Tag: Eureka Math Algebra 2 Module 3

Engage NY Eureka Math Algebra 2 Module 31 Lesson 30 Answer Key

Eureka Math Algebra 2 Module 3 Lesson 31 Mathematical Modeling Exercise Answer Key

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 ne” payment has been made as b.

a. What is the monthly interest rate, i? Approximate i to 5 decimal places.
Answer:
i = \(\frac{0.1999}{12}\) ≈ 0.01666

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?
Answer:
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?
Answer:
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.

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Answer:

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e. Write a recursive formula for the balance b_n in month n in terms of the balance b_{n – 1}.
Answer:
To calculate the new balance, b_n, we compound interest for one month on the previous balance b_{n – 1} and then subtract the payment R:
b_n = b_{n – 1}(1 + i) – R, with b₀ = 1500.

f. Write an explicit formula for the balance b in month n, leaving the expression 1 + i in symbolic form.
Answer:
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
.
.
.
.
b_n = 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).
Answer:
b_n = b₀rⁿ – Rr^{n – 1} -Rr^{n – 2} – Rr – R
= b₀rⁿ – R(1 + r + r² + … + r^{n – 1})

h. What can you say about your formula in part (g)? What term do we use to describe r in this formula?
Answer:
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.
Answer:
b_n = b₀rⁿ – R(1 + r + r² + …. + r^{n – 1})
= b₀rⁿ – R\(\left(\frac{1-r^{n}}{1-r}\right)\)

k. Find the month when your balance is paid off.
Answer:
The balance is paid off when b_n = 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\(\left(\frac{1-r^{n}}{1-r}\right)\) = 0
b₀rⁿ = R\(\left(\frac{1-r^{n}}{1-r}\right)\)
(1 – r)(b₀rⁿ) = R(1 – rⁿ)
(1 – r)(b₀rⁿ) + Rrⁿ = R
rⁿ = \(\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}\)
n log(r) = log\(\left(\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}\right)\)

n = \(\frac{\log \left(\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}\right)}{\log (\boldsymbol{r})}\)

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 18months.
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?
Answer:
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\(\left(\frac{1-r^{n}}{1-r}\right)\)
≈ 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\(\left(b_{0} r^{17}-R\left(\frac{1-r^{17}}{1-r}\right)\right)\)
≈ 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\(\left(b_{0} r^{n}-R\left(\frac{1-r^{n}}{1-r}\right)\right)\)
≈ 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.

Eureka Math Algebra 2 Module 3 Lesson 31 Problem Set Answer Key

Question 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.
Answer:
img 5

b. Calculate the total amount paid over the life of the debt.
Answer:
16.419 . $150 = $2462.85

c. How much money was paid entirely to the interest on this debt?
Answer:
$462.85

Question 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.

a. Find the amount of time it takes to pay off this debt. Give your answer in months and years.
Answer:
img 6

b. Calculate the total amount paid over the life of the debt.
Answer:
14. 676 . $150 = $2,201.40

c. How much money was paid entirely to the interest on this debt?
Answer:
$201.40

Question 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.

a. Find the amount of time it takes to pay off this debt. Give your answer in months and years.
Answer:
img 7

b. Calculate the total amount paid over the life of the debt.
Answer:
14.009 . $150 = $2101.35

c. How much money was paid entirely to the interest on this debt?
Answer:
$101.35

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

Question 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.

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a. Who has the higher initial balance? Explain how you know.
Answer:
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.
Answer:
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 b = 1500(1.01)ⁿ – 100\(\left(\frac{1.01^{n}-1}{0.01}\right)\), 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.

Question 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.

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

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

Question 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 g_n = g_{n – 1}(1.01666) – 200 with g₀ = 2500, and the balance on Helena’s credit card debt can be modeled by the explicit formula h_n = 2000(1.01666)ⁿ – 250\(\left(\frac{1.01666^{n}-1}{0.01666}\right)\) for n ≥ 0.

a. Who has the higher Initial balance? Explain how you know.
Answer:
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.
Answer:
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.
Answer:
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.

Question 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.
Answer:
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.
Answer:
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.
Answer:
Answers will vary.

Question 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.

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Answer:

Eureka Math Algebra 2 Module 3 Lesson 31 Exit Ticket Answer Key

Question 1.
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 b = b₀rⁿ – R(1 + r + r² + …. + r^{n – 1}), where b₀ is the initial balance, b_n 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 j? What is the growth rate, r?
Answer:
The monthly interest rate ils given by i = \(\frac{0.24}{12}\) = 0.02, and r = 1 + i = 1.02.

b. Explain why we can rewrite the given formula as b_n = b₀rⁿ – R\(\left(\frac{1-r^{n}}{1-r}\right)\).
Answer:
Using summation notation and the sum formula for afinite geometric series, we have
1 + r + r² + …. + r^{n – 1} = \(\sum_{k=0}^{n-1} r^{n k}\) = \(\frac{1-r^{n}}{1-r}\).

Then the formula becomes
b_n = b₀rⁿ – R(1 + r + r² + …. + r^{n – 1})
= b₀rⁿ – R\(\frac{1-r^{n}}{1-r}\)

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.
Answer:
When the debt is paid off, b_n ≤ 0. Then b₀rⁿ – R\(\frac{1-r^{n}}{1-r}\) = 0, and b₀rⁿ = R\(\frac{1-r^{n}}{1-r}\) Since b₀ = 10000, R = 250, and r = 1.02, we have
10000(1.02)ⁿ ≤ 250\(\frac{1-1.02^{n}}{1-1.02}\)
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 ≥ \(\frac{log{5}}log{1.02}\)
n ≥ 81.27
It takes 82 months to pay off this debt, which means it takes 6 years and 10 months.