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-The Union of set A, set B and set C has 104 elements.

-The Union of Set A and B has 51 elements

-The Union of Set A and C has 84 elements

-The Union of Set B and C has 97 elements

-The Intersection of Set A and the Union of Set B and C has 17 elements.

-Set C has twice as many elements as set B and three times as many elements as Set A.

How many elements does A have?

Hint: Take C is 6x and solve for x.

This is what I did after drawing a Venn diagram:

A minus B[itex]\cup[/itex]C = 104-97=7

B minus A[itex]\cup[/itex]C = 104-84=20

C minus A[itex]\cup[/itex]B = 104-51=53

If you add up these numbers and subtract them from 104, you'll get that all the intersections of these sets together have 24 elements.

It is given that A[itex]\cap[/itex](B[itex]\cup[/itex]C) = 17.

Therefore, the intersection of B and C without whatever is in A, should be 7.

From this fact the elements of A can be calculated from the fact that A[itex]\cup[/itex]C has 84 elements. To calculate the elements of A take 84 and subtract the elements of C that are not in A, 84 -53 -7= 24

I didn't use the hint nor did I use the size relationship between the sets, so I am not sure if I did this problem right.

How could I solve this problem with the hint (solve as an equation of x) and the given relationship between A, B and C?

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# Problem Involving Counting of Elements in Three Sets

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