lesson 31: credit cards - engageny.org

ALGEBRA II

NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 31

Lesson 31: Credit Cards

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Student Outcomes

Students compare payment strategies for a decreasing credit card balance.

Students apply the sum of a finite geometric series formula to a decreasing balance on a credit card.

Lesson Notes

This lesson develops the necessary tools and terminology to analyze the mathematics behind credit cards and other

unsecured loans. Credit cards can provide flexibility to budgets, but they must be carefully managed to avoid the pitfalls

of bad credit. For young adults, credit card interest rates can be expected to be between 19.99% and 29.99% per year

(29.99% is currently the maximum allowable interest rate by federal law). Adults with established credit can be offered

interest rates around 8% to 14%. The credit limit for a first credit card is typically around $500, but these limits quickly

increase with a history of timely payments.

In this modeling lesson, students explore the mathematics behind calculating the monthly balance on a single credit card

purchase and recognize that the decreasing balance can be modeled by the sum of a finite geometric series (A-SSE.B.4).

We are intentionally keeping the use of rotating credit such as credit cards simple in this lesson. The students make one

charge of $1,500 on this hypothetical credit card and pay down the balance without making any additional charges.

With this simple example, we can realistically ignore the fact that the interest on a credit card is charged based on the

average daily balance of the account; in our example, the daily balance only changes once per month when the payment

is made.

The students need to recall the following definitions from Lesson 29:

SERIES: Let 𝑎1, 𝑎2, 𝑎3, 𝑎4, … be a sequence of numbers. A sum of the form

𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛

for some positive integer 𝑛 is called a series (or finite series) and is denoted 𝑆𝑛. The 𝑎𝑖’s are called the terms of

the series. The number 𝑆𝑛 that the series adds to is called the sum of the series.

GEOMETRIC SERIES: A geometric series is a series whose terms form a geometric sequence.

The sum 𝑆𝑛 of the first 𝑛 terms of the finite geometric series 𝑆𝑛 = 𝑎 + 𝑎𝑟 + ⋯ + 𝑎𝑟𝑛−1 (when 𝑟 ≠ 1) is given

by

𝑆𝑛 = 𝑎 (1 − 𝑟𝑛

1 − 𝑟).

The sum formula of a geometric series can be written in summation notation as

∑ 𝑎𝑟𝑘

𝑛−1

𝑘=0

= 𝑎 (1 − 𝑟𝑛

1 − 𝑟).

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Classwork

Opening (3 minutes)

Assign students to small groups, and keep them in the same groups throughout this lesson. In the first mathematical

modeling exercise, all groups work on the same problem, but in the second mathematical modeling exercise, the groups

are assigned one of three different payment schemes to investigate.

In the previous lesson, you investigated the mathematics needed for a car loan. What if you have decided to

buy a car, but you have not saved up enough money for the down payment? If you are buying through a

dealership, it is possible to put the down payment onto a credit card. For today’s lesson, we investigate the

finances of charging $1 ,500 onto a credit card for the down payment on a car. We investigate different

payment plans and how much you end up paying in total using each plan.

The annual interest rates on a credit card for people who have not used credit in the past tend to be much

higher than for adults with established good credit, ranging between 14.99% and 29.99%, which is the

maximum interest rate allowed by law. Throughout this lesson, we use a 19.99% annual interest rate, and we

explore problems with other interest rates in the Problem Set.

One of the differences between a credit card and a loan is that you can pay as much as you want toward your

credit card balance, as long as it is at least the amount of the “minimum payment,” which is determined by the

lender. In many cases, the minimum payment is the sum of the interest that has accrued over the month and

1% of the outstanding balance, or $25, whichever is greater.

Another difference between a credit card and a loan is that a loan has a fixed term of repayment—you pay it

off over an agreed-upon length of time such as five years—and that there is no fixed term of repayment for a

credit card. You can pay it off as quickly as you like by making large payments, or you can pay less and owe

money for a longer period of time. In the mathematical modeling exercise, we investigate the scenario of

paying a fixed monthly payment of various sizes toward a credit card balance of $1 ,500.

Mathematical Modeling Exercise (25 minutes)

In this exercise, students model the repayment of a single charge of $1,500 to a credit card that charges 19.99% annual

interest. Before beginning the Mathematical Modeling Exercise, assign students to small groups, and assign groups to be

either part of the 50-team, 100-team, or 150-team. The groups in each of the three teams investigate how long it takes

to pay down the $1,500 balance making fixed payments of either $50, $100, or $150 each month.

As you circulate the room while students are working, take note of groups that are working well together on this set of

problems. Select at least one group on each team to present their work at the end of the exercise period.

Mathematical Modeling Exercise

You have charged $𝟏, 𝟓𝟎𝟎 for the down payment on your car to a credit card that charges

𝟏𝟗. 𝟗𝟗% annual interest, and you plan to pay a fixed amount toward this debt each month until it

is paid off. We denote the balance owed after the 𝒏th payment has been made as 𝒃𝒏.

a. What is the monthly interest rate, 𝒊? Approximate 𝒊 to 𝟓 decimal places.

𝒊 =𝟎. 𝟏𝟗𝟗𝟗

𝟏𝟐≈ 𝟎. 𝟎𝟏𝟔𝟔𝟔

Scaffolding:

For struggling students, use an interest rate of 24.00% so that 𝑖 = 0.02 and 𝑟 = 1.02.





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b. You have been assigned to either the 𝟓𝟎-team, the 𝟏𝟎𝟎-team, or the 𝟏𝟓𝟎-team, where the number indicates

the size of the monthly payment 𝑹 you make toward your debt. What is your value of 𝑹?

