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3 years ago

The following costs and useful life data are associated with two new machines being considered at Arun Tech Inc.

Business

2 answers:

Sidana [21]3 years ago

6 0

Answer:

Machine B has a higher NPV therefore should be produced

Explanation:

The machine with the higher Net Present Value (NPV) should be produced .

NPV of Machine A

PV of cash flow

PV of annual profit = A × (1- (1+r)^*(-n)/r

A- 92,000, n- 11, r- 12%

PV = 92,000 × (1- (1.12^(-11)/0.12 = 546268.32

PV of salvage value = 13,000× 1.12^(-11)= 3737.189

NPV = 546268.320 + 3737.189 -250,000 = $300,005.50

NPV of Machine B

A- 103,00, n- 19, r- 12%

PV = 103,000 × (1- (1.12^(-19)/0.12= 758675.0165

Pv of salvage value = 26000× 1.12^(-19)= 3018.776199

NPV =758675.0165 + 3018.77 -460,000 = $301,693.79

Machine B has a higher NPV , therefore should be produced.

xenn [34]3 years ago

6 0

Answer:

While Machine B has an slightly higher net present value his annual worth is much lower than machine A therefore, the company should purchase machine A which yield better annual return

Machine A

Net Present Value: 300,005.00

Anual worth $ 50,525.463

Machine B

Net Present Value: 301.693,8‬

Anual worth $ 40,958.857

Explanation:

We calculate the present value of each machine

and also, the annual worth of each one to get a fair comparison considering their useful life differ

Machine A

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 92,000.00

time 11

rate 0.12

$92000 \times \frac{1-(1+0.12)^{-11} }{0.12} = PV\\$

PV $546,268.3202

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $13,000.00

time 11.00

rate 0.12000

$\frac{13000}{(1 + 0.12)^{11} } = PV$

PV 3,737.1894

Net Present Value:

$546,268.32 + $3,737.19 - $250,000 = 300.005,51‬

Annual worth:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV 300,005.00

time 11

rate 0.12

$300005 \div \frac{1-(1+0.12)^{-11} }{0.12} = C\\$

C $ 50,525.463

Machine B

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 103,000.00

time 19

rate 0.12

$103000 \times \frac{1-(1+0.12)^{-19} }{0.12} = PV\\$

PV $758,675.0165

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $26,000.00

time 19.00

rate 0.12000

$\frac{26000}{(1 + 0.12)^{19} } = PV$

PV 3,018.7762

Net present value

$758,675.02 + $3,018.78 - 460,000 = 301.693,8‬

Annual worth

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV 301,693.80

time 19

rate 0.12

$301693.8 \div \frac{1-(1+0.12)^{-19} }{0.12} = C\\$

C $ 40,958.857

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Answer:

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3 years ago

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devlian [24]

Answer:

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Explanation:

Diane's Designs has two classes of stock authorized: 8%, $10 par value preferred and $1 par value common.

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JOURNAL ENTRIES

Dr. Bank.......................3,000,000

Cr. Common stock.......................200,000

Cr. Additional Paid-in capital..2,800,000

February 6 Issue 1,000 shares of preferred stock for $11 per share.

JOURNAL ENTRIES

Dr. Bank.......................11,000

Cr. Preferred stock.......................10,000

Cr. Additional Paid-in capital.......1,000

October 10 Purchase 10,000 shares of its own common stock for $18 per share.

JOURNAL ENTRIES

Dr. Treasury Stock.......................180,000

Cr. Bank........................................................180,000

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JOURNAL ENTRIES

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Cr. Treasury Stock.......................60,000

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Answer:

See answers below in the explanation

Explanation:

Journal Entries :

Journals

<u>Date Account and Explanation Debit Credit </u>

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Cash 62300

Service Revenue* 62300

(Record earned revenue)

b cash 199100

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c Supplies 62900

Accounts Payable 62900

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Cash 7400

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Cash 14200

Record Payment of previous wages

Wages Expenses 112000

Cash 112000

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f Income Tax payable 15100

cash 15100

(Record Payment of Income taxes

g Accounts Payable 73000

cash 73000

(record payment of account)

h Interest Expense 2700

Interest Payable 2700

Cash*** 5400

(Record Payment of Interest)

i No journal entry required

j Property Taxes Expense 17000

cash 17000

(Record payment of property taxes)

k Dividends 7200

Cash 7200

(Record Payment of dividends)

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3 years ago