Two solutions X and Y contain alcohol and water. Solution X contains 52% alcohol, while solution Y contains 28% alcohol. They are mixed so that the final mixture contains 58% water. Find by what percentage the quantity of solution X is more than the quantity of solution Y in the final mixture.

ATQ, Solution X contains 52% alcohol. Solution Y contains 28% alcohol. Final mixture contains 58% water, so it contains 42% alcohol. Using alligation: Ratio of solution X to solution Y = (42 − 28) : (52 − 42). Ratio of solution X to solution Y = 14 : 10. Ratio of solution X to solution Y = 7 : 5. So, quantity of solution X = 7 parts and quantity of solution Y = 5 parts. Here, X is more than Y, but we are asked by what percentage Y is more than X, so we reverse the comparison. We first find by what percentage X is more than Y: Difference = 7 − 5 = 2 parts. Percentage by which X is more than Y = 2 × 100/5 = 40%.