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Naddika [18.5K]
2 years ago
15

Aborkian Co. is forecasting sales of 75,000 units of product for November. To make one unit of finished product, seven pounds of

raw materials are required. Actual beginning and desired ending inventories of raw materials and finished goods are:
Business
1 answer:
melamori03 [73]2 years ago
4 0

Questions

Aborkian Co. is forecasting sales of 75,000 units of product for November. To make one unit of finished product, seven pounds of raw materials are required. Actual beginning and desired ending inventories of raw materials and finished goods are:

November 1 November 30

(Actual) (Desired)

Raw materials (pounds) 91,400 86,400

Finished goods 8,500 9,600

(a.) Calculate the number of units of product to be produced during November.

(b.) Calculate the number of pounds of raw materials to be purchased during November

Answer:

Number of units to be produced= 76,100  units

Raw materials to be purchased=   527,700 pounds

Explanation:

<em>Units to be produced</em>

<em>Number of units to be produced = sales budget + closing inventory - opening inventory</em>

= 75,000 + 9,600  - 8,500 =  76,100 units

Number of units to be produced= 76,100  units

<em>Raw materials purchase budget</em>

Raw materials to be purchased = Raw materials to be used + closing inventory of raw materials - opening inventory of raw materials

Raw material usage = production units × standard pounds per unit

                               = 76,100× 7 =532700  pounds

Raw materials to be purchased = 532,700  +86,400 - 91,400=527700

Raw materials to be purchased=   527,700 pounds

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P. Jameson Co. sold $500 of merchandise on Master Card credit sales. The net cash receipts from the sale are immediately deposit
Alchen [17]

Answer:

The journal entry would be as follows:

Account                                  Debit           Credit

Cash                                       $480

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The Credit Card Expense corresponds to the 4% fee that Master Card charged P. Jameson Co. ($500 x 20% = $20)

4 0
3 years ago
Inside Incorporated was issued a charter on January 15 authorizing the following capital stock: Common stock, $6 par, 100,000 sh
marin [14]

Answer:  

$1,114,000   -  total equity section

Balance sheet extract

common stock   (120,000 units)                   $720,000

common stock share premium                     $240,000

preference shares (8 000 units)                   $80,000

preference share premium                            $36,000

Profit (net income)                                       <u>     $ 38,000</u>

                                                                          $1,114,000      

Explanation:

common stock account (100,000 + 20,000) x $6 par value = $720,000

common stock premium per unit is calculated $18 minus par value of $6 = $12. total premium is 12 x 20,000 units issued= $240,000

Preference shares account = (5000+3000) x $10 = $80,000

preference share premium (22 minus 10) = $12 per unit

total preference shares premium is $12 x 3000 issued units= $36,000

8 0
3 years ago
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svet-max [94.6K]

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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

<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
Hartong Corporation is contemplating purchasing equipment that would increase sales revenues by $185,000 per year and cash opera
krek1111 [17]

Answer:

The simple rate of return on the investment is closest to: C. 10.6%

Explanation:

In Hartong Corporation:

Increasing net income = Increase sales revenues - Cash operating expenses - Annual depreciation expense = $185,000 - $89,000 - $52,000 = $44,000

This is the net income from the equipment per year

Return on the investment (ROI) is calculated by using following formula:

ROI = (Net income/Cost of investment )x 100%

Cost of investment  = Cost of equipment = $416,000

ROI = ($44,000/$416,000) x 100% = 10.6%

8 0
3 years ago
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