Mathematical Modeling Exercise:<\/p>\n
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\u201d payment has been made as b.<\/p>\n
a. What is the monthly interest rate, i? Approximate i to 5 decimal places.
\nAnswer:
\ni = \\(\\frac{0.1999}{12}\\) \u2248 0.01666<\/p>\n
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?
\nAnswer:
\nStudents will answer 50, 100, or 150 as appropriate.<\/p>\n
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?
\nAnswer:
\nThe minimum payment is 0.01($1500) + 0.01666($1500) = $39. 99. All given values of R are greater than the minimum payment.<\/p>\n
d. Complete the following table to show 6 months of payments.<\/p>\n
img 1<\/p>\n
Answer:<\/p>\n
img 2<\/p>\n
img 3<\/p>\n
img 4<\/p>\n
e. Write a recursive formula for the balance bn<\/sub> in month n in terms of the balance bn – 1<\/sub>. f. Write an explicit formula for the balance b in month n, leaving the expression 1 + i in symbolic form. g. Rewrite your formula in part (f) using r to represent the quantity (1 + i). h. What can you say about your formula in part (g)? What term do we use to describe r in this formula? i. Write your formula from part(g) in summation notation using \u03a3. k. Find the month when your balance is paid off. b0<\/sub>rn<\/sup> – R\\(\\left(\\frac{1-r^{n}}{1-r}\\right)\\) = 0 n = \\(\\frac{\\log \\left(\\frac{\\boldsymbol{R}}{\\left(\\boldsymbol{b}_{0}(\\mathbf{1}-\\boldsymbol{r})+\\boldsymbol{R}\\right)}\\right)}{\\log (\\boldsymbol{r})}\\)<\/p>\n If R = 50, then n \u2248 41.925. The debt is paid off in 42 months. l. Calculate the total amount paid over the life of the debt. How much was paid solely to interest? The total amount paid using monthly payments of $50 is $2,096.37. Of this amount, $596.37 is interest. 100(17) + (1 + i)b17<\/sub> = 1700 + r\\(\\left(b_{0} r^{17}-R\\left(\\frac{1-r^{17}}{1-r}\\right)\\right)\\) Question 1. a. Find the amount of time it takes to pay off this debt. Give your answer in months and years. b. Calculate the total amount paid over the life of the debt. c. How much money was paid entirely to the interest on this debt? Question 2. a. Find the amount of time it takes to pay off this debt. Give your answer in months and years. b. Calculate the total amount paid over the life of the debt. c. How much money was paid entirely to the interest on this debt? Question 3. a. Find the amount of time it takes to pay off this debt. Give your answer in months and years. b. Calculate the total amount paid over the life of the debt. c. How much money was paid entirely to the interest on this debt? Question 4. Question 5. img 8<\/p>\n a. Who has the higher initial balance? Explain how you know. b. Who will pay their debt off first? Explain how you know. Question 6. img 9<\/p>\n img 10<\/p>\n a. What is the annual interest rate on Alan\u2019s debt? Explain how you know. b. Who has the higher initial balance? Explain how you know. c. Who will pay their debt off first? Explain how you know. d. What do your answers to parts (a), (b), and (c) tell you about the interest rate for Emma\u2019s debt? Question 7. a. Who has the higher Initial balance? Explain how you know. b. Who has the higher monthly payment? Explain how you know. c. Who will pay their debt off first? Explain how you know. Question 8. a. Research the median housing price in the county where you live or where you wish to relocate. 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. 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. Question 9. img 11<\/p>\n Answer:<\/p>\n Question 1. a. What is the monthly interest rate j? What is the growth rate, r? b. Explain why we can rewrite the given formula as bn<\/sub> = b0<\/sub>rn<\/sup> – R\\(\\left(\\frac{1-r^{n}}{1-r}\\right)\\). Then the formula becomes 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. 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 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/102325"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=102325"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/102325\/revisions"}],"predecessor-version":[{"id":135277,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/102325\/revisions\/135277"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=102325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=102325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=102325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nAnswer:
\nTo calculate the new balance, bn<\/sub>, we compound interest for one month on the previous balance bn – 1<\/sub> and then subtract the payment R:
\nbn<\/sub> = bn – 1<\/sub>(1 + i) – R, with b0<\/sub> = 1500.<\/p>\n
\nAnswer:
\nWe have the following formulas:
\nb1<\/sub> = b0<\/sub>(1 + i) – R
\nb2<\/sub> = b1<\/sub>(1 + i) – R
\n= [b0<\/sub>(1 + i) – R](1 + i) – R
\n= b0<\/sub>(1 + i)2<\/sup> – R(1 + i) – R
\nb3<\/sub> = b2<\/sub>(1 + i) – R
\n= [b0<\/sub>(1 + i)2<\/sup> – R(1 + i) – R](1 + i) – R
\n= b0<\/sub>(1 + i)3<\/sup> R(1 + i)2<\/sup> – R(1 + i) – R
\n.
