the table shows a schedule of mr. kirovs plan for paying off his credit card balance. if mr. kirov continues…

the table shows a schedule of mr. kirovs plan for paying off his credit card balance. if mr. kirov continues to make monthly payments of $100 and does not make any new purchases, how many more payments will he need to make before the balance is 0?\n| balance | payment | new balance | rate | interest |\n| ---- | ---- | ---- | ---- | ---- |\n| $800.00 | $100 | $700.00 | 0.012 | $8.40 |\n| $708.40 | $100 | $608.40 | 0.012 | $7.30 |\n| $615.70 | $100 | $515.70 | 0.012 | $6.19 |

the table shows a schedule of mr. kirovs plan for paying off his credit card balance. if mr. kirov continues to make monthly payments of $100 and does not make any new purchases, how many more payments will he need to make before the balance is 0?\n| balance | payment | new balance | rate | interest |\n| ---- | ---- | ---- | ---- | ---- |\n| $800.00 | $100 | $700.00 | 0.012 | $8.40 |\n| $708.40 | $100 | $608.40 | 0.012 | $7.30 |\n| $615.70 | $100 | $515.70 | 0.012 | $6.19 |

Answer

Explanation:

Step1: Initial balance and payment

The initial balance is $800.00 and the monthly payment is $100.

Step2: First - month calculation

After the first - month payment of $100 from the initial balance of $800, the balance before interest is $800 - 100=700$. Interest of $0.012\times700 = 8.40$ is added, so the new balance is $700+8.40 = 708.40$.

Step3: Second - month calculation

After the second - month payment of $100 from the balance of $708.40, the balance before interest is $708.40 - 100 = 608.40$. Interest of $0.012\times608.40\approx7.30$ is added, so the new balance is $608.40+7.30 = 615.70$.

Step4: Third - month calculation

After the third - month payment of $100 from the balance of $615.70, the balance before interest is $615.70 - 100 = 515.70$. Interest of $0.012\times515.70\approx6.19$ is added, so the new balance is $515.70+6.19 = 521.89$.

Step5: Keep repeating the process

We keep subtracting the $100$ payment and adding the interest until the balance is $0$. Let $B_n$ be the balance after $n$ months. The formula for the balance after a payment and interest addition is $B_{n}=(B_{n - 1}-100)(1 + 0.012)$ where $B_0 = 800$. We can also do it in a brute - force way: Starting with $B_0 = 800$ After 1st payment: $800-100=700$, $700\times(1 + 0.012)=708.40$ After 2nd payment: $708.40-100 = 608.40$, $608.40\times(1 + 0.012)=615.70$ After 3rd payment: $615.70-100=515.70$, $515.70\times(1 + 0.012)=521.89$ After 4th payment: $521.89-100 = 421.89$, $421.89\times(1 + 0.012)=427.06$ After 5th payment: $427.06-100 = 327.06$, $327.06\times(1 + 0.012)=330.98$ After 6th payment: $330.98-100 = 230.98$, $230.98\times(1 + 0.012)=233.76$ After 7th payment: $233.76-100 = 133.76$, $133.76\times(1 + 0.012)=135.37$ After 8th payment: $135.37-100 = 35.37$, $35.37\times(1 + 0.012)=35.79$ After 9th payment: $35.79<100$, and after paying $35.79$ the balance will be $0$.

Answer:

9