Students will answer 𝟓𝟎, 𝟏𝟎𝟎, or 𝟏𝟓𝟎 as appropriate.

c. Remember that you can make any size payment toward a credit card debt, as long as it is at least as large as

the minimum payment specified by the lender. Your lender calculates the minimum payment as the sum of

𝟏% of the outstanding balance and the total interest that has accrued over the month, or $𝟐𝟓, whichever is

greater. Under these stipulations, what is the minimum payment? Is your monthly payment 𝑹 at least as

large as the minimum payment?

The minimum payment is 𝟎. 𝟎𝟏($𝟏𝟓𝟎𝟎) + 𝟎. 𝟎𝟏𝟔𝟔𝟔($𝟏𝟓𝟎𝟎) = $𝟑𝟗. 𝟗𝟗. All given values of 𝑹 are greater

than the minimum payment.

d. Complete the following table to show 𝟔 months of payments.

Month, 𝒏 Interest Due

(in dollars)

Payment, 𝑹

(in dollars)

Paid to Principal

(in dollars)

Balance, 𝒃𝒏

(in dollars)

𝟎 𝟏, 𝟓𝟎𝟎. 𝟎𝟎 𝟏 𝟐𝟒. 𝟗𝟗 𝟓𝟎 𝟐𝟓. 𝟎𝟏 𝟏, 𝟒𝟕𝟒. 𝟗𝟗 𝟐 𝟐𝟒. 𝟓𝟕 𝟓𝟎 𝟐𝟓. 𝟒𝟑 𝟏, 𝟒𝟒𝟗. 𝟓𝟔 𝟑 𝟐𝟒. 𝟏𝟓 𝟓𝟎 𝟐𝟓. 𝟖𝟓 𝟏, 𝟒𝟐𝟑. 𝟕𝟏 𝟒 𝟐𝟑. 𝟕𝟐 𝟓𝟎 𝟐𝟔. 𝟐𝟖 𝟏, 𝟑𝟗𝟕. 𝟒𝟑 𝟓 𝟐𝟑. 𝟐𝟖 𝟓𝟎 𝟐𝟔. 𝟕𝟐 𝟏, 𝟑𝟕𝟎. 𝟕𝟏 𝟔 𝟐𝟐. 𝟖𝟑 𝟓𝟎 𝟐𝟕. 𝟏𝟕 𝟏, 𝟑𝟒𝟑. 𝟓𝟒


(in dollars)

Payment, 𝑹

(in dollars)

Paid to Principal

(in dollars)

Balance, 𝒃𝒏

(in dollars)

𝟎 𝟏, 𝟓𝟎𝟎. 𝟎𝟎 𝟏 𝟐𝟒. 𝟗𝟗 𝟏𝟎𝟎 𝟕𝟓. 𝟎𝟏 𝟏, 𝟒𝟐𝟒. 𝟗𝟗 𝟐 𝟐𝟑. 𝟕𝟒 𝟏𝟎𝟎 𝟕𝟔. 𝟐𝟔 𝟏, 𝟑𝟒𝟖. 𝟕𝟑 𝟑 𝟐𝟐. 𝟒𝟕 𝟏𝟎𝟎 𝟕𝟕. 𝟓𝟑 𝟏, 𝟐𝟕𝟏. 𝟐𝟎 𝟒 𝟐𝟏. 𝟏𝟖 𝟏𝟎𝟎 𝟕𝟖. 𝟖𝟐 𝟏, 𝟏𝟗𝟐. 𝟑𝟖 𝟓 𝟏𝟗. 𝟖𝟔 𝟏𝟎𝟎 𝟖𝟎. 𝟏𝟒 𝟏, 𝟏𝟏𝟐. 𝟐𝟒 𝟔 𝟏𝟖. 𝟓𝟑 𝟏𝟎𝟎 𝟖𝟏. 𝟒𝟕 𝟏, 𝟎𝟑𝟎. 𝟕𝟕


(in dollars)

Payment, 𝑹

(in dollars)

Paid to Principal

(in dollars)

Balance, 𝒃𝒏

(in dollars)

𝟎 𝟏, 𝟓𝟎𝟎. 𝟎𝟎 𝟏 𝟐𝟒. 𝟗𝟗 𝟏𝟓𝟎 𝟏𝟐𝟓. 𝟎𝟏 𝟏, 𝟑𝟕𝟒. 𝟗𝟗 𝟐 𝟐𝟐. 𝟗𝟏 𝟏𝟓𝟎 𝟏𝟐𝟕. 𝟎𝟗 𝟏, 𝟐𝟒𝟕. 𝟗𝟎 𝟑 𝟐𝟎. 𝟕𝟗 𝟏𝟓𝟎 𝟏𝟐𝟗. 𝟐𝟏 𝟏, 𝟏𝟏𝟖. 𝟔𝟗 𝟒 𝟏𝟖. 𝟔𝟒 𝟏𝟓𝟎 𝟏𝟑𝟏. 𝟑𝟔 𝟗𝟖𝟕. 𝟑𝟑 𝟓 𝟏𝟔. 𝟒𝟓 𝟏𝟓𝟎 𝟏𝟑𝟑. 𝟓𝟓 𝟖𝟓𝟑. 𝟕𝟖 𝟔 𝟏𝟒. 𝟐𝟐 𝟏𝟓𝟎 𝟏𝟑𝟓. 𝟕𝟖 𝟕𝟏𝟖. 𝟎𝟎

e. Write a recursive formula for the balance 𝒃𝒏 in month 𝒏 in terms of the balance 𝒃𝒏−𝟏.