\n.
\n.
\n.
\nbn<\/sub> = b0<\/sub>(1 + i)n<\/sup> – R(1 + i)n – 1<\/sup> – R(1 + i)n – 2<\/sup> ….. -R(1 + i) – R<\/p>\n
\nAnswer:
\nbn<\/sub> = b0<\/sub>rn<\/sup> – Rrn – 1<\/sup> -Rrn – 2<\/sup> – Rr – R
\n= b0<\/sub>rn<\/sup> – R(1 + r + r2<\/sup> + … + rn – 1<\/sup>)<\/p>\n
\nAnswer:
\nThe formula in part (g) contains the sum of a finite geometric series with common ratio r.<\/p>\n
\nAnswer:
\nbn<\/sub> = b0<\/sub>rn<\/sup> – R(1 + r + r2<\/sup> + …. + rn – 1<\/sup>)
\n= b0<\/sub>rn<\/sup> – R\\(\\left(\\frac{1-r^{n}}{1-r}\\right)\\)<\/p>\n
\nAnswer:
\nThe balance is paid off when bn<\/sub> = 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<\/sub>, and R substituted in.<\/p>\n
\nb0<\/sub>rn<\/sup> = R\\(\\left(\\frac{1-r^{n}}{1-r}\\right)\\)
\n(1 – r)(b0<\/sub>rn<\/sup>) = R(1 – rn<\/sup>)
\n(1 – r)(b0<\/sub>rn<\/sup>) + Rrn<\/sup> = R
\nrn<\/sup> = \\(\\frac{\\boldsymbol{R}}{\\left(\\boldsymbol{b}_{0}(\\mathbf{1}-\\boldsymbol{r})+\\boldsymbol{R}\\right)}\\)
\nn log(r) = log\\(\\left(\\frac{\\boldsymbol{R}}{\\left(\\boldsymbol{b}_{0}(\\mathbf{1}-\\boldsymbol{r})+\\boldsymbol{R}\\right)}\\right)\\)<\/p>\n
\nIf R = 100, then n \u2248 17.49. The debt is paid off in 18months.
\nIf R = 150, then n \u2248 11.0296. The debt is paid off in 12 months.<\/p>\n
\nAnswer:
\nFor R = 50: The debt is paid in 41 payments of $50, and the last payment is the amount b41 with interest:
\n50(41) + (1 + i)b41<\/sub> = 2050 + r(b0<\/sub>rn<\/sup> – R\\(\\left(\\frac{1-r^{n}}{1-r}\\right)\\)
\n\u2248 2050 + r(45.61)
\n\u2248 2096.37.<\/p>\n
\nFor R = 100: The debt is paid in 17 payments of $100, and the last payment is the amount b17<\/sup> with interest.<\/p>\n
\n\u2248 1700 + r(40.52)
\n\u2248 1740.52
\nThe total amount paid using monthly payments of $100 is $1,740.52. Of this amount, $240.52 is interest.
\nFor R = 150: The debt is paid in 11 payments of $150, and the last payment is the amount b11<\/sub> with interest.
\n150(11) + (1 + i)b11<\/sub> = 1700 + r\\(\\left(b_{0} r^{n}-R\\left(\\frac{1-r^{n}}{1-r}\\right)\\right)\\)
\n\u2248 1650 +r(4.49)
\n\u2248 1654.49
\nThe total amount paid using monthly payments of $150 is $1,654.49. Of this amount, $154.49 is interest.<\/p>\nEureka Math Algebra 2 Module 3 Lesson 31 Problem Set Answer Key<\/h3>\n
\nSuppose 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.<\/p>\n
\nAnswer:
\nimg 5<\/p>\n
\nAnswer:
\n16.419 . $150 = $2462.85<\/p>\n
\nAnswer:
\n$462.85<\/p>\n
\nSuppose 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.<\/p>\n
\nAnswer:
\nimg 6<\/p>\n
\nAnswer:
\n14. 676 . $150 = $2,201.40<\/p>\n
\nAnswer:
\n$201.40<\/p>\n
\nSuppose 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.<\/p>\n
\nAnswer:
\nimg 7<\/p>\n
\nAnswer:
\n14.009 . $150 = $2101.35<\/p>\n
\nAnswer:
\n$101.35<\/p>\n
\nSummarize the results of Problems 1, 2, and 3.
