Question 672642
From 9 AM to noon is 3 hrs
Let {{{ x }}} = number of hours the main pump
is used pumping by itself
{{{ 3 - x }}} = number of hours that both pumps are pumping
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The main pump's rate of pumping is ( 1 tanker / 4 hrs )
the auxilliary pump's rate is ( 1 tanker / 9 hrs )
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( fraction of tanker pumped in x hrs ) + 
( fraction pumped by both pumps in 3-x hrs ) = 1 whole tanker pumped
{{{ (1/4)*x + ( 1/9 + 1/4 )*( 3-x ) = 1 }}} 
Multiply both sides by {{{ 36 }}}
{{{ 9x + ( 4 + 9)*( 3-x ) = 36 }}}
{{{ 9x + 13*( 3-x ) = 36 }}}
{{{ 9x + 39 - 13x = 36 }}}
{{{ -4x = -3 }}}
{{{ x = .75 }}}
{{{ 3 - x = 2.25 }}}
The auxiliary pump should be started at 9:45
check answer:
In 3/4 of an hour the main pump empties
{{{ (1/4)*(3/4) = 3/16 }}} of the tanker
In 2.25 hrs, both together empty
{{{ ( 1/4 + 1/9 )*(9/4) }}}
{{{ ( 9/36 + 4/36 )*81/36 }}}
{{{ ( 13/36 )*( 81/36 ) }}}
{{{ ( 13*81 ) / ( 36^2 ) }}}
{{{ 1053 / 1296 }}}
{{{ 13/16 }}}
and {{{ 13/16 + 3/16 = 1 }}}
OK