Author: Anonymous

Interest is money paid for the use of borrowed money.

A car’s warranty is 3 years. Upon expiration, annual maintenance starts at $150 and then climbs $25 per year until the car is sold at the end of year 7. Use a 10% interest rate and find the present worth of these expenses

A man is retiring with $250,000 in the bank. He plans on spending $30,000 a year for the rest of his life. What is the interest rate he needs to make this happen?

Leasing or buying a new car decision is considered as an intermediate problem which is suitable for engineering economic analysis.

P= 500, n=3 years, if 6% compounded Weekly then F = 598.55

Two pumps are being considered for purchase. Assume that interest is 8%. Which pump should you select. Pump A Pump B Initial cost $   8,000 $   5,000 End-of-useful life salvage value 2000 1000 Useful life, in years 12 6

A machine will cost $30,000 to purchase. Annual operating and maintenance costs will be $2,000. The machine will save $10,000 per year in labor costs. The salvage value of the machine after 5 years will be $7,000. Calculate the machines present worth for an interest rate of 10%.

The Annual percentage rate (APR) can be > or = the Annual percentage yield (APY)

Feedback process is the last step of decision making process.

How long would it take to double an amount of money P, if it is invested at 8% simple interest? And at 8% compound interest?