SOLUTION: In a school with 92 students, 47 live with their biological father and 68 like with their biological mother. How many children live with both Biological parents, if there are 8

Mathematical formulation to this problem is as follows:


You have an Universal set U with n(U) = 92 elements.


There is a subset F (first letter of the word "Father") with n(F) = 47 elements.

There is another subset M (first letter of the word "Mother") with n(M) = 68 elements.


The part (a) asks about the number of elements n(F and M) of the intersection of these two subsets, F and M.


From elementary set theory, you have THIS EQUATION

    92 - 8 = n(F) + n(M) - n(F and M),

where the last term in the equation is your unknown to find.


Substitute the given values n(F) = 47  and  n(M) = 58 into this equation

    92 - 8 = 47 + 58 - n(F and M).

Then you get the solution

    n(F and M) = 47 + 58 - 84 = 31.


31 is the ANSWER to question (a).


Part (a) is solved.


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(b)  This question asks, how many elements has the set  (F U M) \ (F intersection M).

     The sign " \ " symbolizes the difference of the sets.


     The number of elements in this set is  n(F) + n(M) - 2*n(F and M) = 47 + 68 - 2*31 = 53.    ANSWER




(c)  This question asks, how many elements has the set  F U M.


     The number of elements in this set is  n(F) + n(M) - n(F and M) = 47 + 68 - 31 = 84.    ANSWER

I made it CONSCIOUSLY to leave a room for your thoughts.