A percentage is a way of expressing a number as a fraction of 100 (per cent meaning “per hundred”). It is denoted using percent sign, “%”. This concept is useful for comparison of fractions as all the fractions are indexed to 100 when converted to percentages. Any percentage can be expressed as a fraction or a decimal fraction and vice versa. For example, fractions such as 2/5 can be written as 0.8 or 80% and vice versa.
Percentage Value = Fractional Value x 100
Important Percentages Formulas & Concepts
Here are some very important Percentages formulas and concepts to solve various percentages based problems.
1) Percentages Formulas for Percentage increase or decrease
Percentage increase or decrease of a quantity is the percent representation of the ratio of the actual (absolute) increase and decrease to the original quantity.
Example: If the price of an idly is increased from Rs. 25 to Rs. 35, the percentage increase in the price can be computed as Percentage change = (10/25)*100 = 40%
2) Percentages Formulas for Multiplication Factor
Whenever there is any percentage increase or decrease in a quantity, instead of calculating the actual increase/decrease and then adding to/subtracting from the original quantity, we can directly compute the new quantity by using the concept of the multiplication factor.
For every percentage increase or decrease, a number called multiplication factor can be calculated which when multiplied by the original quantity will give the increased quantity.
For a percentage increase of I%, the multiplication factor is (1 + I/100).
If the number of goals scored by Indian hockey team in 2000 Olympics is 20 and it is increased by 20% in the 2004 Olympics, then the number of goals scored by Indian hockey team in 2004 Olympics can be calculated by directly finding the product of the original quantity (20 in this case) and the multiplication factor which is
(1+I/100) = (1 + 20/100) = 1.2
Hence, the new quantity will be the product of 20 and 1.2, which is 24.
The inverse of the above concept can also be used to good effect in tackling problems which involve calculating the original quantity with data given on the new value and percentage increase. Suppose in the same example, the number of goals in 2004 is given along with the percentage increase from 2000 to 2004, and it is asked to calculate the number of goals in 2000, then it can be computed as,
Number of goals in 2000 = Number of goals in 2004/Multiplication factor = 24/1.2 = 20
Additional Percentages Formulas for calculating Multiplication Factors are:
For percentage decrease of D%, the multiplication factor is (1 - (D/100))
3) General Percentages Formulas
a) Given original value(X) and percentage change (p), the final value (Y) can be computed using the Percentages formula
b) Given final value(Y) and percentage change(p), the original value(X) can be computed using the Percentages formula
4) Percentages Formulas for Successive increase/decrease
There were 400 students in a school in the 1st year. The strength increased by 20% by the 2nd year. By the 3rd year, the strength increased by 25%. How many students are there in the school, by the 3rd year? Also, what is the percentage increase in the number of students?
Therefore there are 600 students by the 3rd year. But what about the percentage increase in the number of students in the 3 years?
Change = 600-400 = 200
Initial (base)value = 400
Thus total percentage change will be 400/200 x 100 = 50%
If there are successive increases of p%, q%, and r% to a value, the effective multiplication factor is
If one or more of p%, q%, and r% are percentages of decrease, then those figures will be taken as negative to arrive at the effective multiplication factor.
5) Product constancy rule based formulas
The product of two variables being constant, when one variable is increased by 1xtimes itself, the other decreases by 1x+1 times itself. This is easy to understand using variables. Let A×B=Constant.
Let A be increased by 1/x times itself and B decreased by a factor K.
Comparing with A x B =Constant,
This means that,
Therefore, keeping the product constant, when A increased by 1/x times itself, B decreased by 1/(x+1) times itself or vice-versa. The concept and its application can be further understood by considering some examples.
A man cycles at 10 km/hr, then he arrives at a certain place at 1 P.M. If he cycles at 15 km/ hr, he will arrive at the same place at 11 A.M. Find the speed at which he must cycle to get there at noon.
The speed increases from 10 to 15km/hr, by 5 km/hr which is an increase of 1/2. Therefore, the time must have decreased by 1/3 times the original time. The decrease in time is given as 2 hours. if 1/3 times the original time is 2 hours, original time = 6 hours. Therefore, the man must have started at 7 A.M. To reach at noon he must take 5 hours which is 1 hour or 1/6 th less than original time. As the time decreases by 1/6 the of the original time, the speed must increase by 1/5 th of the original speed.
Speed = 10 + 10 x (1/5) = 12 Km/hr