To calculate the new balance, 𝒃𝒏, we compound interest for one month on the previous balance 𝒃𝒏−𝟏 and

then subtract the payment 𝑹:

𝒃𝒏 = 𝒃𝒏−𝟏(𝟏 + 𝒊) − 𝑹, with 𝒃𝟎 = 𝟏𝟓𝟎𝟎.





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f. Write an explicit formula for the balance 𝒃𝒏 in month 𝒏, leaving the expression 𝟏 + 𝒊 in symbolic form.

We have the following formulas:

𝒃𝟏 = 𝒃𝟎(𝟏 + 𝒊) − 𝑹

𝒃𝟐 = 𝒃𝟏(𝟏 + 𝒊) − 𝑹

= [𝒃𝟎(𝟏 + 𝒊) − 𝑹](𝟏 + 𝒊) − 𝑹

= 𝒃𝟎(𝟏 + 𝒊)𝟐 − 𝑹(𝟏 + 𝒊) − 𝑹

𝒃𝟑 = 𝒃𝟐(𝟏 + 𝒊) − 𝑹

= [𝒃𝟎(𝟏 + 𝒊)𝟐 − 𝑹(𝟏 + 𝒊) − 𝑹](𝟏 + 𝒊) − 𝑹

= 𝒃𝟎(𝟏 + 𝒊)𝟑 − 𝑹(𝟏 + 𝒊)𝟐 − 𝑹(𝟏 + 𝒊) − 𝑹

⋮

𝒃𝒏 = 𝒃𝟎(𝟏 + 𝒊)𝒏 − 𝑹(𝟏 + 𝒊)𝒏−𝟏 − 𝑹(𝟏 + 𝒊)𝒏−𝟐 − ⋯ − 𝑹(𝟏 + 𝒊) − 𝑹

g. Rewrite your formula in part (f) using 𝒓 to represent the quantity (𝟏 + 𝒊).

𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹𝒓𝒏−𝟏 − 𝑹𝒓𝒏−𝟐 − ⋯ − 𝑹𝒓 − 𝑹

= 𝒃𝟎𝒓𝒏 − 𝑹(𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏)

h. What can you say about your formula in part (g)? What term do we use to describe 𝒓 in this formula?

The formula in part (g) contains the sum of a finite geometric series with common ratio 𝒓.

i. Write your formula from part (g) in summation notation using 𝚺.

𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹(𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏)

= 𝒃𝟎𝒓𝒏 − 𝑹 ∑ 𝒓𝒌𝒏−𝟏

𝒌=𝟎

j. Apply the appropriate formula from Lesson 29 to rewrite your formula from part (g).

Using the sum of a finite geometric series formula,

𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹(𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏)

= 𝒃𝟎𝒓𝒏 − 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓)

Scaffolding:

Ask advanced learners to develop a generic formula for the balance 𝑏𝑛 in terms of the payment amount R and the growth factor 𝑟.





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k. Find the month when your balance is paid off.

The balance is paid off when 𝒃𝒏 = 𝟎. (The final payment is less than a full payment so that the debt is not

overpaid.)

Students will likely do this calculation with the values of 𝒓, 𝒃𝟎, and 𝑹 substituted in.

𝒃𝟎𝒓𝒏 − 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓) = 𝟎

𝒃𝟎𝒓𝒏 = 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓)

(𝟏 − 𝒓)(𝒃𝟎𝒓𝒏) = 𝑹(𝟏 − 𝒓𝒏)

(𝟏 − 𝒓)(𝒃𝟎𝒓𝒏) + 𝑹𝒓𝒏 = 𝑹

𝒓𝒏(𝒃𝟎(𝟏 − 𝒓) + 𝑹) = 𝑹

𝒓𝒏 =𝑹

(𝒃𝟎(𝟏 − 𝒓) + 𝑹)

𝒏 𝐥𝐨𝐠(𝒓) = 𝐥𝐨𝐠 (𝑹

(𝒃𝟎(𝟏 − 𝒓) + 𝑹))

𝒏 = 𝐥𝐨𝐠 (

𝑹(𝒃𝟎(𝟏 − 𝒓) + 𝑹)

)

𝐥𝐨𝐠(𝒓)

If 𝑹 = 𝟓𝟎, then 𝒏 ≈ 𝟒𝟏. 𝟗𝟐𝟓. The debt is paid off in 𝟒𝟐 months.

If 𝑹 = 𝟏𝟎𝟎, then 𝒏 ≈ 𝟏𝟕. 𝟒𝟗. The debt is paid off in 𝟏𝟖 months.

If 𝑹 = 𝟏𝟓𝟎, then 𝒏 ≈ 𝟏𝟏. 𝟎𝟐𝟗𝟔. The debt is paid off in 𝟏𝟐 months.

l. Calculate the total amount paid over the life of the debt. How much was paid solely to interest?

For 𝑹 = 𝟓𝟎: The debt is paid in 𝟒𝟏 payments of $𝟓𝟎, and the last payment is the amount 𝒃𝟒𝟏 with interest:

𝟓𝟎(𝟒𝟏) + (𝟏 + 𝒊)𝒃𝟒𝟏 = 𝟐𝟎𝟓𝟎 + 𝒓 (𝒃𝟎𝒓𝒏 − 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓))

≈ 𝟐𝟎𝟓𝟎 + 𝒓(𝟒𝟓. 𝟔𝟏)

≈ 𝟐𝟎𝟗𝟔. 𝟑𝟕.

The total amount paid using monthly payments of $𝟓𝟎 is $𝟐, 𝟎𝟗𝟔. 𝟑𝟕. Of this amount, $𝟓𝟗𝟔. 𝟑𝟕 is interest.

