Question

A savings plan is set up so that $100 is placed into an account on the...

A savings plan is set up so that $100 is placed into an account on the 1st and 15th of each month. The first payment was made on January 1, 2010 and this continued until December 1, 2013. Then no money was deposited for the rest of December 2013 until March 1, 2014 when the $100 payments were continued until the last payment was made on February 1, 2017. Interest is 2.5% effective annual interest. a. Draw a careful number line and indicate the number of payments that were made before the missed payment, the number of missed payments and the number of payments after the missed payments. Remember that the payments are twice per month. b. Find the annual rate compounded twice per month. c. Find the amount in the account after the payment is made on December 1, 2013. d. Find the amount in the account after the payment is made on March 15, 2014. e. Find the amount in the account on February 15, 2017

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Part (a)

Timeline and payment

Payment no. Date Payment
1 1-Jan-10 100
2 15-Jan-10 100
3 1-Feb-10 100
4 15-Feb-10 100
5 1-Mar-10 100
6 15-Mar-10 100
7 1-Apr-10 100
8 15-Apr-10 100
9 1-May-10 100
10 15-May-10 100
11 1-Jun-10 100
12 15-Jun-10 100
13 1-Jul-10 100
14 15-Jul-10 100
15 1-Aug-10 100
16 15-Aug-10 100
17 1-Sep-10 100
18 15-Sep-10 100
19 1-Oct-10 100
20 15-Oct-10 100
21 1-Nov-10 100
22 15-Nov-10 100
23 1-Dec-10 100
24 15-Dec-10 100
25 1-Jan-11 100
26 15-Jan-11 100
27 1-Feb-11 100
28 15-Feb-11 100
29 1-Mar-11 100
30 15-Mar-11 100
31 1-Apr-11 100
32 15-Apr-11 100
33 1-May-11 100
34 15-May-11 100
35 1-Jun-11 100
36 15-Jun-11 100
37 1-Jul-11 100
38 15-Jul-11 100
39 1-Aug-11 100
40 15-Aug-11 100
41 1-Sep-11 100
42 15-Sep-11 100
43 1-Oct-11 100
44 15-Oct-11 100
45 1-Nov-11 100
46 15-Nov-11 100
47 1-Dec-11 100
48 15-Dec-11 100
49 1-Jan-12 100
50 15-Jan-12 100
51 1-Feb-12 100
52 15-Feb-12 100
53 1-Mar-12 100
54 15-Mar-12 100
55 1-Apr-12 100
56 15-Apr-12 100
57 1-May-12 100
58 15-May-12 100
59 1-Jun-12 100
60 15-Jun-12 100
61 1-Jul-12 100
62 15-Jul-12 100
63 1-Aug-12 100
64 15-Aug-12 100
65 1-Sep-12 100
66 15-Sep-12 100
67 1-Oct-12 100
68 15-Oct-12 100
69 1-Nov-12 100
70 15-Nov-12 100
71 1-Dec-12 100
72 15-Dec-12 100
73 1-Jan-13 100
74 15-Jan-13 100
75 1-Feb-13 100
76 15-Feb-13 100
77 1-Mar-13 100
78 15-Mar-13 100
79 1-Apr-13 100
80 15-Apr-13 100
81 1-May-13 100
82 15-May-13 100
83 1-Jun-13 100
84 15-Jun-13 100
85 1-Jul-13 100
86 15-Jul-13 100
87 1-Aug-13 100
88 15-Aug-13 100
89 1-Sep-13 100
90 15-Sep-13 100
91 1-Oct-13 100
92 15-Oct-13 100
93 1-Nov-13 100
94 15-Nov-13 100
95 1-Dec-13 100
96 15-Dec-13
97 1-Jan-14
98 15-Jan-14
99 1-Feb-14
100 15-Feb-14
101 1-Mar-14 100
102 15-Mar-14 100
103 1-Apr-14 100
104 15-Apr-14 100
105 1-May-14 100
106 15-May-14 100
107 1-Jun-14 100
108 15-Jun-14 100
109 1-Jul-14 100
110 15-Jul-14 100
111 1-Aug-14 100
112 15-Aug-14 100
113 1-Sep-14 100
114 15-Sep-14 100
115 1-Oct-14 100
116 15-Oct-14 100
117 1-Nov-14 100
118 15-Nov-14 100
119 1-Dec-14 100
120 15-Dec-14 100
121 1-Jan-15 100
122 15-Jan-15 100
123 1-Feb-15 100
124 15-Feb-15 100
125 1-Mar-15 100
126 15-Mar-15 100
127 1-Apr-15 100
128 15-Apr-15 100
129 1-May-15 100
130 15-May-15 100
131 1-Jun-15 100
132 15-Jun-15 100
133 1-Jul-15 100
134 15-Jul-15 100
135 1-Aug-15 100
136 15-Aug-15 100
137 1-Sep-15 100
138 15-Sep-15 100
139 1-Oct-15 100
140 15-Oct-15 100
141 1-Nov-15 100
142 15-Nov-15 100
143 1-Dec-15 100
144 15-Dec-15 100
145 1-Jan-16 100
146 15-Jan-16 100
147 1-Feb-16 100

the number of payments that were made before the missed payment = 95

the number of missed payments = 5

and the number of payments after the missed payments = 47

Part (b)

