Solve the problem.
-A tank initially contains 120 gal of brine in which 40 lb of salt are dissolved. A brine containing 4 lb/gal of salt runs into the tank at the rate of 8 gal/min. The mixture is kept uniform by stirring and flows out of the tank at The rate of 5 gal/min. Write, in standard form, the differential equation that models the mixing process.
A) $\frac { d y } { d t } = 32 - \frac { 5 } { 120 + 3 t } y$
B) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 5 } { 120 - 3 \mathrm { t } } \mathrm { y } = 32$
C) $\frac { d y } { d t } + \frac { 5 } { 120 + 3 t } y = 32$
D) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 8 } { 120 + 3 \mathrm { t } } \mathrm { y } = 20$

Question

Solve the problem.
-A tank initially contains 120 gal of brine in which 40 lb of salt are dissolved. A brine containing 4 lb/gal of salt runs into the tank at the rate of 8 gal/min. The mixture is kept uniform by stirring and flows out of the tank at The rate of 5 gal/min. Write, in standard form, the differential equation that models the mixing process. 
A) $\frac { d y } { d t } = 32 - \frac { 5 } { 120 + 3 t } y$
B) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 5 } { 120 - 3 \mathrm { t } } \mathrm { y } = 32$
C) $\frac { d y } { d t } + \frac { 5 } { 120 + 3 t } y = 32$
D) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 8 } { 120 + 3 \mathrm { t } } \mathrm { y } = 20$

Quizplus · Accepted Answer

Let y be the amount of salt in the tank at time t.
The volume of brine in the tank is V = 120 + 3t (since brine is flowing in at a rate of 8gal/min and flowing out at a rate of 5 gal/min, so the net increase in volume per minute is 3 gal/min).
The rate of change of the amount of salt in the tank is given by:
$\frac{dy}{dt} = \frac{	ext{salt in}}{	ext{time in}} - \frac{	ext{salt out}}{	ext{time out}}$

The amount of salt flowing in per minute is:
$$ (4 	ext{ lb/gal})\cdot (8 	ext{ gal/min}) = 32 	ext{ lb/min} $$

The amount of salt in the tank at time t is y, and the concentration of salt in the brine flowing out of the tank is $\frac{y}{V}$, so the amount of salt flowing out per minute is:
$$ (5 	ext{ gal/min})\cdot \frac{y}{V} 	ext{ lb/gal} = (5 	ext{ gal/min})\cdot \frac{y}{120+3t} 	ext{ lb/min} $$

Putting it all together, we get:
$$ \frac{dy}{dt} = 32 - \frac{5y}{120+3t} $$

which is equivalent to choice C.

Solve the Problem dydt=32−5120+3ty\frac { d y } { d t } = 32 - \frac { 5 } { 120 + 3 t } ydtdy​=32−120+3t5​y

Solve the Problem $\frac { d y } { d t } = 32 - \frac { 5 } { 120 + 3 t } y$