MHB Union & Intersection questions

[bala2014]

Sixty six cats signed up for the contest MISS CAT 2013. After the first round 21 cats were eliminated
because they failed to catch a mouse. Of the remaining cats, 27 had stripes and 32 had one black
ear. All striped cats with one black ear got to the final. What is the minimum number of finalists?
(A) 5 (B) 7 (C) 13 (D) 14 (E) 27

 

[Prove It]

66 cats started out, 21 were eliminated, so 45 made it to round two.

In round two, all those that had both a black ear and stripes continued.

Since there were 27 with stripes and 32 with a black ear, but only 45 cats, some had to have both. How many must have had both?

 

[Evgeny.Makarov]

Since there were 27 with stripes and 32 with a black ear, but only 45 cats, some had to have both. How many must have had both?

Out of 45 that passed the first round, let $B$ be the set of cats with a black ear and $S$ be the set of cats with stripes. It's easier to figure out $|B\cap S|$ (the number of elemenst in the intersection) under the assumption that all 45 cats had stripes or a black ear or both. Then the answer is unique. I'll wait for the OP to show some effort on this, and then we can figure out how an upper bound on $|B\cup S|$ turns into a lower bound on $|B\cap S|$.

 

[bala2014]

Out of 45 that passed the first round, let $B$ be the set of cats with a black ear and $S$ be the set of cats with stripes. It's easier to figure out $|B\cap S|$ (the number of elemenst in the intersection) under the assumption that all 45 cats had stripes or a black ear or both. Then the answer is unique. I'll wait for the OP to show some effort on this, and then we can figure out how an upper bound on $|B\cup S|$ turns into a lower bound on $|B\cap S|$.

By keeping $|B\cap S|$ = x , B = 27-x, S = 32-x, we know $|B\cup S|$ s 45. So we can form a equation 27-x+x+32-x = 45 where x= 59-45 = 14.

 

[Evgeny.Makarov]

By keeping $|B\cap S|$ = x , B = 27-x, S = 32-x

Something is not right here. First, $B$ is a set (at least as I defined it) and $27-x$ is a number, so $B$ and $27-x$ are objects of different types and cannot be equal. Second, if you mean $|B|=27-x$, this is also not right. "$B$ [is] the set of cats with a black ear", and the problem statement says that $|B|=32$. What you probably mean is that $|B\setminus (B\cap S)|=|B\setminus S|=32-x$ if $|B\cap S|=x$. Similarly, $|S\setminus (B\cap S)|=|S\setminus B|=27-x$. Now, the sets $B\setminus S$, $B\cap S$ and $S\setminus B$ ar disjoint and their union has 45 element (under the additional assumption that all 45 cats had stripes or a black ear or both). From here we get your equation
\[
(27-x)+x+(32-x) = 45.
\]

I would write a solution as follows.
\[
|B|+|S|-|B\cap S|=|B\cup S|\qquad(*)
\]
from where
\[
|B\cap S|=|B|+|S|-|B\cup S|=32+27-45=14
\]
If we don't make the assumption that $|B\cup S|=45$, then we only know that $|B\cup S|\le 45$, so (*) gives
\[
|B\cap S|\ge |B|+|S|-45=32+27-45=14.
\]