**Percentage**

A value or ratio that may be stated as a fraction of 100 is called a percentage in mathematics. Divide the number by the whole and multiply the result by 100 if we need to get the percentage of a given number. As a result, the percentage denotes a fraction per hundred. The definition of percent is per 100. “%” is the symbol used to represent it.

**Percentage Formula:**

To determine the percentage, we have to divide the value by the total value and then multiply the resultant by 100.

Percentage formula = (Value/Total value) × 100

Example: 6/4 × 100 = 1.5 × 100 = 150 per cent

**How to Calculate Percentage?**

We must use a separate formula, such as this one, to determine the percentage of a number.

P/% of Total = X

where X is the necessary proportion.

In order to represent the aforementioned formulas, we must eliminate the % symbol.

P/100 * Number = X

**Example: Calculate 10% of 60**

Let 10% of 60 = x

10/100*60= x

x = 6

Converting Fractions into Percentage:

A fraction can be represented by a/b, Multiplying and Dividing the fraction by 100, we have

From the definition of percentage, we have;

(1/100) = 1%

Thus, equation (i) can be written as:

(a/b) × 100%

Therefore, a fraction can be converted to a percentage simply by multiplying the given fraction by 100.

**Conversion of Fraction into Percentage**

**Percentage Difference Formula :**

The following formula can be used to determine the percentage difference between two values if we are given two values:

Percentage Increase and Decrease:

The original number is subtracted from a new number, divided by the original number, and then multiplied by 100 to get the percentage increase.

**% increase = [(New number – Original number)/Original number] x 100**

where, increase in number = New number – original number

Comparably, a percentage drop is calculated by deducting a new figure from the base number, dividing the result by the base number, and then multiplying the result by 100.

**% decrease = [(Original number – New number)/Original number] x 100**

Where decrease in number = Original number – New number

So basically if the answer is negative then there is a percentage decrease.

**Percentage Questions**

**Q.1. If 12% of 50% of a number is 12, then find the number?**

**Solution:**

Let the required number be x,

Therefore, as per the given question

(12/100) * (50/100)* x = 12

So, x= ( 10*100*100)/(12*50) = 200

**Q.2. What percentage of 1/5 is 1/35 ?**

**Solution:**

let x% of 1/5 is 1/25

Therefore, [(1/5)/100] * x = 1/25

x= 1/25* 5/1 * 100= 4%

**Q.3. Which number is 60% less than 80?**

**Solution:**

Required number= 40% of 80

(80*40)/100 = 32

Therefore, the number 32 is 60%, less than 80.

**Q.4. The sum of (10% OF 12.5) and (6% of 25.2) is equal to what value?**

**Solution: **

As per the given question,

Sum= (10% of 12.5) + (6% of 25.2)

(12.55*10)/10 + (6* 25.2)/100

= 14.012

**Word Problem**

**Q.1 A lady spends her total income as follows-30% on food, 25% on rent, 16% on travel and 19% on education. After all expenses made, she saves Rs. 8,900. Find the amount spent on education.**

**Solution:**

Given, A lady spends 30% of her income on food, 25% on rent, 16% on travel and 19% on education.

Her total saving: 8,900

Concepts used: Expenditure + Saving+ Income

Calculation:

Let the income earned by lady be Rs. x

Expenditure made by lady on food= 30% of x = 0.30x

Expenditure made by lady on rent= 25% of x= 0.25x

Expenditure made by lady on travel= 16% of x= 0.16x

Expenditure made by man on education = 19% of x = 0.19x

Savings= Rs. 8,900

Income= Expenditure + Savings

x=0.30x+0.25x+0.16x+0.19x+Rs.8,900

x=0.90x+ Rs.8,900

x-0.90x=Rs.8,900

0.10x= Rs.8,900/0.10

x=Rs.89,000

Expenditure made by lady on education= 19% of x

0.19*Rs.89,000= Rs.16,910

**Thus, the expenditure made on education by the lady is equal to Rs.16,910**

**Q.2. On a shelf, the first row contains 15% more books than the second row and the third row contains 15% less books than the second row. If the total number of books contained in all the rows is 900, then find the number of books in the first row.**

**Solution:**

Let the number of books on second row be 100x

Then, 1st row=

Given, the books on first row contains 15% more books than the second row

Thus, 1st row= 100x+15x= 115x

2nd row= 100x

3rd row=

Given, the books on second row contains 15% less books than second row

Thus, 3rd row= 100x-15x= 85x

Total number of books= 115x+100x+85x= 300x

According to the question, 300x=900

x= 3

**So, on 1st row = 115*3= 345 books.**

**Q.3. A television depreciates in value each year at the rate of 10% of its previous value. However, every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of the fourth year , the value of the television stands at Rs. 1,46,205, then the value of the television at the start of the first year.**

**Solution:**

Let the value of television at the start of first year be Rs. x

According to the question, x*90/100*95/100*90/100*95/100= 1,46,205

x= 1,46,205*10,00,000/81*95*95

x= Rs. 2,00,000

**Q.4. In a mixture of 80 litres of mineral oil and edible oil, 25% of the mixture is mineral oil. How much edible oil should be added to the mixture so that mineral oil becomes 20% of the mixture?**

**Solution:**

In a mixture of 80 litres of mineral and edible oil, 25% of the mixture is mineral oil.

Calculation:

Amount of Mineral oil= 80*25/100= 20 litres

Let x be the amount of edible oil added

ATQ,, 20=20% of (80+x)

20= 20/100 (80+x)

100=80+x

x+20 litres= x= 20 lts

**Q.5. There are three solutions, A, B and C containing milk and water. The strength of the three solutions are 80%,90%, and 70% respectively. When 1 litre each of A and B is mixed with C, the strength of C increases to 75%. What would be strength of C if two litres of A and 0.5 litres of B were mixed with C?**

**Solution:**

Let the volume of C be x litres.

As the strengths of A and B are 80% and 90%, by mixing 1 litre each of A and B with C, we are adding 300 ml and 900 ml respectively of alcohol to C. The strenth of C is 70% which increases to 75%.

So, the new strength = 75%

7x/10+ 1.7/x+2 = 75/100

On solving we get x= 4 litres

So, alcohol present in C= 70% of 4 litres= 2.8 litres

In the second case, total alcohol added= (1.6+0.45) ltrs= 2.05 ltrs

Required strength = 2.8+2.05/4=2.5 = 74.6%

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