# Program to find the rate percentage from compound interest of consecutive years

Last Updated : 31 May, 2022

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

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## Javascript

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Output:

`20 %`

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

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

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