SOLUTION: a coin purse is full of $9.65 worth of quarters and dimes. there are 56 coins in the purse. how many quarters and dimes are there?

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Question 1020123: a coin purse is full of $9.65 worth of quarters and dimes. there are 56 coins in the purse. how many quarters and dimes are there?
Found 2 solutions by FrankM, Edwin McCravy:
Answer by FrankM(1040)   (Show Source): You can put this solution on YOUR website!
25Q+10D=965 ( 25 Q is the value 25 times the number of quarters, Q, 10 D is the value 10 times the number of dimes, D)
Q+D=56 ( total number of coins is 56)
10Q+10D=560 (X 2nd equation by 10)
15Q= 405
Q = 27 (quarters)
D = 29 (dimes)
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Let the number of quarters be x
Let the number of dimes be y


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
quarters      x      $0.25       $0.25x
dimes         y      $0.10       $0.10y
-------------------------------------------
TOTALS       56      -----       $9.65

 The first equation comes from the second column.

  

                 x + y = 56

 The second equation comes from the last column.
  

        0.25x + 0.10y = 9.65

Get rid of decimals by multiplying every term by 100:

            25x + 10y = 965

 So we have the system of equations:
           .

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

                x + y = 56
                    y = 56 - x

Substitute (56 - x) for y in 25x + 10y = 965

     25x + 10(56 - x) = 965
      25x + 560 - 10x = 965
            15x + 560 = 965
                  15x = 405
                    x = 27 = the number of quarters.

Substitute in y = 56 - x
              y = 56 - (27)
              y = 29 dimes.


Checking:  27 quarters is $6.75 and 29 dimes is $2.90
            That's 56 coins.
            And indeed $6.75 + $2.90 = $9.65
Edwin
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