查找在给定混合物中要达到目标比例所要添加的量

您将得到一个X升的容器，其中装有葡萄酒和水的混合物。该混合物中包含W％的水。必须添加多少升水才能使水的比例达到Y％?
输入分别包含3个整数X，W和Y。
输出应为浮点格式，最多2个小数点。
例子：

Input : X = 125, W = 20, Y = 25
Output : 8.33 liters
20% of 125 is 25. If we add 8.33 liters, we get 33.33 which is 25% of 133.33.

Input : X = 100, W = 50, Y = 60
Output : 25

Let the amount of water to be added be A liters.
So, the new amount of mixture = (X + A) liters
And the amount of water in the mixture = (old amount + A) = ((W % of X ) + A )
Also, the amount of water in the mixture = new percentage of water in new mixture = Y % of (X + A)
Now, we can write the expression as
———————————
Y % of ( X + A) = W % of X + A
———————————-
Since, both denote the amount of water.
On simplifying this expression, we will get

A = [X * (Y – W)] / [100 – Y]

Illustration :
X = 125, W = 20% and Y = 25%;

So, for the given question, the amount of water to be added = (125 * (25 – 20)) / (100 – 25) = 8.33 liters

下面是上述方法的实现：

C

// C program to find amount of water to
// be added to achieve given target ratio.
#include 
  
float findAmount(float X, float W, float Y)
{
    return (X * (Y - W)) / (100 - Y);
}
  
int main()
{
    float X = 100, W = 50, Y = 60;
    printf("Water to be added = %.2f ", 
                 findAmount(X, W, Y));
    return 0;
}

Java

// Java program to find amount of water to
// be added to achieve given target ratio.
  
public class GFG {
      
    static float findAmount(float X, float W, float Y)
    {
        return (X * (Y - W)) / (100 - Y);
    }
  
      
    // Driver code
    public static void main(String args[])
    {
           float X = 100, W = 50, Y = 60;
           System.out.println("Water to be added = "+ findAmount(X, W, Y));
  
  
    }
    // This code is contributed by ANKITRAI1
}

Python3

# Python3 program to find amount 
# of water to be added to achieve 
# given target ratio. 
def findAmount(X, W, Y):
      
    return (X * (Y - W) / (100 - Y))
  
X = 100
W = 50; Y = 60
print("Water to be added",
       findAmount(X, W, Y))
  
# This code is contributed
# by Shrikant13

C#

// C# program to find amount of water to
// be added to achieve given target ratio.
using System;
class GFG
{
  
public static double findAmount(double X, 
                                double W, 
                                double Y)
{
    return (X * (Y - W)) / (100 - Y);
}
  
// Driver code
public static void Main()
{
    double X = 100, W = 50, Y = 60;
    Console.WriteLine("Water to be added = {0}", 
                           findAmount(X, W, Y));
}
}
  
// This code is contributed by Soumik

输出：

Water to be added = 25.00