SOLUTION: an all day parking meter only takes dimes and quarters. If it contains 100 Coins with a total value of $14.50, how many of each type of coin are in the meter?

Algebra ->  Test -> SOLUTION: an all day parking meter only takes dimes and quarters. If it contains 100 Coins with a total value of $14.50, how many of each type of coin are in the meter?      Log On


   



Question 364798: an all day parking meter only takes dimes and quarters. If it contains 100 Coins with a total value of $14.50, how many of each type of coin are in the meter?
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The 1st method, more or less traditional:
d + q = 100 (100 coins)
10d + 25q = 1450 (the value)
d + q = 100 --> q = 100 - d
Sub for q in the 2nd eqn
10d + 25q = 1450
10d + 25(100-d) = 1450
2500 - 15d = 1450
-15d = - 1050
d = 70
q = 30
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A better method, imo:
If all the coins were dimes, it would be $10, = 1000
Each dime replace by a quarter adds 15 cents.
1450 - 1000 = 450
450/15 = 30 replacements
--> 30 quarters, 70 dimes

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) 10d+%2B+25q+=+1450 (in cents)
(2) d+%2B+q+=+100
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From (1):
(1) 2d+%2B+5q+=+290
Multiply both sides of (2) by 2 and
subtract from (1)
(1) 2d+%2B+5q+=+290
(2) -2d+-+2q+=+-200
3q+=+90
q+=+30
and
(2) d+%2B+q+=+100
d+=+100+-+30
d+=+70
70 dimes and 30 quarters
check:
(1) 2d+%2B+5q+=+290
(1) 2%2A70+%2B+5%2A30+=+290
140+%2B+150+=+290
290+=+290
OK