Question 768816
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Suppose the container holds x cubic feet of water.
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a pipe can fill a container with water in 3 hrs.
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So the pipe's fill rate is x cu.ft. per 3 hours. or {{{"x_cu_ft"/"3_hr"}}} or {{{x/3}}}{{{cu_ft/hr}}}
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However, if water is drawn from it at the same time at the rate of 4 cubic ft per hour,
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That means the new slower fill rate is the pipe's original fill rate {{{"x_cu_ft"/"3_hr"}}}
reduced by {{{4_cu_ft/hr}}} to {{{(x/3-4)}}}{{{cu_ft/hr}}}
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it requires 5 hrs to fill it.
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So this new slower fill rate is also x cu.ft. per 5 hours. or {{{"x_cu_ft"/"5_hr"}}} or {{{x/5}}}{{{cu_ft/hr}}}.

So we equate the two expressions for the new slower fill rate:

{{{(x/3-4)}}}{{{cu_ft/hr}}} = {{{x/5}}}{{{cu_ft/hr}}}. 

{{{x/3}}}{{{""-""}}}{{{4}}} {{{""=""}}} {{{x/5}}}

Multiply thru by LCD of 15

5x - 60 = 3x

     2x = 60

      x = 30 cu ft, which is how much the container holds.
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At what rate is the pipe delivering water?
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Since the pipe's fill rate is {{{x/3}}}{{{cu_ft/hr}}}, that becomes:

{{{30/3}}}{{{cu_ft/hr}}}, or 10{{{cu_ft/hr}}},

Edwin</pre>