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Question Number 34850 by mondodotto@gmail.com last updated on 11/May/18

Answered by MJS last updated on 11/May/18

A.......(1/3) tank/min.  B.......(1/4) tank/min.  ((1/3)+(1/4))×t_1  min.=(1/2) tank  (7/(12))t_1 =(1/2)  t_1 =(6/7) min.  (1/4)×t_2  min.=(1/2) tank  (1/4)t_2 =(1/2)  t_2 =2 min.  t=t_1 +t_2 =((20)/7) min.≈2min51s

A.......13tank/min.B.......14tank/min.(13+14)×t1min.=12tank712t1=12t1=67min.14×t2min.=12tank14t2=12t2=2min.t=t1+t2=207min.2min51s

Answered by candre last updated on 11/May/18

let call the tank capacity T  v_A =T/3  v_B =T/4  v(t)=(v_A +v_B )t   [t min]  ((T/3)+(T/4))t=(T/2)  ((7T)/(12))t=(T/2)  t=(T/2)∙((12)/(7T))=(6/7)  the time B fills the half is  v_B t=(T/2)  t=(T/2)∙(4/T)=2  the total time is  2+(6/7)=((20)/7)≈2,85min  ≈2min51,43s

letcallthetankcapacityTvA=T/3vB=T/4v(t)=(vA+vB)t[tmin](T3+T4)t=T27T12t=T2t=T2127T=67thetimeBfillsthehalfisvBt=T2t=T24T=2thetotaltimeis2+67=2072,85min2min51,43s

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