SOLUTION: A cash register contains a total of $580 in bills. It has 76 bills consisting of $5 bills and $10 bills. How many of each type does he have?

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Question 1038983: A cash register contains a total of $580 in bills. It has 76 bills consisting of $5 bills and $10 bills. How many of each type does he have?
Answer by ikleyn(52781) About Me  (Show Source):
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A cash register contains a total of $580 in bills. It has 76 bills consisting of $5 bills and $10 bills. How many of each type does he have?
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Let  f = # of  $5  bills.
Then the number of $10 bills is (76-f).

 $5 bills contribute 5f        dollars to the total.
$10 bills contribute 10*(76-f) dollars to the total.

The total is 5f + 10*(76-f)  dollars.
From the other side, the total is $580.

Therefore, you have an equation

5f + 10*(76-f) = 580.

Simplify and solve:

5f + 760 - 10f = 580,

-5f = 580 - 760,

-5f = -180.

f = %28-180%29%2F-5%29 = 36.

There are 36 of $5 bills and 76-36 = 40 of $10 bills.

Check.  36*5 + 40*10 = 180 + 400 = 580.   Correct!

Answer.  There are 36 of $5 bills and 40 of $10 bills.