For 𝑹 = 𝟏𝟎𝟎: The debt is paid in 𝟏𝟕 payments of $𝟏𝟎𝟎, and the last payment is the amount 𝒃𝟏𝟕 with interest.

𝟏𝟎𝟎(𝟏𝟕) + (𝟏 + 𝒊)𝒃𝟏𝟕 = 𝟏𝟕𝟎𝟎 + 𝒓 (𝒃𝟎𝒓𝟏𝟕 − 𝑹 (𝟏 − 𝒓𝟏𝟕

𝟏 − 𝒓))

≈ 𝟏𝟕𝟎𝟎 + 𝒓(𝟒𝟎. 𝟓𝟐)

≈ 𝟏𝟕𝟒𝟎. 𝟓𝟐

The total amount paid using monthly payments of $𝟏𝟎𝟎 is $𝟏, 𝟕𝟒𝟎. 𝟓𝟐. Of this amount, $𝟐𝟒𝟎. 𝟓𝟐 is interest.

For 𝑹 = 𝟏𝟓𝟎: The debt is paid in 𝟏𝟏 payments of $𝟏𝟓𝟎, and the last payment is the amount 𝒃𝟏𝟏 with interest.

𝟏𝟓𝟎(𝟏𝟏) + (𝟏 + 𝒊)𝒃𝟏𝟏 = 𝟏𝟕𝟎𝟎 + 𝒓 (𝒃𝟎𝒓𝒏 − 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓))

≈ 𝟏𝟔𝟓𝟎 + 𝒓(𝟒. 𝟒𝟗)

≈ 𝟏𝟔𝟓𝟒. 𝟒𝟗

The total amount paid using monthly payments of $𝟏𝟓𝟎 is $𝟏, 𝟔𝟓𝟒. 𝟒𝟗. Of this amount, $𝟏𝟓𝟒. 𝟒𝟗 is interest.





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Discussion (9 minutes)

Have students from each team present their solutions to parts (k) and (l) to the class. After the three teams have made

their presentations, lead students through the following discussion, which should help them to make sense of the

different results that arise from the different payment values 𝑅.

What happens to the number of payments as you increase the amount 𝑅 of the recurring monthly payment?

As the amount 𝑅 of the payment increases, the number of payments decreases.

What happens to the total amount of interest paid as you increase the amount 𝑅 of the recurring monthly

payment?

As the amount 𝑅 of the payment increases, the number of payments decreases.

What is the largest possible amount of the payment 𝑅? In that case, how many payments are made?

The largest possible payment would be to pay the entire balance in one payment: (1 + 𝑖)$1500 = $1524.99.

Ask students about the formulas that they developed in the Mathematical Modeling Exercise to calculate the balance of

the debt in month 𝑛. Students may use different notations, but they should have come up with a formula similar to

𝑏𝑛 = 𝑏0𝑟𝑛 − 𝑅 (1−𝑟𝑛

1−𝑟). Depending on what notation the students used, you may need to draw the parallel from this

formula to the present value of an annuity formula developed in Lesson 30. If we substitute 𝑏𝑛 = 0 as the future value

of the annuity when it is paid off in 𝑛 payments, and 𝐴𝑝 = 𝑏0 as the present value/initial value of the annuity, then we

have

𝑏𝑛 = 𝑏0𝑟𝑛 − 𝑅 (1 − 𝑟𝑛

1 − 𝑟)

0 = 𝐴𝑝𝑟𝑛 − 𝑅 (1 − 𝑟𝑛

1 − 𝑟)

𝐴𝑝𝑟𝑛 = 𝑅 (1 − 𝑟𝑛

1 − 𝑟)

𝐴𝑝(1 + 𝑖)𝑛 = 𝑅 (1 − (1 + 𝑖)𝑛

1 − (1 + 𝑖))

𝐴𝑝(1 + 𝑖)𝑛 = 𝑅 (1 − (1 + 𝑖)𝑛

−𝑖)

𝐴𝑝 = 𝑅 ((1 + 𝑖)𝑛 − 1

𝑖) ⋅ (1 + 𝑖)−𝑛

𝐴𝑝 = 𝑅 (1 − (1 + 𝑖)−𝑛

𝑖).





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Closing (3 minutes)

Ask students to summarize the main points of the lesson either in writing or with a partner. Some highlights that should

be included are listed below.

Calculating the balance from a single purchase on a credit card requires that we sum a finite geometric series.

We have a formula from Lesson 29 that calculates the sum of a finite geometric series:

∑ 𝑎𝑟𝑘

𝑛−1

𝑘=0

= 𝑎 (1 − 𝑟𝑛

1 − 𝑟).

When you have incurred a credit card debt, you need to decide how to pay it off.

If you choose to make a lower payment each month, then both the time required to pay off the debt

and the total interest paid over the life of the debt increases.

If you choose to make a higher payment each month, then both the time required to pay off the debt

and the total interest paid over the life of the debt decreases.

Exit Ticket (5 minutes)





ALGEBRA II



520



Name Date


Exit Ticket

Suppose that you currently have one credit card with a balance of $10,000 at an annual rate of 24.00% interest. You

have stopped adding any additional charges to this card and are determined to pay off the balance. You have worked

out the formula 𝑏𝑛 = 𝑏0𝑟𝑛 − 𝑅(1 + 𝑟 + 𝑟2 + ⋯ + 𝑟𝑛−1), where 𝑏0 is the initial balance, 𝑏𝑛 is the balance after you have

made 𝑛 payments, 𝑟 = 1 + 𝑖, where 𝑖 is the monthly interest rate, and 𝑅 is the amount you are planning to pay each

month.

a. What is the monthly interest rate 𝑖? What is the growth rate, 𝑟?

b. Explain why we can rewrite the given formula as 𝑏𝑛 = 𝑏0𝑟𝑛 − 𝑅 (1−𝑟

𝑛

1−𝑟).

c. How long does it take to pay off this debt if you can afford to pay a constant $250 per month? Give the

answer in years and months.





