CAT 2022 Slot 1 DILR Set 3: Arrangements (5 Questions) with Solutions

Singers: 6 in all, 4 boys, so 2 girls. Dancers: 10 in all, 4 girls, so 6 boys. No dancer is a singer.

Let there be $B$ boys. Then $0.8B$ boys want a 1-day event and $0.6B$ want a 2-day event, so $B$ is a multiple of 5.

Of the boys interested in a 2-day event, 70% are neither singers nor dancers. The other 30% are singers or dancers, which is $0.3\times0.6B=\dfrac{9B}{50}$ boys. This must be a whole number, so $B$ is a multiple of 50.

Bounds on this count: Clue 4 puts all 4 male singers in the 3-day event, so they are also in the 2-day event. Two dancers are in the 3-day event, and no girl is, so these 2 are boys and are also in the 2-day event. So at least $4+2=6$ boys in the 2-day event are singers or dancers. At most $4+6=10$ can be, since there are 4 male singers and 6 male dancers.

So $6\le \dfrac{9B}{50}\le 10$, which gives $33.3\le B\le 55.5$. The only multiple of 50 in this range is $B=50$.

Singers: 6, of whom 4 are boys, so 2 are girls. Dancers: 10, of whom 4 are girls, so 6 are boys. No dancer is a singer.

Let there be $B$ boys. Boys interested in a 2-day event number $0.6B$, and 70% of them are neither singers nor dancers. So $0.3 \times 0.6B = 0.18B$ boys in the 2-day group are singers or dancers.

Clue 4 says all 4 male singers attend a 3-day event, so they also attend a 2-day event. The boy dancers in the 2-day group are therefore $0.18B - 4$, and this must lie between 0 and 6.

$0.8B$, $0.6B$ and $0.18B$ must all be whole numbers, so $B$ is a multiple of 50. $B = 100$ would give $18 - 4 = 14$ boy dancers, which exceeds 6. So $B = 50$, giving 5 boy dancers in the 2-day group.

Clue 5: let $s$ and $d$ be the girl singers and girl dancers interested in a 2-day event. Singers number $4 + s$ and dancers number $5 + d$. Then $4 + s = 5 + d + 1$, so $s = d + 2$. Since $s \le 2$, we get $s = 2$ and $d = 0$.

Clue 3: 40% of the girls in the 2-day group are singers or dancers, and that is $s + d = 2$ girls. So 5 girls are interested in a 2-day event. None of them is a dancer.

all girls are interested in a 1-day event (clue 1), so all 4 girl dancers are interested in it. This is determined and equals 4.

boys interested in a 1-day event number $0.8 \times 50 = 40$. This includes the 4 male singers and the 5 boy dancers in the 2-day group. The sixth boy dancer may or may not be interested in a 1-day event, and nothing in the clues settles it. So the count of boys who are neither singers nor dancers is 31 or 32, and it cannot be determined.

There are 50 boys (the count that satisfies the integer and bound constraints). So 30 boys want a 2-day event, and 21 of them are neither singers nor dancers. The other 9 are the 4 male singers (all in the 3-day event, hence the 2-day event) plus 5 male dancers.

Let $s$ and $d$ be the numbers of female singers and female dancers interested in a 2-day event. Singers there number $4+s$ and dancers $5+d$.

Clue 5 gives $4+s=5+d+1$, so $s=d+2$. There are only 2 girl singers, so $s\le 2$. Hence $d=0$ and $s=2$.

In Clue 3, 60% of the girls in the 2-day event are neither singers nor dancers, so the 2 female singers are the other 40%. Then $0.4g=2$, so $g=5$ girls want a 2-day event.

Total interested in a 2-day event: $30+5=35$. Class size: $50+15=65$.

Fraction $=\dfrac{35}{65}=\dfrac{7}{13}$.

There are 6 male dancers (10 dancers, 4 of them girls).

Among the 30 boys in the 2-day event, 9 are singers or dancers (70% of 30 are neither, so 21 are). The 4 male singers account for 4, so exactly 5 male dancers are interested in a 2-day event.

Anyone who wants a 2-day event also wants a 1-day event. So at least 5 male dancers are interested in a 1-day event.

The sixth male dancer is not interested in a 2-day event. He may or may not be interested in a 1-day event, and no clue decides this. Of the 50 boys, 10 are outside the 1-day event, so either case is possible.

So the number of male dancers interested in a 1-day event is 5 or 6.

With 50 boys, 30 want a 2-day event, and 9 of them are singers or dancers (30% of 30). The 4 male singers are all in the 3-day event, hence the 2-day event. So 5 male dancers are in the 2-day event.

Let $s$ and $d$ be the numbers of female singers and female dancers in the 2-day event. Singers: $4+s$. Dancers: $5+d$.

Clue 5: $4+s=(5+d)+1$, so $s=d+2$.

There are only 2 girl singers, so $s\le2$. Since $d\ge0$, this forces $d=0$ and $s=2$.