Bunuel wrote:
\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?
(1) \(X = 80\)
(2) \(Y = 2\)
Simple Logical Explanation:
1) X = 80 G. This means that 80G of water was added to 80G of strong acid solution.
In this scenario, the proportion of acid in the final solution will reduce by half i.e. (If initially the proportion was 80% - it will become 40%).
However, Initial proportion cannot be determined.
Not Suff.
2) Y = 2. This means that the proportion of acid in the final solution reduces by half.
For this to happen exactly the same amount of water must be added i.e. 80G
However, Initial proportion cannot be determined.
Not Suff.
1 & 2 Together:
Not Sufficient. Just proves that the initial concentration was halved. Cannot determine initial concentration.
OA: E