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gulaghasi [49]
3 years ago
6

Flynn Industries has three activity cost pools and two products. It expects to produce 2,200 units of Product BC113 and 1,430 of

Product AD908. Having identified its activity cost pools and the cost drivers for each pool, Flynn accumulated the following data relative to those activity cost pools and cost drivers.
Annual Overhead data Expected Use of Cost Drivers per Product
Activity Cost Pools Cost Drivers Estimated Overhead Expected Use of Cost Drivers per Activity Product BC113 Product AD908
Machine setup Setups $19,608 38 21 17
Machining Machine hours 123,650 4,945 1,122 3,824
Packing Orders 33,530 479 192 287
Using the above data, do the following:
Prepare a schedule showing the computations of the activity-based overhead rates per cost driver.
Activity Cost Pools Estimated Overhead Expected Use of Cost Drivers per Activity Activity-Based Overhead Rates
Machine setup $_______ _______setups $_____ per setup
Machining ________ _______machine hours $_____ per machine hr.
Packing ________ _______orders 192 287
$______
Prepare a schedule assigning each activity's overhead cost to the two products.
BC113 AD908
Activity Cost Pools Expected Use of Cost Drivers per Product Activity-Based Overhead Cost Assigned Expected Use of Cost Drivers per Product Activity-Based Overhead Rates Cost Assigned
Machine setup $ $ $ $
Machining $ $
Packing $ $
Total assigned $ $
Compute the overhead cost per unit for each product. Round answers to 2 decimal places.
BC113 AD908
Overhead cost per unit
Business
1 answer:
netineya [11]3 years ago
3 0

Answer:

Results are below.

Explanation:

<u>First, we need to calculate the allocation rate:</u>

Predetermined manufacturing overhead rate= total estimated overhead costs for the period/ total amount of allocation base

Machine setup= 19,608 / 38= $516 per set up

Machining= 123,650 / 4,945= $25 per machine hour

Packing= 33,530 / 479= $70 per  order

<u>Now, we can allocate to each product:</u>

Allocated MOH= Estimated manufacturing overhead rate* Actual amount of allocation base

Product BC113:

Machine setup= 516*21= $10,836

Machining= 25*1,122= $28,050

Packing= 70*192= $13,440

Total= $52,326

Product AD908:

Machine setup= 516*17= $8,772

Machining= 25*3,824= $95,600

Packing= 70*287= $20,090

Total= $124,462

<u>Finally, the unitary cost:</u>

Product BC113= 52,326 / 2,200= $28,79

Product AD908= 124,462 / 1,430= $87.04

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5 0
4 years ago
For product costs associated with a particular product to be reported on the income statement:
RUDIKE [14]

Answer:

a. The product must be sold

Explanation:

Total revenue and total expenses are recorded in the income statement.  

If the total income exceeds than the total expenditure then the company earns net income And if the total income is less than the total expenditure then the company has a net loss.

The product includes direct material cost, direct labor cost ,and the manufacturing overhead cost. If the product cost is not sold then it is shown in the asset side of the balance sheet as an inventory

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3 years ago
Durable goods $3,000
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Answer:

c. $11,450

Explanation:

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6 0
3 years ago
During a conference call with the corporate office, you are told by a senior executive that you will be going abroad in the next
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C.Clarify the situation, and ask specific questions about the overseas company's cultural and ethical practices. Also, ask what your company policies are regarding intercultural ethics.

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6 0
3 years ago
If you have 4.00 gg of H2, how many grams of NH3 can be produced?
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Answer:

22.49 g of NH3

Explanation:

The balanced equation for this would be:

3H₂+N₂ → 2NH₃

So let's take note of this:

We will need 3 moles of H₂ to produce 2 moles of NH₃.

Now let us convert:

First we determine the molar mass of H2:

Element:

     number of atoms   x     molar mass

H   =          2                 x       1.01 g/mol = 2.02 g/mol

Let's see how many moles of H2 there are in 4.00g

4.00 g\times \dfrac{1\;mole\;of\;H_2}{2.02g} = 1.98\;moles\;of\;H_2

Now we can see how many moles of NH₃ we can make given the ratio and convert it again to grams by getting the molar mass of NH₃:

1.98\;moles\;of\;H_2\times\dfrac{2\;moles\;of\;NH_3}{3\;moles\;of\;H_2} = 1.32\;moles\;of\;NH_3

This means that with 1.98 moles of H₂, we produce 1.32 moles of NH₃

So let's get the molar mass of NH₃ so we can convert it to grams:

N      =   1 x 14.01 = 14.01

H      =  3 x 1.01   = 3.03

                             17.04g/mol

1.32\;moles\;of\;NH_3\times\dfrac{17.04g\;of\;NH_3}{1\;mole\;of\;NH_3} = 22.49g\;of\;NH_3

4 0
3 years ago
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