ALGEBRA II



521



Exit Ticket Sample Solutions

Suppose that you currently have one credit card with a balance of $𝟏𝟎, 𝟎𝟎𝟎 at an annual rate of 𝟐𝟒. 𝟎𝟎% interest. You

have stopped adding any additional charges to this card and are determined to pay off the balance. You have worked out

the formula 𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹(𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏), where 𝒃𝟎 is the initial balance, 𝒃𝒏 is the balance after you have

made 𝒏 payments, 𝒓 = 𝟏 + 𝒊, where 𝒊 is the monthly interest rate, and 𝑹 is the amount you are planning to pay each

month.

a. What is the monthly interest rate 𝒊? What is the growth rate, 𝒓?

The monthly interest rate 𝒊 is given by 𝒊 =𝟎.𝟐𝟒𝟏𝟐

= 𝟎. 𝟎𝟐, and 𝒓 = 𝟏 + 𝒊 = 𝟏. 𝟎𝟐.

b. Explain why we can rewrite the given formula as 𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹 (𝟏−𝒓

𝒏

𝟏−𝒓).

Using summation notation and the sum formula for a finite geometric series, we have

𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏 = ∑ 𝒓𝒌𝒏−𝟏

𝒌=𝟎

=𝟏 − 𝒓𝒏

𝟏 − 𝒓.

Then the formula becomes

𝒃𝒏 = 𝒃𝟎𝒓𝒏 − 𝑹(𝟏 + 𝒓 + 𝒓𝟐 + ⋯ + 𝒓𝒏−𝟏)

= 𝒃𝟎𝒓𝒏 − 𝑹 (𝟏 − 𝒓𝒏

𝟏 − 𝒓).

c. How long does it take to pay off this debt if you can afford to pay a constant $𝟐𝟓𝟎 per month? Give the

answer in years and months.

When the debt is paid off, 𝒃𝒏 ≤ 𝟎. Then 𝒃𝟎𝒓𝒏 − 𝑹 (𝟏−𝒓𝒏

𝟏−𝒓) = 𝟎, and 𝒃𝟎𝒓𝒏 = 𝑹 (

𝟏−𝒓𝒏

𝟏−𝒓). Since 𝒃𝟎 = 𝟏𝟎𝟎𝟎𝟎,

𝑹 = 𝟐𝟓𝟎, and 𝒓 = 𝟏. 𝟎𝟐, we have

𝟏𝟎𝟎𝟎𝟎(𝟏. 𝟎𝟐)𝒏 ≤ 𝟐𝟓𝟎 (𝟏 − 𝟏. 𝟎𝟐𝒏

𝟏 − 𝟏. 𝟎𝟐)

𝟏𝟎𝟎𝟎𝟎(𝟏. 𝟎𝟐)𝒏 ≤ −𝟏𝟐𝟓𝟎𝟎(𝟏 − 𝟏. 𝟎𝟐𝒏)

𝟏𝟎𝟎𝟎𝟎(𝟏. 𝟎𝟐)𝒏 ≤ 𝟏𝟐𝟓𝟎𝟎(𝟏. 𝟎𝟐𝒏 − 𝟏)

(𝟏. 𝟎𝟐)𝒏 ≤ 𝟏. 𝟐𝟓(𝟏. 𝟎𝟐)𝒏 − 𝟏. 𝟐𝟓

𝟏. 𝟐𝟓 ≤ 𝟎. 𝟐𝟓(𝟏. 𝟎𝟐)𝒏

𝟓 ≤ 𝟏. 𝟎𝟐𝒏

𝐥𝐨𝐠(𝟓) ≤ 𝒏 𝐥𝐨𝐠(𝟏. 𝟎𝟐)

𝒏 ≥𝐥𝐨𝐠(𝟓)

𝐥𝐨𝐠(𝟏. 𝟎𝟐)

𝒏 ≥ 𝟖𝟏. 𝟐𝟕

It takes 𝟖𝟐 months to pay off this debt, which means it takes 𝟔 years and 𝟏𝟎 months.





ALGEBRA II



522



Problem Set Sample Solutions

Problems 1–4 ask students to compare credit card scenarios with the same initial debt and the same monthly payments

but different interest rates. Problems 5, 6, and 7 require students to compare properties of functions given by different

representations, which aligns with F-IF.C.9 and F-LE.B.5.

The final two problems in this Problem Set require students to do some online research in preparation for Lesson 32, in

which they select a career and model the purchase of a house. Have some printouts of real-estate listings ready to hand

to students who have not brought their own to class. Feel free to add some additional constraints to the criteria for

selecting a house to purchase. The career data in Problem 9 can be found at http://themint.org/teens/starting-

salaries.html. For additional jobs and more information, please visit the U.S. Bureau of Labor Statistics at

http://www.bls.gov/ooh and http://www.bls.gov/ooh/about/teachers-guide.htm. The salary for the “entry-level full-

time” position is based on the projected federal minimum wage in 2016 of $10.10 per hour and a 2,000-hour work year.

1. Suppose that you have a $𝟐, 𝟎𝟎𝟎 balance on a credit card with a 𝟐𝟗. 𝟗𝟗% annual interest rate, compounded

monthly, and you can afford to pay $𝟏𝟓𝟎 per month toward this debt.

a. Find the amount of time it takes to pay off this debt. Give your answer in months and years.

𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏𝟓𝟎 (𝟏 − (𝟏 +

𝟎. 𝟐𝟗𝟗𝟗𝟏𝟐

)𝒏

−𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐

) = 𝟎

𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

= 𝟏𝟓𝟎 ((𝟏 +

𝟎. 𝟐𝟗𝟗𝟗𝟏𝟐

)𝒏

− 𝟏

𝟎. 𝟐𝟗𝟗𝟗𝟏𝟐

)

𝟐𝟗𝟗𝟗

𝟗𝟎𝟎𝟎(𝟏 +

𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

= (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏

(𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

(𝟐𝟗𝟗𝟗

𝟗𝟎𝟎𝟎− 𝟏) = −𝟏

(𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏

(𝟏 −𝟐𝟗𝟗𝟗

𝟗𝟎𝟎𝟎) = 𝟏

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐) + 𝐥𝐨𝐠 (

𝟔𝟎𝟎𝟏

𝟗𝟎𝟎𝟎) = 𝐥𝐨𝐠(𝟏)

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐) = −𝐥𝐨𝐠 (

𝟔𝟎𝟎𝟏

𝟗𝟎𝟎𝟎)

𝒏 = −𝐥𝐨𝐠 (

𝟔𝟎𝟎𝟏𝟗𝟎𝟎𝟎

)

𝐥𝐨𝐠 (𝟏 +𝟎. 𝟐𝟗𝟗𝟗

𝟏𝟐)

𝒏 ≈ 𝟏𝟔. 𝟒𝟏𝟗

So it takes 𝟏 year and 𝟓 months to pay off the debt.

b. Calculate the total amount paid over the life of the debt.

𝟏𝟔. 𝟒𝟏𝟗 ⋅ $𝟏𝟓𝟎 = $𝟐𝟒𝟔𝟐. 𝟖𝟓

c. How much money was paid entirely to the interest on this debt?

$𝟒𝟔𝟐. 𝟖𝟓





http://themint.org/teens/starting-salaries.html

http://themint.org/teens/starting-salaries.html

http://www.bls.gov/ooh

http://www.bls.gov/ooh/about/teachers-guide.htm

ALGEBRA II



523



2. Suppose that you have a $𝟐, 𝟎𝟎𝟎 balance on a credit card with a 𝟏𝟒. 𝟗𝟗% annual interest rate, and you can afford to

pay $𝟏𝟓𝟎 per month toward this debt.


𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏𝟓𝟎 (𝟏 − (𝟏 +

𝟎. 𝟏𝟒𝟗𝟗𝟏𝟐

)𝒏

−𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐

) = 𝟎

𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

= 𝟏𝟓𝟎 ((𝟏 +

𝟎. 𝟏𝟒𝟗𝟗𝟏𝟐

)𝒏

− 𝟏

𝟎. 𝟏𝟒𝟗𝟗𝟏𝟐

)

𝟏𝟒𝟗𝟗

𝟗𝟎𝟎𝟎(𝟏 +

𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

= (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏

(𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

(𝟏𝟒𝟗𝟗

𝟗𝟎𝟎𝟎− 𝟏) = −𝟏

(𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏

(𝟏 −𝟏𝟒𝟗𝟗

𝟗𝟎𝟎𝟎) = 𝟏

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐) + 𝐥𝐨𝐠 (

𝟕𝟓𝟎𝟏

𝟗𝟎𝟎𝟎) = 𝐥𝐨𝐠(𝟏)

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐) = −𝐥𝐨𝐠 (

𝟕𝟓𝟎𝟏

𝟗𝟎𝟎𝟎)


𝟕𝟓𝟎𝟏𝟗𝟎𝟎𝟎

)

𝐥𝐨𝐠 (𝟏 +𝟎. 𝟏𝟒𝟗𝟗

𝟏𝟐)

𝒏 ≈ 𝟏𝟒. 𝟔𝟕𝟔

The loan is paid off in 𝟏 year and 𝟑 months.


𝟏𝟒. 𝟔𝟕𝟔 ⋅ $𝟏𝟓𝟎 = $𝟐, 𝟐𝟎𝟏. 𝟒𝟎


$𝟐𝟎𝟏. 𝟒𝟎





ALGEBRA II



524



3. Suppose that you have a $𝟐, 𝟎𝟎𝟎 balance on a credit card with a 𝟕. 𝟗𝟗% annual interest rate, and you can afford to

pay $𝟏𝟓𝟎 per month toward this debt.


𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏𝟓𝟎 (𝟏 − (𝟏 +

𝟎. 𝟎𝟕𝟗𝟗𝟏𝟐

)𝒏

−𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐

) = 𝟎

𝟐𝟎𝟎𝟎 (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

= 𝟏𝟓𝟎 ((𝟏 +

𝟎. 𝟎𝟕𝟗𝟗𝟏𝟐

)𝒏

− 𝟏

𝟎. 𝟎𝟕𝟗𝟗𝟏𝟐

)

𝟕𝟗𝟗

𝟗𝟎𝟎𝟎(𝟏 +

𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

= (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

− 𝟏

(𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

(𝟕𝟗𝟗

𝟗𝟎𝟎𝟎− 𝟏) = −𝟏

(𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏

(𝟏 −𝟕𝟗𝟗

𝟗𝟎𝟎𝟎) = 𝟏

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐) + 𝐥𝐨𝐠 (

𝟖𝟐𝟎𝟏

𝟗𝟎𝟎𝟎) = 𝐥𝐨𝐠(𝟏)

𝒏 ⋅ 𝐥𝐨𝐠 (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐) = −𝐥𝐨𝐠 (

𝟖𝟐𝟎𝟏

𝟗𝟎𝟎𝟎)


𝟖𝟐𝟎𝟏𝟗𝟎𝟎𝟎

)

𝐥𝐨𝐠 (𝟏 +𝟎. 𝟎𝟕𝟗𝟗

𝟏𝟐)

𝒏 ≈ 𝟏𝟒. 𝟎𝟎𝟗

The loan is paid off in 𝟏 year and 𝟑 months.


