Solution manual intermediate accounting 15th kiesoch11

would be the method that yields the highest accumulated depreciation at the end of Year 3, which is DDB.. sold at the end of Year 4 is the method which will yield the highest book value

CHAPTER 11

SOLUTIONS TO B EXERCISES

E11-1B (15–20 minutes)

10

2

depreciation rate =

100%

X 2 = 20.00%

10

($500,000 – $100,000 – $80,000) X 20.00% = $ 64,000 depreciation, Year 3

E11-2B (20–25 minutes)

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% = $34,000

$34,000 ÷ 0.40 = $85,000 Cost of asset

(b) $85,000 cost [from (a)] – $75,000 total depreciation = $10,000 salvage

value.

double-declining-balance method.

method.

would be the method that yields the highest accumulated depreciation at the end of Year 3, which is DDB.

Computations:

St.-line = $85,000 – ($15,000 + $15,000 + $15,000) = $40,000 book value, end

of Year 3.

S.Y.D = $85,000 – ($25,000 + $20,000 + $15,000) = $25,000 book value, end of Year 3.

D.D.B = $85,000 – ($34,000 + $20,400 + $12,240) = $18,360 book value, end of Year 3.

sold at the end of Year 4 is the method which will yield the highest book value at the end of Year 4, which is the double-declining balance method

in this case.

E11-3B (15–20 minutes)

2 3/4 X 20/210 X ($355,500 – $30,000) = $23,250 for 2014

1/4 X 20/210 X ($355,500 – $30,000) = $ 7,750

E11-3B (Continued)

20

3/4 X 10% X $355,500 = $26,663 for 2014

10% X ($355,500 – $26,663) = $32,884 for 2015

E11-4B (15–25 minutes)

10

X $300,000 X 1/3 = $18,182 55

9

X $300,000 X 2/3 = 32,727 55

[$345,000 – ($345,000 X 20%)] X 20% X 2/3 = 36,800

[May also be computed as 20% of ($345,000 – 2/3 of 20% of $345,000).]

(a) ($235,800 – $25,800) = $21,000/yr = $21,000 X 5/12 = $8,750

10

2014 depreciation—straight-line = $8,750

42,000

2014 depreciation—machine usage = 800 X $5.00 = $4,000

* $38,182 X 5/12 = $15,909

** $38,182 X 7/12 = $22,273

*** $34,364 X 5/12 = $14,318

2015 Depreciation—sum-of-the-years’-digits = $36,591

2015 40% X ($235,800 – $39,300) = $78,600

OR

2014 depreciation = 5/12 X $94,320 = $39,300

2015 depreciation = 7/12 X $94,320 = $55,020

5/12 X $56,592 = 23,580

$78,600

E11-6B (20–30 minutes)

8

3 months—depreciation $6,250 = ($25,000 X 3/12)

20,000 1,500 units X $10.00 = $15,000

10,000

900 hours X $20.00 = $18,000

Allocated to Sum-of-the-years’-digits Total 2014 2015 2016

machine’s life).

(e) Double-declining balance 2005: 1/8 X 2 = 25%

2014: 25% X $250,000 X 3/12 = $15,625

2015: 25% X ($250,000 – $15,625) = $58,594

OR

2014 Depreciation 3/12 X $62,500 = $15,625

2015 Depreciation 9/12 X $62,500 = $46,875

3/12 X $46,875 = 11,719

E11-7B (25–35 minutes)

Methods of Depreciation

Description

Date

Accum Depr.

E11-7B (Continued)

Machine A—Testing the methods

6] X 1/2

6] X 3

6/21 X 5]

($180,000 X 5/21 X 1/2)

($180,000 X 4/21 X 1/2)

($180,000 X 3/21 X 1/2)

Machine B—Computation of the cost

Year 1 depreciation is then equal to depreciable base X 5/15 X 1/2 and year

2 depreciation is equal to the depreciable base X (5/15 X 1/2 + 4/15 X 1/2)

Total depreciation equals depreciable base x (5/30 + 5/30 +4/30)

$46,667 = depreciable base x 14/30, depreciable base = 100,000

Cost is $115,000 ($100,000 + $15,000].

