
^{2}+ 4q) / (3p^{2}- 4q)

**Solution:**

Given p : q = 3 : 4

Let p = 3x and q = 4x

- 2p + 3q = 2(3x) + 3(4x) = 18x

For each different value of 'x', we will get different values for the given expression. Hence a unique value for the given expression 2p + 3q can't determine. - (5p + 3q) / (5p -3q) = (5(3x) + 3(4x)) / (5(3x) - 3(4x)) = 27x / 3x = 9

Hence this expression has a unique value for any different value of x. - (3p
^{2}+ 4q) / (3p^{2}- 4q) = (3(3x)^{2}+ 4(4x)) / (3(3x)^{2}- 4(4x))

= 9x^{2}+ 16x / 9x^{2}- 16x = 9x + 16 / 9x - 16

For this expression also we will get different values for each different value of x. So it’s not possible to find a unique value for this expression.

**Example 3: Ratio in Age Problems.**

The ratio of the present ages of Ajay and Varun is 4:3. Five years ago it was 7:5. Find the present age of Varun.

- 40 years
- 35 years
- 30 years
- 25 years

**Solution:**

And varun's present age = 3x

Five years ago;

Ajay's age = 4x - 5

Varun's age = 3x - 5

Given (4x - 5)/(3x - 5) = 7/5

X = 10

From assumption;

Varun's present age = 3x

= 3 * 10 = 30 years

Ans: C.

**Alternate method:**

Present age of Varun is represented by the ratio value of 3. Therefore normally Varun's age should be a multiple of three. Please note, in almost all the questions related to age problems, very rarely only ages expressed in the form of fractions, rest all in the form of integers. This favourable situation will help the student to reach a conclusion easily through the basic testing of divisibility. Here also our options consisting only one value as a multiple of three.i.e. 30. Hence this could be the possible answer.

If more than one of the options satisfying the basic testing condition (here it is the divisibility of 3), then student needs to make verification from the selected set of options and finally reach the conclusion.

**Example 4: Ratio in Mixtures:**

The ratio of the quantities of milk and water in a solution of 60 liters (consisting milk and water only) is 3:7. How much liters of additional milk should pour in to the mixture to get a resultant mixture which consisting the milk and water in the respective ratio of 2:3?

- 12
- 14
- 10
- 5

Solution:

= 18 liters.

Present quantity of water in the solution = 60 - 18 = 42 liters

Let 'x' liters be the additional required quantity of milk.

Then; (18 + x) /42 = 2/3

Therefore x = 10 liters.

Ans: C

Example 5: **Ratio in geometry**

If the length and breadth of a rectangle are in the ratio of 4:3 and it has an area of 108 sq:c.m. find the length of diagonal in centimeters.

- 12
- 10
- 15
- Can't be determined.

**Solution:**

Let length and breadth of the rectangles are 4x and 3x respectively.

Area = length * breadth

= 12 x

^{2}

12x

^{2}= 108

X = ± 3

X = - 3 doesn't exist, because the dimensions of a rectangle is always a magnitude. Hence it should be non-negative.

So x = 3.

Therefore, length = 4x = 12 c.m.

Breadth = 3x = 9 c.m.

Diagonal of a rectangle = √(length

^{2}+ breadth

^{2})

= √(12

^{2}+ 9

^{2})

= 15 c.m.

Ans: C

**Example 6: income - Expenditure - Savings problems.**

Ratio of the respective incomes of A and B is 2:3 and the ratio of their respective expenditures is 5:8. Find the income of A, if each of them saves an equal amount of Rs.2000.

- Rs. 8,000.
- Rs. 10,000.
- Rs. 12,000.
- Rs. 8,500.

**Solution:**

Let;

Income of A = 2i, Expenditure of A = 5e, Savings of A = 2i - 5e

Income of B = 3i, Expenditure of B = 8e, Savings of B = 3i - 8e

Note: Income and expenditure are basically two different kinds of quantities, so while expressing them algebraically use different variables for income and expenditure. |

Given that each of them saves an equal amount.

