The count of the denominations of the bills of $1, $5, and $10, are <u>15, 10, and 5</u>, respectively.
In the question, we are given that there are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills.
We are asked to find the count of each denomination, given there was altogether $115 in the bag.
We assume the number of $1-bills in the bag to be x.
Total amount in x bills of $1 = x * $1 = $x.
Given that there are 5 fewer $5-bills than $1-bills in the bag, number of $5-bills in the bag = x - 5.
Total amount in (x - 5) bills of $5 = (x - 5) * $5 = $5(x - 5).
Given that there are half as many $10-bills as $5-bills in the bag, number of $10-bills in the bag = (x - 5)/2.
Total amount in (x - 5)/2 bills of $10 = (x - 5)/2 * $10 = $5(x - 5).
Thus, the total amount in the bag = $x + $5(x - 5) + $5(x - 5).
But, the total amount in the bag = $115.
Thus, we get an equation:
$x + $5(x - 5) + $5(x - 5) = $115,
or, x + 5x - 25 + 5x - 25 = 115,
or, 11x = 115 + 50,
or, 11x = 165,
or, x = 165/11,
or, x = 15.
Thus, number of $1-bills = x = 15.
The number of $5-bills = x - 5 = 15 - 5 = 10.
The number of $10-bills = (x - 5)/2 = (15 - 5)/2 = 10/2 = 5.
Thus, the count of the denominations of the bills of $1, $5, and $10, are <u>15, 10, and 5</u>, respectively.
The provided question is incomplete. The complete question is:
There are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills. If there is $115 altogether, find the number of each type of bill in the wallet."
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