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++
#include <bits/stdc++.h>
using namespace std;
float noOfYears( int t1, int n1, int t2)
{
float years = ((t2 - 1) * n1 / ( float )(t1 - 1));
return years;
}
int main()
{
int T1 = 3, N1 = 5, T2 = 6;
cout << noOfYears(T1, N1, T2);
return 0;
}
|
Java
class GFG
{
static float noOfYears( int t1, int n1, int t2)
{
float years = ((t2 - 1 ) * n1 / ( float )(t1 - 1 ));
return years;
}
public static void main(String[] args)
{
int T1 = 3 , N1 = 5 , T2 = 6 ;
System.out.println(noOfYears(T1, N1, T2));
}
}
|
Python3
def noOfYears(t1, n1, t2):
years = (t2 - 1 ) * n1 / (t1 - 1 )
return years
if __name__ = = "__main__" :
T1, N1, T2 = 3 , 5 , 6
print (noOfYears(T1, N1, T2))
|
C#
using System;
class GFG
{
static float noOfYears( int t1, int n1, int t2)
{
float years = ((t2 - 1) * n1 / ( float )(t1 - 1));
return years;
}
public static void Main(String[] args)
{
int T1 = 3, N1 = 5, T2 = 6;
Console.WriteLine(noOfYears(T1, N1, T2));
}
}
|
PHP
<?php
function noOfYears( $t1 , $n1 , $t2 )
{
$years = (( $t2 - 1) * $n1 / ( $t1 - 1));
return $years ;
}
$T1 = 3;
$N1 = 5;
$T2 = 6;
print (noOfYears( $T1 , $N1 , $T2 ));
?>
|
Javascript
<script>
function noOfYears(t1, n1, t2)
{
var years = ((t2 - 1) * n1 / (t1 - 1));
return years;
}
var T1 = 3, N1 = 5, T2 = 6;
document.write( noOfYears(T1, N1, T2));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
09 Jun, 2022
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