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frosja888 [35]
3 years ago
5

On June 10, 20X8, Tower Corporation acquired 100 percent of Brown Company's common stock. Summarized balance sheet data for the

two companies immediately after the stock acquisition are as follows:
Tower Corp. Brown Company
Item Book Value Fair Value
Cash $ 34,000 $ 24,000 $ 24,000
Accounts Receivable 48,000 28,000 28,000
Inventory 98,000 24,000 29,000
Buildings & Equipment (net) 136,000 65,000 85,000
Investment in Brown Stock 175,000
Total $ 491,000 $ 141,000 $ 66,000
Accounts Payable $ 24,000 $ 4,000 $ 4,000
Bonds Payable 139,000 12,000 12,000
Common Stock 41,000 20,000
Retained Earnings 287,000 105,000
Total $ 491,000 $ 141,000 $ 16,000
Required:
a. Prepare the consolidating entries required to prepare a consolidated balance sheet immediately after the acquisition of Brown Company shares. (If no entry is required for a transaction/event, select "No journal entry required" in the first account field.)

Business
1 answer:
Blababa [14]3 years ago
3 0

Answer:

The answer is attached below

Explanation:

You might be interested in
What type of supply chain configuration is most appropriate if your organization needs to minimize inbound transportation costs,
pochemuha

Answer:

centralized suplly chain

Explanation:

Decentralize supply chain processes can be defined as processes that must be performed in the plant because they involve physical interaction with the material. There processes are the ones where the decision-making is ‘localized’.  It involves supply chain managers, planners, manufacturing teams, health and safety team and possibility trade management folks.

3 0
3 years ago
During March, the production department of a process operations system completed and transferred to finished goods 20,000 units
Tju [1.3M]

Answer:

$2.18 per unit

Explanation:

The computation of the direct material cost per equivalent units is shown below:

As we know that

Direct Material cost per equivalent unit is

= Direct material cost ÷ equivalent units

where,

Direct material cost is

= $253,500 + $93,700

= $347,200

And, the number of equivalent units is

Equivalent units of production = Units processed + closing work-in-progress  

= [(20,000 + 100,000) × 100%] + (39,000 × 100%)

= 120,000 + 39,000

= 159,000

Now the cost per equivalent unitis

= $347,200 ÷ 159,200

= $2.18 per unit

7 0
3 years ago
Calculate net worth using the following information. Assets Liabilities car - $3,500 car loan - $3,000 savings account - $1,000
raketka [301]
Ok, I'm going to tell you how to calculate it and the answer.
so what you do is add up your assets and then add up your liabilities.
then you subtract your liabilities from your assets in this case your assets add up to 4,700 and your liabilities add up to 3,500.
then you subtract 4,700 from 3,500 since your liability is a lower number.
And then your answer would be $1,200 dollars hope it helped :D
3 0
3 years ago
Read 2 more answers
At January 1, 2019, Deer Corp. has beginning inventory of 2,000 surfboards. Deer estimates it will sell 10,000 units during the
coldgirl [10]

Answer:

The correct answer is $1,881,600

Explanation:

According to the scenario, the computation of the given data are as follows:

Unit sells = 10,000 units

Growth rate = 12%

Selling price = $150 per unit

Costing = $100 per unit

So, we can calculate the budget sales revenue by using following formula:

Budget sales unit for quarter 3 = (10,000 × 112%) × 112% = 12,544

So, budget sales amount for quarter 3 = 12,544 × $150

= $1,881,600

4 0
3 years ago
Suppose that output (Y ) in an economy is given by the following aggregate production function: Yt = Kt + Nt where Kt is capital
shusha [124]

Answer:

Check the explanation

Explanation:

Yt = Kt + Nt

Taking output per worker, we divide by Nt

Yt/Nt = Kt/Nt + 1

yt = kt + 1

where yt is output per worker and kt is capital per worker.

a) With population being constant, savings rate s and depreciation rate δ.

