Question 980127
I picture these liquids in the same container,
but I am able to separate them from eachother
and recombine them in different amounts
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Let {{{ a}}} = liters of 1st liquid originally
Let {{{ b }}} = liters of 2nd liquid originally
Let {{{ c }}} = liters of 3rd liquid originally
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(1) {{{ a + b + c = 700 }}}
(2) {{{ a/b = 2/3 }}}
(3) {{{ b/c = 4/5 }}}
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It sounds like {{{ a + b }}} is going to stay the sme
and also {{{ c }}} will stay the same
The new {{{ a }}} and {{{ b }}} is {{{ a[1] }}} and {{{ b[1] }}}
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(4) {{{ a + b = a[1] + b[1] }}}
(5) {{{ a[1]/b[1] = 6/5 }}}
(6) {{{ a[1] + b[1] + c = 700 }}}
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(2) {{{ a = (2/3)*b }}}
(3) {{{ b = (4/5)*c }}}
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Substitute (3) into (2)
(2) {{{ a = (8/15)*c }}}
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(1) (8/15)*c + (4/5)*c + c = 700 }}}
(1) {{{ (8/15)*c + 12/15)*c + (15/15)*c = 700 }}}
(1) {{{ (35/15)*c = 700 }}}
(1) {{{ c = (3/7)*700 }}}
(1) {{{ c = 300 }}}
and
(3) {{{ b = (4/5)*c }}}
(3) {{{ b = (4/5)*300 }}}
(3) {{{ b = 240 }}}
and
(2) {{{ a = (8/15)*c }}}
(2) {{{ a = (8/15)*300 }}}
(2) {{{ a = 160 }}}
check:
(1) {{{ a + b + c = 700 }}}
(1) {{{ 160 + 240 + 300 = 700 }}}
(1) {{{ 700 = 700 }}}
OK
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(4) {{{ a + b = a[1] + b[1] }}}
(4) {{{ 160 + 240 = a[1] + b[1] }}}
(4) {{{ a[1] + b[1] = 400 }}}
and
(5) {{{ a[1]/b[1] = 6/5 }}}
(5) {{{ a[1] = (6/5)*b[1] }}}
Plug this result into (4)
(4) {{{ (6/5)*b[1] + b[1] = 400 }}}
(4) {{{ (11/5)*b[1] = 400 }}}
(4) {{{ b[1] = 2000/11 }}}
(4) {{{ b[1] = 181.8182 }}}
and
(5) {{{ a[1] = (6/5)*(2000/11) }}}
(5) {{{ a[1] = 12000/55 }}}
(5) {{{ a[1] = 218.1818 }}}
check:
(4) {{{ a[1] + b[1] = 400 }}}
(4) {{{ 181.8182 + 218.1818 = 400 }}}
(4) {{{ 400 = 400 }}}
OK
check:
(6) {{{ a[1] + b[1] + c = 700 }}}
(6) {{{ 400 + 300 = 700 }}}
(6) {{{ 700 = 700 }}}
OK
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Finally:
{{{ a[1] / b[1] = 218.1818/181.8182 }}}
{{{ a[1]/b[1] = 1.2 }}}
{{{ 1.2 = 12/10 }}}
{{{ 12/10 = 6/5 }}}
OK
So, the amounts of the new {{{a}}} and {{{b}}} are:
218.1818 liters and 181.8182 liters