Question 925607
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These problems are like "D=RT" problems, except that the 'distance' is
replaced by 'job to do', The 'job' here is 'one tank'.
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An inlet pipe can fill up the tank in 6 hours 
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So the inlet pipe's filling rate is 

{{{matrix(1,10,

1,tank,per,6,hours,""="",matrix(1,2,1,tank)/matrix(1,2,6,hours),""="",1/6,tank/hour)}}}
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while an outlet pipe can empty the tank in 9 hours. 
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The outlet's emptying rate is considered as a NEGATIVE filling rate.
So the outlet pipe's NEGATIVE filling rate is

{{{matrix(1,10,

-1,tank,per,9,hours,""="",matrix(1,2,-1,tank)/matrix(1,2,9,hours),""="",-1/9,tank/hour)}}}
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One day, the tank was being filled. When it was 1/3 full, a boy opened the outlet pipe. 
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So the fraction of a job left to do is to fill {{{2/3}}} of a tankful.
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How long did it take to fill the rest of the tank?
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Now the filling rate is the sum of the rates of the two pipes:
{{{matrix(1,2,1/6-1/9=3/18-2/18=1/18,tank/hour)}}}

 and the job is {{{2/3}}} of a tank.

Just as in "D=RT" problems we use "TIME=DISTANCE/RATE" we use here 
"TIME=JOB/RATE".  The job is {{{2/3}}} of a tank and the rate is {{{matrix(1,2,1/18,tank/hour)}}}

So,

{{{matrix(1,10,TIME,""="",JOB/RATE,""="",(2/3)/(1/18),""="",(2/3)(18/1),""="",12,hours)}}} 

Edwin</pre>