𝟏𝟒. 𝟎𝟎𝟗 ⋅ $𝟏𝟓𝟎 = $𝟐𝟏𝟎𝟏. 𝟑𝟓


$𝟏𝟎𝟏. 𝟑𝟓

4. Summarize the results of Problems 1, 2, and 3.

Answers will vary but should include the fact that the total interest paid in each case dropped by about half with

every problem. Lower interest rates meant that the loan was paid off more quickly and that less was paid in total.





ALGEBRA II



525



5. Brendan owes $𝟏, 𝟓𝟎𝟎 on a credit card with an interest rate of 𝟏𝟐%. He is making payments of $𝟏𝟎𝟎 every month

to pay this debt off. Maggie is also making regular payments to a debt owed on a credit card, and she created the

following graph of her projected balance over the next 𝟏𝟐 months.

a. Who has the higher initial balance? Explain how you know.

Reading from the graph, Maggie’s initial balance is between $𝟏, 𝟕𝟎𝟎 and $𝟏, 𝟖𝟎𝟎, and we are given that

Brendan’s initial balance is $𝟏, 𝟓𝟎𝟎, so Maggie has the larger initial balance.

b. Who will pay their debt off first? Explain how you know.

From the graph, it appears that Maggie will pay off her debt between months 𝟏𝟐 and 𝟏𝟒. Brendan’s balance

in month 𝒏 can be modeled by the function 𝒃𝒏 = 𝟏𝟓𝟎𝟎(𝟏. 𝟎𝟏)𝒏 − 𝟏𝟎𝟎 (𝟏.𝟎𝟏

𝒏−𝟏

𝟎.𝟎𝟏), which is equal to zero

when 𝒏 ≈ 𝟏𝟔. 𝟑. Thus, Brendan’s debt will be paid in month 𝟏𝟕, so Maggie’s debt will be paid off first.

6. Alan and Emma are both making $𝟐𝟎𝟎 monthly payments toward balances on credit cards. Alan has prepared a

table to represent his projected balances, and Emma has prepared a graph.

Alan’s Credit Card Balance

Month, 𝒏 Interest Payment Balance, 𝒃𝒏

𝟎 𝟐, 𝟎𝟎𝟎. 𝟎𝟎

𝟏 𝟒𝟏. 𝟔𝟓 𝟐𝟎𝟎 𝟏, 𝟖𝟒𝟏. 𝟔𝟓

𝟐 𝟑𝟖. 𝟑𝟓 𝟐𝟎𝟎 𝟏, 𝟔𝟖𝟎. 𝟎𝟎

𝟑 𝟑𝟒. 𝟗𝟗 𝟐𝟎𝟎 𝟏, 𝟓𝟏𝟒. 𝟗𝟗

𝟒 𝟑𝟏. 𝟓𝟓 𝟐𝟎𝟎 𝟏, 𝟑𝟒𝟔. 𝟓𝟒

𝟓 𝟐𝟖. 𝟎𝟒 𝟐𝟎𝟎 𝟏, 𝟏𝟕𝟒. 𝟓𝟖

𝟔 𝟐𝟒. 𝟒𝟔 𝟐𝟎𝟎 𝟗𝟗𝟗. 𝟎𝟒

𝟕 𝟐𝟎. 𝟖𝟏 𝟐𝟎𝟎 𝟖𝟏𝟗. 𝟖𝟓

𝟖 𝟏𝟕. 𝟎𝟕 𝟐𝟎𝟎 𝟔𝟑𝟔. 𝟗𝟐

𝟗 𝟏𝟑. 𝟐𝟔 𝟐𝟎𝟎 𝟒𝟓𝟎. 𝟏𝟖

𝟏𝟎 𝟗. 𝟑𝟕 𝟐𝟎𝟎 𝟐𝟓𝟗. 𝟓𝟓

𝟏𝟏 𝟓. 𝟒𝟏 𝟐𝟎𝟎 𝟔𝟒. 𝟗𝟔

a. What is the annual interest rate on Alan’s debt? Explain how you know.

One month’s interest on the balance of $𝟐, 𝟎𝟎𝟎 was $𝟒𝟏. 𝟔𝟓, so 𝟒𝟏. 𝟔𝟓 = 𝒊(𝟐𝟎𝟎𝟎). Then the monthly

interest rate is 𝒊 = 𝟎. 𝟎𝟐𝟎𝟖𝟐𝟓, and the annual rate is 𝟏𝟐𝒊 = 𝟎. 𝟐𝟒𝟗𝟗, so the annual rate on Alan’s debt is

𝟐𝟒. 𝟗𝟗%.

Month

Cre

dit

Car

d B

alan

ce





ALGEBRA II



526



b. Who has the higher initial balance? Explain how you know.

From the table, we can see that Alan’s initial balance is $𝟐, 𝟎𝟎𝟎, while Emma’s initial balance is the

𝒚-intercept of the graph, which is above $𝟐, 𝟎𝟎𝟎. Thus, Emma’s initial balance is higher.

c. Who will pay their debt off first? Explain how you know.

Both Alan and Emma will pay their debts off in month 𝟏𝟐 because both of their balances in month 𝟏𝟏 are

under $𝟏𝟎𝟎.

d. What do your answers to parts (a), (b), and (c) tell you about the interest rate for Emma’s debt?

