There are four bottles. Each bottle is known to contain only P or only I. They will be considered to be “collectively ready for despatch” if all of them contain only P. In minimum how many tests, is it possible to ascertain whether these four bottles are “collectively ready for despatch”?
Explanation:
The bottles contain only P or only I.
Let us mix equal quantities from each of these bottles and check for impurities.
Case 1: All bottles contain only P Concentration of I in final mixture = 0+0+0+04 = 0% ∴ Impurities will not be detected.
Case 2: 3 bottles contain P and 1 contains I Concentration of I in final mixture = 0+0+0+1004 = 25% ∴ Impurities will be detected.
Case 3: 2 bottles contain P and 2 contain I Concentration of I in final mixture = 0+0+100+1004 = 50% ∴ Impurities will be detected.
Case 4: 1 bottle contains P and 3 contain I Concentration of I in final mixture = 0+100+100+1004 = 75% ∴ Impurities will be detected.
Case 5: All bottles contain only I Concentration of I in final mixture = 100+100+100+1004 = 100% ∴ Impurities will be detected.
There is only one case where impurities are not detected and that is when all the bottles have only P.
⇒ If we mix equal quantities of all 4 bottles and test for impurities, if impurity is not detected then we can safely accept all 4 bottles.
Only one test is required is such a situation.
Hence, 1.
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