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CAT 2021 Slot 3 — DILR Question 2

Mixed PracticeEasy
Passage / Data

Answer the next 4 questions based on the information given 

Each of the bottles mentioned in this question contains 50 ml of liquid. The liquid in any bottle can be 100% pure content (P) or can have certain amount of impurity (I). Visually it is not possible to distinguish between P and I. There is a testing device which detects impurity, as long as the percentage of impurity in the content tested is 10% or more.

For example, suppose bottle 1 contains only P, and bottle 2 contains 80% P and 20% I. If content from bottle 1 is tested, it will be found out that it contains only P. If content of bottle 2 is tested, the test will reveal that it contains some amount of I. If 10 ml of content from bottle 1 is mixed with 20 ml content from bottle 2, the test will show that the mixture has impurity, and hence we can conclude that at least one of the two bottles has I. However, if 10 ml of content from bottle 1 is mixed with 5 ml of content from bottle 2. the test will not detect any impurity in the resultant mixture.

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”?

Answer & solution

Correct answer: 1

Answer: 1

Solution

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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CAT 2021 Slot 3 DILR Q2: There are four bottles. Each bottle is known to contain only P or only I. They will be considered to be &ldquo — Solution | TheCATExam