EXERCISE 11-2 (20-25 minutes)

(a)               If there is any salvage value and the amount is unknown (as is the case here), the cost would have to be determined by looking at the data for the double-declining balance method.

 100% = 20%; 20% X 2 = 40% 5

Cost X 40% = \$20,000

\$20,000 ¸ .40 = \$50,000 Cost of asset

(b)            \$50,000 cost [from (a)] – \$45,000 total depreciation = \$5,000 salvage value.

(c)        The highest charge to income for Year 1 will be yielded by the double-declining balance method.

(a)               The highest charge to income for Year 4 will be yielded by the straight-line method.

(b)               The method that produces the highest book value at the end of Year 3 would be the method that yields the lowest accumulated depreciation at the end of Year 3, which is the straight-line method.

Computations:

St.-line = \$50,000 – (\$9,000 + \$9,000 + \$9,000) = \$23,000 book value, end of Year 3.

S.Y.D. = \$50,000 – (\$15,000 + \$12,000 + \$9,000) = \$14,000 book value, end of Year 3.

D.D.B. = \$50,000 – (\$20,000 + \$12,000 + \$7,200) = \$10,800 book value, end of Year 3.

(c)               The method that will yield the highest gain (or lowest loss) if the asset is sold at the end of Year 3 is the method which will yield the lowest book value at the end of Year 3, which is the double-declining balance method in this case.

EXERCISE 11-5

 (a) (\$117,900 – \$12,900) = \$21,000/yr. = \$21,000 X 5/12 = \$8,750 5

2004 Depreciation — Straight line = \$8,750

 (b) (\$117,900 – \$12,900) = \$5.00/hr. 21,000

2004 Depreciation — Machine Usage = 800 X \$5.00 = \$4,000

 (c) Machine Allocated to Year Total 2004 2005 1 5/15 X \$105,000 = \$35,000 \$14,583* \$20,417** 2 4/15 X \$105,000 = \$28,000 ______ 11,667*** \$14,583 \$32,084 * \$35,000 X 5/12 = \$14,583 ** \$35,000 X 7/12 = \$20,417 *** \$28,000 X 5/12 = \$11,667

2005 Depreciation — Sum-of-the-Years’-Digits = \$32,084

(d)      2004 40% X (\$117,900) X 5/12 = \$19,650

2005 40% X (\$117,900 – \$19,650) = \$39,300

OR

1st full year (40% X \$117,900) = \$47,160

2nd full year [40% X (\$117,900 – \$47,160)] = \$28,296

 2004 Depreciation = 5/12 X \$47,160 = \$19,650 2005 Depreciation = 7/12 X \$47,160 = \$27,510 5/12 X \$28,296 = 11,790 \$39,300

EXERCISE 11-6 (20-30 minutes)

 (a) 2003 Straight-line \$212,000 – \$12,000 = \$25,000/year 8 3 months — Depreciation \$6,250 = (\$25,000 X 3/12)

 (b 2003 Output \$212,000 – \$12,000 = \$5.00/output unit 40,000 1,000 units X \$5.00 = \$5,000 (c) 2003 Working hours \$212,000 – \$12,000 = \$10.00/hour 20,000 525 hours X \$10.00 = \$5,250 (d) 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 OR n (n + 1) = 8(9) = 36 2 2

 Allocated to Sum-of-the-years’-digits Total 2003 2004 2005 Year 1 8/36 X \$200,000 = \$44,444 \$11,111 \$33,333 2 7/36 X \$200,000 = \$38,889 9,722 \$29,167 3 6/36 X \$200,000 = \$33,333 _______ _______ 8,333 \$11,111 \$43,055 \$37,500

2005:      \$37,500 = (9/12 of 2nd year of machine’s life plus 3/12 of 3rd year of machine’s life)

(e)               Double-declining balance 2004: 1/8 X 2 = 25%.

