# Times required by Simple interest for the Principal to become Y times itself

Given that a certain amount of money becomes T1 times itself in N1 years. The task is to find the number of years i.e. N2 so that the amount becomes T2 times itself at the same rate of simple interest.

Examples:

Input: T1 = 5, N1 = 7, T2 = 25
Output: 42

Input: T1 = 3, N1 = 5, T2 = 6
Output: 12.5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Let us consider the 1st Example where T1 = 5, N1 = 7, T2 = 25
Now, Let P principal becomes 5P i.e (T1 * P) then Simple interest received is 4P.
(As S.I = Amount – P)

Now, in the second case, P has become 25P i.e (T2 * P) then simple interest received is 24P.

Now if we received 4P interest in N1 i.e 7 years then we will get an interest of 24P
in 7 * 6 years i.e in 42 years.

Formula: Below is the implementation of the above approach:

## C++

 `// C++ implementaion of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the no. of years ` `float` `noOfYears(``int` `t1, ``int` `n1, ``int` `t2) ` `{ ` `    ``float` `years = ((t2 - 1) * n1 / (``float``)(t1 - 1)); ` ` `  `    ``return` `years; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `T1 = 3, N1 = 5, T2 = 6; ` ` `  `    ``cout << noOfYears(T1, N1, T2); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementaion of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the no. of years ` `static` `float` `noOfYears(``int` `t1, ``int` `n1, ``int` `t2) ` `{ ` `    ``float` `years = ((t2 - ``1``) * n1 / (``float``)(t1 - ``1``)); ` ` `  `    ``return` `years; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `T1 = ``3``, N1 = ``5``, T2 = ``6``; ` ` `  `    ``System.out.println(noOfYears(T1, N1, T2)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the no. of years  ` `def` `noOfYears(t1, n1, t2):  ` ` `  `    ``years ``=` `(t2 ``-` `1``) ``*` `n1 ``/` `(t1 ``-` `1``)  ` `    ``return` `years  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``T1, N1, T2 ``=` `3``, ``5``, ``6` `    ``print``(noOfYears(T1, N1, T2)) ` `     `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# implementation for above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the no. of years ` `static` `float` `noOfYears(``int` `t1, ``int` `n1, ``int` `t2) ` `{ ` `    ``float` `years = ((t2 - 1) * n1 / (``float``)(t1 - 1)); ` ` `  `    ``return` `years; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `T1 = 3, N1 = 5, T2 = 6; ` ` `  `    ``Console.WriteLine(noOfYears(T1, N1, T2)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## PHP

 ` `

Output:

```12.5
```

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