Question 1054900
Rate of filling with both taps open:
[ 1 tank filled ] / [ 9.375 hrs ]
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Let {{{ t }}} = time in hrs for smaller tap to fill tank
{{{ t - 10 }}} = time in hrs for larger tap to fill tank
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Add the rates of filling of the 2 taps to get
the rate of filling with both taps open
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{{{ 1/t + 1/( t-10 ) = 1/9.375 }}}
Multiply both sides by {{{ t*( t-10 )*9.375 }}}
{{{ 9.375*( t-10 ) + 9.375t = t*( t-10 ) }}}
{{{ 9.375t - 93.75 + 9.375t = t^2 - 10t }}}
{{{ t^2 - 10t - 18.75t = -93.75 }}}
{{{ t^2 - 28.75t + 93.75 = 0 }}}
Use quadratic formula to solve for {{{ t }}}
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -28.75 }}}
{{{ c = 93.75 }}}
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{{{ t = (-(-28.75) +- sqrt( (-28.75)^2 - 4*1*93.75 )) / (2*1) }}}
{{{ t = ( 28.75 +- sqrt(  826.5625 - 4*1*93.75 )) / 2 }}}
{{{ t = ( 28.75 + sqrt( 826.5625 - 375 )) / 2 }}}
{{{ t = ( 28.75 + sqrt( 451.5625 ))/2 }}}
{{{ t = ( 28.75 + 21.25 ) / 2 }}}
{{{ t =( 50/2 ) }}}
{{{ t = 25 }}} hrs
and
{{{ t - 10 = 15 }}}
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The smaller tap takes 25 hrs by itself
The larger tap takes 15 hrs by itself
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check:
{{{ 1/t + 1/( t-10 ) = 1/9.375 }}}
{{{ 1/25 + 1/15 = 1/9.375 }}} 
{{{ .04 + .06667 = .106667 }}}
{{{ .106667 = .106667 }}}
OK