We can solve this problem using the principle of inclusion and exclusion. Let’s define the sets:

• Let $C$ be the set of students who play Cricket.
• Let $H$ be the set of students who play Hockey.
• Let $V$ be the set of students who play Volleyball.

Given dataWe can solve this problem using the principle of inclusion and exclusion. Let’s define the sets:

• Let $C$ be the set of students who play Cricket.
• Let $H$ be the set of students who play Hockey.
• Let $V$ be the set of students who play Volleyball.

Given

• $|C| = 50$ (students who played Cricket),
• $|H| = 50$ (students who played Hockey),
• $|V| = 40$ (students who played Volleyball),
• $|C \cap H| = 5$ (students who played both Cricket and Hockey),
• $|H \cap V| = 10$ (students who played both Hockey and Volleyball),
• $|C \cap V| = 5$ (students who played both Cricket and Volleyball),
• $|C \cap H \cap V| = 10$ (students who played all three games).

Step 1: Total number of students

Using the principle of inclusion and exclusion, the total number of students $N$ is:

$N = |C| + |H| + |V| - |C \cap H| - |H \cap V| - |C \cap V| + |C \cap H \cap V|$

Substituting the values:

$N = 50 + 50 + 40 - 5 - 10 - 5 + 10 = 130 - 20 + 10 = 120$

So, the total number of students is $120$.

Step 2: Students who played only one game

Only Cricket:

Students who played only Cricket are those in $C$ but not in $H$ or $V$. This is calculated as:

$\text{Only Cricket} = |C| - |C \cap H| - |C \cap V| + |C \cap H \cap V|$

Substituting the values:

$\text{Only Cricket} = 50 - 5 - 5 + 10 = 50 - 10 + 10 = 50$

So, $50$ students played only Cricket.

Only Hockey:

Students who played only Hockey are those in $H$ but not in $C$ or $V$. This is calculated as:

$\text{Only Hockey} = |H| - |C \cap H| - |H \cap V| + |C \cap H \cap V|$

Substituting the values:

$\text{Only Hockey} = 50 - 5 - 10 + 10 = 50 - 15 + 10 = 45$

So, $45$ students played only Hockey.

Only Volleyball:

Students who played only Volleyball are those in $V$ but not in $C$ or $H$. This is calculated as:

$\text{Only Volleyball} = |V| - |C \cap V| - |H \cap V| + |C \cap H \cap V|$

Substituting the values:

$\text{Only Volleyball} = 40 - 5 - 10 + 10 = 40 - 15 + 10 = 35$

So, $35$ students played only Volleyball.

Final Answer:

• Total number of students: $120$
• Number of students who played only Cricket: $50$
• Number of students who played only Hockey: $45$
• Number of students who played only Volleyball: $35$

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Would you like further details or have any questions? Here are some related questions you might explore:

1. How many students played exactly two games?
2. How many students played exactly one game?
3. How many students played all three games?
4. If 5 more students played only Hockey, how would the total change?
5. What is the minimum number of students who must have played two games?

Tip: The inclusion-exclusion principle is very useful in counting problems involving multiple overlapping sets.