A cash register has no one dollar bill in it all of the bills are five dollar bills $10 bills or $20 bills there are a total 37 bills in the cash register and the total value of this bills is $310 if the number of $10 bills in the register is twice the number of $20 bills how many five dollar bills are in the register?
x + y + z = 37
5 x + 10 y + 20 z = 310
y = 2 z
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x + 3 z = 37
5 x + 40 z = 310
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5 x + 15 z = 185
5 x + 40 z = 310
----------------- subtract
-25 z = - 125
z = 5
then y = 2 z = 10
so
x + 10 + 5 = 37
x = 22
To solve this problem, we can set up a system of equations using the given information.
Let's define three variables:
- Let "x" represent the number of $5 bills.
- Let "y" represent the number of $10 bills.
- Let "z" represent the number of $20 bills.
Based on the given information, we can establish the following equations:
1) Total number of bills: x + y + z = 37
2) Total value of bills: 5x + 10y + 20z = 310
3) Relation between number of $10 bills and $20 bills: y = 2z
Now, we can solve this system of equations step by step.
First, substitute the value of "y" from equation 3 into equation 1:
x + 2z + z = 37
x + 3z = 37 ---- (Equation 4)
Next, substitute the value of "y" from equation 3 into equation 2:
5x + 10(2z) + 20z = 310
5x + 20z + 20z = 310
5x + 40z = 310
x + 8z = 62 ---- (Equation 5)
Now, we have a system of two equations (Equation 4 and Equation 5) with two variables (x and z).
To solve this system, we can use the substitution method or the elimination method. Let's use the substitution method:
From Equation 4, we can solve for x in terms of z:
x = 37 - 3z
Substitute this value of x into Equation 5:
(37 - 3z) + 8z = 62
37 + 5z = 62
5z = 62 - 37
5z = 25
z = 5
Now that we have the value of z, we can substitute it back into Equation 4 to find x:
x + 3(5) = 37
x + 15 = 37
x = 37 - 15
x = 22
Therefore, there are 22 $5 bills, 10 $10 bills, and 5 $20 bills in the cash register.