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Question Number 135420 by liberty last updated on 13/Mar/21

Algebra

Algebra Pipe A can fill a tank in two hours and pipe B can fill it in half the time it takes pipe C to empty it. When all 3 are opened, it takes 1.5 hours to fill the pool how much time is required for pipe 'C to empty the tank?\n

Answered by john_santu last updated on 13/Mar/21

(1)pipe A ⇒V_1 = (W/2)  (2) pipe C ⇒ V_2  = (W/a)  (3) pipe B ⇒V_3  = (W/(a/2))  (4) When all 3 pipe are opened  ⇒ (W/(V_1 +V_2 +V_3 )) = (3/2)  ⇒(W/((W/2)+(W/a)+((2W)/a))) = (3/2)  ⇒ (1/2)+(3/a) = (2/3) ; (3/a) = (1/6)  we get a = 18 hour

(1)pipeAV1=W2 (2)pipeCV2=Wa (3)pipeBV3=Wa/2 (4)Whenall3pipeareopened WV1+V2+V3=32 WW2+Wa+2Wa=32 12+3a=23;3a=16 wegeta=18hour

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