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Fynjy0 [20]
3 years ago
6

Machinery purchased for $64,200 by Sheridan Co. in 2016 was originally estimated to have a life of 8 years with a salvage value

of $4,280 at the end of that time. Depreciation has been entered for 5 years on this basis. In 2021, it is determined that the total estimated life should be 10 years with a salvage value of $4,815 at the end of that time. Assume straight-line depreciation.
Required:
a. Prepare the entry to correct the prior year's depreciation, if necessary
b. Prepare the entry to record depreciation for 2015
Business
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

Sheridan Co.

a. It is not necessary to correct the prior year's depreciation.  Depreciation is an accounting estimate and does not require the adjustment of prior year's accounts when there is a correction in its estimates.

b. Entry to record depreciation for 2021:

Debit Depreciation Expense $4,387

Credit Accumulated Depreciation $4,387

To record the depreciation expense for the year.

Explanation:

a) Data and Calculations:

Purchase of machinery in 2016 = $64,200

Original estimated useful life = 8 years

Salvage value = $4,280

Depreciation amount = $59,920 ($64,200 - $4,280)

Depreciation expense per year = $7,490 ($59,920/8)

Accumulated depreciation for 5 years = $37,450

Net book value = $26,750 ($64,200 - $37,450)

Remaining estimated useful life = 5 years

Salvage value = $4,815

New depreciable amount = $21,935 ($26,750 - $4,815)

Depreciation expense per year = $4,387 ($21,935/5)

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6 0
3 years ago
Suppose that you have the option to lease a new car, which you otherwise intend to purchase for $21,000. The lease terms: $3000
slava [35]

Answer:

The amount that will be paid to buy the car is $18,539.43.

Explanation:

This can be calculated using the following 3 steps:

Step 1: Calculation of the present of the monthly payment

Since the payments are made at the beginning of each month, this can be calculated using the formula for calculating the present value (PV) of annuity due given as follows:

PVM = P * ((1 - (1 / (1 + r))^n) / r) * (1 + r) .................................. (1)

Where;

PVM = Present value monthly payments = ?

P = Monthly withdraw = $298

r = monthly financing rate = Financing rate / Number of months in a year = 5.4% / 12 = 0.054 / 12 = 0.0045

n = number of months = 48

Substitute the values into equation (1), we have:

PVM = $298 * ((1 - (1 / (1 + 0.0045))^48) / 0.0045) * (1 + 0.0045) = $12,896.55

Step 2: Calculation of the present of the purchase amount at lease expiration

This can be calculated using the present value formula as follows:

PVP = P / (1 + r)^n  .................................. (2)

Where;

PVP = Present value of the purchase amount at lease expiration = ?

P = Purchase amount at lease expiration = $7000

r = monthly financing rate = Financing rate / Number of months in a year = 5.4% / 12 = 0.054 / 12 = 0.0045

n = number of months = 48

Substitute the values into equation (2), we have:

PVP = $7000 / (1 + 0.0045)^48 = $5,642.88

Step 3: Calculation of the amount that will be paid to buy the car

This can be calculated as follows:

Amount to pay to buy car = PVM + PVP ............... (3)

Where:

PVM = Present value monthly payments = $12,896.55

PVP = $5,642.88

Substitute the values into equation (3), we have:

Amount to pay to buy car = $12,896.55 + $5,642.88 = $18,539.43

Therefore, the amount that will be paid to buy the car is $18,539.43.

5 0
2 years ago
End item A requires three component parts: B, C, and D. The BOM indicates that for each completed A, 1 unit of B, 2 units of C,
I am Lyosha [343]

Answer:

30 in total

Explanation:

In order to calculate how many items A we can produce we need to check how many units required we have, in this case, we have:

40 B's

50 C's

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As you can see in our inventory we only have 15 units of D's, meaning that that is our maximum number of items A produced this week, since we already have 15 A items, we can deliver 30 A products this week.

3 0
2 years ago
An oil refinery is located on the north bank of a straight river that is 3km wide. A pipeline is to be constructed from the refi
Mariulka [41]

Answer:

$6,598,076.21

Explanation:

<h2>THE KEY IS TO FIND OUT THE COST FUNCTION, the calculations are very easy!!!</h2><h2></h2><h3>In order to find the cost function, take a look at the drawing attached. </h3>

We can see the river (sort of) that is 3 km wide and the storage tanks on the other side of the river 8 km apart.

<h3 />

Laying pipes under (across) the river costs 1,000,000 the km & laying pipes over land costs 500,000 per km.

<h3 /><h3>So basically the cost function is 1,000,000 multiplied by something plus 500,000 multiplied by another something.</h3><h3 />

The distance across the river can be found by using Pythagoras Theorem. A side is 3 km the other is unknown, so we call it X. And it is equal to:

\sqrt{3^{2} +x^{2}}=\\\sqrt{9 +x^{2}}

And we multiply it by 1,000,000; the cost of laying pipe under the river, the we get:

1000000\sqrt{9+x^{2}

The distance over the land is (8-x), as we can see in the drawing. So we multiply it by its cost, 500,000. And we get 500,000(8-x).

So the cost function f(x) would be:

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)

<h2>From here, we just have to differentiate and the derivative found must be equal to zero in order to minimize cost. </h2><h3>The value of x when the derivative is zero is plugged in the original function to get the cost.</h3><h3 /><h2>LET'S DO THIS</h2>

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(x)=1000000(9+x^{2})^{1/2}+4000000-500000x\\f'(x)=\frac{1}{2} 1000000(9+x^{2})^{-1/2}(2x)-500000\\\\f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000

<h2>f'(x)=0</h2>

f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000=0\\\frac{1000000x}{\sqrt{9+x^2}}  = 500000\\\frac{2x}{\sqrt{9+x^2}}  = 1\\2x={\sqrt{9+x^2}}\\4x^2=9+x^2\\3x^2=9\\x^2=3\\x=\sqrt{3} \\

And we plug square root of 3 in the original cost function  ad we get

f(\sqrt{3} )=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(\sqrt{3})=1000000\sqrt{9+(\sqrt{3} )^{2}} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{9+3} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{12}+500000(6.27)\\f(\sqrt{3})=1000000(3.46)+500000(6.27)\\f(\sqrt{3})=3464101.62+3133974.60\\f(\sqrt{3})=6598076.21\\

<h2>so the minimal cost is $6,598,076.21</h2><h2 /><h3 />

6 0
3 years ago
Perrette Motor Company rebuilds automobile engines that have been damaged or are in need of extensive repair. The rebuilt engine
kow [346]

Answer:

Perrette Motor Company

Computation of the cost per equivalent unit for materials and conversion for the month:

1. Material cost per equivalent unit: $1,100

2. Conversion cost per equivalent unit: $1,350

Explanation:

a) Data and Calculations:

                                Units     % of completion          Equivalent unit

                                          Materials Conversion    Materials Conversion

Beginning WIP        120         50%        50%                 60                60

a) Started and

Completed             530        100%       100%             530               530

b) Ending WIP        220          50%        30%              110                  66

Equivalent units                                                        640                596

Cost of production:

                                       Materials      Conversion

Beginning WIP              $68,000          $73,200

Started & completed    636,000           731,400

Total cost                    $704,000       $804,600

Equivalent units               640                   596

Cost per equivalent unit $1,100             $1,350

b) When calculating the equivalent units under the weighted average process costing method, the units beginning work-in-process are not taken into consideration, but the costs are.

5 0
3 years ago
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