Chapter 10, Problem 3BPSB

Requirement 1-

To determine

To compute:

The table showing the cost incurred and its allocation.

Answer to Problem 3BPSB

Solution:

First we need to calculate apportionment of purchase price-

Allocation of purchase price	Appraised value	Percentage	Allocated costs
Land	795,200	56%	868,000
Building B	482,800	34%	527,000
Land Improvement B	142,000	10%	155,000
Total	1,420,000	100%	1,550,000

Table showing cost incurred-

Details	Land	Building B	Building C	Land Improvement B	Land Improvement C
Purchase price	868,000	527,000		155,000
Demolition	122,000
Land grading	174,500
Improvements (new)					103,500
New building			1,458,000
Total	1,164,500	527,000	1,458,000	155,000	103,500

Explanation of Solution

Given,
Appraised value

• Land - $795,200
• Building B - $482,800
• Land Improvement B - $142,000
  

  $\begin{array}{l}TotalAppraisedvalue=Land+BuildingB+LandImprovementB\hfill \\ TotalAppraisedvalue=\$795,200+\$482,800+\$142,000\hfill \\ TotalAppraisedvalue=\$1,420,000\hfill \end{array}$

Now, we need to calculate the percentage allocated to each asset-

  $\begin{array}{l}\begin{array}{l}PercentageofLand=\frac{Appraisedvalueofland}{TotalAppraisedvalue}\hfill \\ PercentageofLand=\frac{\$795,200}{\$1,420,000}\hfill \end{array}\\ PercentageofLand=56\%\end{array}$

  $\begin{array}{l}\begin{array}{l}PercentageofBuildingB=\frac{AppraisedvalueofBuildingB}{TotalAppraisedvalue}\hfill \\ PercentageofBuildingB=\frac{\$482,800}{\$1,420,000}\hfill \end{array}\\ PercentageofBuildingB=34\%\end{array}$

  $\begin{array}{l}\begin{array}{l}PercentageofLandImprovementB=\frac{AppraisedvalueofLandImprovementB}{TotalAppraisedvalue}\hfill \\ PercentageofLandImprovementB=\frac{\$142,000}{\$1,420,000}\hfill \end{array}\\ PercentageofLandImprovementB=10\%\end{array}$

Allocation of purchase price-
Given,

• Purchase price = $1,550,000

  $\begin{array}{l}PurchasepriceallocatedtoLand=Purchaseprice\times PercentageallocatedtoLand\hfill \\ PurchasepriceallocatedtoLand=\$1,550,000\times 56\%\hfill \\ PurchasepriceallocatedtoLand=\$868,000\hfill \end{array}$

  $\begin{array}{l}PurchasepriceallocatedtoBuildingB=Purchaseprice\times PercentageallocatedtoBuildingB\hfill \\ PurchasepriceallocatedtoBuildingB=\$1,550,000\times 34\%\hfill \\ PurchasepriceallocatedtoBuildingB=\$527,000\hfill \end{array}$

  $\begin{array}{l}PurchasepriceallocatedtoLandImprovementB=Purchaseprice\times PercentageallocatedtoLandImprovementB\hfill \\ PurchasepriceallocatedtoLandImprovementB=\$1,550,000\times 10\%\hfill \\ PurchasepriceallocatedtoLandImprovementB=\$155,000\hfill \end{array}$

Calculation of cost incurred-
Given,

• Purchase price of Land - $868,000 (Calculated above)
• Purchase price of Building B- $527,000 (Calculated above)
• Purchase price of Land Improvement B - $155,000 (Calculated above)
• Demolition − $122,000
• Land grading - $174,500
• Improvements- $103,500
• New building - $1,458,000
Total cost of Land −

  $\begin{array}{l}TotalcostofLand=Purchaseprice+Demolition+Landgrading\hfill \\ TotalcostofLand=\$868,000+\$122,000+\$174,500\hfill \\ TotalcostofLand=\$1,164,500\hfill \end{array}$

Conclusion

Thus, table for allocation of costs incurred is prepared.

Requirement 2:

To determine

To prepare:

Journal entry to record costs incurred.

