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Common Information

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: 2/5

Additional Information for questions 1 and 2:

In Lab 2 as well as Lab 3, the number of bottles of acids in the category XL as a percentage of the total number of bottles of acids in the respective laboratories is not more than 1%.

If the ratio of the number of bottles of acids in Lab 1 and Lab 4 that are in the category XL is 2: 7, then what can be the difference between the number of number of bottles of acids in Lab 1 and Lab 4 that are not in the category XL?

 A.

212

 B.

202

 C.

207

 D.

Either (A) or (C)

 E.

Either (A) or (B)

 Hide Ans

Solution:
Option(D) is correct

Given that In Lab 2 as well as Lab 3, the number of bottles of acids in the category XL as a percentage of the total number of bottles of acids in the respective laboratories is not more than 1%.

Total number of bottles in these laboratories $= 2400$

Let the number of the bottles of acids in Lab 1 and Lab 4 that are in the category XL be $2x$ and $7x$ respectively.

Therefore, the number of bottles of acids in Lab 1 and Lab 4 that are not in the category XL will be:

$500 – 2x$ and $688 – 7x$ respectively.

Also, let the number of bottles of acids in Lab 2 and Lab 3 that are in category XL be $Y$.

Therefore, $Y + 2x + 7x = 720$

and $Y$ cannot be more than $6 + 6 = 12$

So, the value of $x$ can be 80 or 79.

Required difference $500 - 2x - (688 - 7x) = 5x – 188$

If $x$ is equal to 80, then the required difference is 212 and if $x$ is equal to 79, then the required difference is 207.


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