two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 25 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 / 25 = 1 / 7 hence , tank will be filled in 6.98 hour option ( b )"
inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 25)))