You are given a container of X liters containing a mixture of wine and water. The mixture contains W% of water in it. How many liters of water must be added to increase the ratio of water to Y%?
The input includes 3 integers: X, W, and Y respectively.
The output should be in float format up to 2 decimal points.
Examples:
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 the 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.
By 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.
Below is the implementation of the above approach:
C++
#include <iostream>#include <iomanip>using namespace std;float findAmount(float X, float W, float Y) { return (X * (Y - W)) / (100 - Y);}int main() { float X = 100, W = 50, Y = 60; std::cout << "Water to be added = " << fixed << setprecision(2) << findAmount(X, W, Y); return 0;} |
C
// C program to find amount of water to// be added to achieve given target ratio.#include <stdio.h>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 = 100W = 50; Y = 60print("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 codepublic 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 |
Javascript
<script> // Javascript program to find amount of water to // be added to achieve given target ratio. function findAmount(X, W, Y) { return (X * (Y - W)) / (100 - Y); } let X = 100, W = 50, Y = 60; document.write("Water to be added = "+ findAmount(X, W, Y).toFixed(2));</script> |
PHP
<?php// PHP program to find amount of water to// be added to achieve given target ratio.function findAmount($X, $W, $Y){ return ($X * ($Y - $W)) / (100 - $Y);}// Driver Code$X = 100; $W = 50; $Y = 60;echo "Water to be added = " . findAmount($X, $W, $Y);// This code is contributed // by Akanksha Rai?> |
Water to be added = 25.00
Time Complexity: O(1)
Auxiliary Space: O(1)
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