Question 1206529
.
a survey of 220 families showed that: 
63 had a dog; 64 had a cat; 28 had a dog and a cat; 
84 neither had a cat nor a dog and in addition did not have a parakeet; 
6 had a cat, a dog and a parakeet. 
how many families had a parakeet only?
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<pre>
We have a universal set of 220 families and three its basic subsets

    d  of 63 families having a dog;

    c  of 64 families having a cat;

    p  of unknown number of families having a parakeet.


The number of families in the union (d U c) is  63 + 64 - 28 = 99.


The set of families that neither have a cat nor a dog  is the complement to the set (d U c),
and the number of such families is  220 - 99 = 121.


We are told, that the number of families that "in addition" do not have parakeet is 84.


From it, we conclude that  121-84 = 37  families do have a parakeet, but have neither a cat nor a dog.


So, the answer to the problem's question is 37 families.
</pre>

Solved.



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Notice that the last condition in the post

<pre>
    "6 had a cat, a dog and a parakeet. "
</pre>is excessive and unnecessary.  &nbsp;&nbsp;I did not use it in my solution.



This problem is &nbsp;DEFINITELY &nbsp;not to write and/or to use Venn diagram 
and accompanied system of linear equations in 8 unknowns.


This problem's true destination and true purpose is to solve it &nbsp;MENTALLY
and to teach young students to think logically about the sets and their subsets.