Let r be the annual rate compounded twice per month

EAR = 2.5% = (1 + r / 24)24 - 1

Hence, r = [(1 + 2.5%)1/24 - 1] x 24 = 0.1029%

Part (c)

the amount in the account after the payment is made on December 1, 2013 = FV (Rate, Nper, PMT, PV) = FV (0.1029%, 94, -100, -100) = $9,974.64

Part (d)

the amount in the account after the payment is made on March 1, 2014 = $9,974.64 x (1 + 0.1029%)6 + 100 = $10,136.41

Part (e)

the amount in the account on the last payment date of February 1, 2016 = FV (Rate, Nper, PMT, PV) = FV (0.1029%, 46, -100, -10136.41) = $15,335.84

Add a comment
Know the answer?
Add Answer to:
A savings plan is set up so that $100 is placed into an account on the...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Question 6 An educational saving account is created. The account is set up such that the...

    Question 6 An educational saving account is created. The account is set up such that the beneficiary will receive all money in the account (principal as well as accrued interest) after 12 years from the present. Payments into this account are made by depositing $275 to this account every month. Because this saving plan is registered, a matching grant is given by the Government amounting to 25% of the money deposited (with the exception of the last $275 paid at...

  • To calculate the future value of an annuity (savings plan), we use the formula: FV =...

    To calculate the future value of an annuity (savings plan), we use the formula: FV = PMT[(1 + rin) nt) - 1]/(r/n) where PMT is the payment amount that is deposited on a regular basis, r is the APR, n is the number of regular payments made each year and FV is the future value after t years. At the age of 25, Kyle starts an IRA (Individual Retirement Account) to save for retirement. He deposits $200 into the account...

  • JAVA PLEASE! Suppose you save $100 each month into a savings account with the annual interest...

    JAVA PLEASE! Suppose you save $100 each month into a savings account with the annual interest rate 5%. So, the monthly interest rate is 0.05/12 = 0.00417. After the first month, the value in the account becomes 100 * (1 + 0.00417) = 100.417 After the second month, the value in the account becomes (100 + 100.417) * (1 + 0.00417) = 201.252 After the third month, the value in the account becomes (100 + 201.252) * (1 + 0.00417)...

  • *1.2.18 Smith receives a paycheck of 3500 on the last day of each month, and immediately...

    *1.2.18 Smith receives a paycheck of 3500 on the last day of each month, and immediately deposits all but 1000 of it in a bank account. The first deposit is on December 31, 2017. Smith deposits an addition- al 1000 on the 15ª of every month. The account pays an annual in- terest rate of 10%. Find the balance in the account on March 31. 2018, after the deposit is made and interest is credited, in each of the following...

  • QUESTION 6 A loan of L is to be repaid with 40 payments of 100 at...

    QUESTION 6 A loan of L is to be repaid with 40 payments of 100 at the end of each month. Interest on the loan is charged at an annual nominal rate of i, 0 <i< 1, convertible monthly. The outstanding balances immediately after the 8th and 24th payments are 2308.15 and 1345.50, respectively. Calculate the amount of interest repaid in the 15th payment. Round your answer to the nearest whole number.

  • C++ 18. Savings Account Balance Write a program that calculates the balance of a savings account...

    C++ 18. Savings Account Balance Write a program that calculates the balance of a savings account at the end of a three month period. It should ask the user for the starting balance and the annual interest rate. A loop should then iterate once for every month in the period, performing the following A) Ask the user for the total amount deposited into the account during that month. Do not accept negative numbers. This amount should be added to the...

  • 1)You plan to deposit $100 in a savings account at the end of each month for...

    1)You plan to deposit $100 in a savings account at the end of each month for the next 10 years. How much will you have in 10 years, assuming a 2% annual interest rate. 2)You wish to purchase a car that costs $22,000 today. You will put $2,000 down and borrow the remaining $20,000. The interest will be 2.00% and you will make monthly payments (end of month) over 5 years. What will your monthly payment be? I need help...

  • January 31, 2010 1.5 Since June 30, 2014 Smith has been making deposits of 100 pack...

    January 31, 2010 1.5 Since June 30, 2014 Smith has been making deposits of 100 pack into a bank account on the last day of each month. For all of 2014 and 2015 Smith's account earned nominal interest com pounded monthly at an annual rate of 9%. For the first 9 months of 2016 the account earned i(l2) = .105, and since then the ac count has been earning ;(12) = .12. Find the balance in the account on each...

  • You deposit $100 semi-monthly (twice a month) in an account earning 8% interest. You plan to...

    You deposit $100 semi-monthly (twice a month) in an account earning 8% interest. You plan to continue until your retirement in 35 years. How much will you have in your account after 35 years?

  • ASSIGNMENT 2 QUESTIONS b) to $2000.00? 9. Sarah deposited $4500 into a savings account on July...

    ASSIGNMENT 2 QUESTIONS b) to $2000.00? 9. Sarah deposited $4500 into a savings account on July 17, 2018 that earned simple interest of 2.25%. How much interest was earned and paid into her account on December 1, 2018? 10. Find the size of the deposit that must be made to earn $65 in 150 days at 2.50%. 11. What rate of interest makes deposit of $2000 earn $22 in 91 days? 12. If 45 is earned at 2.5% on a...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT