There are 900 students in a college, categorized into games A, B, and C. The ratio of students in only A, only B, and only C is $2 : 3 : 5$ respectively. The number of students in all three categories is 50\% of the number of students in only C. The combined total of students in only B & C but not A and only A & C but not B is 155. The number of students in A & B but not C is 30 less than the number of students in only B.

What is the total number of students who are in at most two categories?

Question

There are 900 students in a college, categorized into games A, B, and C. The ratio of students in only A, only B, and only C is $2 : 3 : 5$ respectively. The number of students in all three categories is 50\% of the number of students in only C. The combined total of students in only B & C but not A and only A & C but not B is 155. The number of students in A & B but not C is 30 less than the number of students in only B.

What is the total number of students who are in at most two categories?

Accepted Answer

**Solution:**

1. Let the number of students in only A, only B, and only C be $2x$, $3x$, and $5x$ respectively.

2. Total students = 900, so:
   $$2x + 3x + 5x + 	ext{(students in exactly two categories)} + 	ext{(students in all three categories)} = 900$$

3. Number of students in all three categories is 50\% of only C:
   $$	ext{All three} = 0.5 	imes 5x = 2.5x$$

4. Sum of students in only B & C but not A and only A & C but not B is 155.

5. Students in A & B but not C are 30 less than only B:
   $$	ext{A & B only} = 3x - 30$$

6. Now, total students in at most two categories means all except those in all three categories:
   $$	ext{At most two categories} = 900 - 2.5x$$

7. Since total students are 900, and the sum of only A, only B, and only C is $2x + 3x + 5x = 10x$, and all three categories are $2.5x$, the total is:
   $$10x + 	ext{(students in exactly two categories)} + 2.5x = 900$$

8. But the problem gives the sum of two specific two-category groups as 155 and the third two-category group as $3x - 30$.

9. Total students in exactly two categories:
   $$155 + (3x - 30) = 125 + 3x$$

10. So total students:
    $$10x + (125 + 3x) + 2.5x = 900$$
    $$10x + 3x + 2.5x + 125 = 900$$
    $$15.5x = 775$$
    $$x = \frac{775}{15.5} = 50$$

11. Therefore, students in all three categories:
    $$2.5x = 2.5 	imes 50 = 125$$

12. Finally, students in at most two categories:
    $$900 - 125 = 775$$

**Answer:** 775

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