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

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++

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// C++ implementaion of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 */

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PHP

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<?php
// PHP implementation for above approach
      
// Function to return the no. of years
function noOfYears($t1, $n1, $t2)
{
    $years = (($t2 - 1) * $n1 / ($t1 - 1));
  
    return $years;
}
  
// Driver code
$T1 = 3;
$N1 = 5;
$T2 = 6;
  
print(noOfYears($T1, $N1, $T2));
  
// This code contributed by mits
?>

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Output:

12.5


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