SOLUTION: Mary has $6.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and eighty-seven more nickels than dimes, how many coins of each type does she have?

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Question 990746: Mary has $6.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and eighty-seven more nickels than dimes, how many coins of each type does she have?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = number of nickels
Let +d+ = number of dimes
Let +q+ = number of quarters
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(1) +5n+%2B+10d+%2B+25q+=+600+ ( in cents )
(2) +d+=+2q+
(3) +n+=+d+%2B+87+
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There are 3 equations and 3 unknowns, so it's solvable
(2) +q+=+d%2F2+
Substitute (2) and (3) into (1)
(1) +5%2A%28+d%2B87+%29+%2B+10d+%2B+25%2A%28d%2F2%29+=+600+
(1) +5d+%2B+435+%2B+10d+%2B+%281%2F2%29%2A25d+=+600+
Multiply both sids by +2+
(1) +10d+%2B+870+%2B+20d+%2B+25d+=+1200+
(1) +55d+=+330+
(1) +d+=+6+
and
(2) +q+=+d%2F2+
(2) +q+=+6%2F2+
(2) +q+=+3+
and
(3) +n+=+d+%2B+87+
(3) +n+=+6+%2B+87+
(3) +n+=+93+
There are 93 nickels
There are 6 dimes
There are 3 quarters
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Check:
(1) +5n+%2B+10d+%2B+25q+=+600+
(1) +5%2A93+%2B+10%2A6+%2B+25%2A3+=+600+
(1) +465+%2B+60+%2B+75+=+600+
(1) +600+=+600+
OK