Answer:
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
