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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.

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?

Solution

✅ Correct Option: 1
Slide 1/2

Understanding the set :-

  1. 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

  1. 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).

  1. 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.

Given that each of the bottles contains a volume of 50 ml each.

If 5 ml from bottle A which contains only P is mixed with 5 ml of bottle B and in the resultant mixture the presence of I was detected.

In order to detect the presence of I in this, there must be at least 10% impurity in the 10 ml which is equivalent to 1 ml. This must be from bottle B.

Hence 5 ml of solution from B must contain at at least 1ml of impurity and since bottle B is of a total volume of 50 ml. This must contain at least 10 ml of impurity.

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