## APTITUDE PREPARATION: Work Proportions

#### 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.