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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 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 = = 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".