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:Â
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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
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Approach:Â
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Let the ratio in which 1 Rs, 50 paise and 25 paise coin is divided is 1:2:4Â
Now,Â
1 Rs coins in a bag is 1x.Â
50 paise coins in a bag is 2x.Â
25 paise coins in a bag is 4x.
Now convert these paise into Rupees.Â
So,Â
x coins of 1 Rs each, the total value is x rupees.Â
2x coins of 50 paise i.e (1 / 2) rupees each, the total value is x rupees.Â
4x coins of 25 paise i.e (1 / 4) rupees each, the total is x rupees.
Therefore,
3x = 1800Â
x = 600
1 rupee coins = 600 * 1 = 600Â
50 paisa coins = 600 * 2 = 1200Â
25 paisa coins = 600 * 4 = 2400
Â
Below is the implementation of the above approach:Â
Â
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;Â
// function to calculate coin.int coin(int totalRupees, int X, int Y, int Z){Â
    float one = 0, fifty = 0, twentyfive = 0,          result = 0, total = 0;Â
    // 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);Â
    total = one + fifty + twentyfive;Â
    result = ((totalRupees) / total);Â
    return result;}Â
// Driver codeint main(){Â Â Â Â int totalRupees = 1800;Â Â Â Â int X = 1, Y = 2, Z = 4;Â
    int Rupees = coin(totalRupees, X, Y, Z);Â
    cout << "1 rupee coins = " << Rupees * 1 << endl;    cout << "50 paisa coins = " << Rupees * 2 << endl;    cout << "25 paisa coins = " << Rupees * 4 << endl;Â
    return 0;} |
Java
// Java implementation of above approach Â
import java.io.*;Â
class GFG {Â Â // function to calculate coin. Â static int coin(int totalRupees, int X, int Y, int Z) { Â
    float one = 0, fifty = 0, twentyfive = 0,         result = 0, total = 0; Â
    // Converting each of them in rupees.     // As we are given totalRupees = 1800     one = X * 1;     fifty = ((Y * 1) / 2);     twentyfive = ((Z * 1) / 4); Â
    total = one + fifty + twentyfive; Â
    result = ((totalRupees) / total); Â
    return (int)result; } Â
// Driver code Â
    public static void main (String[] args) {             int totalRupees = 1800;     int X = 1, Y = 2, Z = 4; Â
    int Rupees = coin(totalRupees, X, Y, Z); Â
    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 approachÂ
# function to calculate coin.def coin(totalRupees, X, Y, Z):Â
    # 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)Â
    return int(result)Â
# Driver codeif __name__=='__main__':Â Â Â Â totalRupees = 1800Â Â Â Â X, Y, Z = 1, 2, 4Â
    Rupees = coin(totalRupees, X, Y, Z)Â
    print("1 rupee coins = ", Rupees * 1)    print("50 paisa coins = ", Rupees * 2)    print("25 paisa coins = ", Rupees * 4)Â
# This code is contributed by# Sanjit_Prasad |
C#
// C# implementation of above approach using System;Â
class GFG {Â
// function to calculate coin. static int coin(int totalRupees, int X, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â int Y, int Z) { Â
    float one = 0, fifty = 0, twentyfive = 0,           result = 0, total = 0; Â
    // Converting each of them in rupees.     // As we are given totalRupees = 1800     one = X * 1;     fifty = ((Y * 1) / 2);     twentyfive = ((Z * 1) / 4); Â
    total = one + fifty + twentyfive; Â
    result = ((totalRupees) / total); Â
    return (int)result; } Â
// Driver code public static void Main () {Â Â Â Â int totalRupees = 1800; Â Â Â Â int X = 1, Y = 2, Z = 4; Â Â Â Â Â Â Â Â Â int Rupees = coin(totalRupees, X, Y, Z); Â Â Â Â Â Â Â Â Â Console.WriteLine( "1 rupee coins = " + Rupees * 1); Â Â Â Â Console.WriteLine( "50 paisa coins = " + Rupees * 2); Â Â Â Â Console.WriteLine( "25 paisa coins = " + Rupees * 4); }}Â
// This code is contributed by inder_verma |
PHP
<?php // PHP implementation of above approachÂ
//function to calculate coinfunction 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);Â
    $total = $one + $fifty + $twentyfive;Â
    $result = (($totalRupees) / $total);Â
    return $result;     } Â
// 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";Â
// This code is contributed // by Shashank_Sharma. ?> |
Javascript
<script>Â
// Javascript implementation of above approachÂ
// function to calculate coin.function coin(totalRupees, X, Y, Z){Â
    var one = 0, fifty = 0, twentyfive = 0,        result = 0, total = 0;Â
    // 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);Â
    total = one + fifty + twentyfive;Â
    result = ((totalRupees) / total);Â
    return result;}Â
// Driver codevar 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>");Â
</script> |
Output
1 rupee coins = 600 50 paisa coins = 1200 25 paisa coins = 2400
Time Complexity: O(1)
Auxiliary Space: O(1)
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