# Percentages Formulas

## Percentages Formulas and Application

The word “Percentage” was coined from the Latin word “Percentum” which means “by hundred”, therefore, it is said that percentages are the fractions with 100 in the denominator. This Page from here on contains Percentages formulas and definition. ### Formulas for Percentages & Definitions

• In mathematics, a percentage is a number or ratio expressed as a fraction whose denominator (bottom) is 100.
Thus, x percent means x hundredths, written as x%.
We express x% as a fraction as $\frac{x}{100}$
• For example 10% = $\frac{10}{100}$ = $\frac{1}{10}$

### Percentages Difference Formula

• If we are given with two values and we need to find the percentage difference between these two values, then it can be done using the formula:
• $Percentage.Difference =\left \| \frac{N1+N2}{\left [ \frac{N1+N2}{2} \right ]} \right \|\times100$

### Formulas & Basic Concept of Percentages

• To calculate a % of b = $\mathbf{\frac{a}{100} × b }$
• To find what percentage of a is b =  $\mathbf{\frac{b}{a} × 100}$
• To calculate percentage change in value
Percentage change = $\mathbf{\frac{Change}{Initial Value }× 100 }$
• Percentage Increase or Decrease
• Percentage increase =  $\mathbf{\frac{R}{100 + R } × 100 }$ %
• Percentage decrease = $\mathbf{\frac{R}{100 – R } × 100 }$ %
• Successive Percentage Change
If there are successive percentage increases of a % and b%, the effective percentage increase is: $\mathbf{ a + b + \frac{ab}{100} }$ %
• Succesive Discount :

If there are successive discount of a % and b%, the effective discount is: $\mathbf{ a + b – \frac{ab}{100} }$ %

• Results on population
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
• Population after n years = $\mathbf{P (1 +\frac{R}{100})^n}$
• Population n years ago = $\mathbf{\frac{P}{(1 +\frac{R}{100})^n}}$
• Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
• Value of the machine after n years = $\mathbf{P (1 – \frac{R}{100})^n}$
• Value of the machine n years ago = $\mathbf{\frac{P}{(1 -\frac{R}{100})^n}}$
• If A is R% more than B, then B is less than A by $\mathbf{\frac{R}{100 + R } × 100 }$ %
• If A is R% less than B, then B is more than A by$\mathbf{\frac{R}{100 – R } × 100 }$ %
• Increase N by S% : $N ( 1 + \frac{S}{100})$.
• Decrease N by S% : $N ( 1 – \frac{S}{100})$

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Question 1 : Peaches are now 50% more expensive at Reliance Smart Point. What percentage reduction in consumption is required to maintain peach expenditure constant?

1. 35 %
2. 35.50 %
3. 33.33 %
4. 32.50 %

Explanation:
If the price of a commodity increases by r%, then the reduction in consumption so as not to increase the expenditure is –

$(\frac{r}{100+r}\times 100)%$

Therefore r is the increased price(i.e 50)

By using the formula:
$(\frac{50}{100+50}\times 100)%=\frac{50}{150}\times100=33.33 %$

Question2 : Jason Statham could set aside 10% of his earnings. However, when his income increased by 20% two years later, he could only save the same amount as before. How much has his spending increased by percentage?

1. $22\frac{2}{9}%$
2. $21\frac{5}{7}%$
3. $22\frac{2}{8}%$
4. $23\frac{1}{9}%$

Explanation:
Let earlier income be 100 Rs
∴ Savings = 10% of 100 = 10 Rs
∴ Expenditure = 90 Rs

New Income = 120 Rs

Savings (same as before) = 10 Rs

∴ Expenditure = 120 – 10 = 110 Rs

∴ Increase in Expenditure = 110 – 90 = 20

$Percentage Increase= \frac{20}{90}\times 100%=22\frac{2}{9}%$

Question 3 : To pass the NDA Recruitment Test, candidates must receive 290 of the aggregate marks to pass. What is the highest total of marks an applicant can receive if they receive 203 and are considered unsuccessful by 12% of their total scores?

1. 750
2. 725
3. 700
4. 675

Explanation:
Let’s maximum aggregate marks = x

203 + 12% of x = 290

12% of x = 290 – 203

$x=\frac{87\times100}{12}=725$.

Question 4: 12. Robert Pattinson is an IT professional who spends 12% of his monthly income on power, 25% of his income on food, 20% of his income on his children’s education, 30% of his income on rent, and the rest 1040 on everything else. What is the person’s monthly salary?

1. 7500 Rs
2. 7700 Rs
3. 7900 Rs
4. 8000 Rs

Explanation:
Let the monthly salary of the Robert be x. then,

Total spends of Robert = (30 + 25 + 20 + 12) = 87%

Remaining % = 100 – 87 = 13%
So, 13% ≡ 1040

100% ≡ x
Using the cross multiplication method,
$x=\frac{1040\times100}{13}=80\times 100= 8000 Rs$

Question 5 : If Yasir’s income is 20% less than Xavier’s, by what percentage does Yasir’s salary fall short of Xavier’s?

1. 16.66 %
2. 15.50 %
3. 14.45 %
4. 13.45 %

Explanation:
Let the salary of Yasir be 100. then,

Salary of Xavier will be 120.

Yasir less than Xavier in %,

$=\frac{120-100}{120}\times 100=\frac{20}{120}\times 100=16.33 %$

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