**Direct and Inverse proportionality Method:**

Price x Quantity = Constant (Profit and Loss)

Time x Speed = Distance (Time Speed and Distance)

Days x Efficiency = Total Work (Time and Work)

**Example 1:**

Due to reduction in the price of mangoes by 30%, A person got 15 mangoes more for the same amount. What is the number of mangoes originally purchased?

*Traditional Method:***Inverse proportionality Method:**

**Example 2:**

*Traditional method:*

*Inverse proportionality Method:***Two more important relations between variables:**

Apart from direct proportion and indirect proportion, there exist another two relations between variables.

1. Direct relation

2. Indirect relation

**1. Direct Relation:**

Here the relation between variables can be best defined as Y = K + mX

Even though x is zero, y is not zero. It take a value of K. But Y varies directly in relation to x. If x increases, y also increases, If x decreases y also decreases.

**Example 3:**

The expenses of organizing a garden party increased from Rs.9000 to Rs.12000 when the number of registered candidates increased from 25 to 40. Find the total cost of organizing the party if there are 50 final registrations.

We know that the expenses of party increases not directly proportional but directly relational.

Assume the fixed component of expense is K rupees and Variable component is M rupees.

Then the total cost is given by

K + 25 x M = 9000 …………(1)

When there are 40 registrations the total cost is

K + 40 x M = 12000 ………(2)

By substracting (1) from (2)

⇒15M=3000⇒M=200⇒15M=3000⇒M=200

So the variable cost per head is Rs.200

To find the fixed cost we can substitute Rs.200 in either (1) or (2), then K = 4000

If there are 50 registrations then the total cost = 4000 + 50 x 200 = Rs.14000

**2. Inverse Relation:**

Here Y = K -mX holds good. If x increases, y decreases and vice versa.

**Example 4:**

The reduction in the speed of an engine is directly proportional to the square of the number of bogies attached to it. The speed of the train is 100 km/hr when there are 4 bogies and 55 km/hr when there are 5 bogies. What is the maximum number of bogies that can be attached to the train so that, even with those many numbers of bogies, it can just move?

We know that S=Smax−K×(bogies)2S=Smax−K×(bogies)2

When there are 4 bogies speed of the train is 100km/hr 100=Smax−K×(4)2100=Smax−K×(4)2 ⇒100=Smax−K×16⇒100=Smax−K×16 …….(1)

When there are 5 bogies speed of the train is 55km/hr

55=Smax−K×(5)255=Smax−K×(5)2 ⇒⇒ 55 = Smax−K×25Smax−K×25 ……..(2)

(2) – (1) gives 9K=45⇒K=59K=45⇒K=5

By substituting K = 5 in either (1) or (2) we can find the maximum speed of the train = 180 km/hr

Assume, for N number of bogies the train does not move.

⇒0=180−5(N)2⇒0=180−5(N)2⇒N=6⇒N=6

If the engine is attached to 6 bogies, it does not move. If we want the train to move we need to attach maximum of 5 bogies