A1,a2,-each of these 30 sets has 5 elements. b1,-each of these n sets has 3 elements. union of a1,=union of b1,=s.
if each elements of s is in 10 a sets and 9 bsets. then n=?

 the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x.  Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set.  if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times.  This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45

If jon had $4.00 then takes away $2.25, then he should have $1.75 left. $4.00 - $2.25 = $1.75 hope this ; )