Question 726700
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Here it is with explanation and also without rounded-off decimals.
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Two pipes can fill up a water tank in 6 hours and 40 minutes. 
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"6 hours and 40 minutes" is {{{6&40/60}}} hours or {{{6&2/3}}} hours. 

Their combined tank-filling rate is 1 tank per {{{6&2/3}}} hours, or

{{{1_tank/6&2/3hr}}} = {{{1/(6&2/3)}}}{{{tank/hr}}} = {{{1}}}{{{"÷"}}}{{{6&2/3}}}{{{tank/hr}}} = {{{1}}}{{{"÷"}}}{{{20/3}}}{{{tank/hr}}} = {{{1*3/20}}}{{{tank/hr}}} = {{{3/20}}}{{{tank/hr}}} 
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Find the time each will take to fill the tank 
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Let x = the number of hours it takes the slower pipe to fill the tank
alone.

The slower pipe's tank-filling rate is 1 tank per x hours, or

{{{1_tank/(x_hr)}}} = {{{1/x}}}{{{tank/hr}}} 

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 if one of the two pipes can fill it is three hours less than the other.
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Then x-3 = the number of houts it takes the faster pie to fill the tank alone.

The faster pipe's tank-filling rate is 1 tank per x-3 hours, or {{{1_tank/(x+3_hr)}}} = {{{1/(x+3)}}}{{{tank/hr}}}

The equation comes from

          {{{(matrix(5,1,
SLOWER,"PIPE'S",TANK,FILLING,RATE))}}}{{{""+""}}}{{{(matrix(5,1,
FASTER,"PIPE'S",TANK,FILLING,RATE))}}}{{{""=""}}}{{{(matrix(5,1,
THEIR,COMBINED,TANK,FILLING,RATE))}}}

                    {{{1/x}}}{{{""+""}{}}{{{1/(x-3)}}}{{{""=""}}}{{{3/20}}}

Multiply through by 20x(x-3)

                    20(x-3) + 20x = 3x(x-3)
                     20x-60 + 20x = 3x²-9x
                           40x-60 = 3x²-9x
                                0 = 3x²-49x+60
                                0 = (x-15)(3x-4)

                                    x-15 =  0;   3x-4 = 0
                                       x = 15;     3x = 4
                                                    x = {{{4/3}}}
                                                    x = {{{1&1/3}}}

We discard the x = {{{1&1/3}}} because the slower's rate cannot be faster 
that the combined rate.

Answer:  The slower one takes 15 hours and the faster one 3 hours less,
or 12 hours.

Edwin</pre>