(d) Using SYD, 2015 Depreciation is $23,333.

($100,000 X 3/15 X 1/2)

Machine C—Using the straight-line method of depreciation

Accumulated Depreciation

Machine D—Computation of Year Purchased

$29,600 Thus the asset must have been purchased on February 12, 2014

E11-8B (20–25 minutes)

Old Machine

Annual depreciation charge: ($69,000 – $4,000) ÷ 10 = $6,500

On October 1, 2013, debit the old machine for $7,000; the revised total cost

is $76,000 ($69,000 + $7,000); thus the revised annual depreciation charge is: ($76,000 – $4,000 – $19,500) ÷ 7 = $7,500.

E11-8B (Continued)

Book value, old machine, October 1, 2016:

Fair value (22,000)

Cost of removal 75

Total loss $12,075

(Note to instructor: The above computation is done to determine whether there is a gain or loss from the exchange of the old machine with the new machine and to show how the cost of removal might be reported Also, if a gain occurs, the gain is not deferred (1) because the exchange has commercial substance and (2) the cash paid exceeds 25% of the total value of the property received.)

New Machine

Installation cost 3,000

Depreciation for the year beginning October 1, 2016 = ($89,000 – $10,000) ÷ 10

= $7,900.

E11-9B (15–20 minutes)

Estimated Salvage

Depreciable Cost

Estimated Life

Depreciation per Year

Composite life = $404,100 ÷ $48,900, or 8.26 years

Composite rate = $48,900 ÷ $457,800, or approximately 10.7%

(b) Depreciation Expense—Plant Assets 48,900

Accumulated Depreciation—Plant Assets 48,900

(c) Cash 14,400

Plant Assets 57,000

E11-10B (10–15 minutes)

2 Using Y to stand for the years of remaining life:

Y/36 X ($860,000 – $140,000) = $120,000

Multiplying both sides by 36:

Y = $4,320,000 ÷ $720,000

Y = 6

The year in which there are 6 remaining years of life at the beginning of that given year is 2013.

E11-11B (10–15 minutes)

handled in the current and prospective periods.

Book value as of 1/1/2015 [$100,000 – ($8,000 X 5)] = $60,000 Remaining useful life, 4 years (9 years – 5 years)

Revised salvage value, $6,000 ($60,000 – $6,000) ÷ 4 = $13,500 Depreciation Expense—Equipment 13,500

E11-12B (20–25 minutes)

$141,250/yr.

Building

Salvage value 200,000

1,805,000 Remaining useful life 29 years Annual depreciation $ 62,241

*$95,000 X 21 years = $1,995,000

Addition

Salvage value 20,000

832,500 Remaining useful life ÷ 29 years Annual depreciation $ 28,707

**$46,250 X 14 years = $647,500 Annual depreciation expense—building ($62,241 + $28,707) $90,948

(a) $3,000,000 ÷ 40 = $75,000

(b) Loss on Disposal of Plant Assets 100,000

Accumulated Depreciation—Building

($200,000 X 20/40) 100,000

Building 200,000 Building 500,000

Cash 500,000

a loss on disposal, because the cost of old roof is given Another alternative would be to debit Accumulated Depreciation on the theory that the replacement extends the useful life of the building The entry in this case would be as follows:

Cash 500,000

As indicated, this approach does not seem as appropriate as the first approach.

1,900,000 Remaining useful life 25 years

OR (Assuming the cost of new roof is debited to

accumulated depreciation:)

Book value of building prior to the replacement of

Cost of new roof 500,000

$2,000,000 Remaining useful life 25 years

E11-14B (20–25 minutes)

(a) Repair Expense 1,250

Equipment 1,250

Add: New equipment:

Purchases $81,000 Freight 500 Installation 3,000

84,500

(1) Straight-line: $276,500 ÷ 10 = $27,650

(2) Sum-of-the-years’-digits: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

For equipment purchased in 2013: $192,000 ($212,000 – $20,000) of the cost of equipment purchased in 2013, is still on hand.

