Concepts of Percentages and its Important Shortcuts

'Percentage' is a very important concept, applicable in almost all the topics of Quantitative Aptitude. To help you all score better in the exams, we are sharing some important concepts and shortcut tricks that will help you solve questions based on percentages easily and within less time. 

Concept 1

1. A+B+AB/100 When A and B both are a positive change
2. A-B-AB/100 When A is positive change and B is a negative change
3. -A+B-AB/100 When A is negative change and B is a positive change.
4. -A-B+AB/100 When A and B both are a negative change.
Important: There is no need to remember the above formulas, you have to just remember
±A ± B ±AB/100 and put the sign of change, if negative, then (-) and positive then, (+) but keep in mind that sign of AB is the product of signs of A and B.
Example1: The price of a book is reduced by 10% and the sale of the book is increased by 15%. Find the net effect on revenue.

Example 2: If length and breadth of a rectangle is increased by 5% and 8% respectively. Find the % change in area of the rectangle.

Concept 2:

New solution × new % = old solution × old %
This formula is applicable for the commodity which is constant in the solution or mixture, its quantity doesn’t change after mixing in solution.
Example3:  A mixture of sand and water contains 20% sand by weight. Of it 12 kg of water is evaporated and the mixture now contains 30% sand.find the original quantity and quantity of sand and water initially ? 
Solution: In this sand is constant in the mixture. So we will apply this formula on sand not on water.
(a)Find the original mixture.
Let the original mixture is P kg, So new mixture = (P-12) kg
old% = 20 and new % = 3
new solution × new % = old solution × old %
 (P-12)× 30% = P × 20%
 3P – 36 = 2P
 P = 36 Kg.
(b) Find the quantity of sand and water in the original mixture.
     Quantity of sand in original mixture = 20% of 36 = 7.2 Kg
     Quantity of water in original mixture = 80% 0f 36 = 28.8 Kg
OR = Quantity of mixture – quantity of sand = 36 -7.2 = 28.8 Kg

Example 4: 30 litres of a mixture of alcohol and water contains 20% alcohol. How many litres of water must be added to make the alcohol 15% in the new mixture?
Solution: only water is added to the mixture so, there is no change in alcohol. We will apply the above formula on alcohol. Let water added is P litres.
Old mixture = 30 litres, old % of alcohol = 20%
New mixture = 30+P litres, new % of alcohol = 15
using, new solution × new % = old solution × old %
(30+P) × 15% = 30 × 20%
P = 10 litres , hence 10 litres of water is added.