Let's break down and solve the problem step by step. --- ### **Step 1: Define Variables** Let \( y(t) \) = amount of salt (in kg) in the tank at time \( t \) (in minutes). Initial salt: \( y(0) = 60 \) kg. Initial volume: \( V_0 = 1000 \) L. Inflow rate: 10 L/min (pure water, no salt). Outflow rate: 5 L/min. --- ### **Step 2: Volume of Solution in Tank** Since more water enters than leaves, the volume in the tank is increasing: \[ V(t) = 1000 + (10 - 5)t = 1000 + 5t \quad \text{(L)} \] --- ### **Step 3: Rate of Change of Salt** - **Salt in:** \( 0 \), since inflow is pure water. - **Salt out:** Salt leaves with the outflow. The concentration is \( \frac{y(t)}{V(t)} \), and outflow is 5 L/min: \[ \text{Salt out per minute} = 5 \times \frac{y(t)}{V(t)} \] So, \[ \frac{dy}{dt} = 0 - 5 \frac{y(t)}{V(t)} = -5 \frac{y(t)}{1000 + 5t} \] --- ### **Step 4: Solve the Differential Equation** \[ \frac{dy}{dt} = -\frac{5y}{1000 + 5t} \] This is a first-order linear ODE. Separate variables: \[ \frac{dy}{y} = -\frac{5}{1000 + 5t} dt \] \[ \frac{dy}{y} = -\frac{1}{200 + t} dt \] Integrate both sides: \[ \int \frac{1}{y} dy = -\int \frac{1}{200 + t} dt \] \[ \ln|y| = -\ln|200 + t| + C \] \[ \ln|y| + \ln|200 + t| = C \] \[ \ln|y(200 + t)| = C \] \[ y(200 + t) = A \] Let’s solve for \(A\) using the initial condition \(y(0) = 60\): \[ 60 \times 200 = A \implies A = 12000 \] So, \[ y(t) = \frac{12000}{200 + t} \] --- ### **Step 5: Find Amount of Salt After 5 Hours** 5 hours = 300 minutes. \[ y(300) = \frac{12000}{200 + 300} = \frac{12000}{500} = 24 \text{ kg} \] --- ## **Final Answer** \[ \boxed{24 \text{ kg}} \] is the amount of salt in the tank after 5 hours.