8/55 X $192,000 = $27,927

Total $43,291

2010

2011–2015 Incl 2016 Total

1 $300,000 – $25,000 = $275,000

$275,000 ÷ 10 = $27,500 per yr.

265*/365 of $27,500 = $19,966

2011–2015 Include (5 X $27,500) $137,500

166/365 of $27,500 = $12,507 $169,973

2 0 137,500 27,750 165,000

3 27,500 137,500 0 165,000

4 13,750 137,500 13,750 165,000

5 8/12 of $27,500 18,333

2011–2015 Inc 137,500

6/12 of $27,500 13,750 169,583

6 27,500 137,500 0 165,000

*(20 + 31+ 30 + 31 + 31 + 30 + 31 + 30 + 31)

assumed that straight-line depreciation is satisfactory Reasonable accuracy is normally given by 2, 3, or 4 The simplest of the applications are 6, 2, 3, 4, 5, and 1, in about that order Methods 2, 3, and 4 combine reasonable accuracy with simplicity of application.

E11-16B (10–15 minutes)

Loss on Impairment 2,400,000

E11-16B (Continued)

Depreciation Expense 900,000

E11-17B (15–20 minutes)

(a) Loss on Impairment 2,415,000

Fair value $3,975,000

Less: Cost of disposal 15,000 $3,960,000 Carrying amount 3,585,000

(a) December 31, 2014

Loss on Impairment 300,000

be highlighted as an unusual item in a separate section It is not reported as an extraordinary item.

evaluate this step, management does a recoverability test The recoverability test estimates the future cash flows expected from use of that asset and its eventual disposition If the sum of the expected future net cash flows (undiscounted) is less than the carrying amount of the asset, an impairment results If the recoverability test indicates that an impairment has occurred, a loss is computed The impairment loss is the amount by which the carrying amount of the asset exceeds its fair value E11-19B (15–20 minutes)

30 years

Cost of timber sold: $2,000 – $500 = $1,500

$1,500 X 10,000 acres = $15,000,000 of value of timber ($15,000,000 ÷ 2,000,000 bd ft.) X 700,000 bd ft = $5,250,000

$9,750,000 + $100,000 = $9,850,000 ($9,850,000 ÷ 4,000,000 bd ft.) X 900,000 bd ft = $2,216,250

roads are expensed each period and are not part of the depletion base.

E11-20B (10–15 minutes)

Cost per barrel of oil:

500,000

32,000

500,000

E11-21B (15–20 minutes)

$1,000 X 8,000 acres

X 1,000,000 bd ft = 5,000 bd ft X 8,000 acres

$8,000,000

X 1,000,000 bd ft = $200,000 40,000,000 bd ft.

40,000,000 bd ft.

next time timber is harvested.

Depletion base: $2,000,000 + $200,000 – $300,000 + $500,000 = $2,400,000

Depletion rate: $2,400,000 ÷ 50,000 = $48/ton

E11-23B (15–20 minutes)

50,000

*Note to instructor: The $400,000 should be depleted because it is an asset retirement obligation.

30,000 tons extracted X $112 = $3,360,000 depletion for 2014

2014

E11-24B (15–20 minutes)

$2,820

= 1.932 times

$1,436 + $1,484

2

$1,436 + $1,484

2

E11-24B (Continued)

$138

= 4.89%

$2,820

rate of return on assets computed for Brinker International as follows:

Note the answer 9.45% is the same as the rate of return on assets computed in (b) above.

*E11-25B (20–25 minutes)

(a) Revenues $600,000 $600,000

(b) Revenues $600,000 $600,000

Depreciation* 10,800 17,280 Taxable income $149,200 $142,720

(a) 1 ($21,000 – $1,000) X 1/8 X 6/12 = $1,250 depreciation expense for

book purposes.

purposes.