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Let's solve the two questions visible in the image:

### Q24: 
A pipe can fill a tank in 15 hours, while a leak at three-fifths the height of the tank can empty that part in 4 hours. If both are opened simultaneously, and the tank is initially full, when will the tank be three-fifths full?

#### Solution:
1. The pipe can fill the tank in 15 hours. Therefore, in 1 hour, it fills \(\frac{1}{15}\) of the tank.
2. The leak empties three-fifths of the tank in 4 hours. This means it can empty \(\frac{3}{5}\) in 4 hours. So, in 1 hour, the leak empties \(\frac{3}{5} \div 4 = \frac{3}{20}\) of the tank.
3. The net filling rate (when both the pipe and the leak are working) is the filling rate of the pipe minus the emptying rate of the leak:  
   \[
   \text{Net rate} = \frac{1}{15} - \frac{3}{20}
   \]
   To calculate this, find a common denominator:
   \[
   \frac{1}{15} = \frac{4}{60}, \quad \frac{3}{20} = \frac{9}{60}
   \]
   Thus, the net rate is:
   \[
   \frac{4}{60} - \frac{9}{60} = -\frac{5}{60} = -\frac{1}{12}
   \]
   The negative value means the tank is losing water at a rate of \(\frac{1}{12}\) per hour.

4. Initially, the tank is full, and we want to know when it will be three-fifths full. The change in the tank’s fullness is:
   \[
   \frac{1 - \frac{3}{5}}{ \frac{1}{12}} = \frac{2}{5} \div \frac{1}{12} = \frac{2}{5} \times 12 = \frac{24}{5} = 4.8 \text{ hours}.
   \]

Hence, the tank will be three-fifths full after **4.8 hours**.

---

### Q25:
A train 80 m long is running at a speed of 54 km/h and passes a building. What is the time taken by the train to pass the building?

#### Solution:
1. The length of the train is 80 meters, and the speed of the train is 54 km/h. Convert the speed to meters per second (m/s):
   \[
   54 \, \text{km/h} = 54 \times \frac{1000}{3600} = 15 \, \text{m/s}
   \]
2. The time to pass a building is the time the train takes to cover a distance equal to its own length at its given speed. So the time \( t \) is:
   \[
   t = \frac{\text{Distance}}{\text{Speed}} = \frac{80}{15} = 5.33 \text{ seconds}
   \]

Thus, the time taken by the train to pass the building is approximately **5.3 seconds**.

---

Would you like further details on these calculations, or do you have any questions? Here are 5 additional questions related to this topic:

1. How would the time change in Q24 if the pipe filled the tank in 10 hours instead of 15?
2. In Q24, what would happen if the leak could empty the full tank in 6 hours?
3. In Q25, how would the time change if the train was 100 m long?
4. What if the train in Q25 was moving at 72 km/h instead of 54 km/h? How long would it take to pass the building?
5. How would you solve Q25 if the building had a width, and the train had to pass the entire structure?

**Tip:** Always remember to convert units consistently when working with rates and speeds, especially in questions involving time and distance.

Q24: A pipe can fill a tank in 15 hours, while a leak which is at three-fifth the height of the tank (from bottom) can empty up to that part in 4 hours. If both are opened simultaneously and initially the tank is full, when will it be three-fifths full? Q25: A 80 m long train is running at a speed of 54 km/h and passes a building. What is the time taken by the train to pass by the building?

Learn how to solve two key math problems: one involving a tank being filled by a pipe while a leak empties part of it, and another problem where a train passes by a building at a given speed. This solution covers concepts such as time, work rate, speed, and distance, using step-by-step solutions. Suitable for Grades 8-10.

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