SOLUTION: A jar of change contains 64 coins consisting only quarters and dimes. The total value of the coins in the jar $13.60. Let q= the number of quarters and d= the number of dimes. Writ

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Question 1034811: A jar of change contains 64 coins consisting only quarters and dimes. The total value of the coins in the jar $13.60. Let q= the number of quarters and d= the number of dimes. Writes two equations that describe the information given above. how many of each type of coin is in the jar?
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A jar of change contains 64 coins consisting only quarters and dimes. The total value of the coins in the jar $13.60. Let q= the number of quarters and d= the number of dimes. Writes two equations that describe the information given above. how many of each type of coin is in the jar?
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q + d = 64
25q + 10d = 1360
===================
Here's another method:
64 dimes = 640 cents
Each dime replaced by a quarter adds 15 cents.
1360 - 640 = 720
720/15 = 48 quarters replacing dimes
--> 16 dimes

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Let the number of quarters be q
Let the number of dimes be d


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
quarters      q      $0.25       $0.25q
dimes         d      $0.10       $0.10d
-------------------------------------------
TOTALS       64      -----       $13.60

 The first equation comes from the number of coins column.

  %28matrix%283%2C1%2CNumber%2Cof%2Cquarters%29%29%22%22%2B%22%22%28matrix%283%2C1%2CNumber%2Cof%2Cdimes%29%29%22%22=%22%22%28matrix%284%2C1%2Ctotal%2Cnumber%2Cof%2Ccoins%29%29
                 q + d = 64

 The second equation comes from the last column.
  %28matrix%284%2C1%2CValue%2Cof%2CALL%2Cquarters%29%29%22%22%2B%22%22%28matrix%284%2C1%2CValue%2Cof%2CALL%2Cdimes%29%29%22%22=%22%22%28matrix%285%2C1%2CTotal%2Cvalue%2Cof%2CALL%2Ccoins%29%29

           0.25q + 0.10d = 13.6

Get rid of decimals by multiplying every term by 100:

              25q + 10d = 1360

 So we have the system of equations:
           system%28q+%2B+d+=+64%2C25q+%2B+10d+=+1360%29.

We solve by substitution.  Solve the first equation for d:

             q + d = 64
                 d = 64 - q

Substitute (64 - q) for d in 25q + 10d = 1360

    25q + 10(64 - q) = 1360
     25q + 640 - 10q = 1360
           15q + 640 = 1360
                 15q = 720
                   q = 48 = the number of quarters.

Substitute in d = 64 - q
              d = 64 - (48)
              d = 16 dimes.

Checking:  48 quarters is $12.00 and 16 dimes is $1.60
            That's 64 coins.
            And indeed $12.00 + $1.60 = $13.60
Edwin