Finding the cardinal number for the intersection of two sets

[JC2000]

Homework Statement

A survey shows that $63 %$ Americans like cheese where as $76 %$ like apples. If $x %$ like both, find $x$.

Relevant Equations

Since $A \cap B \subset A$ and $A \cap B \subset B$ :
$n(A \cap B) \leq n(A)$ and $n(A \cap B) \leq n(B)$
i.e $n(A \cap B) \leq 63$
Also, $n(A\cap B ) \geq 39$ since $n(A\cap B) = n(A)+ n(B)-n(A\cup B)$ but $n(A\cup B) \leq 100$

Thus : $39 \leq x \leq 63$

My Question :

1.Why are the inequalities considered? Why not simply use $n(A\cap B) = n(A)+ n(B)-n(A\cup B)$ to get $n(A\cap B) = 39$ ?
2. The way I interpret this is : If the set for people liking cheese was to be a subset of the set for people who like apples then the most number of people to like both would be 63. But I still fail to understand why the minimum value should be 39 (Why : $P(A) + P(B) - P(A \cup B) < 1$)?

 

[Mark44]

My Question :

1.Why are the inequalities considered? Why not simply use $n(A\cap B) = n(A)+ n(B)-n(A\cup B)$ to get $n(A\cap B) = 39$ ?

Because it's not given how many like both cheese and apples.

 

[JC2000]

Because it's not given how many like both cheese and apples.

Yes but we are given the remaining variables from which $n(A \cap B)$ can be found (?).

 

[Mark44]

Yes but we are given the remaining variables from which $n(A \cap B)$ can be found (?).

No, since you aren't given $n(A \cup B)$

 

[JC2000]

No, since you aren't given $n(A \cup B)$

Can't it be assumed to be 100? (Also could you shed some light on Q2?) Thanks!

 

[Mark44]

Can't it be assumed to be 100? (Also could you shed some light on Q2?) Thanks!

All you are given is that $n(A \cup B) \le 100$, which doesn't imply that it equals 100.
I need to take off in a bit, so maybe somebody else can take a crack at your other question.

 

[WWGD]

JC2000: I assume your numbers are given as percentages, right, so that it should be 76% and 63%?Edit. As Mark44 wrote, you don't have all the data you need. But you are correct that the percentage that like both is at most 63( Since it is a subset of those who like cheese) and the percentage that likes either is at most 100

 

[JC2000]

Oh yes! Thanks!

 

[WWGD]

As Mark44 pointed out, notice that $n(A\cup B)$ is _at most_ 100, but not necessarily 100. Can you see why?

 

[JC2000]

As Mark44 pointed out, notice that $n(A\cup B)$ is _at most_ 100, but not necessarily 100. Can you see why?

Is it because there may be a percentage of people that like neither? So this would mean $n(A) + n(B)-n(A\cap B) \leq n(A\cup B)$?

 

[WWGD]

Correct. Good job.

 

[JC2000]

Thanks a lot!