How does this work?

[allegansveritatem]

In looking for a video on the solution of certain types of word problem I came upon one that presented this problem:


I tried to solve this by reasoning--because I couldn't think of a formula to apply. If the tank when it is intact takes two hours to fill and when it is leaking takes two hour and another 1/3rd of an hour to fill, then every hour it loses 1/6th for an hour's worth of water, so, after 12 hours it will have lost 2 hours worth of water and the tank will be empty. But, that is not correct, at least the way the problem was worked out by the person in the video. Here is how she worked it out:


I am having a hard time getting this. The teacher presented this as a kind of black box into which you plugged in certain data and out comes the answer. She did not explain why you take the reciprocal of the time it takes to fill with a leak and subtract it from the hourly rate of the filling for the intact tank. This move does not seem intuitive to me. What happened here?

 

[Dr.Peterson]

The "= 1" in the equation is nonsense. One way to make sense of this is to replace it with 1/x, where x is the time for the leak alone to drain the tank.

In each hour, the pump is putting in 1/2 tank (that is, its rate is 1/2 tank per hour), but the amount in the tank is actually increasing by 1/(2 1/3) of a tank because it loses 1 tank in 2 1/3 hour (that is, the net increase is 3/7 of a tank per hour). So the amount removed by the leak is the difference of these two, yielding a rate of 1/14 tank per hour, so the leak alone removes the entire tankful in 14 hours. (No, that last sentence is not obvious to me, either, though with thought I see it is true.)

There are several common ways to work out these problems that make more sense, so you might want to find a different source to learn from.

One way, similar to what you saw but easier to follow, is that we know the pump adds 1/2 tank per hour and the leak removes 1/x tank per hour; the net is the difference, 1/2 - 1/x. But we know the net rate is 1/(7/3) = 3/7, so we can solve the equation 1/2 - 1/x = 3/7. This is equivalent to your 1/2 - 3/7 = 1/x.

One advantage of algebra is that we can separate the process of describing the situation mathematically from actually solving it, which makes things easier to get our minds around.

 

[Romsek]

The effective rate that the tank is filled at is \(\displaystyle r=r_{fill}-r_{drain}\)

\(\displaystyle r_{fill} = \dfrac 1 2 ~tank/hr\)

\(\displaystyle r = \dfrac{1}{2\frac 1 3}~tank/hr = \dfrac 3 7~tank/hr\)

\(\displaystyle r_{drain}=r_{fill}-r = \dfrac 1 2 - \dfrac 3 7 = \dfrac{1}{14}~tank/hr\)

\(\displaystyle \text{The leak can drain the tank in $\dfrac{1~tank}{\frac{1}{14}~tank/hr}=14~hr$}\)

It really behooves you to keep track of units. If nothing else you will quickly spot errors because units fail to match.

 

[allegansveritatem]

The "= 1" in the equation is nonsense. One way to make sense of this is to replace it with 1/x, where x is the time for the leak alone to drain the tank.

In each hour, the pump is putting in 1/2 tank (that is, its rate is 1/2 tank per hour), but the amount in the tank is actually increasing by 1/(2 1/3) of a tank because it loses 1 tank in 2 1/3 hour (that is, the net increase is 3/7 of a tank per hour). So the amount removed by the leak is the difference of these two, yielding a rate of 1/14 tank per hour, so the leak alone removes the entire tankful in 14 hours. (No, that last sentence is not obvious to me, either, though with thought I see it is true.)

There are several common ways to work out these problems that make more sense, so you might want to find a different source to learn from.

One way, similar to what you saw but easier to follow, is that we know the pump adds 1/2 tank per hour and the leak removes 1/x tank per hour; the net is the difference, 1/2 - 1/x. But we know the net rate is 1/(7/3) = 3/7, so we can solve the equation 1/2 - 1/x = 3/7. This is equivalent to your 1/2 - 3/7 = 1/x.

One advantage of algebra is that we can separate the process of describing the situation mathematically from actually solving it, which makes things easier to get our minds around.

My brain is going through its nightly near death experience but I think I see what you are saying here...the 1/(2 1/3) is the rate of defective fill for one hour and the 1/2 is the hourly rate when the tank was intact, so subtracting the one from the other gives the rate of loss, i.e., 1/14 of a tank. Thanks for the explanation.
As for the teacher in the video...I think she is more a purveyor of tricks than a dedicated educator. She is aiming her stuff at young students who just want a quick way to get a grade without too much sweat. I can't fault them for that...that was my school days' mode of operation too. But, as the old Dylan song goes, "I was so much older then, I'm younger than that now".

 

[allegansveritatem]

The effective rate that the tank is filled at is \(\displaystyle r=r_{fill}-r_{drain}\)

\(\displaystyle r_{fill} = \dfrac 1 2 ~tank/hr\)

\(\displaystyle r = \dfrac{1}{2\frac 1 3}~tank/hr = \dfrac 3 7~tank/hr\)

\(\displaystyle r_{drain}=r_{fill}-r = \dfrac 1 2 - \dfrac 3 7 = \dfrac{1}{14}~tank/hr\)

\(\displaystyle \text{The leak can drain the tank in $\dfrac{1~tank}{\frac{1}{14}~tank/hr}=14~hr$}\)

It really behooves you to keep track of units. If nothing else you will quickly spot errors because units fail to match.

Thanks for that tip. I will take your advice.