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.