A tank contains 4L of water at first. Tap A is turned on and water from tap A flows into the container at the rate of 8L per minute. After 3 minutes, Tap B is turned on. Both taps are turned off together after another 5 minutes. If the total amount of water in the container is 77L after both taps are turned off, what is the rate of the water flowing from Tap B?
A's rate --- 8 L/min
B's rare --- x+8 L/min
combined rate = (x+8) L/min
for first 3 minutes, amount of water = 24 L
next 5 minutes, amount of water = 5(x+8) or 5x+40 L
so , 4 + 24 + 4x+40 = 77
4x = 9
x = 9/4 L/min or 2.25 L/min
B's rate is 2.25 L/min
B's rate ---- x L/min
To find the rate of water flowing from Tap B, we need to determine the amount of water Tap A contributed before Tap B was turned on and subtract that from the total amount of water in the tank after both taps were turned off.
1. Calculate the amount of water Tap A contributed:
- Tap A was turned on for 3 minutes, so it delivered 8 liters/min * 3 min = 24 liters of water.
2. Calculate the total amount of water in the tank after Tap B was turned off:
- Tap A was on for a total of 3 minutes and Tap B was on for an additional 5 minutes, so the total time both taps were on is 3 min + 5 min = 8 minutes.
- The total amount of water in the tank after both taps were turned off is 77 liters.
3. Subtract the amount of water Tap A contributed from the total amount of water to find the amount of water Tap B contributed:
- Tap B contributed 77 liters - 24 liters = 53 liters of water.
4. Calculate the rate of water flowing from Tap B:
- Tap B was turned on for 5 minutes, so the rate of water flowing from Tap B is 53 liters / 5 min = 10.6 liters per minute.
Therefore, the rate of water flowing from Tap B is 10.6 liters per minute.