Answer to Problem 3BPSB

Solution:

Date	Accounts titles and Description	Debit ($)	Credit ($)
Jan.1,2015	Land	1,164,500
	Building B	527,000
	Building C	1,458,000
	Land Improvement B	155,000
	Land Improvement C	103,500
	Cash		3,408,000
	(To record plant assets purchased in cash)

Explanation of Solution

Amount of total cash payment is calculated below-

  $\begin{array}{l}\begin{array}{l}Totalcashpayment=Land+BuildingB+BuildingC+LandImprovementB+LandImprovementC\hfill \\ Totalcashpayment=\$1,164,500+\$527,000+\$1,458,000+\$155,000+\$103,500\hfill \end{array}\\ Totalcashpayment=\$3,408,000\end{array}$

In the above entry, cash account is credited as there is reduction in current assets and purchase of assets is increase in assets so accounts are debited.

Conclusion

Thus, journal entry to record all costs incurred is passed.

Requirement 3-

To determine

To prepare:

Adjusting journal entry to record depreciation.

Answer to Problem 3BPSB

Solution:

S. No.	Date	Accounts titles and Description	Debit ($)	Credit ($)
1	Dec.31,2015	Depreciation-Building B	28,500
		Accumulated Depreciation		28,500
		(To record depreciation on Building B)
2	Dec.31,2015	Depreciation-Building C	60,000
		Accumulated Depreciation		60,000
		(To record depreciation on Building C)
3	Dec.31,2015	Depreciation-Land Improvement B	31,000
		Accumulated Depreciation		31,000
		(To record depreciation on Land Improvement B)
4	Dec.31,2015	Depreciation-Land Improvement C	10,350
		Accumulated Depreciation		10,350
		(To record depreciation on Land Improvement C)

Explanation of Solution

Entry 1- Calculation of Depreciation expense-
Given,

• Scrap value = $99,500
• Estimated useful life = 15 years
• Cost = $527,000 (calculated in Req.1)

  $\begin{array}{l}Depreciation=\frac{\left(Cost–ScrapValue\right)}{Estimatedusefullife}\hfill \\ Depreciation=\frac{\left(\$527,000-\$99,500\right)}{15years}\hfill \\ Depreciation=\frac{\$427,000}{15years}\hfill \\ Depreciation=\$28,500\hfill \end{array}$

Entry 2- Calculation of Depreciation expense-
Given,

• Scrap value = $258,000
• Estimated useful life = 20 years
• Cost = $1,458,000 (calculated in Req.1)

  $\begin{array}{l}Depreciation=\frac{\left(Cost–ScrapValue\right)}{Estimatedusefullife}\hfill \\ Depreciation=\frac{\left(\$1,458,000-\$258,000\right)}{\text{20 }years}\hfill \\ Depreciation=\frac{\$1,200,000}{\text{20 }years}\hfill \\ Depreciation=\$60,000\hfill \end{array}$

Entry 3- Calculation of Depreciation expense-
Given,

• Estimated useful life = 5 years
• Cost = $155,000 (calculated in Req.1)
• It has no scrap value

  $\begin{array}{l}Depreciation=\frac{\left(Cost–ScrapValue\right)}{Estimatedusefullife}\hfill \\ Depreciation=\frac{\left(\$155,000-\$0\right)}{\text{5 }years}\hfill \\ Depreciation=\frac{\$155,000}{\text{5 }years}\hfill \\ Depreciation=\$31,000\hfill \end{array}$

Entry 4- Calculation of Depreciation expense-
Given,

• Estimated useful life = 10 years
• Cost = $103,500 (calculated in Req.1)
• It has no scrap value

  $\begin{array}{l}Depreciation=\frac{\left(Cost–ScrapValue\right)}{Estimatedusefullife}\hfill \\ Depreciation=\frac{\left(\$103,500-\$0\right)}{\text{10 }years}\hfill \\ Depreciation=\frac{\$103,500}{10years}\hfill \\ Depreciation=\$10,350\hfill \end{array}$

Conclusion

Thus, depreciation expense has been calculated and entries have been passed.