Question 651706
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Hi,
49 coins...
five more dimes{{{highlight(d)}}} than all the nickels{{{highlight(n)}}} and quarters{{{highlight(q)}}} combined
 q + n + d = 49      || {{{d = (q+n)+5}}}
 q + n + ((q+n)+5) = 49
       2q + 2n = 44
        q + n = 22   0r {{{n = (22-q)}}}  And... the number of dimes = 27  
Question states*** value of $5.20
CENTS makes sense
 25q + 10d + 5n = 520
 25q + 10·27 + 5(22-q) = 520
 25q + 10·27 + 5·22- 5q = 520
                20q = 140
                  q = 7, the number of quaters.  Number of Nickels = 15
and...
{{{25*7 + 10*27 + 5*15 = 520}}}