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Answer the questions on the basis of the information given below.

In a school, there are four chemical laboratories namely Lab 1, Lab 2, Lab 3 and Lab 4. There are only six types of acids that are available in these laboratories. The following table provides information about the number of bottles of each type of acid in each of these laboratories.

Common information image for Pie Charts, Data Interpretation:1673-1

Every bottle of acid in the chemical laboratories is further categorized on the basis of its capacity under one or the other of the five different categories namely ‘S’, ‘M’, ‘L’, ‘XL’ and ‘XXL’.
The following chart provides information about the number of bottles of acids in each of the mentioned categories (on the basis of its capacity) as a percentage of the total number of bottles of acids in these laboratories.

Common information image for Pie Charts, Data Interpretation:1673-2

Q.

Common Information Question: 4/5

Additional Information for questions 3 and 4:

All the bottles containing one or the other of the three acids namely Sulphuric, Nitric and Nitrous are in one or the other of the three categories S, M and L only. Also, the total number of bottles of Benzoic, Hydrochloric and Salicylic acid that are in one or the other of three categories S, M and L are ‘a’, ‘b’ and ‘c’ respectively.

If the value of $a$ is maximum possible, then the number of bottles of Benzoic acid in Lab 2 that do not belong to any of the three categories S, M and L cannot be less than:

 A.

76

 B.

75

 C.

78

 D.

77

 E.

79

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Solution:
Option(A) is correct

Given that all the bottles containing one or the other of the three acids namely Sulphuric, Nitric and Nitrous are in one or the other of the three categories S, M and L.

Total number of bottles containing one or the other of three acids namely Sulphuric, Nitric and Nitrous acid is:

$=384 + 367 + 402$

$= 1153$

Total number of bottles of acids in the three categories S, M and L:

$= 360 + 600 + 240$

$= 1200$

Difference:
$= 1200 - 1153$

$= 47$

$⇒ a + b + c = 47$

Maximum possible value of $a$ could be 47.

So, in order to minimize the number of bottles of Benzoic acid in Lab 2 that do not belong to any of the three categories S, M and L we need to maximize the number of bottles of Benzoic acid that belong to one or the other of three categories S, M and L.

So, required answer is:

$=123 - 47$

$= \textbf{76}$


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