CAT 2021 Slot 3DILR Question 1

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.

5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, detects the presence of I. If it is known that bottle A contains only P, what BEST can be concluded about the volume of I in bottle B?

Answer & solution

  • 10 ml or more

  • B

    Less than 1 ml

  • C

    10 ml

  • D

    1 ml

Solution

Since impurity is detected in the final mixture, hence the mixture has at least 10% impurities.

∴ % impurity in the mixture = 5×0%+5×P%5+5 ≥ 10

⇒ 5P ≥ 100

⇒ P ≥ 20%

∴ Volume of impurities in bottle B is at least 20% of 50 ml i.e., ≥ 10 ml.

Hence, option (a).

CAT 2021 Slot 3 DILR Q1: 5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, — Solution | TheCATExam