orites Tools Help The following information applies to the q

orites Tools Help (The following information applies to the questions displayed below In January 2017, Mitzu Co. pays $2.750.000 for a tract of land with two buildings on it. It plans to demolish Building 1 and build a new store in its place. Building 2 will be a company office: it is appraised at $570.000. with a useful life of 20 years and a $75,000 salvage value. A lighted parking lot near Building 1 has improvements (Land Impr $600.000 that are expected to last another 20 years with no salvage value. Without the buildings and im tract of land is valued at $1,830.000. The company also incurs the following additional costs: s 1 valued at ? 340,400 Cost to demolish Building 1 Cost of additional land grading Cost to construct new building (Building 3), having a 195, 400 2,302, 000 173, 000 useful 1ife of 25 years and a $402,000 salvage value Cost of new land improvements (Land Improvenents 2) near Building 2 having a 20-year useful life and no salvage value 1. Allocate the costs incurred by Mitzu to the sppropriate columns and total each colum

Solution

Solution: 1. Allocation of Purchase Price Appraised Value Percent of Total Appraised Value x Total Cost of Acquisition = Apportioned Cost Land $1,830,000 61% x $2,750,000 = $1,677,500 Building 2 570,000 19% x $2,750,000 = $522,500 Land Improvements 1 600,000 20% x $2,750,000 = $550,000 Totals 3000000 100% 2750000 Land Building 2 Building 3 Land Improvements 1 Land Improvements 2 Purchase Price $1,677,500 $522,500 $0 $550,000 $0 Demolition 340,400 0 0 0 0 Land grading 195,400 0 0 0 0 New building (Construction cost) 0 0 2,302,000 0 0 New improvements cost 0 0 0 0 173,000 Totals $2,213,300 $522,500 $2,302,000 $550,000 $173,000 Working Notes: Percent of total appraised value = Appraised value / total appraised value a b c=a x b Allocation of Purchase Price Appraised Value Percent of Total Appraised Value x Total Cost of Acquisition = Apportioned Cost Land $1,830,000 61% x $2,750,000 = $1,677,500 Building 2 570,000 19% x $2,750,000 = $522,500 Land Improvements 1 600,000 20% x $2,750,000 = $550,000 Totals 3000000 100% 2750000 Land Building 2 Building 3 Land Improvements 1 Land Improvements 2 Purchase Price $1,677,500 $522,500 $0 $550,000 $0 [apportioned cost calculated in above table] Demolition 340,400 0 0 0 0 Land grading 195,400 0 0 0 0 New building (Construction cost) 0 0 2,302,000 0 0 New improvements cost 0 0 0 0 173,000 Totals $2,213,300 $522,500 $2,302,000 $550,000 $173,000 2. Date General Journal Debit Credit Jan 01 Land 2,213,300 Building 2 522,500 Building 3 2,302,000 Land improvements 1 550,000 Land improvements 2 173,000 Cash 5,760,800 Notes: Used data from second table 3. Date General Journal Debit Credit Dec 31 Depreciation expense—Building 2 22,375 Accumulated depreciation—Building 2 22,375 Notes: Depreciation expense = (cost -salvage )/ life Building 2 = (522,500 - 75,000)/20 =22,375 Dec 31 Depreciation expense—Building 3 76,000 Accumulated depreciation—Building 3 76,000 Notes: Depreciation expense = (cost -salvage )/ life Building 3 = (2,302,000 - 402,000)/25 =76,000 Dec 31 Depreciation expense—Land improvements 1 27,500 Accumulated depreciation—Land improvements 1 27,500 Notes: Depreciation expense = (cost -salvage )/ life Land improvements 1 = (550,000 - 0)/20 =76,000 Dec 31 Depreciation expense—Land improvements 2 8,650 Accumulated depreciation—Land improvements 2 8,650 Notes: Depreciation expense = (cost -salvage )/ life Land improvements 2 = (173,000 - 0)/20 =8,650 Notes: Cost value taken in above entry is of totals we got in table 2. Notes are just for your understandings Please feel free to ask if anything about above solution in comment section of the question.