If 3 "A" cans can hold 4 "B" cans and 3 "B" cans hold 5 "C" cans than how many "A" cans are need to hold 20 "C" cans ?

3a=4b

3b=5c

20c = 4(5c) = 4(3b) = 12b = 12(3a/4) = 9a