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++ implementation of the approach #include <bits/stdc++.h> 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 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 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# 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 // PHP implementation of the approach // Function to return the // required rate percentage function Rate( $N1 , $N2 )
{ $rate = ( $N2 - $N1 ) * 100 / $N1 ;
return $rate ;
} // Driver code $N1 = 100;
$N2 = 120;
echo Rate( $N1 , $N2 ), "%" ;
// This code is contributed by AnkitRai01 ?> |
<script> // javascript implementation of the approach // Function to return the
// required rate percentage
function Rate(N1 , N2) {
var rate = (N2 - N1) * 100 / N1;
return parseInt( rate);
}
// Driver code
var N1 = 100, N2 = 120;
document.write(Rate(N1, N2) + " %" );
// This code contributed by Rajput-Ji </script> |
20 %
Time Complexity: O(1), as there is only basic arithmetic operation.
Auxiliary Space: O(1), as no extra space has been taken.