Analyzing School Student Participation Data

The question asks us to find the total number of students participating in Chess from classes B, C, and E combined, based on the provided table.

The table provides the following information for each class:

• Total Number of Students
• Number of Students without participation in both games (neither Chess nor Carrom)
• Ratio of Students participating in Chess to Carrom (among those who play either game)

To find the number of students playing Chess in a specific class, we need to first determine the number of students who participate in *either* Chess or Carrom. These are the students who are not in the 'neither' category. Then, we use the given ratio to distribute these students into Chess and Carrom players.

Step-by-Step Calculation for Each Class

Class B

• Total students in Class B: 330
• Students participating in neither Chess nor Carrom: 88
• Number of students participating in Chess or Carrom = Total Students - Students participating in neither
• Number of students participating in Chess or Carrom in Class B = $330 - 88 = 242$
• Ratio of Chess to Carrom participants in Class B: $7 \ratio; 4$
• The sum of the ratio parts is $7 + 4 = 11$.
• These 11 parts represent the 242 students who play either Chess or Carrom.
• Value of one ratio part = $\frac{\text{Total students playing Chess or Carrom}}{\text{Sum of ratio parts}} = \frac{242}{11} = 22$.
• Number of students participating in Chess in Class B = Ratio part for Chess $\times$ Value of one part = $7 \times 22 = 154$.

Class C

• Total students in Class C: 240
• Students participating in neither Chess nor Carrom: 110
• Number of students participating in Chess or Carrom = Total Students - Students participating in neither
• Number of students participating in Chess or Carrom in Class C = $240 - 110 = 130$
• Ratio of Chess to Carrom participants in Class C: $8 \ratio; 5$
• The sum of the ratio parts is $8 + 5 = 13$.
• These 13 parts represent the 130 students who play either Chess or Carrom.
• Value of one ratio part = $\frac{\text{Total students playing Chess or Carrom}}{\text{Sum of ratio parts}} = \frac{130}{13} = 10$.
• Number of students participating in Chess in Class C = Ratio part for Chess $\times$ Value of one part = $8 \times 10 = 80$.

Class E

• Total students in Class E: 390
• Students participating in neither Chess nor Carrom: 130
• Number of students participating in Chess or Carrom = Total Students - Students participating in neither
• Number of students participating in Chess or Carrom in Class E = $390 - 130 = 260$
• Ratio of Chess to Carrom participants in Class E: $8 \ratio; 5$
• The sum of the ratio parts is $8 + 5 = 13$.
• These 13 parts represent the 260 students who play either Chess or Carrom.
• Value of one ratio part = $\frac{\text{Total students playing Chess or Carrom}}{\text{Sum of ratio parts}} = \frac{260}{13} = 20$.
• Number of students participating in Chess in Class E = Ratio part for Chess $\times$ Value of one part = $8 \times 20 = 160$.

Calculating Total Chess Participants from B, C, and E

To find the total number of students participating in Chess from B, C, and E together, we sum the number of Chess participants from each of these classes:

Total Chess Participants (B, C, E) = (Chess participants in B) + (Chess participants in C) + (Chess participants in E)

Total Chess Participants (B, C, E) = $154 + 80 + 160$

Total Chess Participants (B, C, E) = $394$

Therefore, the total number of students participating in Chess from B, C, and E together is 394.

Summing the Chess participants from B, C, and E:

$154 + 80 + 160 = 394$

Revision Table: Key Calculations

Additional Information: Understanding Participation Data

This type of data interpretation question requires careful reading of the table columns and understanding the relationships between the different categories of students. The key steps involve:

1. Identifying the total relevant population (students playing either game).
2. Using the given ratio to divide this population into specific activity groups (Chess or Carrom).
3. Summing the numbers for the requested classes or categories.

The column "Number of Students without participation in both games" is crucial because it tells us how many students are left who *must* be participating in at least one of the two games mentioned (Chess or Carrom). The problem states that the "remaining students play either chess or carrom," which simplifies the problem by implying no student plays both or plays a different game not listed.

The ratio provided applies only to this remaining group of students who play at least one of the two games. Understanding which group the ratio applies to is vital for correct calculation.