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garik1379 [7]
3 years ago
15

QS 11-9 Recording warranty repairs LO P4 On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty

that covers parts. Warranty expense is estimated at 10% of sales. On July 24, 2017, the mower is brought in for repairs covered under the warranty requiring $34 in materials taken from the Repair Parts Inventory. Prepare the September 11, 2016, entry to record the mower sale, and the July 24, 2017, entry to record the warranty repairs. (Round your answers to 2 decimal places.)
Business
1 answer:
prisoha [69]3 years ago
7 0

Answer:

<em>On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty that covers parts</em>

<u>Recording of revenue:</u>

Cash $490 (debit)

Revenue $490 (credit)

<u>Recording of Warranty granted :</u>

Assurance Warranty expense $49.00 (debit)

Warranty Provision $49.00  (credit)

$490 × 10% = $49.00

<em>On July 24, 2017, the mower is brought in for repairs covered under the warranty requiring $34 in materials taken from the Repair Parts Inventory</em>

<u>When warranty is subsequently received:</u>

Warranty Provision $ 34 (debit)

Repair Parts Inventory $ 34 (credit)

Explanation:

<em>On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty that covers parts</em>

<u>Recording of revenue:</u>

Cash $490 (debit)

Revenue $490 (credit)

<em>We Recognise Revenue to depict transfer of control of mower</em>

<u>Recording of Warranty granted :</u>

Assurance Warranty expense $49.00 (debit)

Warranty Provision $49.00  (credit)

$490 × 10% = $49.00

<em>There is no option for customer to take the warranty or not, so this is a service warranty.The warranty is measured at the best estimate of expenditure required to settle the obligation that is at 10% of sales.</em>

<em>On July 24, 2017, the mower is brought in for repairs covered under the warranty requiring $34 in materials taken from the Repair Parts Inventory</em>

<u>When warranty is subsequently received:</u>

Warranty Provision $ 34 (debit)

Repair Parts Inventory $ 34 (credit)

<em>Utilise the Warranty Provision when the warranty claim is subsequently received</em>

<em></em>

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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

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

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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

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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

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since Yt/Nt = yt

g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)

at kt* = s/(δ + n -s), yt* = kt* + 1

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