Discussion

Explanation:

Let the bottles be A, B, C and D.

We first mix equal quantities of A and B and test the mixture.

Case 1: If any one of A or B contains 20% impurities, test will detect the impurity.
Then we check A for impurity. If impurity is detected then A contains I else B contains I.
∴ 2 tests are required to detect I.

Case 2: If none of A or B contains impurities, test will not detect the impurity.
This means one of C or D contains I.
Then we check C for impurity. If impurity is detected then C contains I else D contains I.
∴ 2 tests are required to detect I.

In both cases 2 tests are required to detect I.

Hence, 2.

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