PERCENTAGE
Per means ‘/’ & Cent means ‘100‘ so the term ‘Per Cent’ means ‘For every hundred’. The term "per cent" is derived from the Latin 'per centum' meaning "by the hundred".
A fraction whose denominator is 100 is called Percentage, and the numerator of the fraction is called the Rate per cent. A percentage is a dimensionless number.
⇒ X/100 = X%
A percentage can be written as fraction, ratio or decimal.
For example 35% = 35/100 or 35:100 or 0.35
⇒ X% = X/100
Here 80% dots are green and Total dots are 40.
⇒ Per 100 dots = 80 dots are green
⇒ Per 40 dots = 80/100 x 40 dots are green.
⇒ 32 dots are green
⇒ Per 100 dots = 80 dots are green
⇒ Per 40 dots = 80/100 x 40 dots are green.
⇒ 32 dots are green
To express any fraction or Ratio X/Y as a percentage: (X/Y) x 100
NOTE: From above diagrams one thing is very clear - Everything is complete or 100% in itself.
General formula
Percentage = (Value obtained/Total value) x 100
Or
Percentage = (Quantity which is compared/ Quantity to whom first quantity is compared) x 100
METHODS
Or
Percentage = (Quantity which is compared/ Quantity to whom first quantity is compared) x 100
METHODS
Relation between two quantities in term of percentage
1. If A is x% more than that of B then B is less than that of A by: [x/(100 + x)] x 100%
Example 1: If A’s income is 20% more than that of B’s income, then how much percent is B’s income less than that of A’s income?
Solution: B’s income = [20/(100 + 20)] x 100 = 16.66% of A’s income
Solution: B’s income = [20/(100 + 20)] x 100 = 16.66% of A’s income
2. If A is x% less than that of B then B is more than that of A by: [x/(100 - x)] x 100%
Example 2: If A’s income is 20% less than that of B’s income, then how much percent is B’s income more than A’s income?
Solution: B’s income = [20/(100 - 20)] x 100 = 25% of A’s income
Solution: B’s income = [20/(100 - 20)] x 100 = 25% of A’s income
3. If A is X% of C and B is Y% of C, then A = (X/Y) x 100 % of B
Example 3: If A is 20% of C and B is 25% of C then what percentage is A of B?
Solution: A = (20/25) x 100% = 80% of B
Solution: A = (20/25) x 100% = 80% of B
4. If A is x% more than C and B is y% more than C, then A = 100 x [(100 + x)/(100 + y)]% of B
Example 4: Two numbers are respectively 20% and 50% more than the third. What percentage is the first of the second?
Solution: First number = 100 x (120/150)% of second number = 80% of second number
Population Formula
If the present population of a town be P and the population changes in R% per annum, then:
1. Population after n years: P x (1 + R/100)n
2. Population n years ago: P/(1 + R/100)n
1. Population after n years: P x (1 + R/100)n
2. Population n years ago: P/(1 + R/100)n
Note: R will be positive if population increase, if population decrease R will be negative.
Example: The population of a town increase at the rate of 20% annually due to excessive migration. If present population is 144000 find population two years ago?
Solution: Population 2 years ago = 144000/(1 + 20/100)2 = 100000
Solution: Population 2 years ago = 144000/(1 + 20/100)2 = 100000
Increment or Decrement in Consumption
1. If price of a commodity increased by P%, then reduction in consumption so as not to increase the expenditure will be: 100 x [P/(100 + P)] %
2. If price of a commodity decreased by P%, then increase in consumption so as not to decrease expenditure will be: 100 x [P/(100 - P)]%
Example: If the price of a commodity be raised by 20%, find by how much percent must a householder reduce his consumption of that commodity so as to not increase his expenditure?
Solution: Reduction = 100 x (20/120) = (50/3)%
Successive Percentage Change
1. If a number is changed successively by x% and y%, then net percentage change will be given by: x + y + xy/100 %
Note: For percentage increase x and y will be positive, for percentage decrease x and y will be negative.
Example: If price of a commodity first increased by 20% then decreased by 10% then what is the net change price of commodity?
Solution: 20 – 10 – (10 x 20)/100 = 8%
Solution: 20 – 10 – (10 x 20)/100 = 8%
2. If any quantity A is changed by X%, Y% and Z% successively then final value of A will be:
Final Value = A x (1 + X/100)(1 + Y/100)(1 + Z/100)
Final Value = A x (1 + X/100)(1 + Y/100)(1 + Z/100)
Note: X, Y, Z will be positive for increment and negative for decrement.
Example: The income of Ramesh increase by 10%, 20% and 30% successively in three years, find final income if present income is Rs 150000?
Solution: Final Income = 150000 x (1 + 10/100)(1 + 20/100)(1 + 30/100) = Rs 257400
Solution: Final Income = 150000 x (1 + 10/100)(1 + 20/100)(1 + 30/100) = Rs 257400
Same Increase and Decrease
If the value of any quantity first increased by x% and then decreased by x%, the net change is always decrease which is equal to: x2/100
Increased, Decreased & Original price
1. If after reduction in the price of any commodity by x%, a person can buy ‘y’ kg more for Rs ‘z’, then:
Reduced price = [xz/100y]
Original price = [xz/(100 - x)y]
2. If after increment in the price of any commodity by x%, a person can buy ‘y’ kg less for Rs ‘z’, then:
Increased price = [xz/100y]
Original Price = [xz/(100 + x)y]
Reduced price = [xz/100y]
Original price = [xz/(100 - x)y]
2. If after increment in the price of any commodity by x%, a person can buy ‘y’ kg less for Rs ‘z’, then:
Increased price = [xz/100y]
Original Price = [xz/(100 + x)y]
Example: Due to an increase of 20% in the price of eggs, 2 eggs less are available for Rs 24. The present rate of eggs per dozen is?
Solution: Increased price = 20 x 24/(100 x 2) = Rs 2.4 per egg
Present rate per dozen = 12 x 2.4 = Rs 28.80
Solution: Increased price = 20 x 24/(100 x 2) = Rs 2.4 per egg
Present rate per dozen = 12 x 2.4 = Rs 28.80
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