On January 1 of this year, Shannon Company completed the following transactions (assume a 10% annual interest rate): (FV of $1, PV of $1, FVA of $1, and PVA of $1) (Use the appropriate factor(s) from the tables provided.)

a. Bought a delivery truck and agreed to pay $60,000 at the end of three years.
b. Rented an office building and was given the option of paying $10,000 at the end of each of the next three years or paying $28,000 immediately.
c. Established a savings account by depositing a single amount that will increase to $90,000 at the end of seven years.
d. Decided to deposit a single sum in the bank that will provide 10 equal annual year-end payments of $40,000 to a retired employee (payments starting December 31 of this year).

1. What is the cost of the truck that should be recorded at the time of purchase? (Round your answer to nearest whole dollar.)
2. Which option for the office building results in the lowest present value?
Pay in single installment or Pay in three installments?
3. What single amount must be deposited in this account on January 1 of this year? (Round your answer to nearest whole dollar.)
4. What single sum must be deposited in the bank on January 1 of this year? (Round your answer to nearest whole dollar.)

1. Amount to be paid at the end of three years (A) = $60,000

Rate of interest (r) = 10% = 0.10

Number of years (n) = 3 years.

Cost of truck that should be recorded at the time of purchase = A÷(1+r)n = 60,000÷(1+0.10)3 = 60,000÷1.13 = 60,000÷1.331 = $45,078

2. Annual payment (P) = 10,000

Number of years (n) = 3 years

Rate of interest (r) = 10% = 0.10

Present value of annual payment = P×[1-(1+r)-n]÷r = 10,000×[1-(1+0.10)-3]÷0.10 = 10,000×[1-1.1-3]÷0.10 = 10,000×[1-0.7513]÷0.10 = 10,000×0.2487÷0.10 = $24,870.

Single installment payment is $28,000 and the present value of $10,000 paid annually for 3 years is $24,870, which means annual payment for three years will be the better option because the present value is less than the single installment.

Pay in three installment will be the better option.

3. Amount at the end of 7 years (A) = $90,000

Number of years (n) = 7 years

Rates of interest (r) = 10% = 0.10

Single amount to be deposited in this account on January 1 of this year = A÷(1+r)n = 90,000÷(1+0.10)7 = 90,000÷1.17 = 90,000÷1.9487 = $46,184

4. Annual payment (P) = $40,000

Number of years (n) = 10 years

Rate of interest (r) = 10% = 0.10

Single sum to be deposited in the bank on January 1 of this year = P×[1-(1+r)-n]÷r = 40,000×[1-(1+0.10)-10]÷0.10 = 40,000×[1-1.1-10]÷0.10 = 40,000×[1-0.3855]÷0.10 = 40,000×0.6145÷0.10 = $245782.6842