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2 years ago

Vista Company installed a standard cost system on January 1. Selected transactions for the month of January are as follows.

Business

1 answer:

nika2105 [10]2 years ago

8 0

Answer:

1. Dr Raw Materials Inventory $69,920

Dr Materials Price Variance $1,840

Cr Accounts Payable $71,760

2. Dr Work in Process Inventory $68,780

Dr Materials Quantity Variance $1,140

Cr Raw Materials Inventory $69,920

3. Dr Factory Labor $73,600

Cr Labor Price Variance $8,000

Cr Factory Wages Payable $65,600

4. Dr Work in Process Inventory $74,474

Cr Labor Quantity Variance $874

Cr Factory Labor $73,600

5. Dr Work in Process Inventory $143,254

Cr Manufacturing Overhead $143,254

Explanation:

Preparation of the anuary transactions

1. Dr Raw Materials Inventory $69,920

(18,400*$3.80)

Dr Materials Price Variance $1,840 [18,400 x ($3.90 - $3.80)]

Cr Accounts Payable $71,760

($69,920+$1,840)

2. Dr Work in Process Inventory $68,780

(18,100*$3.80)

Dr Materials Quantity Variance $1,140 [$3.80 x (18,400 - 18,100)]

Cr Raw Materials Inventory $69,920

(18,400*$3.80)

3. Dr Factory Labor $73,600

($16,000*$4.60)

Cr Labor Price Variance $8,000

[16,000 x ($4.10 - $4.60)]

Cr Factory Wages Payable $65,600

(16,000*$4.10)

4. Dr Work in Process Inventory $74,474

(16,190*$4.60)

Cr Labor Quantity Variance $874 [$4.60 x (16,000 - 16,190)]

Cr Factory Labor $73,600

($8,000+$65,600)

5. Dr Work in Process Inventory $143,254

($68,780+$74,474)

Cr Manufacturing Overhead $143,254

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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

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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*

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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.

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3 years ago

2. The following table provides information about the production possibilities frontier of a Country.(4)

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If it produces 20 more computers, it will move to the point where it can produce 30 computers and 360 kg of wheat. Opportunity cost would be:

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2 years ago

Assume that you purchase a 6-year, 8% certificate of deposit for $1,000. If interest is compounded annually, what will be the va

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Answer:

$ 1,586.8743

Explanation:

Calculation to determine what will be the value of the certificate when it matures

Compounded annually

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Period n = 6

Using this formula

A = P (1+r)^n

Let plug in the formula

1000 (1.08)^6

= 1586.8743

Therefore what will be the value of the certificate when it matures is $1586.8743

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