Engineering Economics A young engineer borrowed P 10,000 at 12% interest and paid P 2,000 per annum for the last 4 years. What does he have to pay at the end of the fifth year in order to pay off his loan? P 6,999.39 P 6,222.39 P 6,922.93 P 6,292.93 P 6,999.39 P 6,222.39 P 6,922.93 P 6,292.93 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A sum of P1,000 is invested now and left for eight years, at which time the principal is withdrawn. The interest has accrued is left for another eight years. If the effective annual interest rate is 5%, what will be the withdrawal amount at the end of the 16th year? P702.15 P693.12 P700.12 P705.42 P702.15 P693.12 P700.12 P705.42 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics An amount of P1,000 becomes P1,608.44 after 4 years compounded bimonthly. Find the nominal interest. 0.1189 0.12 0.1208 0.1232 0.1189 0.12 0.1208 0.1232 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics What is defines as the analysis and evaluation of the monetary consequences by using the theories and principles of economics to engineering applications, designs and projects? Design cost analysis Engineering economy Engineering cost analysis Economic Analysis Design cost analysis Engineering economy Engineering cost analysis Economic Analysis ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics As applied to capitalized asset, the distribution of the initial cost by a periodic changes to operation as in depreciation or the reduction of a debt by either periodic or irregular prearranged programs is called ______. Annuity factor Amortization Annuity Capital recovery Annuity factor Amortization Annuity Capital recovery ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics What is the term for an annuity with a fixed time span? Annuity due Ordinary annuity Perpetuity Annuity certain Annuity due Ordinary annuity Perpetuity Annuity certain ANSWER DOWNLOAD EXAMIANS APP
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