See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let equal masses be m g. Then amount of ‘X’ left out 30/10 m 1 10 10 2         3 1 m 2        and amount of ‘Y’ left out 30/t m 1 20 20 2         30/t 1 m 2        Then 3 30/t 1 m mass of 'X' 2 mass of 'Y' 1 m 2                                          3 3 30/t 30/t 1 m 4 1 1 1 2 1 4 2 2 1 m 2 5 30/t 1 1 2 2                t = 6 days.