· Those students who are interested in attending a 3-day event are also interested in attending a 2-day event; those who are interested in attending a 2-day event are also interested in attending a 1-day event.

This statement means that:

Number of students in 1-day event ≥ Number of students in 2-day event ≥ Number of students in 3-day event.

· Given is number of girls = 15. Let the number of boys = 100x.

· Total singers = 6, Boy-Singers=4, then Girl-Singers=2.

· Total dancers = 10, Boy-Dancers=6, then Girl-Dancers=4.

· It is also given that ‘No dancer in the class is a singer’. This means both dancer and singer=0, and neither dancer and singer may exist.

· Neither D&S among boys = 100x – 10.

Neither D&S among girls = 15 – 6 = 9.

· We can use point(1) as follows:

§ All the girls are interested in attending the 1-day event. This implies, 2 singers, 4 dancers, and 9 neither sinder or dancer are interested in 1-day event.

§ 80% of the boys are interested in 1-day event (=80x), and 60% of the boys are interested in 2-day event (=60x)

· Point(3)

§ 70% of the boys attending 2-day event will be neither dancers or singers (=60x * 70% = 42x).

§ Let the total number of girls attending 2-day event be 10a. So that 60% of them (=6a) will be neither dancers or singers.

· Point(4) can be used as:

§ If no girl is interested in attending 3-day event, then it will be 0 in all three categories of dancers and singers.

§ All male singers (=4) are interested in attending the 3-day event. This also means all male singers are also interested in 2-day and 1-day events.

§ 2 dancers are also interested in 3-day event. But, we also know that no girl is interested in 3-day events. Therefore, these 2 dancers are boys.

		Not-interested	1-day event	2-day event	3-day event
Boys	Singer = 4		4	4	4
	Dancer = 6				2
	Neither = 100x - 10			42x
	Total = 100x		80x	60x
Girls	Singer = 2		2		0
	Dancer = 4		4		0
	Neither = 9		9	6a	0
	Total = 15		15	10a	0

· Using point(2):

It says that ‘Some of the girls are interested in attending a 1-day event, but not a 2-day event; some of the other girls are interested in attending both.’

This means only 1-day value exists and only 2-day value is 0. Also, both 1-day and 2-day also exists. There are 0 interested girls for 3-day event, therefore, 2-day value must be < 1-day value.

⟹ 10a<15.

· Value of x must be in form of , where ‘z’ must LCM of 100, 80, 60, or 42, this is because number of persons must be in whole numbers. ‘z’ can be 1, 2 only.

§ If z=1, then x=1, and total boys for 2-day event = 60x = 60.

Boy dancers for 2-day event = 60 – 42x – 4 = 60 – 42 – 4 = 14. This is not possible, as Total boy dancers = 6.

§ If z=2, then x=, and total boys for 2-day event = 60x =30.

Boy dancers for 2-day event = 30 – 42x – 4 = 30 – 21 – 4 = 5.

		Not-interested	1-day event	2-day event	3-day event
Boys	Singer = 4		4	4	4
	Dancer = 6			60x – 42x – 4 = 5	2
	Neither = 100x – 10 = 40			42x = 21
	Total = 100x = 50		80x = 40	60x = 30
Girls	Singer = 2		2		0
	Dancer = 4		4		0
	Neither = 9		9	6a	0
	Total = 15		15	10a	0

Let Girl singers in 2-day event = b, and Girl dancers in 2-day event = 10a – 6a – b = 4a – b.

Last point(5) says that: 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.

§ It means that 4+b = 5 + 4a – b + 1

⟹ b = 2a+1.

‘a’ can be 0, 1 or only, so that b’s value is whole number, and 10a<15.

Since b’s value cannot be more than 2, therefore, a’s value must be , then b = 2a+1 = 2.

Girl dancers in 2-day event = 4a – b = 2 – 2 = 0.

		Not-interested	1-day event	2-day event	3-day event
Boys	Singer = 4		4	4	4
	Dancer = 6			60x – 42x – 4 = 5	2
	Neither = 100x – 10 = 40			42x = 21
	Total = 100x = 50		80x = 40	60x = 30
Girls	Singer = 2		2	b = 2	0
	Dancer = 4		4	4a-b = 0	0
	Neither = 9		9	6a = 3	0
	Total = 15		15	10a = 5	0

Fraction of the class interested in attending a 2-day event = $\frac{Total\ 2-day\ interested\ persons}{Total\ number\ of \ persons}$ = $\frac{30+5}{50+15}$ = $\frac{35}{65}$ = $\frac{7}{13}$

Therefore, answer is option(c): 7/13.

CAT