in township k , 1 / 5 of the housing units are equipped with cable television . if 1 / 10 of the housing units , including 1 / 3 of those that are equipped with cable television , are equipped with videocassette recorders , what fraction of the housing units have neither cable television nor videocassette recorders ?
let total housing units = 5 x . as per question , equipped with cable tv = x equipped with vcr = x / 2 , which includes x / 3 ctv . so , only vcr = x / 2 - x / 3 = x / 6 . so , cable tv + vcr = x + x / 6 = 7 x / 6 no cable tv and vcr = 5 x - 7 x / 6 = 23 x / 6 . fraction = 23 x / ( 6 * 5 x ) = 23 / 30 . hence , a
add(subtract(subtract(const_1, divide(1, 5)), divide(1, 10)), multiply(divide(1, 5), divide(1, 3)))