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.

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

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