CAT 2021 Slot 3 · DILR · Question 3

CAT 2021 Slot 3 previous year question on Caselet DI, with the answer and a step-by-step solution.

Caselet DIData InterpretationType in the answer (TITA)Medium

Set 1: Testing Bottles for Impurity — this question belongs to a set. Read the full set with all 4 questions →

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. It is known that three of these bottles contain only P, while the remaining one contains 80% P and 20% I. What is the minimum number of tests required to definitely identify the bottle containing some amount of I?

Answer & solution

Answer: 2

Solving the set

Key idea: the device detects impurity only when it is at least 10% of the tested liquid, so by choosing how much of each bottle to mix, one test can answer a yes/no question about several bottles at once.

Question	Situation	Best test plan	Tests
Volume of I in B	5 ml A (pure) + 5 ml B shows I	Impurity ≥ 10% of 10 ml ⇒ ≥ 1 ml I in 5 ml of B ⇒ ≥ 20% of B ⇒ ≥ 10 ml of I in B	—
All pure?	Each bottle is only P or only I	Mix equal amounts of all four: any I bottle makes ≥ 25% impurity	1
Find the 20% bottle	Three pure, one 20% I	Mix two bottles 1 : 1 (10% if the impure one is there); then test one bottle of the indicated pair	2
One or two pure?	Rest 15% I	Mix equal amounts of all four: 3 impure ⇒ 11.25% (detected), 2 impure ⇒ 7.5% (not detected)	1

This question

A single test has only two outcomes, so it cannot pick out one of four bottles. Two tests suffice: test bottles 1 + 2 mixed 1 : 1 (impurity 10% if one of them is the bad bottle), then test one bottle of the indicated pair.

Hence, the answer is 2.

CAT 2021 Slot 3 · DILR · Question 3

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