A cashier has a total of 34 bills, made up of ones, fives, and twenties. The number of fives is one less than the number of twenties. The total value of the money is $237. How many of each denomination of bills are there? (Hint: Let x = the number of ones, y = the number of fives, and z = the number of twenties).
Not going to follow the hint, why use 3 unknowns?
I think the following way is easier:
number of twenties -- x
number of fives ------ x-1
number of ones = 34 - x - (x-1)
= 35 - 2x
20x + 5(x-1) + 35-2x = 237
23x = 207
x = 9
So we have 9 twenties, 8 fives, and 17 ones
check:
they do add up to 34
value: 9(20) + 8(5) + 17 = 237
All is good!
^ where do you get the 35 and 2x?
34 - x - (x-1)
34 - x + x + 1
34 - 2x + 1
35 - 2x
correct?
we have to have three equations. as this is a system of equations and inequalities problem. I have this so far:
x=# of 1's
y=# of 5's
z=# 20's
and I have the following equations.
1x+5y+20z=237
x+y+z=34
and I need another equation. but I am having trouble doing it. the teacher stated to use the hint as an equation but I am having trouble putting it together
To solve this problem, let's assign variables and create equations based on the given information.
Let x be the number of ones, y be the number of fives, and z be the number of twenties.
From the information given, we can create the following equations:
1) x + y + z = 34 (since the total number of bills is 34)
2) y = z - 1 (since the number of fives is one less than the number of twenties)
3) x + 5y + 20z = 237 (since the total value of the money is $237)
Now, we have a system of three equations:
x + y + z = 34
y = z - 1
x + 5y + 20z = 237
We can solve this system of equations by substitution or elimination.
Let's use substitution:
Substitute equation 2 into equations 1 and 3 to solve for x and z.
Starting with equation 1:
Replace y with z - 1 to get:
x + (z - 1) + z = 34
Simplifying:
x + 2z - 1 = 34
x + 2z = 35 (equation 4)
Now, substitute equation 2 into equation 3:
Replace y with z - 1 to get:
x + 5(z - 1) + 20z = 237
Simplifying:
x + 5z - 5 + 20z = 237
x + 25z = 242 (equation 5)
Now, we have two equations:
x + 2z = 35 (equation 4)
x + 25z = 242 (equation 5)
We can proceed to solve these equations simultaneously.
Multiply equation 4 by 25 and equation 5 by 2 to eliminate x:
25x + 50z = 875
2x + 50z = 484
Subtract equation 4 from equation 5:
(2x + 50z) - (25x + 50z) = 484 - 875
-23x = -391
x = 17
Substitute the value of x back into equation 4:
17 + 2z = 35
2z = 35 - 17
2z = 18
z = 9
Since y = z - 1, substitute the value of z into this equation:
y = 9 - 1
y = 8
Therefore, the number of ones is 17, the number of fives is 8, and the number of twenties is 9.