Understanding the Data and the Question

The table provides information about students in five different classes (A, B, C, D, E) of a school. For each class, we know:

• The total number of students.
• The number of students who do not participate in either Chess or Carrom.
• The ratio of students participating in Chess to those participating in Carrom, among the students who play either game.

The question asks us to compare two quantities: the number of students participating in either Chess or Carrom from Class E, and the total number of students participating in Chess from Class C and Class D combined. We need to find the percentage difference.

First, we need to calculate the number of students who participate in either Chess or Carrom for each class. This is done by subtracting the number of students who play neither game from the total number of students.

Then, for classes C, D, and E, we will use the given ratio to find the specific number of students playing Chess or the total playing either, as required by the question.

Step-by-Step Calculation

1. Calculate Students Playing Either Chess or Carrom per Class

Students playing either Chess or Carrom = Total Students - Students playing Neither.

• Class A: \(420 - 119 = 301\)
• Class B: \(330 - 88 = 242\)
• Class C: \(240 - 110 = 130\)
• Class D: \(125 - 45 = 80\)
• Class E: \(390 - 130 = 260\)

The number of students participating in Chess and Carrom (meaning either Chess or Carrom) from Class E is 260.

2. Calculate Students Playing Chess in Class C and Class D

For the students playing either game, we use the given ratio to find the number playing Chess.

Number of Chess players = (Total playing either) × \(\frac{\text{Chess Ratio Part}}{\text{Total Ratio Parts}}\)

• Class C:
  • Total playing either = 130
  • Ratio Chess : Carrom = 8 : 5
  • Total Ratio Parts = \(8 + 5 = 13\)
  • Number of Chess players in Class C = \(130 \times \frac{8}{13} = 10 \times 8 = 80\)
• Class D:
  • Total playing either = 80
  • Ratio Chess : Carrom = 2 : 3
  • Total Ratio Parts = \(2 + 3 = 5\)
  • Number of Chess players in Class D = \(80 \times \frac{2}{5} = 16 \times 2 = 32\)

The number of students participating in Chess from Class C is 80.

The number of students participating in Chess from Class D is 32.

3. Calculate Total Chess Players from Class C and D Together

Total Chess players from C and D = (Chess players in C) + (Chess players in D)

Total Chess players from C and D together = \(80 + 32 = 112\)

4. Calculate the Percentage Difference

We need to find what percent the number of students participating in Chess and Carrom from Class E (which is 260) is more or less than the number of students participating in Chess from C and D together (which is 112).

Since \(260 > 112\), the number for Class E is more than the number for Classes C and D combined.

Percentage Difference = \(\frac{|\text{Value 1} - \text{Value 2}|}{\text{Value 2}} \times 100\)

Here, Value 1 = Students in E (either) = 260

Value 2 = Students in C and D (Chess) = 112

Difference = \(260 - 112 = 148\)

Percentage More = \(\frac{148}{112} \times 100\)

Simplify the fraction \(\frac{148}{112}\):

Both are divisible by 4: \(\frac{148 \div 4}{112 \div 4} = \frac{37}{28}\)

Now calculate \(\frac{37}{28} \times 100 = \frac{3700}{28}\)

Perform the division:

\(\frac{3700}{28} = \frac{1850}{14} = \frac{925}{7}\)

Convert the improper fraction to a mixed number:

\(925 \div 7\)

• \(925 = 7 \times 100 + 225\)
• \(225 = 7 \times 30 + 15\)
• \(15 = 7 \times 2 + 1\)

So, \(925 = 7 \times 100 + 7 \times 30 + 7 \times 2 + 1 = 7 \times (100 + 30 + 2) + 1 = 7 \times 132 + 1\)

Thus, \(\frac{925}{7} = 132 \frac{1}{7}\)

The percentage is \(132 \frac{1}{7}\) %.

Since the number from Class E (260) is greater than the number from Classes C and D Chess combined (112), it is \(132 \frac{1}{7}\)% more.

Comparing Quantities

• Students in Class E playing either Chess or Carrom = 260
• Students in Class C and D playing Chess = 112

The number of students from Class E is 260, and the number from Classes C and D together is 112. The difference is \(260 - 112 = 148\). To find the percentage difference compared to the number from Classes C and D, we use the formula: \(\frac{\text{Difference}}{\text{Base}} \times 100\).

\(\text{Percentage} = \frac{148}{112} \times 100 = \frac{37}{28} \times 100 = \frac{3700}{28} = \frac{925}{7} = 132 \frac{1}{7}\) %.

This is \(132 \frac{1}{7}\)% more.

Revision Table: Key Calculations Summary

Value 1 (E either) = 260

Value 2 (C Chess + D Chess) = \(80 + 32 = 112\)

Percentage More = \(\frac{260 - 112}{112} \times 100 = \frac{148}{112} \times 100 = 132 \frac{1}{7}\) %.

Additional Information: Percentage Change Concepts

When calculating percentage change, it's crucial to identify the 'base' value correctly. The question asks "what percent more or less than" a specific value. The value following "than" is typically the base for the calculation.

• Percentage Increase: If Value 1 > Value 2, Percentage Increase = \(\frac{\text{Value 1} - \text{Value 2}}{\text{Value 2}} \times 100\).
• Percentage Decrease: If Value 1 < Value 2, Percentage Decrease = \(\frac{\text{Value 2} - \text{Value 1}}{\text{Value 2}} \times 100\).

In this problem, Value 1 (260) is greater than Value 2 (112), so we calculate percentage increase (or 'more'). The base is 112.