Question

The cash drawer in a market stall contains $227 in bills. There are six more $5 bills than the number of $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

To solve this problem, equations are set up for each bill denomination based on provided information. After defining variables and creating a total amount equation, we find that there are 5 $10 bills, 11 $5 bills, and 122 $1 bills in the cash drawer.

Step-by-step explanation:

The question asks us to find out how many bills of each denomination are there in a market stall's cash drawer which contains $227 in total. We are given that there are 6 more $5 bills than $10 bills and that the number of $1 bills is two more than 24 times the number of $10 bills.

To solve this problem, we will set up an equation for each denomination based on the information provided.:

Let the number of $10 bills be x.

The number of $5 bills would then be x + 6.

The number of $1 bills would be 24x + 2.

Now, we can set up an equation for the total amount of money:

10x + 5(x + 6) + (24x + 2) = 227

Simplifying this equation, we get:

10x + 5x + 30 + 24x + 2 = 227

39x + 32 = 227

39x = 227 - 32

39x = 195

x = 5

So, there are 5 $10 bills, 5 + 6 = 11 $5 bills, and 24(5) + 2 = 122 $1 bills.

Answer 2

X = number of $10 bills

X + 6 = number of $5 bills

24X + 2 = number of $1 bills

10X + 5(X + 6) + 1(24X + 2) = 227 hope it helps!!!!!!!