SOLUTION: A study of 115 people shows that 85 eat breakfast, 58 use dental floss regularly, and 27 subscribe to the morning paper. Among those who eat breakfast, 52 floss regularly and 15

The number of those who eat breakfast, or floss regularly, or get the morning paper 
(who do any (~ at least one) of these three morning activity) is


    115 - 14 = 101.

Let B is the set of those who eat breakfast   (and the number of people in this set, at the same time, which is 85 - given !).

Let F is the set of those who floss regularly (and the number of people in this set, at the same time, which is 58 - given !).

Let P is the set of those who subscribe paper (and the number of people in this set, at the same time, which is 27 - given !).


Let BF is the intersection of the sets B and F (and the number of people in this intersection, at the same time, which is 52 - given !).

Let BP is the intersection of the sets B and P (and the number of people in this intersection, at the same time, which is 15 - given !).


Let BFP is the intersection of the sets B, F and P (and the number of people in this intersection, at the same time, which is 10 - given !).



The number of those who do any of these morning activities is 115 - 14 = 101.



This number, which is the number of people in the union of the sets B, F and P, can be calculated in other way, too, using the formula


   101 = n(B U F U P) = B + F + P - BF - BP - FP + BFP.    *()


In this equation, we know every term of the right side, except FP (which is under the question b) ).

So, we substitute all known values into equation (*), and we get


    101 = 85 + 58 + 27 - 52 - 15 - FP + 10.


Now we can easily find the value of FP as


    FP = 85 + 58 + 27 -52 - 15 + 10 - 101 = 12.


Answer.  The number of those who use dental floss regularly and also get the morning paper is 12.