A cashier has a total of 33 bills consisting of ones, fives and twenties. The number of twenties is 14 less than the number of ones. The total value of money is $164. How many of each denomination of bill are there?
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To solve this problem, we can set up a system of equations based on the given information. Let's use the variables:
- x: number of ones
- y: number of fives
- z: number of twenties
From the problem statement, we can deduce the following information:
1) The cashier has a total of 33 bills, so:
x + y + z = 33
2) The number of twenties is 14 less than the number of ones, so:
z = x - 14
3) The total value of the money is $164, which can be expressed as:
1x + 5y + 20z = 164
Now, we have a set of three equations. We can solve these equations simultaneously to find the values of x, y, and z.
First, let's simplify equation 3 by substituting the value of z from equation 2:
1x + 5y + 20(x - 14) = 164
Simplifying this equation gives us:
x + 5y + 20x - 280 = 164
21x + 5y = 444
Now, we have two equations with two variables (equations 1 and the simplified version of equation 3). We can use any method of solving simultaneous equations, such as substitution or elimination.
Let's solve by substitution. Rearrange equation 1 to express x in terms of y:
x = 33 - y
Substitute this value of x into the second equation:
21(33 - y) + 5y = 444
693 - 21y + 5y = 444
693 - 444 = 21y - 5y
249 = 16y
y = 15.56
We need to have a whole number for y, so we round y to the nearest whole number:
y ≈ 16
Now substitute the value of y back into equation 1 to solve for x:
x + 16 + z = 33
x + z = 17
Substitute the value of x into equation 2:
z = x - 14
Using the value of y:
16 + z = 17
z ≈ 1
Now we can find x using equation 3 (or the simplified version of it):
21x + 5y = 444
21x + 5(16) = 444
21x + 80 = 444
21x = 364
x ≈ 17.33
Again, rounding x to the nearest whole number:
x ≈ 17
So, approximately, there are:
17 ones, 16 fives, and 1 twenty.