Solve the engineering economics question below:

SOAL 4: To operate, maintain, and repair a motor grader costs $1,000 the first year and increases $500 per-year thereafter, thus the second year's costs are $1,500, the third year's costs $2,000, and so forth, for five years. The interest rate i is 8%. Find the following: Present worth of these costs Annual worth of these costs - Future worth of these costs SOAL 5: Find which of two alternative sewer designs has the lowest present worth. The expected life of each is 20 years and i = 8% per year. All annual payments are end-of-year. Design A: Gravity flow, deep installation Cost new, $780,000 Operations and Maintenance costs $1,000 per year Salvage value, none Design B: Lift station, shallow installation Cost new, $500,000 Operations and Maintenance $12,000 for years 1 through 5, increasing by $1,200 year per year thereafter (in other words, $13,200 @EOY 6, $14,400 @ EOY 7, etc.) Salvage value, $100,000 at EOY 20 SOAL 1: $2000 is invested at 5% p.a. for 20 years. What sum will be received at the end of the investment period? SOAL 2: What are the monthly repayments for a loan of $100,000 if the annual interest rate is 6% p.a. (computed monthly) and the loan is over 50 years? SOAL 3: Find the unknown quantity in each of the following: a) 30 10 10 10 n=6 1 N 3 4 5 3 i=5% 6 =? b) F = ? A 8004 I=7%, n=5 1000 5000