L.C.M. of 36 and 45 = 180. The total work = 180 parts. The number of parts filled by pipe P in one minute = 5 The number of parts filled by pipe Q in one minute = 4 Let the number of minutes, pipes P and Q together opened be M, and the number of minutes pipe P only opened be N. M(5+4)+N×5=180 –––(1) M+N=27 –––(2) By solving (1) and (2), we get M=11