A cash register at a store contains $277 bills. There are six more $5 bills than $10 bills. The number of $1bills is two more than 24 times the number of $10 bills. How many bills of each kind are there

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Step-by-step explanation:

Your question isn't well written, the store contains $227 and not $277.

let c be the number of $1 bills

let f be the number of $5 bills

let t be the number of $10 bills.

Therefore, the equation will be:

1c + 5f + 10t = 227

Since there are six more $5 bills than $10 bills, therefore,

f = t+6

Also, the number of $1 bills is two more than 24 times the number of $10 bills and this will be:

c = 24t + 2

Now, we substitute the equation for f and c into the main equation and this will be:

1c + 5f + 10t = 277

1(24t + 2) + 5(t + 6) + 10t = 227

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

39t + 32 = 227

39t = 227 - 32

39t = 195

t = 195/39

t = 5

Therefore, there are 5 $10 bills

Since c = 24t + 2

c = 24(5) + 2

c = 120 + 2

c = 122

Therefore, there are 122 $1 bills

Since f = t+6

f = 5+6

f = 11

Therefore, there are 11 $5 bills.