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A tank contains 60 gallons of salt water with 120 lb of dissolved salt. salt water with 3 lb of salt per gallon flows into the tank at the rate of 2 gallons/min.,and the mixture ,kept uniform by stirring flows out at the same rate. how long will it be(in min.) before there are 135 lb of salt in the tank?

My attempt—

Let at any time t the quantity of salt in the tank be x lb. now in a infinitesimal time dt let a amount dv of water enter into the tank.the salt content now will become x+dv*3,now as per question dv equals 2dt,so the equation becomes x+dx=x+6dt,which gives dx=6dt.now as the mixture is continuously stirred,the salt concentration after adding this volume of water is $\frac{x+6dt}{60+2dt}$.so the amount of salt leaving the tank equals $\frac{x+6dt}{60+2dt}$*2dt.hence the actual increase in the salt content is 6dt -$\frac{x+6dt}{60+2dt}$*2dt.this is not a good equation and I couldnot go ahead with it.please help me out.thanks in advance.

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  • $\begingroup$ Sure, an amount of 3dv=6dt arrives during the time interval (t,t+dt), but also, an amount of (x/60)dv=(x/30)dt leaves, hence dx=(6-(x/30))dt, that is, x'(t)=6-(x(t)/30). Can you continue? $\endgroup$ – Did Sep 4 '16 at 8:25
  • $\begingroup$ @Did how can an amount of (x/60)dv=(x/30)dt leave it should be (x+6dt/60+dv)dv=(x/30)dt,because we should also consider the water that has entered when deceiding what should leave $\endgroup$ – Navin Sep 4 '16 at 8:31
  • $\begingroup$ Nope -- the total quantity of salt is still x(t), neglecting an infinitesimal hence, assuming uniform mixture, the quantity that leaves is indeed (x/60)dv, up to higher order terms which will not play into the differential equation. $\endgroup$ – Did Sep 4 '16 at 8:34
  • $\begingroup$ @Did thanks.i got it.i was confused in that only. $\endgroup$ – Navin Sep 4 '16 at 8:43

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