ΔKt = It - δKt

dividing by Nt, we get

ΔKt/Nt = It/Nt - δKt/Nt ..... [1]

for kt = Kt/Nt, taking derivative

d(kt)/dt = d(Kt/Nt)/dt ... since Nt is a constant, we have

d(kt)/dt = d(Kt/Nt)/dt = (dKt/dt)/Nt = ΔKt/Nt = It/Nt - δKt/Nt = it - δkt

thus, Capital accumulation Δkt = i – δkt

In steady state, Δkt = 0

That is I – δkt = 0

S = I means that I = s.yt

Thus, s.yt – δkt = 0

Then kt* = s/δ(yt) = s(kt+1)/(δ )

kt*= skt/(δ) + s/(δ)

kt* - skt*/(δ) = s/(δ)

kt*(1- s/(δ) = s/(δ)

kt*((δ - s)/(δ) = s/(δ)

kt*(δ-s)) = s

kt* = s/(δ -s)

capital per worker is given by kt*

b) with population growth rate of n,

d(kt)/dt = d(Kt/Nt)/dt =

= \frac{\frac{dKt}{dt}Nt - \frac{dNt}{dt}Kt}{N^{2}t}

= \frac{dKt/dt}{Nt} - \frac{dNt/dt}{Nt}.\frac{Kt}{Nt}

= ΔKt/Nt - n.kt

because (dNt/dt)/Nt = growth rate of population = n and Kt/Nt = kt (capital per worker)

so, d(kt)/dt = ΔKt/Nt - n.kt

Δkt = ΔKt/Nt - n.kt = It/Nt - δKt/Nt - n.kt ......(from [1])

Δkt = it - δkt - n.kt

at steady state Δkt = it - δkt - n.kt = 0

s.yt - (δ + n)kt = 0........... since it = s.yt

kt* = s.yt/(δ + n) =s(kt+1)/(δ + n)

kt*= skt/(δ + n) + s/(δ + n)

kt* - skt*/(δ + n) = s/(δ + n)

kt*(1- s/(δ + n)) = s/(δ + n)

kt*((δ + n - s)/(δ + n)) = s/(δ + n)

kt*(δ + n -s)) = s

kt* = s/(δ + n -s)

.... is the steady state level of capital per worker with population growth rate of n.

3. a) capital per worker. in steady state Δkt = 0 therefore, growth rate of kt is zero

b) output per worker, yt = kt + 1

g(yt) = g(kt) = 0

since capital per worker is not growing, output per worker also does not grow.

c)capital.

kt* = s/(δ + n -s)

Kt*/Nt = s/(δ + n -s)

Kt* = sNt/(δ + n -s)

taking derivative with respect to t.

d(Kt*)/dt = s/(δ + n -s). dNt/dt

(dNt/dt)/N =n (population growth rate)

so dNt/dt = n.Nt

d(Kt*)/dt = s/(δ + n -s).n.Nt

dividing by Kt*

(d(Kt*)/dt)/Kt* = s/(δ + n -s).n.Nt/Kt* = sn/(δ + n -s). (Nt/Kt)

\frac{sn}{\delta +n-s}.\frac{Nt}{Kt}

using K/N = k

\frac{s}{\delta +n-s}.\frac{n}{kt}

plugging the value of kt*

\frac{sn}{\delta +n-s}.\frac{(\delta + n -s)}{s}

n

thus, Capital K grows at rate n

d) Yt = Kt + Nt

dYt/dt = dKt/dt + dNt/dt = s/(δ + n -s).n.Nt + n.Nt

using d(Kt*)/dt = s/(δ + n -s).n.Nt from previous part and that (dNt/dt)/N =n

dYt/dt = n.Nt(s/(δ + n -s) + 1) = n.Nt(s+ δ + n -s)/(δ + n -s) = n.Nt((δ + n)/(δ + n -s)

dYt/dt = n.Nt((δ + n)/(δ + n -s)

dividing by Yt

g(Yt) = n.(δ + n)/(δ + n -s).Nt/Yt

since Yt/Nt = yt

g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)

at kt* = s/(δ + n -s), yt* = kt* + 1

so yt* = s/(δ + n -s) + 1 = (s + δ + n -s)/(δ + n -s) = (δ + n)/(δ + n -s)

thus, g(Yt) = n.(δ + n)/(δ + n -s) (1/yt) =  n.(δ + n)/(δ + n -s) ((δ + n -s)/(δ + n)) = n

therefore, in steady state Yt grows at rate n.

5 0
3 years ago
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