Janice had a water tank with 2 different sized taps. if she used tap a the tank would empty in 6 hours. if she used tap b the tank would empty in 3 hours. how long would it take the tank to empty if both taps were open
Let's call the time it takes for both taps to empty the tank "x".
When only tap A is open, it empties the tank in 6 hours. So in 1 hour, it empties 1/6th of the tank.
When only tap B is open, it empties the tank in 3 hours. So in 1 hour, it empties 1/3rd of the tank.
When both taps are open, they work together to empty the tank. So in 1 hour, they empty 1/6th + 1/3rd = 1/2 of the tank.
Therefore, to empty the whole tank (which is 1), it would take 2 hours when both taps are open.
To find out how long it would take the tank to empty if both taps were open, you need to determine the combined rate at which taps A and B empty the tank.
Tap A empties the tank in 6 hours, so its emptying rate is 1/6 of the tank per hour (1 tank/6 hours = 1/6 tank/hour).
Tap B empties the tank in 3 hours, so its emptying rate is 1/3 of the tank per hour (1 tank/3 hours = 1/3 tank/hour).
To find the combined emptying rate when both taps A and B are open, you sum up their individual emptying rates:
1/6 + 1/3 = 2/6 + 6/6 = 8/6 = 4/3 tank/hour.
So, the combined emptying rate of both taps A and B is 4/3 tank per hour. To find the time it takes for the tank to empty, we need to invert this rate:
(1/ (4/3)) hours = 3/4 hour.
Therefore, if both taps A and B were open, it would take 3/4 hour or 45 minutes for the tank to empty.