Question

Two friends sold many pieces of furniture and made ​$1550 during their garage sale. They had fifteen more​ $10 bills than​ $50 bills. They had 4 more than three times as many​ $20 bills as​ $50 bills. How many of each denomination did they​ have? The number of​ $50 bills is nothing​, the number of​ $10 bills is nothing​, and the number of​ $20 bills is 550.

Answer 1

The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.

Explanation:

Let n₁ = number of $ 10 bills,

Let n₂ = number of $ 20 bills and

Let n₃ = number of $ 50 bills

Given that 10n₁ + 20n₂ + 50n₃ = 1550 (1)

Also, since we have fifteen more​ $10 bills than​ $50 bills, n₁ = n₃ + 15 (2)

and we have 4 more than three times as many​ $20 bills as​ $50 bills. n₃ = 3n₂ + 4. (3)

substituting equations (2) and (3) into (1), we have

10(n₃ + 15) + 20n₂ + 50n₃ = 1550

expanding the bracket, we have

10n₃ + 150 + 20n₂ + 50n₃ = 1550

collecting like terms, we have

60n₃ + 150 + 20n₂ = 1550

inserting equation (3), we have

60(3n₂ + 4) + 150 + 20n₂ = 1550

expanding the bracket, we have

180n₂ + 240 + 150 + 20n₂ = 1550

collecting like terms, we have

200n₂ + 390 = 1550

subtracting 390 from both sides, we have

200n₂ = 1550 - 390

200n₂ = 1160

dividing both sides by 200, we have

n₂ = 1160/200

n₂ = 5.8

n₂ ≅ 6 since it cannot be a fraction.

Substituting this into (3), we have

n₃ = 3n₂ + 4 = 3(6) + 4 = 18 + 4 = 22

substituting n₃ into (2), we have

n₁ = n₃ + 15 = 22 + 15 = 37

So, the number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.