Question 807710
Add the rates of filling to get the rate of all 3
pipes filling together
Let {{{ R[1] }}} = the rate of filling of the largest pipe
Let {{{ R[2] }}} = the rate of filling of the middle size pipe
Let {{{ R[3] }}} = the rate of filling of the smallest pipe
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If three pipes are all opened, the total rate is 
( 1 pool filled ) / ( 3 hrs ) = {{{ 1/3 }}}, so
(1) {{{ R[1] + R[2] + R[3] = 1/3 }}}
(2) {{{ R[3] = 3*R[1] }}}
( The largest pipe's rate is 3 times the smallest pipe )
(3) {{{ R[3] = 2*R[2] }}}
( The largest pipe's rate is 2 times the medium size pipe )
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(2) {{{ R[1] = (1/3)*R[3] }}}
and
(3) {{{ R[2] = (1/2)*R[3] }}}
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Substitute these results into (1)
(1) {{{ R[1] + R[2] + R[3] = 1/3 }}}
(1) {{{ (1/3)*R[3] + (1/2)*R[3] + R[3] = 1/3 }}}
(1) {{{ R[3]*( 1/3 + 1/2 + 1 ) = 1/3 }}}
(1) {{{ R[3]*( 2/6 + 3/6 + 6/6 ) = 1/3 }}}
(1) {{{ (11/6)*R[3] = 1/3 }}}
(1) {{{ R[3] = (6/11)*(2/6) }}}
(1) {{{ R[3] = 2/11 }}}
and, since
(2) {{{ R[1] = (1/3)*R[3] }}}
(2) {{{ R[1] = (1/3)*(2/11) }}}
(2) {{{ R[1] = 2/33 }}}
and
(3) {{{ R[2] = (1/2)*R[3] }}}
(3) {{{ R[2] = (1/2)*(2/11) }}}
(3) {{{ R[2] = 1/11 }}}
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The largest pipe can fill ( 2 pools ) / ( 33 hrs ), so
it can fill ( 1 pool ) / ( 16.5 hrs )
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The medium size pipe can fill ( 1 pool ) / ( 11 hrs )
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The smallest pipe can fiill ( 2 pools ) / ( 11 hrs )
It can fill ( 1 pool ) / ( 5.5 hrs )
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Check the answer:
(1) {{{ R[1] + R[2] + R[3] = 1/3 }}}
(1) {{{ 1/16.5 + 1/11 + 1/5.5 = 1/3 }}}
Multiply both sides by {{{ 5.5 }}}
(1) {{{ 1/3 + 1/2 + 1 = (1/3)*5.5 }}}
(1) {{{ 2/6 + 3/6 + 6/6 = 11/6 }}}
(1) {{{ 11/6 = 11/6 }}}
OK