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Question:

How many boys are there in the class?

Explanation:

There are boys and girls in the class. Each of the boy or girl is either a dance or singer or none of these but not both.

There are 6 singers in the class, 4 of them are boys. 
⇒ There are 4 male singers and 2 female singers

There are 10 dancers in the class, 4 of them are girls. No dancer in the class is a singer
⇒ There are 6 male dances and 4 female dancers

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From (4): No girl is interested in attending a 3-day event. All male singers and 2 of the dancers are interested in attending a 3-day event.

Since all 4 male singers are interested in a 3-day event, they all must also be interested in a 2-day and a 1-day event.

Since 2 male dancers are interested in a 3-day event, male dancers interested in a 2-day event must be greater than or equal to 2.

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From (1): All the girls and 80% of the boys are interested in attending a 1-day event. 60% of the boys are interested in attending a 2-day event.

From (3): 70% of the boys who are interested in attending a 2-day event are neither singers nor dancers. Hence, 30% of those who are interested in a 2-day event are dancers or singers.

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Now, (2 + b) = 30% of 60% of (10 + x)

⇒ 2 + b + 4 = $\frac{3}{10}$ × $\frac{3}{5}$ × (10 + x)

⇒ 6 + b = $\frac{9}{50}$ × (10 + x)

Here b has to be an integer, hence (10 + x) should be completely divisible by 50.
∴ x should be 40 or 90 or 140 and so on.
But since b cannot be greater than 4, x should be 40 and b = 3.

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From (5): The number of singers interested in attending a 2-day event is one more than the number of dancers interested in attending a 2-day event.

This is only possible when 2 female singers while no female dance is interested in a 2-day event.

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From (3): 60% of the girls who are interested in attending a 2-day event are neither singers nor dancers.

⇒ 0 + 2 = $\frac{2}{5}$ × (0 + 2 + females interested in a 2-day event who are neither singer nor dancer)

females interested in a 2-day event who are neither singer nor dancer = 3

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∴ There are 50 boys in the class.

Hence, 50.