Final answer:
The question involving a teller's cash drawer with a mix of $10 and $20 bills can be solved by setting up a system of equations. There are 11 $10 bills in the teller's cash drawer.
Step-by-step explanation:
The question asks how many $10 bills are present in a teller's cash drawer if there are a total of 49 $10 and $20 bills with a cumulative value of $870. To solve this, we can set up a system of equations. Let x represent the number of $10 bills and y represent the number of $20 bills.
The first equation comes from the total number of bills:
x + y = 49
The second equation comes from the total value of the bills:
10x + 20y = 870
We can solve this system using substitution or elimination. If we multiply the first equation by 10 and then subtract it from the second equation, we eliminate x and can find y:
10x + 20y = 870
(10)(x + y) = (10)(49)
10x + 20y - 10x - 10y = 870 - 490
10y = 380
y = 38
Now, we substitute y back into the first equation and solve for x:
x + 38 = 49
x = 49 - 38
x = 11
Thus, there are 11 $10 bills in the teller's cash drawer.