\nAnswer:
\nAnswers 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.<\/p>\n
\nBrendan 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.<\/p>\n
\nAnswer:
\nReading from the graph, Maggie\u2019s initial balance is between $1,700 and $1,800, and we are given that Brendan\u2019s initial balance is $1, 500, so Maggie has the larger initial balance.<\/p>\n
\nAnswer:
\nFrom the graph, it appears that Maggie will pay off her debt between months 12 and 14. Brendan\u2019s balance in month n can be modeled by the function b = 1500(1.01)n<\/sup> – 100\\(\\left(\\frac{1.01^{n}-1}{0.01}\\right)\\), which is equal to zero when n \u2248 16.3. Thus, Brendan\u2019s debt will be paid in month 17, so Maggie\u2019s debt will be paid off first.<\/p>\n
\nAlan 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.<\/p>\n
\nAnswer:
\nOne month\u2019s 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\u2019s debt is 24.99%.<\/p>\n
\nAnswer:
\nFrom the table, we can see that Alan\u2019s initial balance is $2,000, while Emma\u2019s initial balance is the y-intercept of the graph, which is above $2, 000. Thus, Emma\u2019s initial balance is higher.<\/p>\n
\nAnswer:
\nBoth Alan and Emma will pay their debts off in month 12 because both of their balances in month 11 are under $100.<\/p>\n
\nAnswer:
\nBecause 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%.<\/p>\n
\nBoth Gary and Helena are paying regular monthly payments to a credit card balance. The balance on Gary\u2019s credit card debt can be modeled by the recursive formula gn<\/sub> = gn – 1<\/sub>(1.01666) – 200 with g0<\/sub> = 2500, and the balance on Helena\u2019s credit card debt can be modeled by the explicit formula hn<\/sub> = 2000(1.01666)n<\/sup> – 250\\(\\left(\\frac{1.01666^{n}-1}{0.01666}\\right)\\) for n \u2265 0.<\/p>\n
\nAnswer:
\nGary has the higher initial balance. Helena\u2019s initial balance is $2,000, and Gary\u2019s is $2,500.<\/p>\n
\nAnswer:
\nHelena has the higher monthly payment. She is paying $250 every month while Gary is paying $200.<\/p>\n
\nAnswer:
\nHelena 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.<\/p>\n
\nIn 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.<\/p>\n
\nAnswer:
\nAnswers will vary.<\/p>\n
\nAnswer:
\nAnswers will vary.<\/p>\n
\nAnswer:
\nAnswers will vary.<\/p>\n
\nSelect 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.<\/p>\nEureka Math Algebra 2 Module 3 Lesson 31 Exit Ticket Answer Key<\/h3>\n
\nSuppose 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 = b0<\/sub>rn<\/sup> – R(1 + r + r2<\/sup> + …. + rn – 1<\/sup>), where b0<\/sub> is the initial balance, bn<\/sub> 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.<\/p>\n
\nAnswer:
\nThe monthly interest rate ils given by i = \\(\\frac{0.24}{12}\\) = 0.02, and r = 1 + i = 1.02.<\/p>\n
\nAnswer:
\nUsing summation notation and the sum formula for afinite geometric series, we have
\n1 + r + r2<\/sup> + …. + rn – 1<\/sup> = \\(\\sum_{k=0}^{n-1} r^{n k}\\) = \\(\\frac{1-r^{n}}{1-r}\\).<\/p>\n
\nbn<\/sub> = b0<\/sub>rn<\/sup> – R(1 + r + r2<\/sup> + …. + rn – 1<\/sup>)
\n= b0<\/sub>rn<\/sup> – R\\(\\frac{1-r^{n}}{1-r}\\)<\/p>\n
\nAnswer:
\nWhen the debt is paid off, bn<\/sub> \u2264 0. Then b0<\/sub>rn<\/sup> – R\\(\\frac{1-r^{n}}{1-r}\\) = 0, and b0<\/sub>rn<\/sup> = R\\(\\frac{1-r^{n}}{1-r}\\) Since b0<\/sub> = 10000, R = 250, and r = 1.02, we have
\n10000(1.02)n<\/sup> \u2264 250\\(\\frac{1-1.02^{n}}{1-1.02}\\)
\n10000(1.02)n<\/sup> \u2264 -12500(1 – 1.02n<\/sup>)
\n10000(1.02)n<\/sup> \u226412500(1.02n<\/sup> – 1)
\n(1.02)n<\/sup> \u2264 1.25(1.02)n<\/sup> – 1.25
\n1.25 \u2264 0.25(1.02)n<\/sup>
\n5 \u2264 1.02n<\/sup>
\nlog(5) \u2264 n log(1.02)
\nn \u2265 \\(\\frac{log{5}}log{1.02}\\)
\nn \u2265 81.27
\nIt takes 82 months to pay off this debt, which means it takes 6 years and 10 months.<\/p>\n","protected":false},"excerpt":{"rendered":"