Therefore 2i - 5e= 3i - 8e

i = 3e

Substitute this relation in the previous assumption.

Income of A = 2i= 2(3e) = 6e, Expenditure of A = 5e, Savings of A = 6e - 5e = e

Income of B = 3i= 3(3e) = 9e, Expenditure of B = 8e, Savings of B = 9e - 8e = e

Given that their individual savings is Rs.2,000.

I.e. e = Rs. 2,000

Income of A = 6e

= 6 * 2000 = Rs.12,000.

Ans: C

**Alternate Method:**

For this logical and numerical arrangement method require your numerical skills in the form of catching the mutual relations between values. Here it's given that each of them saves an equal amount. So we have to think that, which kind of a change in the ratio values will help us to make the difference of corresponding incomes and expenditures are equal.

In this initial step of arrangement we are not bothering about the original constant values which are available in the data. Finally with the help of available constant values and through a unitary method we can easily reach the solution. By a thorough practice will help the student make able to do these type of problems by mind, without doing rough works.

A | B | |
---|---|---|

Income | 2 | 3 |

Multiply the above ratio values by 3 | 6 | 9, These are the new ratio values for their respective incomes. |

Expenditure | 5 | 8 |

Savings = Income - Expenditure | 1 | 1 |

1 → Rs.2,000.

Therefore; 6 → 6 * 2000 = Rs. 12,000.

**Method of algebraic constant:**

It is a very effective method for the above type of problems.

Two quantities are in the ratio of a:b, when a constant 'x' is add/subtract from both the quatities, the resultants are in the ratio of c:d. Let us take the quantities are 'ak' and 'bk', where 'k' is the multiplicative constant. And, k = (c ∼ d) x / (ad ∼ bc) |

Ratio of incomes = 2 : 3

Ratio of Expenditures = 5 : 8

Savings of each person = Rs. 2000

i.e. a:b = 2 : 3

c : d = 5 : 8

x = 2000

Hence, the multiplicative constant, k =

Therefore, Income of A = 2k = 2 * 6000 = Rs. 12000

In addition to the concept of ratio, the concepts of Proportion and Variation are also the fundamentals of arithmetic. Most of the arithmetic situations have a basic thread of proportion or variation. For example, the topics Time & Work and Time & Distance are basically called the 'rate problems', just because of its unavoidable relation with the concept of variations.

## Combining Ratios.

### Case 1: combining two ratios:

**Example:**

Note: For combining ratios, make it sure that the consequent (second value in ratio) of the first ratio should be algebraically same as the antecedent (first value in the ratio) of the second ratio. Otherwise arrange the ratio values as per the requirement. |

**Case 2: Combining three ratios.**

**Example:**

If a:b = 2:3, b:c = 4:5 and c:d = 3:2, find a:b:c:d.

Here 3 is a common factor in all the products. Hence for finding the ratio, we can avoid the presence of one 3 from each product.

Therefore, a:b:c:d = 2*4 : 3*4 : 3*5 : 5*2 = 8:12:15:10

**Case 3: Combining four ratios.**

**Example:**

If p:q = 1:2, q:r = 3:2, r:s = 5:3 and s:t = 4:1, find p:q:r:s:t.

4 is a common factor in all the products, hence we can remove 4 from all the products.

p:q:r:s:t= 15:30:20:12:3

**Example 1: **

A is 3/4^{th} of B and B is 2/5^{th} of C. find A:B:C ?

**Solution:**

i.e. A : B = 3:4

B = (2/5) C

B/C = 2/5

i.e. B:C = 2:5

A:B:C = (2*3) : (2*4) : (4*5)

'2' is a common factor in all the values, so remove 2 from each product.

A:B:C = 3:4:10

**Example 2:**

Anil's salary is 80% of that of Kesav. Kesav's salary is 30% lesser than that of Ram. Find tha ratio of the respective salaries of Anil, Kesav and Ram.

- 8:7:10
- 7:8:10
- 28:35:50
- 35:28:50

**Solution:**

If Kesav's salary is 100 then Anil's salary is 80% of this, i.e. 80.

