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d1i1m1o1n [39]
3 years ago
6

PB13.

Business
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

Journal entry for each transaction is given below.

Materials were purchased on account for $5,429.

Debit Material Account      $5,429

Credit Payable                    $5,429

Materials were requisitioned to begin work on Job C15 in the amount of $2,500.

Debit WIP JOB C15 Account      $2,500

Credit Material Account             $2,500

Direct labor expense for Job C15 was $4,250.

Debit WIP JOB C15 Account     $4,250

Credit Payroll Account              $4,250

Actual overhead was incurred on account for $5,385.

Debit Factory Overhead Control Account   $5,385

Credit Expense payable/cash                      $5,385

Factory overhead was charged to Job C15 at the rate of 200% direct labor.

Debit WIP JOB C15 Account                                     $8,500

Credit Applied Factory overhead Account              $8,500

Job C15 was transferred to finished goods at $15,250.

Debit Finished Good Account    $15,250

Credit WIP JOB C15 Account      $15,250

Job C15 was sold on account for $28,000

(2 entries will be passed at this stage)

Debit Cost of Good Sold                $15,250

Credit Finished Good Account      $15,250

Debit Receivable/Cash Account   $28,000

Credit Sales                                    $28,000    

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A 9-year annuity of 18 $10,400 semiannual payments will begin 11 years from now, with the first payment coming 11.5 years from n
hichkok12 [17]

The value of the annuity in 9 years is $204,112.77

The value of the annuity in 7 years is $174,993.80

The present value is  $102,107.20

An annuity can be described as a cash flow at regular periods. Here, this annuity provides cash flows semi-annually.

The present value of the annuity has to be determined first.

Present value is the value of an annuity at time zero. It is calculated by discounting cash flows with the discount rate.

Present value would be determined with the aid of a  Present Value of an Ordinary Annuity Table. Please find attached an image of the table.

<em><u>Annuity information </u></em>

  • payments = $10,400
  • years of payments = 9
  • Number of payments = 18
  • Start date = year 11.5
  • End date = year 20

How to use the table : the present value of annuity factor is found at where 20 (end date of the annuity) and 8% (discount rate) meet. This is 9.818

Present value =  the present value of annuity factor x semi-annual payment

$10,400 x 9.818 = $102,107.20

Value of the annuity in 9 years = $102,107.20 x (1.08)^9 = $204,112.77

Value of the annuity in 7 years = $102,107.20 x (1.08)^7 = $174,993.80

A similar question was solved here: brainly.com/question/13405140?referrer=searchResults

5 0
3 years ago
Ocean water contains 0.9 ounces of gold per ton. Method A costs $550 per ton of water processed and will recover 90% of the meta
qwelly [4]

Answer:

Method A should be recommended, because it produces a profit of $61.73 more than Method B

Explanation:

To determine, the recommended, method, let us calculate the amount needed to extract 1 ounce of gold using each method, then subtract these from the selling price to get the profit when each method is used.

Method A:

Recovery rate of metal = 90% = 90/100 = 0.9

Hence for 1 ton of water processed, amount of gold that can be recovered

= 0.9 × 0.9 = 0.81 ounces of gold.

Therefore, to produce 1 ounce of gold, we will solve as follows:

0.81 ounce of gold = 1 ton of water

∴ 1 ounce of gold = 1/0.81 = 1.2345679 ounces of water

Next, we are told that 1 ton of water costs $550 to process

∴ 1.2345679 tons of water = 550 × 1.2345679 = $679.01

Therefore, for method A, the effective amount in dollars used to extract 1 ounce of gold = $679.01

Calculating net income from this method is as follows

profit per ounce = selling price per ounce -  cost price per ounce

profit per ounce = 1,750 - 679.01 = $1,070.99

Method B:

recovery rate of metal = 60% = 60/100 = 0.6

Hence for 1 ton of water processed, amount of gold that can be recovered

= 0.6 × 0.9 = 0.54 ounces of gold.

Therefore, to produce 1 ounce of gold, we will solve as follows:

0.54 ounce of gold = 1 ton of water

∴ 1 ounce of gold = 1/0.54 = 1.8518519 ounces of water

Next, we are told that 1 ton of water costs $400 to process

∴ 1.8518519 tons of water = 400 × 1.2345679 = $740.7

Therefore, for method B, the effective amount in dollars used to extract 1 ounce of gold = $740.74

Calculating net income from this method is as follows

profit per ounce = selling price per ounce -  cost price per ounce

profit per ounce = 1,750 - 740.74 = $1,009.26

Since the net income from method A ($1070.99) is more than the net income from method B ($1,009.25), method A is recommended

5 0
3 years ago
LO 3.2A company has wants to earn an income of $60,000 after-taxes. If the tax rate is 32%, what must be the company’s pre-tax
Keith_Richards [23]

Answer:

$88,235

Explanation:

The computation of the pre tax income is shown below:

We know that

Income after tax = Income before tax (1 - tax rate)

$60,000 = Income before tax × (1 - 0.32)

$60,000 =  Income before tax × 0.68

So, the income before tax would be

= $60,000 ÷ 0.68

= $88,235

If we consider the tax rate than we can easily compute the income after tax

4 0
3 years ago
What is the rationale behind the ceiling when applying the lower-of-cost-or-market method to inventory?
Taya2010 [7]

Answer:

The correct answer is letter "D": Prevents overstatement of the value of obsolete or damaged inventories.

Explanation:

The lower-of-cost-or-market method values assets according to the lowest value possible choosing between the market value or the asset's historical cost. This accounting principle is useful to avoid exaggerating the value of obsolete or damaged assets the firm might have in stock.

5 0
3 years ago
Meena Chavan​ Corp.'s computer chip production process yields DRAM chips with an average life of 1 comma 800 hours and sigma ​=
Lena [83]

Answer:

Cp= 1.33

Explanation:

Giving the following information:

Meena Chavan​ Corp.'s computer chip production process yields DRAM chips with an average life of 1,800 hours and sigma ​= 100 hours. The tolerance upper and lower specification limits are 2,400 hours and 1,600 ​hours, respectively.

Cp= (upper specification - lower specification)/6*sigma

Cp= (2400 - 1600)/6*100= 1.33

3 0
3 years ago
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