A cashier has a total of 37 bills consisting of ones, fives and twenties. The number of twenties is 15 less than the number of ones. The total value of money is $191. How many of each denomination of bill are there?
There are ___ ones, ___fives, ____twenties.
number of ones --- x
number of twenties --- x-15
number of fives = 37 - x - (x-15) = 52 - 2x
x + 5(52-2x) + 20(x-15) = 191
solve for x, then sub back into my definitions.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the number of ones is represented by x.
According to the problem, the number of twenties is 15 less than the number of ones, so we can represent the number of twenties as (x - 15).
Given that the cashier has a total of 37 bills, we can create the equation:
x + (x - 15) + y = 37,
where y represents the number of fives.
Additionally, we know that the total value of money is $191. Since each one is worth $1, each five is worth $5, and each twenty is worth $20, we can create the following equation:
x + 5y + 20(x - 15) = 191.
Now we have a system of two equations with two variables. We can solve this system using substitution or elimination.
Let's solve by substitution method:
From the first equation, we can express y in terms of x:
y = 37 - (2x - 15)
y = 22 - 2x
Substituting this value of y into the second equation:
x + 5(22 - 2x) + 20(x - 15) = 191
x + 110 - 10x + 20x - 300 = 191
11x - 190 = 191
11x = 191 + 190
11x = 381
x = 381 / 11
x = 34.64
Since the number of ones cannot be fractional, we need to round it to the nearest whole number.
Therefore, the number of ones is 35.
Substituting the value of x back into the first equation to find the number of twenties:
35 - 15 = 20
Finally, substituting x = 35 into y = 22 - 2x to find the number of fives:
y = 22 - (2 * 35)
y = 22 - 70
y = -48
However, since we cannot have a negative number of fives, it means our initial assumption for x was incorrect. Let's go back and try a different approach.
Let's assume the number of twenties is represented by x.
According to the problem, the number of ones is 15 more than the number of twenties, so we can represent the number of ones as (x + 15).
Now we can create the equations based on these assumptions.
The first equation based on the total number of bills:
(x + 15) + x + y = 37.
The second equation based on the total value of money:
(x + 15) + 5y + 20x = 191.
Simplifying the first equation:
2x + y = 22.
Simplifying the second equation:
21x + 5y = 176.
Now we have a new system of equations with two variables. Let's solve this system.
Multiplying the first equation by 5 and subtracting it from the second equation:
(21x + 5y) - 5(2x + y) = 176 - 5(22)
21x + 5y - 10x - 5y = 176 - 110
11x = 66
x = 66 / 11
x = 6
Substituting the value of x back into the first equation to find y:
2(6) + y = 22
12 + y = 22
y = 22 - 12
y = 10
Therefore, there are 6 twenties, 10 ones, and 10 fives.
In conclusion, there are 10 ones, 10 fives, and 6 twenties.