Let's first assign variables to the unknown quantities. Let's call the number of $1 bills "x," the number of $5 bills "y," and the number of $10 bills "z."
From the first statement, we know that there are 5 fewer $5 bills than $1 bills.
So, the number of $5 bills is y = x - 5.
From the second statement, we know that there are half as many $10 bills as $5 bills.
So, the number of $10 bills is z = (1/2)y = (1/2)(x - 5) = (x - 5)/2.
From the third statement, we know that there is altogether $115.
The total value of the $1 bills is 1 * x = x.
The total value of the $5 bills is 5 * y = 5 * (x - 5) = 5x - 25.
The total value of the $10 bills is 10 * z = 10 * (x - 5)/2 = 5(x - 5).
So the equation is x + 5x - 25 + 5(x - 5) = 115.
This simplifies to x + 5x - 25 + 5x - 25 = 115.
Combining like terms, we get 11x - 50 = 115.
Adding 50 to both sides, we get 11x = 165.
Dividing both sides by 11, we get x = 15.
So, there are x = <<15=15>>15 $1 bills in the wallet.
Therefore, there are y = x - 5 = 15 - 5 = <<15-5=10>>10 $5 bills in the wallet.
And there are z = (x - 5)/2 = (15 - 5)/2 = <<15-5=10>>10/2 = <<10/2=5>>5 $10 bills in the wallet. Answer: \boxed{15, 10, 5}.
There are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5 dollar bills than $1 dollar bills. There are half as many $10 dollar bills as $5 dollar bills. If there is $115 altogether, find the number of each type of bill in the wallet.
3 answers
No.
Apologies for the incorrect solution. Let's try again.
Let's assign variables to the unknown quantities. Let's call the number of $1 bills "x," the number of $5 bills "y," and the number of $10 bills "z."
From the first statement, we know that there are 5 fewer $5 bills than $1 bills.
So, the number of $5 bills is y = x - 5.
From the second statement, we know that there are half as many $10 bills as $5 bills.
So, the number of $10 bills is z = (1/2)y = (1/2)(x - 5) = (x - 5)/2.
From the third statement, we know that there is altogether $115.
The total value of the $1 bills is 1 * x = x.
The total value of the $5 bills is 5 * y = 5 * (x - 5) = 5x - 25.
The total value of the $10 bills is 10 * z = 10 * (x - 5)/2 = 5(x - 5).
So the equation is x + 5x - 25 + 5(x - 5) = 115.
This simplifies to x + 5x - 25 + 5x - 25 = 115.
Combining like terms, we get 11x - 50 = 115.
Adding 50 to both sides, we get 11x = 165.
Dividing both sides by 11, we get x = 15.
So, there are x = 15 $1 bills in the wallet.
Therefore, there are y = x - 5 = 15 - 5 = 10 $5 bills in the wallet.
And there are z = (x - 5)/2 = (15 - 5)/2 = 10/2 = 5 $10 bills in the wallet.
Answer: 15 $1 bills, 10 $5 bills, and 5 $10 bills.
Let's assign variables to the unknown quantities. Let's call the number of $1 bills "x," the number of $5 bills "y," and the number of $10 bills "z."
From the first statement, we know that there are 5 fewer $5 bills than $1 bills.
So, the number of $5 bills is y = x - 5.
From the second statement, we know that there are half as many $10 bills as $5 bills.
So, the number of $10 bills is z = (1/2)y = (1/2)(x - 5) = (x - 5)/2.
From the third statement, we know that there is altogether $115.
The total value of the $1 bills is 1 * x = x.
The total value of the $5 bills is 5 * y = 5 * (x - 5) = 5x - 25.
The total value of the $10 bills is 10 * z = 10 * (x - 5)/2 = 5(x - 5).
So the equation is x + 5x - 25 + 5(x - 5) = 115.
This simplifies to x + 5x - 25 + 5x - 25 = 115.
Combining like terms, we get 11x - 50 = 115.
Adding 50 to both sides, we get 11x = 165.
Dividing both sides by 11, we get x = 15.
So, there are x = 15 $1 bills in the wallet.
Therefore, there are y = x - 5 = 15 - 5 = 10 $5 bills in the wallet.
And there are z = (x - 5)/2 = (15 - 5)/2 = 10/2 = 5 $10 bills in the wallet.
Answer: 15 $1 bills, 10 $5 bills, and 5 $10 bills.