SOLUTION: suppose A and B are sets such that A Union B has 20 elements, A intersection B has 7 elements and the number of elements in B is twice that of A.what is the number of elements in:

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Question 1073560: suppose A and B are sets such that A Union B has 20 elements, A intersection B has 7 elements and the number of elements in B is twice that of A.what is the number of elements in: A and B
Answer by ikleyn(52803) About Me  (Show Source):
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suppose A and B are sets such that A Union B has 20 elements, A intersection B has 7 elements and the number of elements in B
is twice that of A.what is the number of elements in highlight%28each_of_the_sets%29 A and B.
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Let a = n(A)  (the number of elements in the set A);
    b = n(B)   (same for B).

Then you have these equations

20 = a + b - 7,     (1)
b = 2a.             (2)

Regarding equation (1), it is the particular case of the general equation

n(A U B) = n(A) + n(B) - n(A n B).   (*)  (See my comments at the END)

To solve the system (1), (2), simply substitute the expression (2) into (1). You will get

20 = a + 2a - 7  --->  3a = 27  --->  a = 27%2F3 = 9.

Then b = 2*a = 2*9 = 18.

Answer.  a = n(A) = 9,   b = n(B) = 19.

Check.   n(A) + n(B) - 7 = 9 + 18 - 7 = 20.


Regarding the equation (*), its proof is OBVIOUS:

count elements of A. Then add the elements of B. Elements in the intersection are counted twice.
Therefore, subtract n(a n B) from the sum n(A) + n(B), and you will get n(a U B).


See the lesson
    - Counting elements in sub-sets of a given finite set
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Miscellaneous word problems".