SOLUTION: John has 27 coins totaling $5.70. If he only has dimes and quarters, how many of each coin does he have?

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: John has 27 coins totaling $5.70. If he only has dimes and quarters, how many of each coin does he have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 884303: John has 27 coins totaling $5.70. If he only has dimes and quarters, how many of each coin does he have?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a) He can't have an odd number of quarters,
since that would make the total look like @@.@5
------------------------------------------
(b) He can't have 23 or more quarters, since
that is $5.75 or greater
-----------------------
Let +d+ = number of dimes he has
Let +q+ = number of quarters he has
-------------
(1) +d+%2B+q+=+27+
(2) +10d+%2B+25q+=+570+ ( in cents )
--------------------------------
Multiply both sides of (1) by +10+
and subtract (1) from (2)
(2) +10d+%2B+25q+=+570+
(1) +-10d+-+10q+=+-270+
--------------------------
+15q+=+300+
+q+=+20+
and, since
(1) +d+%2B+q+=+27+
(1) +d+%2B+20+=+27+
(1) +d+=+7+
He has 7 dimes and 20 quarters
check:
(2) +10%2A7+%2B+25%2A20+=+570+
(2) +70+%2B+500+=+570+
(2) +570+=+570+
OK