Answer:
The number of $ 10 bills is 37, the number of $ 20 is 6 and the number of $ 50 bills is 22.
Step-by-step 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.
, Area of traingle = 40