Because Emma had the higher initial balance, and they made the same number of payments, Emma must

have a lower interest rate on her credit card than Alan does. In fact, since the graph decreases apparently

linearly, this implies that Emma has an interest rate of 𝟎%.

7. Both Gary and Helena are paying regular monthly payments to a credit card balance. The balance on Gary’s credit

card debt can be modeled by the recursive formula 𝒈𝒏 = 𝒈𝒏−𝟏(𝟏. 𝟎𝟏𝟔𝟔𝟔) − 𝟐𝟎𝟎 with 𝒈𝟎 = 𝟐𝟓𝟎𝟎, and the balance

on Helena’s credit card debt can be modeled by the explicit formula 𝒉𝒏 = 𝟐𝟎𝟎𝟎(𝟏. 𝟎𝟏𝟔𝟔𝟔)𝒏 − 𝟐𝟓𝟎 (𝟏.𝟎𝟏𝟔𝟔𝟔

𝒏−𝟏

𝟎.𝟎𝟏𝟔𝟔𝟔)

for 𝒏 ≥ 𝟎.

a. Who has the higher initial balance? Explain how you know.

Gary has the higher initial balance. Helena’s initial balance is $𝟐, 𝟎𝟎𝟎, and Gary’s is $𝟐, 𝟓𝟎𝟎.

b. Who has the higher monthly payment? Explain how you know.

Helena has the higher monthly payment. She is paying $𝟐𝟓𝟎 every month while Gary is paying $𝟐𝟎𝟎.

c. Who will pay their debt off first? Explain how you know.

Helena will pay her debt off first since she starts at a lower balance and is paying more per month.

Additionally, they appear to have the same interest rates.

8. In the next lesson, we will apply the mathematics we have learned to the purchase of a house. In preparation for

that task, you need to come to class prepared with an idea of the type of house you would like to buy.

a. Research the median housing price in the county where you live or where you wish to relocate.

Answers will vary.

b. Find the range of prices that are within 𝟐𝟓% of the median price from part (a). That is, if the price from part

(a) was 𝑷, then your range is 𝟎. 𝟕𝟓𝑷 to 𝟏. 𝟐𝟓𝑷.

Answers will vary.

c. Look at online real estate websites, and find a house located in your selected county that falls into the price

range specified in part (b). You will be modeling the purchase of this house in Lesson 32, so bring a printout

of the real estate listing to class with you.

Answers will vary.





ALGEBRA II



527



9. Select a career that interests you from the following list of careers. If the career you are interested in is not on this

list, check with your teacher to obtain permission to perform some independent research. Once it has been

selected, use the career to answer questions in Lesson 32 and Lesson 33.

Occupation Median Starting Salary Education Required

Entry-level full-time

(wait staff, office clerk,

lawn care worker, etc.)

$𝟐𝟎, 𝟐𝟎𝟎 High school diploma or GED

Accountant $𝟓𝟒, 𝟔𝟑𝟎 𝟒-year college degree

Athletic Trainer $𝟑𝟔, 𝟓𝟔𝟎 𝟒-year college degree

Chemical Engineer $𝟕𝟖, 𝟖𝟔𝟎 𝟒-year college degree

Computer Scientist $𝟗𝟑, 𝟗𝟓𝟎 𝟒-year college degree or more

Database Administrator $𝟔𝟒, 𝟔𝟎𝟎 𝟒-year college degree

Dentist $𝟏𝟑𝟔, 𝟗𝟔𝟎 Graduate degree

Desktop Publisher $𝟑𝟒, 𝟏𝟑𝟎 𝟒-year college degree

Electrical Engineer $𝟕𝟓, 𝟗𝟑𝟎 𝟒-year college degree

Graphic Designer $𝟑𝟗, 𝟗𝟎𝟎 𝟐- or 𝟒-year college degree

HR Employment Specialist $𝟒𝟐, 𝟒𝟐𝟎 𝟒-year college degree

HR Compensation Manager $𝟔𝟔, 𝟓𝟑𝟎 𝟒-year college degree

Industrial Designer $𝟓𝟒, 𝟓𝟔𝟎 𝟒-year college degree or more

Industrial Engineer $𝟔𝟖, 𝟔𝟐𝟎 𝟒-year college degree

Landscape Architect $𝟓𝟓, 𝟏𝟒𝟎 𝟒-year college degree

Lawyer $𝟏𝟎𝟐, 𝟒𝟕𝟎 Law degree

Occupational Therapist $𝟔𝟎, 𝟒𝟕𝟎 Master’s degree

Optometrist $𝟗𝟏, 𝟎𝟒𝟎 Master’s degree

Physical Therapist $𝟔𝟔, 𝟐𝟎𝟎 Master’s degree

Physician—Anesthesiology $𝟐𝟓𝟗, 𝟗𝟒𝟖 Medical degree

Physician—Family Practice $𝟏𝟑𝟕, 𝟏𝟏𝟗 Medical degree

Physician’s Assistant $𝟕𝟒, 𝟗𝟖𝟎 𝟐 years college plus 𝟐-year program

Radiology Technician $𝟒𝟕, 𝟏𝟕𝟎 𝟐-year degree

Registered Nurse $𝟓𝟕, 𝟐𝟖𝟎 𝟐- or 𝟒-year college degree plus

Social Worker—Hospital $𝟒𝟖, 𝟒𝟐𝟎 Master’s degree

Teacher—Special Education $𝟒𝟕, 𝟔𝟓𝟎 Master’s degree

Veterinarian $𝟕𝟏, 𝟗𝟗𝟎 Veterinary degree





lesson 31: credit cards - engageny.org

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