2003:            25% X \$212,000 X 3/12 = \$13,250

2004:            25% X (\$212,000 – \$13,250) = \$49,688

OR

1st full year (25% X \$212,000) = \$53,000

2nd full year [25% X (\$212,000 – \$53,000)] = \$39,750

2003 Depreciation 3/12 X \$53,000 = \$13,250

2004 Depreciation 9/12 X \$53,000 = \$39,750

3/12 X \$39,750 =     9,938

\$49,688

EXERCISE 11-16

 (a) December 31, 2004 Loss on Impairment................................................................ 3,200,000 Accumulated Depreciation—Equipment.................. 3,200,000

 Cost \$9,000,000 Accumulated depreciation 1,000,000 Carrying amount 8,000,000 Fair value 4,800,000 Loss in impairment \$3,200,000 (b) December 31, 2005 Depreciation Expense........................................................... 1,200,000 Accumulated Depreciation—Equipment.................. 1,200,000 New carrying amount \$4,800,000 Useful life 4 years Depreciation per year \$1,200,000

(c)        No entry necessary. Restoration of any impairment loss is not permitted.

EXERCISE 11-17

 (a) Loss on Impairment................................................................ 3,220,000 Accumulated Depreciation—Equipment.................. 3,220,000 Cost \$9,000,000 Accumulated depreciation 1,000,000 Carrying amount 8,000,000 Less: Fair value 4,800,000 Plus: Cost of disposal 20,000 Loss on impairment \$3,220,000

(a)               No entry necessary. Depreciation not taken on assets intended to be sold.

 (c) Accumulated Depreciation—Equipment............................ 500,000 Recovery of Loss on Impairment............................ 500,000 Fair value \$5,300,000 Less: Cost of disposal 20,000 5,280,000 Carrying amount 4,780,000 Recovery of impairment loss \$   500,000

EXERCISE 11-23

 (a) \$970,000 + \$170,000 + \$40,000* – \$100,000 = .09 depletion per unit 12,000,000

*(Note to instructor: The \$40,000 should be depleted because it is a cost of the mine. This cost is incurred to get the land back to its original value of \$100,000.)

2,500,000 units extracted X \$.09 = \$225,000 depletion for 2004

(b)            2,100,000 units sold X \$.09 = \$189,000 charged to cost of goods sold for 2004

EXERCISE 11-24

(a)        Asset turnover ratio:

 \$13,234 = .96 times \$14,212 + \$13,362 2

(a)               Rate of return on assets:

 \$76 = .55% \$14,212 + \$13,362 2

(b)               Profit margin on sales:

 \$76 = .57% \$13,234

(c)               The asset turnover ratio times the profit margin on sales provides the rate of return on assets computed for Eastman Kodak as follows:

 Profit margin on sales X Asset Turnover Return on Assets .57% X .96 = .55%

Note the answer .55% is the same as the rate of return on assets computed in (b) above.

 PROBLEM 11-3

 (a) Depreciation Expense—Asset A 2,900 Accumulated Depreciation—Asset A 2,900 (5/55 X [\$35,000 – \$3,100]) Accumulated Depreciation—Asset A 26,100 Asset A (\$35,000 – \$13,000) 22,000 Gain on Disposal of Plant Assets 4,100 (b) Depreciation Expense—Asset B 6,720 Accumulated Depreciation—Asset B 6,720 ([\$51,000 – \$3,000] ¸ 15,000 X 2,100) (c) Depreciation Expense—Asset C 6,000 Accumulated Depreciation—Asset C 6,000 ([\$80,000 – \$15,000 – \$5,000] ¸ 10) (d) Asset E 22,000 Retained Earnings 22,000 Depreciation Expense—Asset E 4,400* Accumulated Depreciation—Asset E 4,400 *(\$22,000 X .20) PROBLEM 11-8

(a)               The amounts to be recorded on the books of Selig Sporting Goods Inc. as of December 31, 2004, for each of the properties acquired from Starks Athletic Equipment Company are calculated as follows:

Cost Allocations to Acquired Properties

 Appraisal Value Remaining Purchase Price Allocations Renovations Capitalized Interest Total (1) Land \$280,000 \$280,000 (2) Building \$   84,0001 \$100,000 \$21,6002 205,600 (3) Machinery _______ 36,0001 _______ ______ 36,000 Totals \$280,000 \$120,000 \$100,000 \$21,600 \$521,600

Supporting Calculations

 1Balance of purchase price to be allocated. Total purchase price \$400,000 Less land appraisal 280,000 Balance to be allocated \$120,000 Appraisal Values Ratios Allocated Values Building \$105,000 105/150 = .70 X \$120,000 \$  84,000 Machinery 45,000 45/150 = .30 X \$120,000 36,000 Totals \$150,000 1.00 \$120,000

2Capitalizable interest.

