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Assume that Beaver uses the periodic system, and the end of period ending inventory for January is 110 units. a. Prepare all nec

aleksandr82 [10.1K]

Answer:

jan 4

Debit ; Accounts Receivable (80 x $8.00) $640

Credit : Revenue $640

jan 11

Debit ; Purchases (150 x $6) $900

Credit : Accounts Payable $900

jan 13

Debit ; Accounts Receivable (120 x $8.75) $1,050

Credit : Revenue $1,050

jan 20

Debit ; Purchases (160 x $7) $1,120

Credit : Accounts Payable $1,120

jan 27

Debit ; Accounts Receivable (100 x $9.00) $900

Credit : Revenue $900

jan 31

Debit ; Cost of Sales (100 x $5 + 150 x $6 + 160 x $7) $2,520

Credit : Inventory $2,520

<em>Gross Profit = Sales - Cost of Sales</em>

Sales = ( 80 x $8.00 + 120 x $8.75 + 100 x $9.00) = $2,590

Cost of Sales = (100 x $5 + 150 x $6 + 160 x $7) = $2,520

Therefore,

Gross Profit = $2,590 - $2,520

= $70

jan 4

Debit ; Accounts Receivable (80 x $8.00) $640

Debit : Cost of Sales (80 x $5.00) $400

Credit : Revenue (80 x $8.00) $640

Credit : Inventory (80 x $5.00) $400

jan 11

Debit ; Purchases (150 x $6) $900

Credit : Accounts Payable $900

jan 13

Debit ; Accounts Receivable (120 x $8.75) $1,050

Debit : Cost of Sales (20 x $5.00 + 100 x $6) $700

Credit : Revenue (120 x $8.75) $1,050

Credit : Inventory (20 x $5.00 + 100 x $6) $700

jan 20

Debit ; Purchases (160 x $7) $1,120

Credit : Accounts Payable $1,120

jan 27

Debit ; Accounts Receivable (100 x $9.00) $900

Debit : Cost of Sales (50 x $6.00 + 50 x $7) $650

Credit : Revenue (100 x $9.00) $900

Credit : Inventory (50 x $6.00 + 50 x $7) $650

<em>Gross Profit = Sales - Cost of Sales</em>

Sales = ( 80 x $8.00 + 120 x $8.75 + 100 x $9.00) = $2,590

Cost of Sales = ($400 + $700 + $650) = $1,750

Therefore,

Gross Profit = $2,590 - $1,750

= $840

Explanation:

<em>Hie, see the attached the full question as images below</em>

Note that the question in this part requires us to use the Periodic Inventory System. In Periodic Inventory system, Inventory Valuation and calculation of Cost of Goods Sold is done at the <em>end of the Period</em>, in this case at the end of the month of January.

Again it is important to note that the question in this part requires us to use the Perpetual Inventory System. In Perpetual Inventory system, Inventory Valuation and calculation of Cost of Goods Sold is done at the <em>after each and every transaction made</em>.

<u>Overall Comment</u>

The Company use of FIFO should be considered in both the Periodic Inventory System in Part 1 and Perpetual Inventory System in Part 2. FIFO method assumes that the first goods received by the business will be the first ones to be delivered to the final customer.

That said, Cost of Sales for Part 1 are determined and recognized at the end of the period and Cost of Sales for Part 2 are determined and recognized after every sale transaction made

4 0

3 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>

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

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

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

A firm is considering a project with annual cash flows of $300,000. The project would have a five-year life, and the company use

stealth61 [152]

Answer:

$1,081,434

Explanation:

<em>At indifference point, the present value of cash outflow equals present value of cash inflow.</em>

Present value of cash inflow = Annual cash inflow * PV annuity factor (12%, 5 years)

Present value of cash inflow = $300,000*3.60478

Present value of cash inflow = $1,081,434

So, the amount at which the firm would be indifferent between accepting or rejecting the investment is $1,081,434.

7 0

3 years ago

According to Michael Porter, what are three effective competitive positioning strategies? a. focus, differentiation, and middle-

Norma-Jean [14]

Answer: D) overall cost leadership, differentiation, and focus

Explanation:

3 0

3 years ago

Josephine is a teenager who works part-time to make money to pay for the upkeep of her car, gasoline, and car insurance. she als

tino4ka555 [31]

Answer:

The correct answer is: in retail.

Explanation:

To begin with, if Josephine is a teenager who works part-time to make money for her own expenses and also she is like most of american youth who hold part-time jobs then the most probable place to work is in retail, where many teenagers are required due to the fact that the majority of the jobs given in that area are for people who do not need many skills or abilities and therefore that the common action to take is to hire young people to pay low salaries for part-time jobs.

5 0

3 years ago

Read 2 more answers