in the junior basketball league there are 18 teams , 2 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ?
otal teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . answer : e
add(multiply(18, divide(const_1, const_2)), const_1)