Ratio of the salaries of Anil and Kesav = 80:100 = 4:5

If Ram's salary is 100 then Kesav's salary is 30% less than this, i.e. 70.

Ratio of the salaries of Kesav and Ram = 70:100 = 7:10

Ans: C.

**Example 3:**

In a hundred meter race A beats B by 20 meters and B beats C by 30 meters. Find the ratio of speeds of A, B and C.

- 14:20:25
- 25:20:14
- 10:8:7
- 7:8:10

**Solution:**

In a particular time interval A reached at the 100^{th} meter and B reached at the 80^{th} meter.

When time is a constant then the ratio of their speed is same as the ratio of the distances they covered.

Therefore ratio of the speeds of A and B is 100:80 = 5:4

In another time interval (note this time interval is not same as that for the first time interval) B reached at the 100^{th} meter and C reached at the 70^{th} meter.

Ratio of the speeds of B and C is 100:70 = 10:7

A:B:C = (5 * 10) : (10 * 4) : (4 * 7) = 50:40:28 = 25:20:14

Ans: B.

## Proportion

If two ratios are equal then the four quantities involved in the ratios are said to be in a proportion. i.e. if a:b = c:d the a,b,c and d in proportion.

The exact mathematical expression for the proportion of four quantities is a:b :: c:d and it is read as ' **a is to b as c is to d**'.

Here, a and d are called the '**extremes**' and b and c are called the '**means**'.

**Product of extremes = product of means.**

i.e. ad = bc.

**Example:** 2:8 = 3:12 hence 2 * 12 = 8 * 3.

In a proportion a:b :: c:d, a is called the first proportional, b is the second proportional c is the third proportional and d is the fourth proportional.

**Example:** Find the first, second, third and fourth proportional of 3, 9 and 27.

**First proportional:**

Let us consider 'a' as the first proportional, then a: 3 = 9:27

Therefore; a * 27 = 3 * 9

Hence a = 1.

**Second proportional:**

Let 'b' be the second proportional, then 3:b = 9:27

Therefore 3*27 = b*9, hence b = 9.

**Third proportional:**

Let 'c' be the third proportional, then 3:9 = c:27

Therefore 3*27 = 9* b, hence b = 9.

**Fourth proportional:**

Let 'd' be the fourth proportional, then 3:9 = 27: d.

Therefore 3 * d = 9 * 27, hence d = 81.

**Continued proportion**

If three quantities a,b and c such that a:b = b:c, then a,b and c are in continued proportion.

a → first proportional

b → mean proportional

c → third proportional

**Results **

If x and y are two quantities then we can arrange them in different orders to get the first, mean and third proportionals.

**First proportional:**

Let 'a' be the first proportional, then a:x = x:y

Therefore a = x^{2} / y = square of the first quantity / second quantity

**Mean proportional:**

Let 'b' be the mean proportional, then x:b = b:y

Therefore, b = √xy = square root of the product of both the quantities.

**Third proportional:**

Let 'c' be the third proportional, then x:y = y:c

Therefore c = y^{2} / x = square of the second quantity / first quantity

**Example:**

Find the first, mean and third proportionals of 4 and 16.

First proportional = 4^{2 }/ 16 = 1

Second proportional = √(4 * 16) = 8

Third proportional = 16^{2} / 4 = 64

**Re-arrangement of proportionals:**

If a:b = c:d, then we can rearrange the values in the following manners.

- b : a = d : c ( this arrangement is called
**Alternendo**) - a : c = b : d (this arrangement is called
**Invertendo**) - (a+b) : b = (c+d) : d ( this is called
**Componendo**) - (a - b) : b = (c - d) :d ( this is called
**Dividendo**) - (a + b) : (a -b) = (c + d) : (c + d) : (c - d) ( this is called
**Componendo-Dividendo**).

**Example:**

The ratio of the ages of Ananad and Bharat is same as the ratio of the ages of Chandru and Deepak. The ratio of the sum of the ages Anand and Bharat to that of Chandru and Deepak is 2:3. Find the ratio of the difference of the ages of Anand and Bharat to that of Chandru and Deepak.

