## Percentage

A percentage is a fraction whose denominator is 100 and the numerator of the fraction is called the rate percent.

Percent is denoted by the sign **'%'. **

The term percent means for every hundred.

If we have to find x % of a number, then x % of number = | times of that number | |

100 |

For example: 20% of 700 = |
× 700 = 20 × 7 = 140 | |

100 |

### Some fast result :-

10% of a number = | |

10 |

11.11% of a number = | |

9 |

12.50% of a number = | |

8 |

14.28% of a number = | |

7 |

6.66% of a number = | |

6 |

20% of a number = | |

5 |

25% of a number = | |

4 |

33.33% of a number = | |

3 |

50% of a number = | |

2 |

**Example**: Find 16.66% of 252.

Solution: 16.66% of 252 = |
× 252 = 42 | |

6 |

**Example:**Find 14.28% of 504.

Solution: 14.28% of 504 = |
× 504 = 72 | |

7 |

**Example:**Find 33.33% of 369.

Solution: 33.33% of 369 = |
× 369 = 123 | |

3 |

### Conversion of Percent into Number :-

To convert any percent into number we should divided by 100. **Example :** Express 20% in fraction.

Solution: 20% = |
= | ||

100 | 5 |

**Example :**Express 75 % in fraction.

Solution: 75% = |
= | ||

100 | 4 |

**Example :**- Express 64% in decimal.

Solution: 64% = |
= 0.64 | |

100 |

### Conversion of Fraction into Percentage :-

To convert fraction into per cent we should multiply by 100 .

Example : Convert |
into percent. | |

5 |

Solution: |
× 100 = 60% | |

5 |

**Example :**Express 3

^{1}/

_{4}in percent.

Solution: 3^{1}/_{4} × 100 = |
× 100 = 13 × 25 = 325 | |

4 |

#### Expressing one quantity as a percent with respect to other .

× 100 % | |||

other quantity |

**Note :**For this formula, both the quantity must be in same metric unit.

**Example:**50 kg is what per cent of 250 kg?

Solution: |
× 100 % | |||

other quantity |

= | × 100 % = 20% | |

250 |

**Example:** 20 g is what per cent of 2 kg?

**Solution**: First quantity = 20 g and second quantity = 2 kg = 2000 g

∴ | × 100 % | |||

other quantity |

= | × 100 % = 1% | |

2000 |

### Increasing percent and Decreasing percent :-

Percentage increase = | × 100 | ||

original value |

**Increase = New value - Original value**

**Example:**Rent of the house is increased from ₹ 8000 to ₹ 10000. Express the increase in price as a percentage of the original rent.

**Solution**: Given, Initial rent = ₹ 8000 , New rent = ₹ 10000

Increase = New rent - Original rent

= 10000 - 8000 = ₹ 2000

Percentage increase = | × 100 | ||

original value |

Percentage increase = | × 100 = 25% | ||

8000 |

Percentage decrease = | × 100 | ||

original value |

**Decrease = Original value - New value**

**Example:** The cost of a bike last year was ₹ 50000. Its cost this year is ₹ 45000. Find the percent decrease in its cost.**Solution**: Given, last year cost = ₹ 50000

This year cost = ₹ 45000

Decrease = Original cost - New cost

Decrease = 50000 - 45000 = 5000

Percentage decrease = | × 100 | ||

original value |

Percentage decrease = | × 100 = 10% | ||

50000 |

## Use of Percent :-

**Trick: Successive change **:-

Net Percentage Change = | A + B + | % | |||

100 |

**Example:** If the salary of a person increased by 10% of income. And again increased by 10% in next year. Find net percent change in his total salary. **Solution**: Let A = 10% and B = 10%

Net Percentage Change = | A + B + | % | |||

100 |

Net percent change in salary = 10 + 10 + | = 20 + 1 = 21% | ||

100 |

**Example:** The price of an article is first increased by 20% and later on the price were decreased by 25% due to reduction in sales. Find the net percentage change in final price of article. **Solution**:

Let A = 20% and B = -25%

Net Percentage Change = | A + B + | % | |||

100 |

Net percent change in final price = 20 - 25 + | = -5 - 5 = -10% | ||

100 |

**Trick**: To solve "Expenditure Constant" questions

Reduction in consumption = | × 100 | ||||

100 + x |

Increase in consumption = | × 100 | ||||

100 - x |

Use '-'sign for decrease

**Example:** Price of sugar is increased by 10%. Find by how much % the consumer should the consumption of sugar to keep the expenditure constant. **Solution**: Here, x = 10%

Using the above given formula , we have

Reduction in consumption = | × 100 = | × 100 | |||

100 + 10 | 110 |

Reduction in consumption = | = 9.09% | ||

11 |

**Example:**Price of rice is decreased by 20 %. Find by how much per cent the consumption should be increased to keep the expenditure constant.

**Solution**: Here x = 20%

Using the above given formula , we have

Increase in consumption = | × 100 = | × 100 = 25% | |||

100 - 20 | 80 |

**Trick**: If the population of a town is **P **and it increases ( or decreases ) at the rate of **r%** per year, then

Population after n year = P | 1 + | ^{n} |
||||

100 |

Population n year ago = P | 1 - | ^{n} |
||||

100 |

Use '-' sign for decrement

**Example:**The population of a town is 472300. If it increases at the rate of 10 % per year, them what will be its population after 2 yr.

**Solution**:- given, P = 473200 , r = 10% , n = 2 yr

Population after 2 year = P | 1 + | ^{n} |
||||

100 |

Population after 2 year = 472300 | 1 + | ^{2} |
||||

100 |

Population after 2 year = 472300 | 1 + | ^{2} |
||||

10 |

Population after 2 year = 472300 | ^{2} |
||||

10 |

**Trick**: If the present population of a city is P and there is a increment or decrement of **r**_{1}**%**,** r**_{2}**%** and **r**_{3}**%** in first, second and third year respectively, then

Population of city after 3 yr = P | 1 + | _{1} |
1 + | _{2} |
1 + | _{3} |
|||||||

100 | 100 | 100 |

**Example:**Population of a city in 2010 was 200000. If in 2011 there is an increment of 10%, in 2012 there is a decrement of 20% and in 2013 there is increment of 30%, then find the population of city at the end of year 2013.

**Solution**:- given, P = 200000, r

_{1}= 10%, r

_{2}= 20%, r

_{3}= 30%

Using above given formula ,

Population of city at the end of year 2013 = P | 1 + | 1 - | 1 + | ||||||||||

100 | 100 | 100 |

Population of city at the end of year 2013 = 200000 | × | × | × | ||||

10 | 5 | 10 |