Problem: -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?