two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 20 hr . if all pipes are opened simultaneously , then the cistern will be filled in
"solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 − 1 / 20 = 2 / 15 hence , tank will be filled in 15 / 2 = 7.5 hour option ( a )"
inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 20)))