Percent value: \(p\%\)

Real numbers: \(A\), \(B\)

Initial balance (principal): \({S_0}\)

Final balance: \(S\)

Real numbers: \(A\), \(B\)

Initial balance (principal): \({S_0}\)

Final balance: \(S\)

Periodic interest rate: \(r\%\)

Number of investment periods: \(n\)

Number of investment years: \(t\)

Number of investment periods: \(n\)

Number of investment years: \(t\)

- One percent means one in a hundred. It is denoted using the percent sign \(\%.\)
- Finding the percent of a number

Given a number \(A.\) It is known that the number \(B\) is \(p\%\) of the number \(A\). Then the number \(B\) is equal to

\(B = A \cdot p/100\) - Finding what percent one number is of another

Given two numbers \(A\) and \(B\). The percentage of the number \(A\) to the number \(B\) is equal to

\(p\% = A/B \cdot 100\% \) - Finding a number given its percent of another number

It is known that the number \(B\) is \(p\%\) of the number \(A\). Then the number \(A\) is

\(A = B \cdot 100/p\) - Percentage increase

Given a number \(A\). The number \(B\) is \(p\%\) greater than \(A\). Then the number \(B\) is equal to

\(B = A + A \cdot p/100 =\) \( A\left( {1 + p/100} \right)\) - Percentage decrease

Given a number \(A\). The number \(B\) is \(p\%\) less than \(A\). Then the number \(B\) is equal to

\(B = A – A \cdot p/100 =\) \( A\left( {1 – p/100} \right)\) - Finding what percent one number is greater than another

Given two numbers \(A\) and \(B\) \(\left({A \gt B}\right).\) The number \(A\) is \(p\%\) greater than \(B\), where

\(p\% = \left( {A – B} \right)/B \cdot 100\% \) - Finding what percent one number is less than another

Given two numbers \(A\) and \(B\) \(\left({A \lt B}\right).\) The number \(A\) is \(p\%\) less than \(B\), where

\(p\% = \left( {B – A} \right)/B \cdot 100\% \) - Simple interest (in banking and finance) is calculated or payable only on the principal. The compound interest includes both the principal and accrued interest.
- Simple interest formula

Suppose that the initial sum (principal amount) is \({S_0}\). The interest rate for a period is \(r\%\). The final sum \(S\) at the end of \(n\) periods is determined by the the formula

\(S = {S_0}\left( {1 + n \cdot r/100} \right)\) - Compound interest formula

Suppose that the initial sum is \({S_0}\) and the interest rate for a period is \(r\%\). The interest earned for each period is reinvested. The final sum \(S\) at the end of \(n\) periods is given by the formula

\(S = {S_0}{\left( {1 + r/100} \right)^n}\) - Finding the interest rate from the compound interest formula

Let the initial (principal) sum be \({S_0}\) and the final sum be \(S\). The number of periods is \(n\). The interest rate \(r\%\) for a period is given by

\(r\% =\) \( \left[ {{{\left( {S/{S_0}} \right)}^{1/n}} – 1} \right] \cdot 100\% \) - Finding the number of periods from the compound interest formula

Let the initial (principal) sum be \({S_0}\) and the final sum be \(S\). The interest rate for a period is \(r\%\). Then the number of periods \(n\), needed to reach the final amount \(S\) is determined by the expression

\(n = {\log _{\left( {1 + r/100} \right)}}\left( {S/{S_0}} \right)\) - Generalized compound interest formula

Suppose that the initial sum is \({S_0}\) and the annual interest rate is \(r\%\). Each year consists of \(n\) equal periods. The profit is reinvested at the end of each period, i.e. \(n\) times per year. The final amount in \(t\) years is determined by the formula

\(S = {S_0}{\left[ {1 + r/\left( {100n} \right)} \right]^{nt}}\) - Continuous compound interest formula

In the limiting case as \(n \to \infty,\) the generalized compound interest formula is represented by the exponential function

\(S = {S_0}\exp \left( {rt} \right)\)