 Dates of loans in 2004 Amounts Periods Outstanding Interest At 12% 1/1 \$  50,000 X 12/12 X .12 \$  6,000 4/1 130,000 X 9/12 X .12 11,700 10/1 130,000 X 3/12 X .12 3,900 12/31 190,000 X X .12 _______ Totals \$500,000 \$21,600

Note t: If the interest is allocated between the building and the machinery, \$15,120 (\$21,600 X 105/150) would be allocated to the building and \$6,480 (\$21,600 X 45/150) would be allocated to the machinery.

(b)               Selig Sporting Goods Inc.’s 2004 depreciation expense, for book purposes, for each of the properties acquired from Starks Athletic Equipment Company is as follows:

 1. Land: No depreciation. 2. Building: Depreciation rate = 1.50 X 1/15 = .10 2004 depreciation expense = Cost X Rate X 1/2 year = \$205,600 X .10 X 1/2 = \$10,280 3. Machinery: Depreciation rate = 2.00 X 1/5 = .40 2004 depreciation expense = Cost X Rate X 1/2 = \$36,000 X .40 X 1/2 = \$7,200

(c)               Arguments for the capitalization of interest costs include the following.

(1)               Diversity of practices among companies and industries called for standardization in practices.

(2)               Total interest costs should be allocated to enterprise assets and operations, just as material, labor, and overhead costs are allocated. That is, under the concept of historical costs, all costs incurred to bring an asset to the condition and location necessary for its intended use should be reflected as a cost of that asset.

Arguments against the capitalization of interest include the following:

(1)        Interest capitalized in a period would tend to be offset by amortiza­tion of interest capitalized in prior periods.

(2)        Interest cost is a cost of financing, not of construction.

 PROBLEM 11-9

(a)               Carrying value of asset: \$8,000,000 – \$2,000,000 = \$6,000,000.

Future cash flows (\$5,300,000) < Carrying value (\$6,000,000)

Impairment entry:

Loss on Impairment.. 1,600,000*.....

Accumulated Depreciation                                    1,600,000

*\$6,000,000 – \$4,400,000

(b)            Depreciation Expense.......             1,100,000**

Accumulated Depreciation                         1,100,000

**(4,400,000 ¸ 4)

(c)        No depreciation is recorded on impaired assets to be disposed of. Recovery of impairment losses are recorded.

Accumulated Depreciation........... 200,000

Recovery of Impairment Loss               200,000

 PROBLEM 11-10

 (1) \$82,000 Allocated in proportion to appraised values (1/10 X \$820,000). (2) \$738,000 Allocated in proportion to appraised values (9/10 X \$820,000). (3) Forty years Cost less salvage (\$738,000 – \$40,000) divided by annual depreciation (\$17,450). (4) \$17,450 Same as prior year since it is straight-line depreciation. (5) \$91,000 [Number of shares (2,500) times fair value (\$30)] plus demolition cost of existing building (\$16,000). (6) None No depreciation before use. (7) \$30,000 Fair market value. (8) \$4,500 Cost (\$30,000) times percentage (1/10 X 150%). (9) \$3,825 Cost (\$30,000) less prior year’s depreciation (\$4,500) equals \$25,500. Multiply \$25,500 times 15%. (10) \$150,000 Total cost (\$164,900) less repairs and maintenance (\$14,900). (11) \$32,000 Cost less salvage (\$150,000 – \$6,000) times 8/36. (12) \$9,333 Cost less salvage (\$150,000 – \$6,000) times 7/36 times one-third of a year.

 (13) \$52,000 Annual payment (\$6,000) times present value of annuity due at 8% for 11 years (7.710) plus down payment (\$5,740). This can be found in an annuity due table since the payments are at the beginning of each year. Alternatively, to convert from an ordinary annuity to an annuity due factor, proceed as follows: For eleven payments use the present value of an ordinary annuity for 11 years (7.139) times 1.08. Multiply this factor (7.710) times \$6,000 annual payment to obtain \$46,260, and then add the \$5,740 down payment. (14) \$2,600 Cost (\$52,000) divided by estimated life (20 years).