two taps F And E are connected to a water tank.tap F can fill the tank in 2hours while E can empty it in 3hours.one day when the tank was a third full of water, the taps were opened at the same time.how long did it take to fill the tank?
each hour, 1/2 - 1/3 = 1/6 of the tank is filled.
That should help some...
Nice
Well, it sounds like Tap F is a real overachiever, filling the tank in just 2 hours. Meanwhile, Tap E seems to have a talent for emptying the tank in 3 hours. Quite the odd couple, huh?
Now, let's think about this. If the taps were opened at the same time and the tank was a third full, that means two-thirds of the tank still needed to be filled. Tap F is doing its thing, filling up the tank, while Tap E is trying to rain on Tap F's parade by emptying it.
Since Tap F can fill the tank in 2 hours, we can say that it fills one unit of the tank per hour. On the other hand, Tap E can empty one-third of the tank per hour because it empties the entire tank in 3 hours.
Now, if we add their rates together, we get the combined rate of filling up the tank. Tap F's rate of 1 unit per hour minus Tap E's rate of -1/3 units per hour (since it empties) gives us a combined rate of 2/3 units per hour.
Since we need to fill two-thirds of the tank, and the combined rate is 2/3 units per hour, it will take exactly 2 hours to fill the remaining two-thirds.
Therefore, it will take a grand total of 2 hours to fill the tank completely and have Tap E wondering why it even bothered trying to empty it in the first place.
To solve this problem, we'll calculate the rates at which the taps fill or empty the tank.
Let's assume that the capacity of the tank is 1 unit (this makes it easier to work with percentages).
Tap F can fill the tank in 2 hours, which means it can fill 1/2 of the tank per hour (since 1 hour is half of 2 hours).
Tap E can empty the tank in 3 hours, which means it can empty 1/3 of the tank per hour (since 1 hour is one-third of 3 hours).
Now, let's consider the scenario where the tank was one-third full when both taps were opened simultaneously.
This means that there was (1/3) * 1 unit = 1/3 unit of water in the tank initially.
Since tap F can fill 1/2 unit per hour and tap E can empty 1/3 unit per hour, their combined rate of filling the tank would be (1/2 - 1/3) unit per hour.
To calculate how long it takes to fill the tank completely, we'll divide the initial amount of water (1/3 unit) by the rate at which the combined taps fill the tank ((1/2 - 1/3) unit per hour):
(1/3 unit) / ((1/2 - 1/3) unit per hour) = (1/3) / (1/6) = (1/3) * (6/1) = 2 hours.
Therefore, it will take 2 hours to fill the tank completely when both taps F and E are opened.
To solve this problem, we need to calculate the rate at which the taps fill or empty the tank. Once we have the rates, we can determine how long it will take to fill the tank by considering the combined effect of both taps.
First, let's determine the rate at which each tap fills or empties the tank:
- Tap F can fill the tank in 2 hours, so its filling rate is 1 tank / 2 hours = 1/2 tank per hour.
- Tap E can empty the tank in 3 hours, so its emptying rate is 1 tank / 3 hours = 1/3 tank per hour.
Since Tap E empties the tank, we'll consider its rate as negative. Thus, the combined filling rate when both taps are open is:
Combined rate = Filling rate of Tap F - Emptying rate of Tap E
= 1/2 tank per hour - 1/3 tank per hour
= (3/6) tank per hour - (2/6) tank per hour
= 1/6 tank per hour
By considering the combined rate, we can determine how long it will take to fill the tank to one-third of its capacity. Since the tank was initially one-third full, we need to fill an additional two-thirds of the tank.
Time required to fill two-thirds of the tank = (2/3) tank / (1/6) tank per hour
= (2/3) divided by (1/6)
= (2/3) * (6/1)
= 4 hours
Therefore, it will take 4 hours to fill the tank completely when both taps are open.