SOLUTION: In one evening, a pizza parlor collected 700 coins, for a total of $145. If the coins are all dimes and quarters, how many of each kind are there?

Algebra ->  Systems-of-equations -> SOLUTION: In one evening, a pizza parlor collected 700 coins, for a total of $145. If the coins are all dimes and quarters, how many of each kind are there?      Log On


   



Question 556071: In one evening, a pizza parlor collected 700 coins, for a total of $145. If the coins are all dimes and quarters, how many of each kind are there?
Found 2 solutions by TutorDelphia, josmiceli:
Answer by TutorDelphia(193) About Me  (Show Source):
You can put this solution on YOUR website!
Dimes and quarters add together to equal 700 coins:
d+q=700
Dimes are worth 10 cents each and quarters 25 and if we multiply how many we have times their value and add it together we'll get the amount of money we have
.10d+.25q=145
Solve the first equation for d
d=700-q
Sub 700-q in for d in the second equation:
.10*(700-q)+.25q=145
Distribute
70-.10q+.25q=145
combine like terms
70+.15q=145
subtract 70 from both sides
.15q=75
divide .15 from both sides
q=500
Plug in 500 for q
d+500=700
d=200

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let d = number of dimes
Let q = number of quarters
given:
(1) +d+%2B+q+=+700+
(2) +10d+%2B+25q+=+14500+ ( in cents )
---------------
Multiply both sides of (1) by 10
and subtract (1) from (2)
(2) +10d+%2B+25q+=+14500+
(1) +-10d+-+10q+=+-7000+
+15q+=+7500+
+q+=+500+
and, since
(1) +d+%2B+q+=+700+
(1) +d+%2B+500+=+700+
(1) +d+=+200+
There are 200 dimes and 500 quarters
check answer:
(2) +10d+%2B+25q+=+14500+
(2) +10%2A200+%2B+25%2A500+=+14500+
(2) +2000+%2B+12500+=+14500+
(2) +14500+=+14500+
OK