**Direct proportion:**

Two variables are said to be in directly proportional, if an increase in one variable also increases the other variable by the same extent and vice-versa.

#### Example:

Distance covered is directly proportional to the velocity for the same time

**Inverse proportion:**

Two variables are said to be in inversely proportional, if an increase in one variable decreases the other variable by the same extent and vice-versa.

#### Example:

Time taken is directly inversely proportional to the speed of travel for the same distance

We will explain the concept with solved examples.

**Example 1:**

If 9 kids can eat 9 pizzas in 9 minutes, how many minutes will 15 kids take to eat 15 pizzas?

#### Solution:

Keep the unknown in the last column.

Work is same.

Kids Pizzas Minutes

9 9 9

15 15 x

Comparing column 1(Kids) and Column 3(Minutes), keeping Column 2(Pizzas) constant;

If number of kids increases (from 9 to 15), the number of minutes to eat the same number of pizzas will DECREASE.

Take SMALLER FRACTION =(9/15)

Comparing column 2(Pizzas) and Column 3(Minutes), keeping Column 1(Kids) constant;

If number of pizzas to be eaten increases (from 9 to 15) by the same number of kids, number of minutes will INCREASE.

Take LARGER FRACTION = (15/9)

x = 9x(9/15)x(15/9) = 9 minutes.

**Example 2:**

If 28 data entry operators can type 28 lines in 28 minutes, how many lines can 84 operators type in 84 minutes? Assume speed of typing is same for all the operators.

#### Solution:

Keep the unknown in the last column.

Work is same.

Operators Minutes Lines

28 28 28

84 84 x

Comparing column 1(Operators) and Column 3(Lines), keeping Column 2(Minutes) constant;

If number of operators increases (from 28 to 84), no of lines types will INCREASE.

Take LARGER FRACTION =(84/28)

Comparing column 2(Minutes) and Column 3(Lines), keeping Column 1(Operators) constant;

If number of minutes typed increases (from 28 to 84), number of lines will INCREASE.

Take LARGER FRACTION = (84/28)

x = 28x(84/28)x(84/28) = 252 lines.

**Example 3:**

If 12 men working for 14 days can dig 5 wells, in how many days will 28 men dig 10 wells?

#### Solution:

Keep the unknown in the last column.

Men Wells Days

12 5 14

28 10 x

Comparing column 1(Men) and Column 3(Days), keeping Column 2(Wells) constant;

If number of men increases (from 12 to 28), to dig the same number of wells, number of days will DECREASE.

Take SMALLER FRACTION =(12/28)

Comparing column 2(Wells) and Column 3(Days), keeping Column 1(Men) constant;

If number of wells to be dug increases (from 5 to 10) by the same number of men, number of days also will INCREASE.

Take LARGER FRACTION = (10/5)

x = 14x(12/28)x(10/5) = 12 days.

**Example 4:**

12 persons working 8 hours a day can construct a shed in 15 days. Find the required number of days for 18 persons working daily 5 hours to construct the same shed.

#### Solution:

Keep the unknown in the last column.

Work is same.

Persons Hours per day Days

12 8 15

18 5 x

Comparing column 1(Persons) and Column 3(Days), keeping Column 2(Hours) constant;

If number of persons increases (from 12 to 18), no of days will DECREASE to construct the shed.

Take SMALLER FRACTION =(12/18)

Comparing column 2(Hours) and Column 3(Days), keeping Column 1(Persons) constant;

If number of working hours decreases (from 8 to 5), number of days will INCREASE.

Take LARGER FRACTION = (8/5)

x = 15x(12/18)x(8/5) = 16 days.

**Example 5:**

30 carpenters working 8 hours a day can make 24 chairs in 21 days. If 35 carpenters want to make 18 such chairs in 12 days, how many hours they should work daily?

#### Solution:

Carpenters Days Chairs Hours

30 21 24 8

35 12 18 x

Comparing column 1(Carpenters) and Column 4(Hours), keeping other columns constant:

If number of carpenters increases (from 30 to 35), no of hours to be worked will DECREASE.

Take SMALLER FRACTION =(30/35)

Comparing column 2(Days) and Column 4(Hours), keeping other columns constant:

If number of days worked decreases (from 21 to 12), number of hours required to work will INCREASE.

Take LARGER FRACTION = (21/12)

Comparing column 3(Chairs) and Column 4(Hours), keeping other columns constant:

If number chairs are to be made decreases (from 24 to 18), number of hours required to work will also DECREASE.

Take SMALLER FRACTION = (18/24)

x = 8x(30/35)x(21/12)x(18/24) =.9 hours.

**About Author**

#### Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics