Program to find the count of coins of each type from the given ratio

Given the total number of Rupees in a bag and the ratio of coins. The bag contains only 1 Rs, 50 paise, 25 paise coins in X, Y, Z, ratio. The task is to determine the number of 1 Rs coins, 50 Paise coins, and 25 paise coins So that after summation of all these we again get the total rupees given.
Examples:

Input: totalRupees = 1800, X = 1, Y = 2, Z = 4
Output:  1 rupees coins = 600
         50 paisa coins = 1200
         25 paisa coins = 2400

Input: totalRupees = 2500, X = 2, Y = 4, Z = 2
Output:  1 rupees coins = 555
         50 paisa coins = 1110
         25 paisa coins = 2220

Approach:

Let the ratio in which 1 Rs, 50 paise and 25 paise coin is divided is 1:2:4 
Now, 
1 Rs coins in a bag is 1x. 
50 paise coins in a bag is 2x. 
25 paise coins in a bag is 4x.
Now convert these paise into Rupees. 
So, 
x coins of 1 Rs each, the total value is x rupees. 
2x coins of 50 paise i.e (1 / 2) rupees each, the total value is x rupees. 
4x coins of 25 paise i.e (1 / 4) rupees each, the total is x rupees.
Therefore,
3x = 1800 
x = 600
1 rupee coins = 600 * 1 = 600 
50 paisa coins = 600 * 2 = 1200 
25 paisa coins = 600 * 4 = 2400

Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// function to calculate coin.
int coin(int totalRupees, int X, int Y, int Z)
{

    float one = 0, fifty = 0, twentyfive = 0,
          result = 0, total = 0;

    // Converting each of them in rupees.
    // As we are given totalRupees = 1800
    one = X * 1;
    fifty = ((Y * 1) / 2.0);
    twentyfive = ((Z * 1) / 4.0);

    total = one + fifty + twentyfive;

    result = ((totalRupees) / total);

    return result;
}

// Driver code
int main()
{
    int totalRupees = 1800;
    int X = 1, Y = 2, Z = 4;

    int Rupees = coin(totalRupees, X, Y, Z);

    cout << "1 rupee coins = " << Rupees * 1 << endl;
    cout << "50 paisa coins = " << Rupees * 2 << endl;
    cout << "25 paisa coins = " << Rupees * 4 << endl;

    return 0;
}

Java

// Java implementation of above approach

import java.io.*;

class GFG {

// function to calculate coin.
 static int coin(int totalRupees, int X, int Y, int Z)
{

    float one = 0, fifty = 0, twentyfive = 0,
        result = 0, total = 0;

    // Converting each of them in rupees.
    // As we are given totalRupees = 1800
    one = X * 1;
    fifty = ((Y * 1) / 2);
    twentyfive = ((Z * 1) / 4);

    total = one + fifty + twentyfive;

    result = ((totalRupees) / total);

    return (int)result;
}

// Driver code

    public static void main (String[] args) {

    int totalRupees = 1800;
    int X = 1, Y = 2, Z = 4;

    int Rupees = coin(totalRupees, X, Y, Z);

    System.out.println( "1 rupee coins = " + Rupees * 1);
    System.out.println( "50 paisa coins = " + Rupees * 2);
    System.out.println( "25 paisa coins = " + Rupees * 4);
    }
}
//This code is contributed by  inder_verma.

Python3

# Python3 implementation of above approach

# function to calculate coin.
def coin(totalRupees, X, Y, Z):

    # Converting each of them in rupees.
    # As we are given totalRupees = 1800
    one = X * 1
    fifty = ((Y * 1) / 2.0)
    twentyfive = ((Z * 1) / 4.0)
    total = one + fifty + twentyfive
    result = ((totalRupees) / total)

    return int(result)

# Driver code
if __name__=='__main__':
    totalRupees = 1800
    X, Y, Z = 1, 2, 4

    Rupees = coin(totalRupees, X, Y, Z)

    print("1 rupee coins = ", Rupees * 1)
    print("50 paisa coins = ", Rupees * 2)
    print("25 paisa coins = ", Rupees * 4)

# This code is contributed by
# Sanjit_Prasad

C#

// C# implementation of above approach
using System;

class GFG
{

// function to calculate coin.
static int coin(int totalRupees, int X,
                int Y, int Z)
{

    float one = 0, fifty = 0, twentyfive = 0,
          result = 0, total = 0;

    // Converting each of them in rupees.
    // As we are given totalRupees = 1800
    one = X * 1;
    fifty = ((Y * 1) / 2);
    twentyfive = ((Z * 1) / 4);

    total = one + fifty + twentyfive;

    result = ((totalRupees) / total);

    return (int)result;
}

// Driver code
public static void Main ()
{
    int totalRupees = 1800;
    int X = 1, Y = 2, Z = 4;

    int Rupees = coin(totalRupees, X, Y, Z);

    Console.WriteLine( "1 rupee coins = " + Rupees * 1);
    Console.WriteLine( "50 paisa coins = " + Rupees * 2);
    Console.WriteLine( "25 paisa coins = " + Rupees * 4);
}
}

// This code is contributed by inder_verma

PHP

<?php
// PHP implementation of above approach

//function to calculate coin
function coin($totalRupees, $X, $Y, $Z)
{
    $one = 0;
    $fifty = 0;
    $twentyfive = 0;
    $result = 0;
    $total = 0;

    // Converting each of them in rupees.
    // As we are given totalRupees = 1800
    $one = $X * 1;
    $fifty = (($Y * 1) / 2.0);
    $twentyfive = (($Z * 1) / 4.0);

    $total = $one + $fifty + $twentyfive;

    $result = (($totalRupees) / $total);

    return $result;

}

// Driver Code
$totalRupees = 1800;
$X = 1;
$Y = 2;
$Z = 4;
$Rupees = coin($totalRupees, $X, $Y, $Z);
echo "1 rupee coins = ", $Rupees * 1, "\n";
echo "50 paisa coins = ", $Rupees * 2, "\n";
echo "25 paisa coins = ", $Rupees * 4, "\n";

// This code is contributed
// by Shashank_Sharma.
?>

Javascript

<script>

// Javascript implementation of above approach

// function to calculate coin.
function coin(totalRupees, X, Y, Z)
{

    var one = 0, fifty = 0, twentyfive = 0,
        result = 0, total = 0;

    // Converting each of them in rupees.
    // As we are given totalRupees = 1800
    one = X * 1;
    fifty = ((Y * 1) / 2.0);
    twentyfive = ((Z * 1) / 4.0);

    total = one + fifty + twentyfive;

    result = ((totalRupees) / total);

    return result;
}

// Driver code
var totalRupees = 1800;
var X = 1, Y = 2, Z = 4;
var Rupees = coin(totalRupees, X, Y, Z);
document.write( "1 rupee coins = " + Rupees * 1 + "<br>");
document.write( "50 paisa coins = " + Rupees * 2 + "<br>");
document.write( "25 paisa coins = " + Rupees * 4 + "<br>");

</script>

Output

1 rupee coins = 600
50 paisa coins = 1200
25 paisa coins = 2400

Time Complexity: O(1)
Auxiliary Space: O(1)