An asset for drilling was purchased and placed in service by a petroleum product

Question

An asset for drilling was purchased and placed in service by a petroleum production company. Its cost basis is ?$65000 and it has an estimated MV of $10,000 at the end of an estimated useful life of 16 years. Compute the depreciation amount in the second year and the BV at the end of the third year of life by each of these ?methods:

a. The SL method.

b. The 200?% DB method with switchover to SL.

c. The GDS.

d. The ADS.

An asset for drilling was purchased and placed in service by a petroleum production company. Its cost basis is S55.000, and it has an estimated MV of $10,000 at the end of an estimated useful ife of 17 years. Compute the depreciation amount in the second year and the BV at the end of the fourth year of life by each of these methods a. The SL method b. The 200% DB method with switchover to SL. c. The GDS d. The ADS More Info Click the icon to view the partial listing of depreciable assets used in business GDS Recovery Rates (rfor the Six Personal Property Classes Click the icon to view the GDS Recovery Rates (). a. Using the SL method the depreciation amount in the second year is (Round to the nearest dollar.) Using the SL method the BV at the end of the fourth year of life is (Round to the nearest dollar.) b. Using the 200% DB method the depreciation amount in the second year is . (Round to the nearest dollar.) Using the 200% DB method the BV at the end ofthe fourth year of life is SD (Round to the nearest dollar ) c. Using the GDS the depreciation amount in the second year is od to the nearest dollar.) Using the GDs the BV at the end of the fourth year of life is (Round to the nearest dollar) d. Using the ADS the depreciation amount in the second year is $Round to the nearest dollar) Using the ADS the BV at the end of the fourth year of if s (Round to the nearest dollar) Recovery Period (and Property Class) Year 3-year 5-year7-year10-year 15-year 20-year 0.0375 00722 0.0668 0.0618 0.0571 0.0528 0.0489 0.0452 0.0447 0.0447 0.0446 0.0446 0.0446 0.0446 0.0446 0.0446 0.0446 0.1429 0 2449 0.1000 0.1800 0.3333 0 4445 0.1481 0.0741 0.2000 0.3200 0.0500 0.0950 0.1152 0.1152 0.0576 0.1249 0.0893 0.0892 0.0893 0.0446 0.0922 0.0737 0.0655 0.0655 0.0656 0.0655 0.0328 0.0770 0.0693 0.0623 0.0590 0.0590 0.0591 0.0590 0.0591 0.0590 0.0591 0.0590 0,0591 0.0295 15 More Info MACRS Class Lives and Recovery Periods Re covery Periods Print Done t Class Descriptions of Assets Class Life GDS ADS Office fumture and equipment Information systems, including computers Light general purpose trucks Heavy general purpose trucks Tractor units for use over the road 10 10 Enter your answer in Production of petroleum and natural gas

Explanation / Answer

Computation of Depreciation in Second year and BV at the end of third year :

a) SL methods

Straight line Depreciation for second year = cost of asset-scrap value/useful life=( 65000-10000)/16=3437.5

Book value at the end third year = 65000 - (3437.5*3) = 65000 - 10312.5 = 54687.5

b) 200 % DB Method

Straight line Depreciation Rate= 6.25%

DB Method Rate = 6.25%*2=12.5%

Depreciation for first year=65000*12.5%=8125

Depreciation for Second year=56875*12.5%=7109.37

Depreciation for Third year=49765.63*12.5%=6220.70

Book value at the end third year = 65000 - (8125+7109.37+6220.70) = 43544.93

c) The GDS

Depreciation for 1st year = 65000*14.29% = 9288.5

Depreciation for 2nd year = 65000*24.49% = 15918.5

Depreciation for 3rd year = 65000*17.49% = 11368.5

Book value at the end third year = 65000 - (9288.5+15918.5+11368.5) = 28424.5

d) The ADS

Depreciation for 1st year = 65000*5% = 3250

Depreciation for 2nd year = 65000*9.5% = 6175

Depreciation for 3rd year = 65000*8.55% = 5557.5

Book value at the end third year = 65000 - (3250+6175+5557.5) = 50017.5

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