Practice Question - 55 (Maxima Minima) | 100 DILR Questions for CAT Preparation PDF Download

Ans: (d)
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 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.

Ans: 1
The bottles contain either P(pure) or I(impure). The possible cases here are :

1- (P, P, P, P), 2 - (P,P,P,I), 3 - (P,P,I,I), 4 - (P,I,I,I), 5 - (I,I,I,I).

In the first case if all the four solutions are pure then taking equal volumes of all the four bottles will get the result to dispatch or not to dispatch.

In the second case if 3 bottles are pure and one impure taking equal volumes of all four bottles and testings will confirm the impurity and hence cannot be dispatched.

In the third case if 2 bottles are pure and two are impure taking equal volumes of all four bottles and testing will confirm the impurity and hence cannot be dispatched.

In the fourth case when only one bottle is pure taking equal volumes of all four bottles will confirm the impurity and hence cannot be dispatched.

In the fifth case if all four bottles are impure taking equal volumes of the four bottles will confirm the impurity and hence cannot be dispatched.

In all the cases a single test is enough to determine if the lot is to be dispatched or not.

Ans: 2
The percentage concentration of the impure solution is 80 percent.

When equal volumes of all four solutions are mixed. 

Considering 10 ml of each we have impurity to be 2ml/40ml. The impurity concentration is less than 10 percent and hence cannot be recognized.

Similarly when equal volumes of one impure and 2 pure solutions are mixed. 

The impurity in the solution is 2ml/30ml which is less than 10 percent and hence cannot be recognized.

Hence for detecting the impure solution we must use equal volumes of 2 solutions at a time.

Considering the three pure solutions to be P and the impure solution to be I. 

P, P, P, I.

Considering equal volumes of solution from the bottle one bottle of P, and I. Testing this would recognize the impurity. 

After this consider one bottle among the other 2 P bottles which are left and test this with one among the previously tested P, I.

If the one considered is I it will detect the impurity and confirms the bottle to be I.

If the one considered is P it will fail to detect the impurity and hence the other bottle will be I.

Hence a minimum of two tests are required to identify the bottle with the impurity.

Ans: (d)

The bottles could possibly be :

Case - 1 Pure, Impure, Impure, Impure.

Case-2, Pure , Pure, Impure, Impure.

Since the concentration in the impure bottle is 85 percent.

In case 1 when equal volumes from all the bottles are considered and mixed. The test result detects the impurity..

Since the overall concentration of impurity is greater than 10 percent.

Considering 10 ml from all four bottles.

The impure concentration is 4.5ml/40ml which is greater than 10.(15ml * 3 = 4.5ml)  (Impurity is detected)

For case 2 when all four bottles are considered. The case here has 2 pure and 2 impure bottles.

When equal volumes from all four bottles are mixed. The resultant concentration of impurity when 10 ml from each of the four solutions is considered :

The impure concentration is 3ml/40ml which is less than 10 percent.. (1.5ml * 2 = 3ml). (Impurity is not detected.)

Hence in one possibility the impurity is detected and not detected in the other case. A single test is enough based on the results of which the number of pure and the number of impure bottles can be identified.