SOLUTION: A parking meter will accept dimes and quarters only. At the end of the day, the meter attendants collected 55 coins whose total value was $10.75. How many coins of each type where

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: A parking meter will accept dimes and quarters only. At the end of the day, the meter attendants collected 55 coins whose total value was $10.75. How many coins of each type where       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 508966: A parking meter will accept dimes and quarters only. At the end of the day, the meter attendants collected 55 coins whose total value was $10.75. How many coins of each type where there?
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
d = number of dimes
10d = value of the dimes in cents
q = number of quarters
24q = value of the quarters in cents
.
d + q = 55
so
d = 55-q
.
10d + 25q = 1075 cents
.
substitute d = 55-q
.
10(55-q) + 25q = 1075
550 -10q + 25q = 1075
15q = 525
q = 35
.
d = 55-q
d = 55-35
d = 20
.
Check to determine if the value is correct.
25(35) = 875 cents
10(20) = 200 cents
875 + 200 = 1075 cents
Correct.
.
Answer: The meter had 35 quarters and 20 dimes.
.
Done.