Logical Reasoning

Math Puzzles

Hard

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 2is tested, the test will reveal that it contains some amount of I. If 10 ml of content from bottle 1is 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?

Entered answer:

✅ Correct Answer: 2

Slide 1/2

Understanding the set :-

This is a mixture and allegations set :

Mixtures and allegations is mixing two things with different prices or strengths to get a mixture with a middle value.

For eg -

Milk costs ₹6 per litre, water is free (₹0).

We want a mixture costing ₹4 per litre.

Step 1: Values → High = 6, Low = 0, Mean = 4

Step 2: Subtract diagonally:

6 - 4 = 2

4 - 0 = 4

Ratio = 2 : 4 = 1 : 2

So, milk : water = 1 : 2

In this question there are 50 ml bottles

also, iquid in any bottle can be 100% pure content (P) or can have certain amount of impurity (I).

There is a testing device which detects impurity, but that detects, only if impurity is more then 10%.

Now, the set will be solved according to each question separately.

Since three bottles contains only P

and one contains 80% P and 20%

Taking equal quantity from each bottle say 10 ml and mixing them I% = $2/40 \times 100 = 5\% < 10\%$ ,

so only one test is not sufficient to detect the bottle Let the bottles be B1, B2, B3 and B4 in any order Now consider any two bottles (say B1 and B2) and take equal quantity from each bottle say 10 ml and mix them,

Case I, if I% = 0,

then both the bottles are only P

and now take any one bottle from either B1 or B2 and mix with either B3 or B4 if I% is still = 0,

then the remaining bottle contains 20% I

and if I% = $(0 + 20)/2 = 10$ ,

then the newly mixed bottle contains 20% I

Case II, if I% = 10,

then remaining B3 and B4 are only P,

now take either B1 or B2 and mix with B3 or B4 If I% = 0, then the other bottle among B1 or B2 contains 20% I And if I% is still = 10,

then the bottle taken among B1 or B2 contains 20% I

Hence, minimum of 2 tests required in either of the cases

Question 17

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?

Question 18

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”?.

Question 20

There are four bottles. It is known that either one or two of these bottles contain(s)only P, while the remaining ones contain 85% P and 15% I. What is the minimum number of tests required to ascertain the exact number of bottles containing only P?

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?

Solution

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