Given two integers **N1** and **N2** which is the **Compound Interest** of two consecutive years. The task is to calculate the rate percentage.**Examples:**

Input:N1 = 660, N2 = 720Output:9.09091 %Input:N1 = 100, N2 = 120Output:20 %

**Approach:** The rate percentage can be calculated with the formula **((N2 – N1) * 100) / N1** where **N1** is the compound interest of some year and **N2** is the compound interest for the next year.

Let us consider the 1st Example:

The difference between the Compound interest in the two consecutive years is because of the interest received on the previous year interest. Therefore,

–> N2 – N1 = N1 * (Rate / 100)

–> 720 – 660 = 660 * (Rate / 100)

–> (60 / 660) * 100 = Rate

–> Rate = (100 / 11) = 9.09% (Approx)

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the` `// required rate percentage` `float` `Rate(` `int` `N1, ` `int` `N2)` `{` ` ` `float` `rate = (N2 - N1) * 100 / ` `float` `(N1);` ` ` `return` `rate;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N1 = 100, N2 = 120;` ` ` `cout << Rate(N1, N2) << ` `" %"` `;` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG ` `{` ` ` `// Function to return the` ` ` `// required rate percentage` ` ` `static` `int` `Rate(` `int` `N1, ` `int` `N2)` ` ` `{` ` ` `float` `rate = (N2 - N1) * ` `100` `/ N1;` ` ` `return` `(` `int` `)rate;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{` ` ` `int` `N1 = ` `100` `, N2 = ` `120` `;` ` ` `System.out.println(Rate(N1, N2) + ` `" %"` `);` ` ` `}` `}` `// This code has been contributed by 29AjayKumar` |

## Python 3

`# Python 3 implementation of the approach` `# Function to return the` `# required rate percentage` `def` `Rate( N1, N2):` ` ` `rate ` `=` `(N2 ` `-` `N1) ` `*` `100` `/` `/` `(N1);` ` ` `return` `rate` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N1 ` `=` `100` ` ` `N2 ` `=` `120` ` ` `print` `(Rate(N1, N2) ,` `" %"` `)` `# This code is contributed by ChitraNayal ` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function to return the` ` ` `// required rate percentage` ` ` `static` `int` `Rate(` `int` `N1, ` `int` `N2)` ` ` `{` ` ` `float` `rate = (N2 - N1) * 100 / N1;` ` ` `return` `(` `int` `)rate;` ` ` `}` ` ` `// Driver code` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `N1 = 100, N2 = 120;` ` ` `Console.WriteLine(Rate(N1, N2) + ` `" %"` `);` ` ` `}` `}` `// This code has been contributed by ajit.` |

## PHP

`<?php` `// PHP implementation of the approach ` `// Function to return the ` `// required rate percentage ` `function` `Rate(` `$N1` `, ` `$N2` `) ` `{ ` ` ` `$rate` `= (` `$N2` `- ` `$N1` `) * 100 / ` `$N1` `; ` ` ` `return` `$rate` `; ` `} ` `// Driver code ` `$N1` `= 100;` `$N2` `= 120; ` `echo` `Rate(` `$N1` `, ` `$N2` `), ` `"%"` `; ` `// This code is contributed by AnkitRai01` `?>` |

## Javascript

`<script>` `// javascript implementation of the approach` ` ` `// Function to return the` ` ` `// required rate percentage` ` ` `function` `Rate(N1 , N2) {` ` ` `var` `rate = (N2 - N1) * 100 / N1;` ` ` `return` `parseInt( rate);` ` ` `}` ` ` `// Driver code` ` ` ` ` `var` `N1 = 100, N2 = 120;` ` ` `document.write(Rate(N1, N2) + ` `" %"` `);` `// This code contributed by Rajput-Ji ` `</script>` |

**Output:**

20 %

**Time Complexity:** O(1), as there is only basic arithmetic operation.

**Auxiliary Space:** O(1), as no extra space has been taken.