A can fill in 5min and B in 20min.if both taps are opened then due to a leakage it took 30min more to fill tank.if tank is full,how long will it take for the leakage alone to empty the tank?
A&B together take 1/(1/5 + 1/20) = 4 minutes to fill the tank.
So, if the leak takes x minutes to empty the tank,
1/4 - 1/x = 1/(30+4)
x = 68/15 = 4 8/15 minutes
sir answer 36 hr. but i dont know how?
Hmmm. I'm not sure I agree. Since A&B can fill the tank so fast, and it takes about 9 times as long to fill it with the leak, the leak must be pretty fast in relation to the fill speeds.
You sure that 5 minutes is right?
To solve this problem, we need to determine the rate at which each tap fills the tank and the rate at which the leakage empties the tank.
Let's start by finding the filling rate of tap A. We know that tap A can fill the tank in 5 minutes, so its filling rate is 1/5 of the tank per minute.
Similarly, tap B can fill the tank in 20 minutes, so its filling rate is 1/20 of the tank per minute.
Now, let's determine the combined filling rate of taps A and B. Since they are both open, their filling rates will add up. Therefore, the combined filling rate is 1/5 + 1/20 = 1/4 of the tank per minute.
We know that due to the leakage, it took an additional 30 minutes to fill the tank. This means that in those 30 minutes, the combined filling rate of taps A and B was cancelled out by the leakage rate.
To find the rate at which the leakage empties the tank, we can calculate the negative filling rate. Since it took 30 minutes for the leakage to cancel out the filling rate, the leakage rate must be equal to 1/4 of the tank per minute (negative rate).
Therefore, it will take the leakage alone 4 minutes to empty the tank completely.