SOLUTION: a bank teller counted P100 and P50 bills. this totaled to 117 bills worth P9950. how many bills of each type did she have?

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Question 1004829: a bank teller counted P100 and P50 bills. this totaled to 117 bills worth P9950. how many bills of each type did she have?
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Let the number of P100's be x
Let the number of P50's be y


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
bill        bills      bill      bills
-------------------------------------------
P100s        x        P100      P100x
P50s         y         P50       P50y
-------------------------------------------
TOTALS      117      -----      P9950

 The first equation comes from the second column.

  
                 x + y = 117

 The second equation comes from the last column.

  

           100x + 50y = 9950

 So we have the system of equations:
           .

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

           x + y = 117
               y = 117 - x

Substitute (117 - x) for y in 100x + 50y = 9950

   100x + 50(117 - x) = 9950
    100x + 5850 - 50x = 9950
           50x + 5850 = 9950
                  50x = 4100
                    x = 82 = the number of P100s.

Substitute in y = 117 - x
              y = 117 - (82)
              y = 35 P50s.

Checking:  82 P100s is P8200 and 35 P50s is P1750
            That's 117 bills.
            And indeed P8200 + P1750 = P9950
Edwin
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
The problem can also be done using only one
unknown or variable:

Let the number of P100s be x
Then the number of P50s, using
ONE PART = TOTAL MINUS OTHER PART,
is 117-x.
                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
bill        bills      bill      bills
-------------------------------------------
P100s        x        P100     P100x
P50s       117-x       P50     P50(117-x)
-------------------------------------------
TOTALS      117      -----     P9950

 The equation comes from the column on the right

  

           100x + 50(117-x) = 9950

           100x + 50(117-x) = 9950

          100x + 5850 - 50x = 9950

                 50x + 5850 = 9950

                        50x = 4100

                     x = 82 = the number of P100s.

The number of P50s is 117-x or 117-82 or 35 P50s.

Checking:  82 P100s is P8200 and 35 P50s is P1750

            That's 117 coins.

            And indeed P8200 + P1750 = P9950
Edwin
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