Question 1074537
Let {{{ r }}} = the rate in gal/min of the NEW pump
{{{ r - 10 }}} = the rate in gal/min of the OLD pump
Let {{{ t }}} = time in minutes to to fill the tank with NEW pump
{{{ t + 15 }}} = time in minutes to fill the tank with OLD pump
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[ gallons ] / [ rate in gal/min ] = [ time in minutes ]
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Equation for old tank:
(1) {{{ 500 / ( r - 10 ) = t + 15 }}}
Equation for the new tank:
(2) {{{ 500 / r = t }}}
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Substitute (2) into (1)
(1) {{{ 500 / ( r - 10 ) = 500/r + 15 }}}
Multiply both sides by {{{ r*( r - 10 ) }}}
(1) {{{ 500r = 500*( r - 10 ) + 15*r*( r - 10 ) }}}
(1) {{{ 500r = 500r - 5000 + 15r^2 - 150r ) }}}
(1) {{{ 15r^2 - 150r - 5000 = 0 }}}
(1) {{{ 3r^2 - 30r - 1000 = 0 }}}
{{{ r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 3 }}}
{{{ b = -30 }}}
{{{ c = -1000 }}}
{{{ r = ( -(-30) +- sqrt( (-30)^2-4*3*(-1000) ))/(2*3) }}} 
{{{ r = ( 30 +- sqrt( 900 + 12000 ))/ 6 }}} 
{{{ r = ( 30 +- sqrt( 12900 ))/ 6 }}} 
{{{ r = ( 30 + 113.58 ) / 6 }}}
{{{ r = 143.58 / 6 }}}
{{{ r = 23.93 }}}
and
{{{ r - 10 = 13.93 }}}
The speed of the old pump is 13.93 gal/min
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check answer:
(2) {{{ 500 / r = t }}}
(2) {{{ t = 500/23.93 }}}
(2) {{{ t = 20.89 }}}
and
(1) {{{ 500 / ( r - 10 ) = t + 15 }}}
(1) {{{ 500 / 13.93 = t + 15 }}}
(1) {{{ 35.89 = t + 15 }}}
(1) {{{ t = 20.89 }}}
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Looks good, but check all my math and
get another opinion if you like