# Formula for calculating rate drain

A question which I am having difficulty with is:

A tank is 3/4 full. Fillpipe A can fill the tank in 12 min. Drainpipe B can empty the tank in 8 min. If both pipes are open, how long will it take to empty the tank?

Is there a specific formula for such questions? Since both the pipes are open I am confused on how to work through this.

Here is what i tried so far: 3/4 of tank fills in 9 minutes. 3/4 of tank drain in 6 minutes. so (1/9) + (1/6) = 5/18 Thus 5/18 part of the job (fill and drain together.. make sense?) is done in one minute and one job is done in : 18/5 = 3.6 minutes..

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Hint: Calculate the rate $a$ at which $A$ is filling the tank and the rate $b$ at which $B$ is emptying the tank, in units of tanks per minute. Then you can pretend that the two pipes $A$ and $B$ are a single pipe that is filling the tank at a rate of $a-b$ tanks per minute. –  MJD Jun 3 '12 at 5:41
Corrected the mistake. –  Rajeshwar Jun 3 '12 at 5:41
@MarkDominus Let me try that.. –  Rajeshwar Jun 3 '12 at 5:42
One error you made is to add together the filling pipe and the emptying pipe as if they were working in the same direction. But they are working against each other, so you should have subtracted. –  MJD Jun 3 '12 at 5:58
Your other error was to calculate the rate of filling and emptying in units of $\frac{\rm minutes}{\rm 3/4\ tank}$, instead of in units of $\frac{\rm minutes}{\rm tank}$. This is likely to trip you up in future problems, although in this case it works okay: $\frac19 - \frac16 = -\frac1{18}$ of the job is done in one minute and the whole job is done in 18 minutes. –  MJD Jun 3 '12 at 6:00

The trick with all such questions is to convert the information about the fill and empty rates from the form you get, which is in units of minutes per tank, or in general $\frac{\rm time}{\rm volume}$, to the reciprocal units of $\frac{\rm volume}{\rm time}$, in this case tanks per minute, because such units can be added and subtracted.

For example, if you have one man who digs a hole in two days and another who digs a hole in three days, you convert the units of $\frac{\rm days}{\rm holes}$ to the reciprocal $\frac{\rm holes}{\rm day}$: we are given that the first man digs at a rate of $\frac{\rm 2\ days}{\rm 1\ hole}$, and when we take the reciprocal we get that the first man digs $\frac{\rm1\ hole}{\rm2\ days} = \frac12\frac{\rm holes}{\rm day}$. Similarly the second man digs $\frac13$ holes per day. Working together, they can dig $\frac12 + \frac13 = \frac 56$ holes per day, and therefore they take $\frac65$ days to dig one hole together.

You can easily convert this technique into a formula for adding rates, but I find it more natural to just convert the units.

Spoiler below.

$A$ fills at a rate of $\frac1{12}$ tanks per minute; $B$ empties at a rate of $\frac18$ tanks per minute. The net rate of filling is $A-B = \frac1{12} - \frac18 = -\frac1{24}$ tanks per minute. If the tank were full, it would empty in 24 minutes, but it is only $\frac34$ full, so it empties in only $\frac34\cdot 24 = 18$ minutes.

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Just edited my question.. Did apply the work formula (rate of work which is job/time) still am stuck.. –  Rajeshwar Jun 3 '12 at 5:52
Thanks for the spoiler. I kind of tried to do this approach but got confused in the way. Thanks for the tip and off course the working!!.. –  Rajeshwar Jun 3 '12 at 5:56
out of curiosity (1/12)-(1/8)=(-1/24). What does the negative sign represents in -1/24. Does it show that rate of drain is faster than fill rate ?? –  Rajeshwar Jun 3 '12 at 6:00
Yes, exactly so. –  MJD Jun 3 '12 at 6:01
I wish I could double-upvote your answer. An excellent explanation of transition to a more natural additive units, plus spoiler-protected answer! –  valdo Jun 3 '12 at 7:32