This problem can be solved using the principles of set theory and a Venn diagram representation. We are given information about the number of students playing three different games: snooker, chess, and tennis. All 95 students play at least one game, meaning the union of the sets of students playing each game is 95.

Let's define the sets:

• S: Set of students who play snooker.
• C: Set of students who play chess.
• T: Set of students who play tennis.

We are given the following values:

• Total students \(|S \cup C \cup T| = 95\)
• Number of students who play snooker \(|S| = 42\)
• Number of students who play tennis \(|T| = 49\)
• Number of students who play chess \(|C| = 43\)
• Number of students who play any and only two games = 29
• Number of students who play all three games \(|S \cap C \cap T| = 5\)
• Number of students who play only snooker and tennis \(|S \cap T|\) only = 11
• Number of students who play only snooker and chess \(|S \cap C|\) only = 6
• The number of students who play only snooker and only chess is equal. This is likely meant to mean the number of students playing only snooker (\(|S|\) only) is equal to the number playing only chess (\(|C|\) only).

Let's denote the number of students in each disjoint region of the Venn diagram:

• \(|S|\) only: Students playing only snooker (let's call this \(n_S\)).
• \(|C|\) only: Students playing only chess (let's call this \(n_C\)).
• \(|T|\) only: Students playing only tennis (let's call this \(n_T\)).
• \(|S \cap C|\) only: Students playing only snooker and chess = 6.
• \(|S \cap T|\) only: Students playing only snooker and tennis = 11.
• \(|C \cap T|\) only: Students playing only chess and tennis (let's call this \(n_{CT}\)).
• \(|S \cap C \cap T|\): Students playing all three games = 5.

We are given that the total number of students who play any and only two games is 29. This means:

\(|S \cap C|\) only + \(|S \cap T|\) only + \(|C \cap T|\) only = 29

Substituting the known values:

\(6 + 11 + n_{CT} = 29\)

\(17 + n_{CT} = 29\)

\(n_{CT} = 29 - 17\)

\(n_{CT} = 12\)

So, 12 students play only chess and tennis.

We are given that the number of students who play only snooker (\(n_S\)) and only chess (\(n_C\)) is equal. So, \(n_S = n_C\).

Now, let's use the total number of students for each game. The total number of students playing snooker is the sum of students in all regions within the snooker circle:

\(|S|\) only + \(|S \cap C|\) only + \(|S \cap T|\) only + \(|S \cap C \cap T| = |S|\)

\(n_S + 6 + 11 + 5 = 42\)

\(n_S + 22 = 42\)

\(n_S = 42 - 22\)

\(n_S = 20\)

Since \(n_S = n_C\), the number of students who play only chess is also 20.

\(n_C = 20\)

Let's check this against the total number of students playing chess:

\(|C|\) only + \(|S \cap C|\) only + \(|C \cap T|\) only + \(|S \cap C \cap T| = |C|\)

\(n_C + 6 + n_{CT} + 5 = 43\)

\(20 + 6 + 12 + 5 = 43\)

\(43 = 43\)

This confirms our values for \(n_S\) and \(n_C\) are consistent.

Finally, we need to find the number of students who play only tennis (\(n_T\)). The total number of students playing tennis is the sum of students in all regions within the tennis circle:

\(|T|\) only + \(|S \cap T|\) only + \(|C \cap T|\) only + \(|S \cap C \cap T| = |T|\)

\(n_T + 11 + n_{CT} + 5 = 49\)

\(n_T + 11 + 12 + 5 = 49\)

\(n_T + 28 = 49\)

\(n_T = 49 - 28\)

\(n_T = 21\)

Alternatively, we can use the fact that the total number of students (95) is the sum of students in all disjoint regions:

\(|S|\) only + \(|C|\) only + \(|T|\) only + \(|S \cap C|\) only + \(|S \cap T|\) only + \(|C \cap T|\) only + \(|S \cap C \cap T| = 95\)

\(n_S + n_C + n_T + 6 + 11 + n_{CT} + 5 = 95\)

Substitute the known values: \(n_S = 20\), \(n_C = 20\), \(n_{CT} = 12\)

\(20 + 20 + n_T + 6 + 11 + 12 + 5 = 95\)

\(74 + n_T = 95\)

\(n_T = 95 - 74\)

\(n_T = 21\)

Both methods give the same result.

Summary of students in each region:

The number of students who play only tennis is 21.

Revision Table: Understanding Set Theory in Word Problems

Additional Information: Solving Venn Diagram Problems

Venn diagrams are visual tools that help represent the relationships between sets. For problems involving three sets, drawing a Venn diagram with 8 distinct regions is often helpful:

1. The region only in the first set.
2. The region only in the second set.
3. The region only in the third set.
4. The region only in the intersection of the first and second sets.
5. The region only in the intersection of the first and third sets.
6. The region only in the intersection of the second and third sets.
7. The region common to all three sets.
8. The region outside all three sets (elements not in any of the sets) - in this problem, this region is empty as all students play at least one game.

When solving these problems, it's usually easiest to start filling in the diagram from the innermost region (the intersection of all sets) and work outwards. If the 'only' regions are given, you can fill those in directly. If total set sizes are given, you'll need to use equations like the ones in the solution above to find the sizes of the disjoint regions.

Always check your final numbers by summing all the disjoint regions to ensure they equal the total number of elements in the universal set (or the union, if not all elements are in the sets).