Question

The cash drawer of the market contains $227. There are six more $5 bills than $10 bills. The number of $1 bills is two more than 24 times the number of $10 bills. How many bills of each kind are there?

Answer 1

Answer:

There are 122 one dollar bills, 11 five dollar bills and 5 ten dollar bills.

Step-by-step explanation:

There are bills of one dollar, five dollars and ten dollars on the cash drawer, therefore the sum of all of them multiplied by their respective values must be equal to the total amount of money on the drawer. We will call the number of one dollar bills, five dollar bills and ten dollar bills, respectively "x","y" and "z", therefore we can create the following expression:

$x + 5*y + 10*z = 227$

We know that there are six more 5 dollar bills than 10 dollar bills and that the number of 1 dollar bills is two more than 24 times the number of 10 dollar bills, therefore:

$y = z + 6\\x = 2 + 24*z$

Applying these values on the first equation, we have:

$2 + 24*z + 5*(z + 6) + 10*z = 227\\2 + 24*z + 5*z + 30 + 10*z = 227\\39*z + 32 = 227\\39*z = 227 - 32\\39*z = 195\\z = \frac{195}{39}\\z = 5$

Applying z to the formulas of y and x, we have:

$y = 5 + 6 = 11\\x = 2 + 24*5 = 122$

There are 122 one dollar bills, 11 five dollar bills and 5 ten dollar bills.