A survey shows that $63\% $ of Americans like cheese whereas $76\% $ like apples. If $x\% $ of the Americans like both cheese and apples, find the value of $x$.
Hint: First we have to define what the terms we need to solve the problem are. Since we need to find the value of the $x$ which is both the apples and the cheese is liked by the Americans, and now values are less liked by the Americans and mostly liked by apples. So, with this information we will find both cheese and apples lovers.
Complete step-by-step solution: We need one formula to proceed which is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ which is the union of the two sets are be written as the adding of the separate plus the intersection of the negative, Fix A is the cheese that means $n(A) = 63$ and B is the apples then $n(B) = 76$ and both the cheese and apples are liked by value of $x$ that means we have $n(A \cap B) = x$ Substitute all in the general equation we get $n(A \cup B) = 63 + 76 - x \Rightarrow 139 - x$ But since there are total only $100\% $ that means the value will not exceed hundred thus $n(A \cup B) \leqslant 100$; thus, comparing these equations we get; $139 - x \leqslant 100$$ \Rightarrow x \geqslant 39$(after cancelling) Also, in generally we know that $n(A \cap B) \leqslant n(A)$(always less than or equals) And hence we get $x \leqslant 63$ Hence, we have two different values and compare them we get $39 \leqslant x \leqslant 63$ is the range of the $x$ that both apples and cheese is liked by the Americans. Minimum $39$ and maximum $63$
Note: less than or equal means the values will not exceed the given points or else equals ,$n(A \cup B)$ is the totals values of the cheese lover and apples lover and also subtracts the both common lovers. $n(A \cup B) \leqslant 100$(since the most value is hundred at percent that will not exceed the hundred)