Engineering Economics If ‘P’ is principal amount, ‘i’ is the rate of interest and ‘n’ is the number of periods in years, then the interest factor is: None of these ni (1 + ni) (ni - 1) None of these ni (1 + ni) (ni - 1) ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A person buys a piece of lot for P 100,000 downpayment and 10 deferred semi-annual payments of P 8,000 each, starting three years from now. What is the present value of the investment if the rate of interest is 12% compounded semi-annually? P 142,189.67 P 143,104.89 P 143,999.08 P 142,999.08 P 142,189.67 P 143,104.89 P 143,999.08 P 142,999.08 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics The ability to convert assets to cash quickly is known as ______. Leverage Liquidity Insolvency Solvency Leverage Liquidity Insolvency Solvency ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics What type of depreciation is due to the reduction in the demand for the function that the equipment or asset was designed to render? Functional depreciation Demand depreciation Design depreciation Physical depreciation Functional depreciation Demand depreciation Design depreciation Physical depreciation ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A loan of P5,000 is made for a period of 15 months, at a simple interest rate of 15%, what future amount is due at the end of the loan period? 5873.2 5937.5 5690.12 5712.4 5873.2 5937.5 5690.12 5712.4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A manufacturer produces certain items at a labor cost of P 115 each, material cost of P 76 each and variable cost of P 2.32 each. If the item has a unit price of P 600, how many units must be manufactured each month for the manufacturer to break even if the monthly overhead is P428,000 1,037 1,053 1,043 1,033 1,037 1,053 1,043 1,033 ANSWER DOWNLOAD EXAMIANS APP
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