{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch.10_Solution_Manual_Ed.1_v14_

# Ch.10_Solution_Manual_Ed.1_v14_ - Exercises 10.1 Identify...

This preview shows pages 1–4. Sign up to view the full content.

Exercises 10.1 Identify the type of annuity and calculate the number of payments during the term in the following problems: Exercise 10.1, Solution 1: Payments are made at the end of every month and compounding period (quarterly) = payment period (quarterly). Therefore, this is an ordinary simple annuity. Payment interval is 3 months. In one year, Ronda would have to make 3 12 = 4 payments Therefore, number of payments in 5 years, 9 months (5.75 years) = 4 × 5.75 = 23 payments. Exercise 10.1, Solution 3: Payments are made at the end of every month and compounding period (semi-annually) payment period (monthly). Therefore, this is an ordinary general annuity. Payment interval is one month. In one year, Mike would receive 1 12 = 12 payments Therefore, number of payments in 15 years = 12 × 15 = 180 payments. Exercise 10.1, Solution 5: Payments are made at the beginning of every month and compounding period (quarterly) = payment period (quarterly). Therefore, this is a simple annuity due. Payment interval is 3 months. In one year, Mary would have to make 3 12 = 4 payments Therefore, number of payments in 2 years = 4 × 2 = 8 payments. Exercise 10.1, Solution 7: Payments are made at the beginning of every month and compounding period (annually) payment period (monthly). Therefore, this is a general annuity due. Payment interval is one month. Therefore, number of payments in 3 years, 5 months (3 × 12 + 5 = 41 months) is 1 × 41 = 41 payments. Exercise 10.1, Solution 9: This forms two different annuities. 1st annuity for 6 years with beginning-of-month payments (PMT = \$280) Payment period (monthly) ≠ compounding period (annually) Therefore, it is a general annuity due. Payment interval is one month. Therefore, number of payments in 6 years (6 × 12 = 72 months) is 1 × 72 = 72 payments. 2nd annuity for 2 years with month-end payments (PMT = \$580) Payment period (monthly) ≠ compounding period (semi-annually)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Therefore, it is an ordinary general annuity. Payment interval is one month. Therefore, number of payments in 2 years (2 × 12 = 24 months) is 1 × 24 = 24 payments. 2
Exercises 10.2 Exercise 10.2, Solution 1: a. This is an ordinary simple annuity because: Payments are made at the end of each payment period (monthly) Compounding period (monthly) = payment period (monthly) n = 12 payments/year × 5 years = 60 monthly payments j = 6% = 0.06, m = 12 m j i = = 12 06 . 0 = 0.005 per month - + = i i PMT FV n 1 ) 1 ( - + = 005 . 0 1 ) 005 . 0 1 ( 50 60 = \$3488.501525… = \$3488.50 N I/Y P/Y C/Y PV PMT FV 60 6 12 12 0 50 ? From the calculator computations shown, we get the FV = 3488.501525 Therefore, the accumulated value of her money at the end of the period in her savings account is \$3488.50. b. Interest I = FV n ( PMT ) I = 3488.50 – 60 × 50 = 3488.50 – 3000 = \$488.50 Therefore, the interest earned over the period is \$488.50. Exercise 10.2, Solution 3: This is an ordinary simple annuity because: Payments are made at the end of each payment period (monthly) Compounding period (monthly) = payment period (monthly) n = 12 payments/year × 10 years = 120 monthly payments j = 3.5% = 0.035, m = 12 m j i = = 0.035 12 = 0.002916...

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 68

Ch.10_Solution_Manual_Ed.1_v14_ - Exercises 10.1 Identify...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online