In the previous section we saw the relationship between percentage and ratios. In this section, we will see Increase in percentage.
In day to day life we see statements such as:
• The price of printers increased by 12%.
This same statement can be written in another form:
• There is a 12% increase in the price of printers.
Both means the same.
• The price of sugar increased by 7%.
This same statement can be written in another form:
• There is a 7% increase in the price of sugar.
Both means the same.
• The number of fever patients increased by 5%.
This same statement can be written in another form:
• There is a 5% increase in the number of fever patients.
Both means the same.
When we read such statements on Newspaper, or hear them on TV, we will want to know the 'situation after the increase'. For example, we will want to know the new prices after the increase. Or we will want to know the number of fever patients after the increase. Let us try to find the new values:
In the above statements, increase is given as per cent. That is., per 100. Same as: 'for every 100'.
So, in the first example, for every hundred, there is an increase of 12. That is every 100 in the original price becomes 112.
• Let the original price be ₹2700/-
• How many hundreds are there in 2700? It is 2700⁄100 =27
• So there are 27 hundreds in the original price of ₹2700.
• Each of these hundreds become 112.
• So the new price is 27 × 112 = ₹3024
The above calculations can be written in one step:
• Divide the original price by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 112]
• So we can write the calculation in just one line: 2700⁄100 × 112
■ This can be rearranged as:
Second example:
• For every 100 there is an increase of 7. That is., every 100 in the original price becomes 107.
• Let the original price be ₹40 per kilogram.
• How many hundreds are there in 40? It is 40⁄100 = 0.40
• So there are 0.40 hundreds in 40
• Each of these hundreds become 107
• So the new price is 0.4 × 107 = 42.8
The above calculations can be written in one step:
• Divide the original price by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 107]
• So we can write the calculation in just one line: 40⁄100 × 107
■ This can be rearranged as:
Third example:
• For every 100 there is an increase of 5. That is., every 100 in the original number becomes 105.
• Let the original number be 87.
• How many hundreds are there in 87? It is 87⁄100 = 0.87
• So there are 0.87 hundreds in 87.
• Each of these hundreds become 105.
• So the new number is 0.87 × 105 = 91.35
The above calculations can be written in one step:
• Divide the original number by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 105]
• So we can write the calculation in just one line: 87⁄100 × 105
■ This can be rearranged as:
From the above 3 examples we get a general method. We will write it as:
Eq.8.1:
Where 'New value of 100' = 100 + 'increase in percent' with out the % sign
We will now see some solved examples:
Solved example 8.23
It was heard on TV that, the price of computers will increase by 3%. If the present price of a computer is ₹27500, what will be the new price?
Solution:
We have
• Given Original value = ₹27500
• Increase in percent = 3
• ∴ New value of 100 = 100 +3 =103
• Substituting these in the equation, we get: New value = 27500 × 103⁄100 = 27500 × 1.03 = 28325
• So the new price = ₹28325
Solved example 8.24
A library has 15250 books. The owner of the library wants to increase the number of books by 15%. How many more books does he have to obtain?
Solution:
We have
• Given Original value = 15250
• Increase in percent = 15
• ∴ New value of 100 = 100 +15 =115
• Substituting these in the equation, we get: New value = 15250 × 115⁄100 = 15250 × 1.15 = 17537.5
• So the new total number of books = 17537.5
∴ the additional number of books = 17537.5 - 15250 = 2287.5
• So the owner has to obtain an additional 2288 books
Solved example 8.25
The population of a district increases by 4% every year. If the present population is 1,20,000, then (i) what will be the population next year? (ii) What was the population the previous year?
Solution:
(a) We have
• Given Original value = 1,20,000
• Increase in percent = 4
• ∴ New value of 100 = 100 + 4 =104
• Substituting these in the equation, we get: New value = 120000 × 104⁄100 = 120000 × 1.04 = 124800
• So the population next year = 124800
(b) Here we can use the same Eq.8.1. But we have to find the original value
• Let original value = x
• New value = 1,20,000
• Increase in percent = 4
• ∴ New value of 100 = 100 + 4 =104
• Substituting these in the equation, we get: 120000 = x × 104⁄100 ⇒ 120000 × 100 = 104x
∴ x = 12000000⁄104 = 115384.6
• So the population in the previous year = 115385
So we have seen how to obtain the new value when the percentage increase is given. We must be able to do the reverse also. That is., when the original value, as well as the new value are given, we must be able to calculate the percentage increase. This can be analysed as follows:
• Let the percentage increase be x%.- - - (1)
• Then, every 100 in the original value will become (100 + x). That is., the 'new value of 100' = (100 + x)
• How many hundreds are there in the original value? It is equal to: original value⁄100
• When we multiply the 'number of hundreds' with the 'new value of 100', we get the 'new value'
• That is., we have to multiply (original value⁄100) by (100 + x) to get the 'new value'. So we can write:
In the last step we obtain 'x'. But 'x' is the percentage increase as mentioned in (1) above. So we can write the equation:
Eq.8.2:
We must remember that, what we get from Eq.8.2 is the percentage increase. So, when we get the result from Eq.8.2, we must immediately put a ‘%’ sign after it.
We will now see some solved examples based on the above discussion:
Solved example 8.26
The price of Rice this year is ₹42. Previous year, it was ₹38. What is the percentage increase in the price?
Solution:
We have:
• New value = ₹42
• Original value = ₹38
• Substituting these values in the equation, we get:
• Percentage increase = (42 - 38)⁄38 × 100 = 4⁄38 × 100 = 10.53%
In the next section we will see 'Decreasing percentage'.
In day to day life we see statements such as:
• The price of printers increased by 12%.
This same statement can be written in another form:
• There is a 12% increase in the price of printers.
Both means the same.
• The price of sugar increased by 7%.
This same statement can be written in another form:
• There is a 7% increase in the price of sugar.
Both means the same.
• The number of fever patients increased by 5%.
This same statement can be written in another form:
• There is a 5% increase in the number of fever patients.
Both means the same.
When we read such statements on Newspaper, or hear them on TV, we will want to know the 'situation after the increase'. For example, we will want to know the new prices after the increase. Or we will want to know the number of fever patients after the increase. Let us try to find the new values:
In the above statements, increase is given as per cent. That is., per 100. Same as: 'for every 100'.
So, in the first example, for every hundred, there is an increase of 12. That is every 100 in the original price becomes 112.
• Let the original price be ₹2700/-
• How many hundreds are there in 2700? It is 2700⁄100 =27
• So there are 27 hundreds in the original price of ₹2700.
• Each of these hundreds become 112.
• So the new price is 27 × 112 = ₹3024
The above calculations can be written in one step:
• Divide the original price by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 112]
• So we can write the calculation in just one line: 2700⁄100 × 112
■ This can be rearranged as:
• For every 100 there is an increase of 7. That is., every 100 in the original price becomes 107.
• Let the original price be ₹40 per kilogram.
• How many hundreds are there in 40? It is 40⁄100 = 0.40
• So there are 0.40 hundreds in 40
• Each of these hundreds become 107
• So the new price is 0.4 × 107 = 42.8
The above calculations can be written in one step:
• Divide the original price by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 107]
• So we can write the calculation in just one line: 40⁄100 × 107
■ This can be rearranged as:
• For every 100 there is an increase of 5. That is., every 100 in the original number becomes 105.
• Let the original number be 87.
• How many hundreds are there in 87? It is 87⁄100 = 0.87
• So there are 0.87 hundreds in 87.
• Each of these hundreds become 105.
• So the new number is 0.87 × 105 = 91.35
The above calculations can be written in one step:
• Divide the original number by 100. Multiply the result by 'the new value of 100'. [The new value of 100 is 105]
• So we can write the calculation in just one line: 87⁄100 × 105
■ This can be rearranged as:
From the above 3 examples we get a general method. We will write it as:
Eq.8.1:
Where 'New value of 100' = 100 + 'increase in percent' with out the % sign
We will now see some solved examples:
Solved example 8.23
It was heard on TV that, the price of computers will increase by 3%. If the present price of a computer is ₹27500, what will be the new price?
Solution:
We have
• Given Original value = ₹27500
• Increase in percent = 3
• ∴ New value of 100 = 100 +3 =103
• Substituting these in the equation, we get: New value = 27500 × 103⁄100 = 27500 × 1.03 = 28325
• So the new price = ₹28325
Solved example 8.24
A library has 15250 books. The owner of the library wants to increase the number of books by 15%. How many more books does he have to obtain?
Solution:
We have
• Given Original value = 15250
• Increase in percent = 15
• ∴ New value of 100 = 100 +15 =115
• Substituting these in the equation, we get: New value = 15250 × 115⁄100 = 15250 × 1.15 = 17537.5
• So the new total number of books = 17537.5
∴ the additional number of books = 17537.5 - 15250 = 2287.5
• So the owner has to obtain an additional 2288 books
Solved example 8.25
The population of a district increases by 4% every year. If the present population is 1,20,000, then (i) what will be the population next year? (ii) What was the population the previous year?
Solution:
(a) We have
• Given Original value = 1,20,000
• Increase in percent = 4
• ∴ New value of 100 = 100 + 4 =104
• Substituting these in the equation, we get: New value = 120000 × 104⁄100 = 120000 × 1.04 = 124800
• So the population next year = 124800
(b) Here we can use the same Eq.8.1. But we have to find the original value
• Let original value = x
• New value = 1,20,000
• Increase in percent = 4
• ∴ New value of 100 = 100 + 4 =104
• Substituting these in the equation, we get: 120000 = x × 104⁄100 ⇒ 120000 × 100 = 104x
∴ x = 12000000⁄104 = 115384.6
• So the population in the previous year = 115385
So we have seen how to obtain the new value when the percentage increase is given. We must be able to do the reverse also. That is., when the original value, as well as the new value are given, we must be able to calculate the percentage increase. This can be analysed as follows:
• Let the percentage increase be x%.- - - (1)
• Then, every 100 in the original value will become (100 + x). That is., the 'new value of 100' = (100 + x)
• How many hundreds are there in the original value? It is equal to: original value⁄100
• When we multiply the 'number of hundreds' with the 'new value of 100', we get the 'new value'
• That is., we have to multiply (original value⁄100) by (100 + x) to get the 'new value'. So we can write:
Eq.8.2:
We must remember that, what we get from Eq.8.2 is the percentage increase. So, when we get the result from Eq.8.2, we must immediately put a ‘%’ sign after it.
We will now see some solved examples based on the above discussion:
Solved example 8.26
The price of Rice this year is ₹42. Previous year, it was ₹38. What is the percentage increase in the price?
Solution:
We have:
• New value = ₹42
• Original value = ₹38
• Substituting these values in the equation, we get:
• Percentage increase = (42 - 38)⁄38 × 100 = 4⁄38 × 100 = 10.53%
In the next section we will see 'Decreasing percentage'.
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