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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# 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 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 */



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?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 


?> 



















                        chevron_right







                        filter_none











Output:


12.5







Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Naman_Garg
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.