Program to solve the Alligation Problem

Write a program to find the ratio in which a shopkeeper will mix two types of rice worth Rs. $X$ kg and Rs. $Y$ kg, so that the average cost of the mixture is Rs. $Z$ kg.

Examples:

Input : X = 50, Y = 70, Z = 65
Output : Ratio = 1:3

Input : X = 1000, Y = 2000, Z = 1400
Output : Ratio = 3:2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

According to Alligation rule, the ratio of the weights of two items mixed will be inversely proportional to the deviation of attributes of these two items from the average attribute of the resultant mixture.

w1 / w2 = (d - m) / (m - c)

Below program illustrate the above approach:

C++

#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the ratio of two mixtures 
void alligation(float x, float y, float m) 
{ 
    // Find the cheaper among x and y 
    float c = (x <= y) ? x : y; 
    // Find the dearer among x and y 
    float d = (x >= y) ? x : y; 
  
    // Find ratio r1:r2 
    int r1 = d - m; 
    int r2 = m - c; 
  
    // Convert the ration into simpler form 
    int gcd = __gcd(r1, r2); 
  
    cout << r1 / gcd << ":" << r2 / gcd; 
} 
  
// Driver code 
int main() 
{ 
    float x, y, z; 
    x = 50; 
    y = 70; 
    z = 65; 
  
    alligation(x, y, z); 
  
    return 0; 
} 

Java

// Java implementation of the  
// above approach. 
import java.util.*; 
  
class solution 
{ 
  
static float __gcd(float a, float b) 
{ 
    float dividend,divisor; 
      
    // a is greater or equal to b 
    if(a>=b) 
    dividend = a; 
    else
    dividend = b; 
      
    // b is greater or equal to a 
    if(a<=b) 
    divisor = a; 
    else
    divisor = b; 
      
while(divisor>0) 
{ 
float remainder = dividend % divisor; 
dividend = divisor; 
divisor = remainder; 
  
} 
return dividend; 
} 
// Function to find the ratio of two mixtures 
static void alligation(float x, float y, float m) 
{ 
    // Find the cheaper among x and y 
    float c; 
    if (x <= y) 
    c = x; 
    else
    c = y; 
    // Find the dearer among x and y 
    float d ; 
    if (x >= y) 
    d = x; 
    else
    d = y; 
  
    // Find ratio r1:r2 
    float r1 = d - m; 
    float r2 = m - c; 
  
    // Convert the ration into simpler form 
    float gcd = __gcd(r1, r2); 
  
    System.out.println((int)(r1 / gcd)+":"+(int)(r2 / gcd)); 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    float x, y, z; 
    x = 50; 
    y = 70; 
    z = 65; 
  
    alligation(x, y, z); 
} 
} 
  
// This code is contribuuted by 
// Shashank_sharma 

Python3

# Python 3 implementation of the  
# above approach. 
from math import gcd 
  
# Function to find the ratio  
# of two mixtures 
def alligation(x, y, m): 
      
    # Find the cheaper among x and y 
    if (x <= y): 
        c = x 
    else: 
        c = y 
          
    # Find the dearer among x and y 
    if (x >= y): 
        d = x 
    else: 
        d = y 
  
    # Find ratio r1:r2 
    r1 = d - m 
    r2 = m - c 
  
    # Convert the ration into simpler form 
    __gcd = gcd(r1, r2) 
  
    print(r1 // __gcd, ":", r2 // __gcd) 
  
# Driver code 
if __name__ == '__main__': 
    x = 50
    y = 70
    z = 65
  
    alligation(x, y, z) 
  
# This code is contributed by 
# Surendra_Gangwar 

C#

// C# implementation of the  
// above approach. 
using System; 
  
class GFG 
{ 
    // Recursive function to return  
    // gcd of a and b  
    static int __gcd(int a, int b)  
    {  
        // Everything divides 0  
        if (a == 0)  
            return b;  
        if (b == 0)  
            return a;  
          
        // base case  
        if (a == b)  
            return a;  
          
        // a is greater  
        if (a > b)  
            return __gcd(a - b, b);  
        return __gcd(a, b - a);  
    }  
      
    // Function to find the ratio of  
    // two mixtures 
    static void alligation(float x,  
                           float y, float m) 
    { 
        // Find the cheaper among x and y 
        float c = (x <= y) ? x : y; 
          
        // Find the dearer among x and y 
        float d = (x >= y) ? x : y; 
      
        // Find ratio r1:r2 
        int r1 = (int)(d - m); 
        int r2 = (int)(m - c); 
      
        // Convert the ration into  
        // simpler form 
        int gcd = __gcd(r1, r2); 
      
        Console.Write(r1 / gcd + ":" +  
                      r2 / gcd); 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        float x, y, z; 
        x = 50; 
        y = 70; 
        z = 65; 
      
        alligation(x, y, z); 
    } 
} 
  
// This code is contributed  
// by Akanksha Rai 

PHP

<?php 
// PHP implementation of the  
// above approach. 
function __gcd($a, $b) 
{ 
    $dividend; $divisor; 
      
    // a is greater or equal to b 
    if($a >= $b) 
        $dividend = $a; 
    else
        $dividend = $b; 
      
    // b is greater or equal to a 
    if($a <= $b) 
        $divisor = $a; 
    else
        $divisor = $b; 
      
    while($divisor > 0) 
    { 
        $remainder = $dividend % $divisor; 
        $dividend = $divisor; 
        $divisor = $remainder; 
    } 
    return $dividend; 
} 
  
// Function to find the ratio of  
// two mixtures 
function alligation($x, $y, $m) 
{ 
    // Find the cheaper among x and y 
    if ($x <= $y) 
        $c = $x; 
    else
        $c = $y; 
      
    // Find the dearer among x and y 
    if ($x >= $y) 
        $d = $x; 
    else
        $d = $y; 
  
    // Find ratio r1:r2 
    $r1 = $d - $m; 
    $r2 = $m - $c; 
  
    // Convert the ration into 
    // simpler form 
    $gcd = __gcd($r1, $r2); 
  
    echo (int)($r1 / $gcd) . ":" .  
         (int)($r2 / $gcd); 
} 
  
// Driver code 
$x = 50; 
$y = 70; 
$z = 65; 
  
alligation($x, $y, $z); 
  
// This code is contributed by 
// Mukul Singh 
?> 

Output:

1:3

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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