SOLUTION: You find 6 coins in an old desk drawer, consisting entirely of nickels, dimes, and quarters, with a face value of 85 cents. However, the coins all date from 1889 and are worth much

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: You find 6 coins in an old desk drawer, consisting entirely of nickels, dimes, and quarters, with a face value of 85 cents. However, the coins all date from 1889 and are worth much      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 718478: You find 6 coins in an old desk drawer, consisting entirely of nickels, dimes, and quarters, with a face value of 85 cents. However, the coins all date from 1889 and are worth much more than their face value, and you were able to sell all the coins to a coin shop for a total of $31. Suppose you received $5 for each nickel, $4 for each dime, and $7 for each quarter. How many of the coins were quarters?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +q+ = number of quarters
Let +d+ = number of dimes
Let +n+ = number of nickels
given:
(1) +n+%2B+d+%2B+q+=+6+
(2) +5n+%2B+10d+%2B+25q+=+85+ ( in cents )
(3) +500n+%2B+400d+%2B+700q+=+3100+ ( in cents )
------------------------------
There are 3 equations and 3 unknowns, so
it's solvable
Multiply both sides of (1) by +400+
and subtract (1) from (3)
---------------------
(3) +500n+%2B+400d+%2B+700q+=+3100+
(1) +-400n+-+400d+-+400q+=+-2400+
+100n+%2B+300q+=+700+
+n+%2B+3q+=+7+
--------------
Multiply both sides of (1) by +10+ and
subtract (1) from (2)
(2) +5n+%2B+10d+%2B+25q+=+85+
(1) +-10n+-+10d+-+10q+=+-60+
+-5n+%2B+15q+=+25+
+-n+%2B+3q+=+5+
----------------
Add the results of these 2 operations:
+n+%2B+3q+=+7+
+-n+%2B+3q+=+5+
+6q+=+12+
+q+=+2+
Two of the coins were quarters
check answer:
+-n+%2B+3q+=+5+
+-n+%2B+3%2A2+=+5+
+-n+=+-1+
+n+=+1+
and
(1) +n+%2B+d+%2B+q+=+6+
(1) +1+%2B+d+%2B+2+=+6+
(1) +d+=+3+
-------------
(3) +500n+%2B+400d+%2B+700q+=+3100+
(3) +5n+%2B+4d+%2B+7q+=+31+
(3) +5%2A1+%2B+4%2A3+%2B+7%2A2+=+31+
(3) +5+%2B+12+%2B+14+=+31+
(3) +31+=+31+
OK