- A) 1:2 B) 2:5 C) 3:5 D) 2:3

**Solution:**

Then, (A+B) :( A - B)= (C+D) : (C - D) {Componendo Dividendo}

And , (A+B) :( C+D)= (A - B) : (C - D) {Alternendo}

Given (A+B) :( C+D ) = 2:3

Therefore (A - B) : (C - D) = 2:3

Ans:D.

## Variation

If two quantities are dependent up on each other, then a change in one of the quantities make a corresponding change in the other quantity. Here one quantity varies as the other or the quantities are in proportion.

**Direct Variation (Direct proportion)**

One quantity X is said to be in vary directly as another quantity Y, if the two quantities depends up on each other, in such a manner that if Y is increased ( or decreased) in a certain ratio, X also increases (or decreases) in the same ratio.

It is expressed as X ∝ Y (X varies directly as Y or X is directly proportional to Y)

If X ∝ Y, then X = kY, where k is the proportionality constant => k = X/Y.

If both the quantities X and Y have two sets of parameters such as x_{1} corresponding to y_{1} and x_{2} corresponding to y_{2}, then their direct proportionality can be expressed in the following way.

x_{1}/x_{2} = y_{1}/y_{2} V x_{1}/y_{1} = x_{2}/y_{2}

**Illustrated example:**

Cost of 15 books → Rs. 75.

Therefore;

Cost of 10 books → Rs. 50

Here number of books (P) is directly vary with the total cost (C).

P ∝ C, i.e.

P_{1}/P_{2} = C_{1}/C_{2} → 15/10 = 75/50

**Inverse variation (Inverse proportion) **

A quantity X is said to be vary inversely as another quantity Y, if the two quantities depends up on each other, in such a manner that if Y is increased (or decreased) in a certain ratio, then X will decrease (or increase) in the same ratio.

More simply;

If X increase → Y decrease.

If X decrease → Y increase.

It is expressed as X ∝ 1/Y (X varies inversely as Y or X is inversely proportional to Y).

i.e. X = k/Y, where k is proportionality constant, and k = XY.

Then, x_{1}/x_{2} ≠ y_{1}/y_{2} but x_{1}/x_{2} = y_{2}/y_{1}

ie. x_{1} * y_{1} = x_{2} * y_{2}

**Example:**

If a car travels at a rate of 30 kmph then it will take 4 hours to cover a certain distance. How long the car will take to cover the same distance at a constant rate of 40 kmph?

**Solution:**

Here basically two quantities, speed and time.

Distance is a constant value. Therefore speed and time are depends up on each other. It is clear that if the speed is increasing then the time taken for journey should reduce. That means speed is inversely vary with time.

Therefore

S_{1} / S_{2} = T_{2} / T_{1} → 30/40 = T_{2}/4 → T_{2} → = (30 * 4)/40 = 3hours

**Joint Variation:**

Three quantities A, B and C, such that A varies with B when C is a constant and A varies with C when B is a constant, then A is jointly vary with B and C.

If A ∝ B when C is a constant and A ∝ C when B is a constant, then A ∝ BC. Then A = k (BC), where k is the proportionality constant and k = A/ (BC).

If A ∝ B when C is a constant and A ∝ 1/C when B is a constant, then A ∝ B / C. Then A = k(B/C), where k is the proportionality constant and k = (AC)/B.

**Example:**

If X varies directly with the square of Y and inversely vary with Z. when x = 5, y = 2 and z = 4. Find x when y = 3 and z = 3.

Solution:

X ∝ Y^{2}/z

Then, x_{1} / x_{2} = y_{1}^{2}/z_{1} / y_{2}^{2}/z_{1}

y_{1}^{2} * z_{2} / y_{2}^{2} * z_{1}

_{1}= 5, y

_{1}= 2, y

_{2}= 3, z

_{1}= 4 and z

_{2}= 3.

And

x_{1} / x_{2} = y_{1}^{2} * z_{2} / y_{2}^{2} * z_{1} → 5/x_{2} = 2^{2} * 3 / 3^{2} * 4

Therefore, x_{2} = 15.