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Given problem is a typical mixing problem
Let y(t) is the salt in tank at time t
The water entering has no salt, rate in =0
The thoroughly mixed solution has concentration of 10L/min . The tank has 1119 L of brine with 15 kg of salt
Rate out =( y/1119) kg/ L * 10 L/min = 10 y/ 1119 kg/mim
Hence, differential equation is
at t=0, y= 15
15= C
c= 15
Therefore, y=
After 20 minutes, t= 20
y= 12.54496 kg