A tank with a capacity of 74 liters, initially contains 20 kilograms of salt dissolved in 50 liters of water. Another solution, containing 3 kilograms of dissolved salt per liter, runs into the tank at 4 liters every minute and that the mixture (kept uniform by stirring) runs out of the tank at the rate of 1 liter per minute. What is the concentration of the solution in the tank at point of overflow? Intermediate values should be rounded only to two decimal places. Round off the final answer to two decimal places.

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Let y(t)kg be amount of salt and V(t)L be the volume of mixture in the tank at any time t min. ThenV(0)=50L" and "y(0)=20kgIn flow{:[" in flow ",(" out flow ")/(y(t))kg//L],[c_("in ")=3kg//L,C_("out ")=(y(t))/(V(t))],[" Rate in "_("in ")=4L//min," Rate out "=1L//min]:}Rate of change of Voleme of mixture =V_("olume Rate ") in -V_("olume Rateout "){:[(dv)/(dt)=4-1],[=>dv=3dt],[=>int dv=int3dt quad" [Integating to both sides] "],[=>v=3t+c]:}Applying initial Condidion50=V(0)=3xx0+c=>c=50Hence, V(t)=3t+50Now, Rate of change of a mount of salt =C_("in ")R_("ate ") in -C_("out ")R_("ate out "){:[=>(dy(t))/(dt)=3xx4-(y(t))/(V(t))xx1],[=>(dy)/(dt)=12-(y)/(3t+50)quad[" using (1) ] "]:}=>(dy)/(dt)+(1)/(3t+50)y=12This is linear first order differention equation.of form (dy)/(dt)+P(t)y=Q(t)Here P(t)=(1)/(3t+50),Q(t)=12Integrating factor, mu(t)=e^(int p(t)dt)=e^(int(1)/(3t+50))dt{:[=e^((ln(3t+50))/(3))],[=e^(ln(3t+50)^(1//3))[a ln x=ln(x^(a))]],[=(3t+50)^(1//3)[e^(ln x)=x]]:}Multiplying by mu(t) to the differential equation gives{:[(3t+50)^(1//3)(dy)/(dt)+(3t+50)^(1//3)*(1)/(3t+50)y=(3t+50)^(1//3)12],[=>quad(d)/(dt)[(3t+50)^(1//3)y]quad=12(3t+50)^(1//3)]:}Integrating to both sides with respect to t gives{:[(3t+50)^(1//3)y=int12(3t+50)^(1//3)dt],[=>(3t+50)^(1//3 ... See the full answer