2. A different aquarium should contain 15,000 liters of water when filled correctly. It w overflow if it gets to 17,600 liters. One day there is an accident and the tank cracks in 4 places. Water flows out of each crack at a rate of \( \frac{1}{2} \) liter per hour. An emergency pump can re-fill the tank at a rate o liters per minute. How many minutes must the pump run to replace the water lost hour?

To determine how long the emergency pump needs to run to replace the water lost in one hour, let's break down the problem step by step. ### Given: 1. **Water Loss:** - The tank has **4 cracks**. - **Each crack leaks at a rate of** \( \frac{1}{2} \) **liter per hour**. - **Total water lost per hour**: \[ 4 \text{ cracks} \times \frac{1}{2} \text{ liter/hour per crack} = 2 \text{ liters/hour} \] 2. **Pump Refill Rate:** - The emergency pump can refill the tank at a rate of **\( o \) liters per minute**. ### Objective: Find the **time (\( t \)) in minutes** that the pump must run to replace the **2 liters** lost in one hour. ### Calculation: To replace **2 liters** of water: \[ \text{Time} \times \text{Pump Rate} = \text{Water Lost} \] \[ t \times o = 2 \text{ liters} \] Solving for \( t \): \[ t = \frac{2}{o} \text{ minutes} \] ### Conclusion: The emergency pump must run for \( \frac{2}{o} \) minutes to replace the water lost in one hour. **Final Answer:** The pump must run for 2 divided by o minutes. Thus, the required time is 2 ⁄ o minutes.