Find amount to be added to achieve target ratio in a given mixture
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.
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]
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:
Water to be added = 25.00
Time Complexity: O(1)
Auxiliary Space: O(1)