We can break down this problem step-by-step by assigning variables for the number of students in each group.

Let’s assign:

• $x$: Number of students who play only Football.
• $y$: Number of students who play only Badminton.
• $z$: Number of students who play only Cricket.
• $ABC$: Number of students who play all three games together.
• $AB$: Number of students who play both Badminton and Football but not Cricket.
• $BC$: Number of students who play both Badminton and Cricket but not Football.
• $AC$: Number of students who play both Football and Cricket but not Badminton.

Now, based on the problem's information, we will derive relationships:

1. Relation between students who play only Badminton and only Football:

"Students who play only Badminton are 50% of students who play only Football." This gives us the equation: $y = \frac{1}{2}x$

2. Ratio of students who play only Cricket to those who play only Badminton:

"The ratio of students who play only Cricket to students who play only Badminton is 5:2." This gives us the equation: $\frac{z}{y} = \frac{5}{2} \quad \text{or} \quad z = \frac{5}{2}y$

3. Students who play all three games:

"Students who play all 3 games together are 50% of students who play only Badminton." This gives us: $ABC = \frac{1}{2}y$

4. Relation between students playing two games (Cricket + Badminton and Football + Badminton):

"Students who play both Cricket and Badminton together are equal to students who play both Football and Badminton but not Cricket." This gives: $BC = AB$

5. Relation involving Cricket and Football:

"Students who play both Cricket and Football but not Badminton are 50% more than students who play all 3 games together." This gives us: $AC = 1.5 \times ABC = 1.5 \times \frac{1}{2}y = \frac{3}{4}y$

6. Another relation between Cricket + Football and Football + Badminton:

"Students who play both Cricket and Football but not Badminton are 50% of students who play both Football and Badminton together." This gives us: $AC = 0.5 \times AB$

Now we can use the above relations to solve step by step:

1. From the last equation $AC = 0.5 \times AB$, we can substitute $AC = \frac{3}{4}y$: $\frac{3}{4}y = 0.5 \times AB$ Solving for $AB$: $AB = \frac{3}{4}y \times 2 = \frac{3}{2}y$

2. Since $BC = AB$, we have $BC = \frac{3}{2}y$.

Now, let’s summarize the relationships and expressions in terms of $y$:

• $x = 2y$
• $z = \frac{5}{2}y$
• $ABC = \frac{1}{2}y$
• $AB = BC = \frac{3}{2}y$
• $AC = \frac{3}{4}y$

This system of variables describes the relationships between students playing each game.

Would you like to see the total number of students or any other detailed explanation about these relationships?

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Here are some related questions you might find interesting:

1. How many students play only one game?
2. What percentage of students play all three games?
3. What is the total number of students playing both Cricket and Football?
4. If the number of students who play only Football is 60, how many students play all three games?
5. How many students play at least two games?

Tip: Visualizing the problem with a Venn diagram can often help clarify relationships between groups in such problems.