Question 586013
The way I do it is:
Let {{{a}}} = number of $1 bills
Let {{{b}}} = number of $5 bills
Let {{{c}}} = number of $10 bills
Let {{{d}}} = number of $20 bills
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given:
(1) {{{ a/b = 5/4 }}}
(2) {{{ c/d = 2/1 }}}
(3) {{{ 1*a + 5*b + 10*c + 20*d = 6795 }}}
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This is 4 unknowns and only 3 equations,
so it isn't completely solvable, but the 
information can still be there
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(1) {{{ 4a = 5b }}}
(1) {{{ b = (5/4)*a }}}
and
(2) {{{ c = 2d }}}
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The possible combinations of $1s and $5s is:
5 $1s
4 $5s
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10 $1s
8 $5s
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15 $1s = $15
12 $5s = $60
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Whatever combination I choose, the remainder of the money
must be divisible by $40 ( 1 $20 and 2 $10s )
{{{ 6795 - 75 = 6720 }}}
{{{ 6720/40 = 168 }}}
 this seems to work
168 $20s = $3360
336 $10s = $3360
12 $5s = $60
15 $1s = $15
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sum = $6795
I don't know if another combination will work.
This one seems to.